Chapter 2 Fractions & Decimals

Solution:

Total quantity of sweets Savita has = $1\frac{3}{4}$ kg.

Converting the mixed fraction to an improper fraction:

$1\frac{3}{4} = \frac{(1 \times 4) + 3}{4} = \frac{4+3}{4} = \frac{7}{4}$ kg.

Number of friends among whom the sweets are divided = 7.

To find how much each friend receives, we need to divide the total quantity of sweets by the number of friends.

Quantity of sweets each friend receives = $\frac{\text{Total quantity of sweets}}{\text{Number of friends}}$

= $\frac{\frac{7}{4}}{7}$ kg

= $\frac{7}{4} \div 7$ kg

Dividing by 7 is the same as multiplying by the reciprocal of 7, which is $\frac{1}{7}$.

= $\frac{7}{4} \times \frac{1}{7}$ kg

Now, we can multiply the fractions:

= $\frac{\cancel{7}^{1}}{4} \times \frac{1}{\cancel{7}_{1}}$ kg

= $\frac{1}{4} \times \frac{1}{1}$ kg

= $\frac{1}{4}$ kg

Thus, each friend receives $\frac{1}{4}$ kg of sweets.

Comparing this result with the given options:

(a) $\frac{3}{4}$ kg

(b) $\frac{1}{4}$ kg

(c) $\frac{1}{2}$ kg

(d) $\frac{3}{28}$ kg

The calculated amount matches option (b).

Answer:

The correct option is (b).

Solution:

Let the unknown number be $x$.

According to the given information, $\frac{3}{4}$ of the number is equal to 12.

We can write this as an equation:

$\frac{3}{4} \times x = 12$

To find the value of $x$, we need to multiply both sides of the equation by the reciprocal of $\frac{3}{4}$, which is $\frac{4}{3}$.

$x = 12 \times \frac{4}{3}$

We can simplify the expression:

$x = \frac{\cancel{12}^{4}}{1} \times \frac{4}{\cancel{3}_{1}}$

$x = 4 \times 4$

$x = 16$

So, the number is 16.

Let's verify the answer:

$\frac{3}{4}$ of 16 = $\frac{3}{4} \times 16 = \frac{3}{\cancel{4}_{1}} \times \cancel{16}^{4} = 3 \times 4 = 12$.

This confirms our answer is correct.

Comparing our result with the given options:

(a) 9

(b) 16

(c) 18

(d) 32

The number 16 matches option (b).

Answer:

The correct option is (b).

Solution:

To find the product of two fractions, we multiply the numerators together and the denominators together.

Let the two fractions be $\frac{a}{b}$ and $\frac{c}{d}$. Their product is given by:

$\frac{a}{b} \times \frac{c}{d} = \frac{a \times c}{b \times d}$

In this example, the fractions are $\frac{2}{7}$ and $\frac{5}{9}$.

Applying the rule for multiplication of fractions:

Product = $\frac{2}{7} \times \frac{5}{9}$

= $\frac{2 \times 5}{7 \times 9}$

This expression matches the format given in option (d).

Comparing with the given options:

(a) $\frac{2 \;×\; 5}{7 \;+\; 9}$ (Incorrect: Denominators are added)

(b) $\frac{2 \;+\; 5}{2 \;+\; 9}$ (Incorrect: Numerators and denominators are added)

(c) $\frac{2 \;×\; 9}{5 \;×\; 7}$ (Incorrect: Incorrect cross-multiplication form)

(d) $\frac{2 \;×\; 5}{7 \;×\; 9}$ (Correct: Numerators multiplied over denominators multiplied)

Answer:

The correct option is (d).

Solution:

We are given that $p, q, r, s$ are integers such that $0 < p < q < r < s$. We need to find the smallest among the four given fractions:

(a) $\frac{p \;+\; q}{r \;+\; s}$

(b) $\frac{p \;+\; s}{q \;+\; r}$

(c) $\frac{q \;+\; s}{p \;+\; r}$

(d) $\frac{r \;+\; s}{p \;+\; q}$

Since $p, q, r, s$ are positive integers, the sums in the numerators and denominators are all positive.

Let's compare the relative sizes of the fractions by considering whether they are greater than or less than 1.

For fraction (c), $\frac{q \;+\; s}{p \;+\; r}$:

Given $q > p$ and $s > r$. Adding these inequalities, we get $q+s > p+r$.

Since $p+r$ is a sum of positive integers, $p+r > 0$.

Therefore, $\frac{q \;+\; s}{p \;+\; r} > 1$.

For fraction (d), $\frac{r \;+\; s}{p \;+\; q}$:

Given $r > p$ and $s > q$. Adding these inequalities, we get $r+s > p+q$.

Since $p+q$ is a sum of positive integers, $p+q > 0$.

Therefore, $\frac{r \;+\; s}{p \;+\; q} > 1$.

For fraction (a), $\frac{p \;+\; q}{r \;+\; s}$:

Given $p < r$ and $q < s$. Adding these inequalities, we get $p+q < r+s$.

Since $r+s$ is a sum of positive integers, $r+s > 0$.

Therefore, $\frac{p \;+\; q}{r \;+\; s} < 1$.

For fraction (b), $\frac{p \;+\; s}{q \;+\; r}$:

It is not immediately clear from the given inequalities whether $p+s$ is greater than or less than $q+r$. For example, if $p=1, q=2, r=3, s=4$, then $p+s = 1+4=5$ and $q+r=2+3=5$, so the fraction is 1. If $p=1, q=2, r=3, s=10$, then $p+s=1+10=11$ and $q+r=2+3=5$, so the fraction is $\frac{11}{5} > 1$. If $p=1, q=5, r=6, s=7$, then $p+s=1+7=8$ and $q+r=5+6=11$, so the fraction is $\frac{8}{11} < 1$. Thus, this fraction can be less than, equal to, or greater than 1.

Since fractions (c) and (d) are both strictly greater than 1, and fraction (a) is strictly less than 1, fraction (a) must be smaller than fractions (c) and (d).

Now, we need to compare fraction (a) $\frac{p \;+\; q}{r \;+\; s}$ with fraction (b) $\frac{p \;+\; s}{q \;+\; r}$.

We can compare them by cross-multiplication, which is valid as all terms are positive. We compare $(p+q)(q+r)$ and $(p+s)(r+s)$.

$(p+q)(q+r) = pq + pr + q^2 + qr$

$(p+s)(r+s) = pr + ps + sr + s^2$

Let's consider the difference: $(p+s)(r+s) - (p+q)(q+r)$

$= (pr + ps + sr + s^2) - (pq + pr + q^2 + qr)$

$= ps + sr + s^2 - pq - q^2 - qr$

Rearranging the terms:

$= (s^2 - q^2) + (ps - pq) + (sr - qr)$

$= (s-q)(s+q) + p(s-q) + r(s-q)$

Factor out the common term $(s-q)$:

$= (s-q)(s+q+p+r)$

Given that $0 < p < q < r < s$. Since $s > q$, we have $s-q > 0$.

Also, since $p, q, r, s$ are positive integers, their sum $s+q+p+r$ is also positive, i.e., $s+q+p+r > 0$.

The product $(s-q)(s+q+p+r)$ is the product of two positive numbers, which is positive.

So, $(p+s)(r+s) - (p+q)(q+r) > 0$.

This implies $(p+s)(r+s) > (p+q)(q+r)$.

Since the denominators $(r+s)(q+r)$ are positive, dividing both sides by this term maintains the inequality sign:

$\frac{(p+s)(r+s)}{(r+s)(q+r)} > \frac{(p+q)(q+r)}{(r+s)(q+r)}$

$\frac{p+s}{q+r} > \frac{p+q}{r+s}$

This shows that fraction (a) $\frac{p \;+\; q}{r \;+\; s}$ is smaller than fraction (b) $\frac{p \;+\; s}{q \;+\; r}$.

Since fraction (a) is smaller than fraction (b) and also smaller than fractions (c) and (d) (which are greater than 1), the smallest fraction among the four is $\frac{p \;+\; q}{r \;+\; s}$.

Answer:

The correct option is (a).

Solution:

The given pattern is 60, 30, 15, ______.

Let's observe the relationship between consecutive terms:

The second term (30) is obtained by dividing the first term (60) by 2:

$30 = \frac{60}{2}$

The third term (15) is obtained by dividing the second term (30) by 2:

$15 = \frac{30}{2}$

This pattern suggests that each term is obtained by dividing the previous term by 2.

This is a geometric progression with a common ratio of $\frac{1}{2}$.

To find the next number in the pattern, we need to divide the last given term (15) by 2.

Next number = $\frac{15}{2}$

Comparing our result with the given options:

(a) 10

(b) 5

(c) $\frac{15}{4}$

(d) $\frac{15}{2}$

The calculated value $\frac{15}{2}$ matches option (d).

Answer:

The correct option is (d).

Solution:

We are given an amount in terms of rupees and paise, and we need to express it entirely in rupees using decimal form.

The amount is 8 rupees and 8 paise.

We know the relationship between rupees and paise:

1 Rupee = 100 Paise

To convert paise into rupees, we divide the number of paise by 100.

1 Paise = $\frac{1}{100}$ Rupee

So, 8 paise can be converted to rupees as follows:

8 paise = $8 \times \frac{1}{100}$ Rupees

= $\frac{8}{100}$ Rupees

= 0.08 Rupees

Now, we add the rupees part and the converted paise part to get the total amount in rupees:

Total amount = 8 Rupees + 8 paise (in Rupees)

= 8 Rupees + 0.08 Rupees

= $8 + 0.08$ Rupees

= 8.08 Rupees

Thus, 8 rupees 8 paise is equal to 8.08 rupees.

Comparing our result with the given options:

(a) 8.8

(b) 8.08

(c) 8.008

(d) 88.0

The calculated decimal expression matches option (b).

Answer:

The correct option is (b).

Solution:

A regular hexagon is a polygon with 6 equal sides and 6 equal angles.

The perimeter of a polygon is the total length of its boundary, which is the sum of the lengths of all its sides.

Given:

The shape is a regular hexagon.

Number of sides of a hexagon = 6.

Length of each side = 3.5 cm.

The formula for the perimeter of a regular polygon is:

Perimeter = Number of sides $\times$ Length of each side

Substituting the given values:

Perimeter = $6 \times 3.5$ cm

Calculating the product:

$6 \times 3.5 = 21.0$

So, the perimeter is 21 cm.

Comparing our result with the given options:

(a) 17.5 cm

(b) 21 cm

(c) 18.3 cm

(d) 20 cm

The calculated perimeter 21 cm matches option (b).

Answer:

The correct option is (b).

Solution:

We need to calculate the value of $2.5 \div 1000$.

Dividing a decimal number by a power of 10 (like 10, 100, 1000, etc.) involves shifting the decimal point to the left.

The number 1000 has three zeros.

When we divide a number by 1000, we move the decimal point three places to the left.

Starting with the number 2.5, the decimal point is between 2 and 5.

To divide by 1000, we move the decimal point 3 places to the left:

Original: 2.5

Move 1 place left: 0.25

Move 2 places left: 0.025

Move 3 places left: 0.0025

We add zeros as placeholders in the empty places to the left of the digits.

Therefore, $2.5 \div 1000 = 0.0025$.

Comparing our result with the given options:

(a) 0.025

(b) 0.0025

(c) 0.2500 (which is the same as 0.25)

(d) 25000

The calculated value 0.0025 matches option (b).

Answer:

The correct option is (b).

Solution:

We need to evaluate the value of each option and then compare them to find the smallest value.

Option (a):

The value is given directly as a decimal.

Value (a) = 0.0002

Option (b):

The value is given as a fraction. We convert it to a decimal by dividing 2 by 1000.

Value (b) = $\frac{2}{1000}$

Dividing by 1000 moves the decimal point 3 places to the left.

Value (b) = 0.002

Option (c):

The value is given as an expression involving squaring and division.

Value (c) = $\frac{\left( 0.2 \right)^2}{2}$

First, calculate the square of 0.2:

$(0.2)^2 = 0.2 \times 0.2 = 0.04$

Now, divide this result by 2:

Value (c) = $\frac{0.04}{2} = 0.02$

Option (d):

The value is given as a division of a fraction by a decimal.

Value (d) = $\frac{2}{100} \;÷\; 0.01$

Convert the fraction to a decimal:

$\frac{2}{100} = 0.02$

Now, perform the division: $0.02 \div 0.01$.

Dividing by 0.01 is the same as dividing by $\frac{1}{100}$, which is equivalent to multiplying by 100.

Value (d) = $0.02 \div 0.01 = \frac{0.02}{0.01} = \frac{0.02 \times 100}{0.01 \times 100} = \frac{2}{1} = 2$

Now, let's list the values of all options in decimal form for comparison:

Value (a) = 0.0002

Value (b) = 0.002

Value (c) = 0.02

Value (d) = 2

To easily compare the decimal values, we can write them with the same number of decimal places as the value with the most decimal places (which is 0.0002, having four decimal places):

Value (a) = 0.0002

Value (b) = 0.0020

Value (c) = 0.0200

Value (d) = 2.0000

Comparing these values, we can see that 0.0002 is the smallest value.

The smallest value corresponds to option (a).

Answer:

The correct option is (a).

Solution:

To find the largest value, we need to calculate the value of each expression given in the options.

Option (a): $\frac{32}{0.05}$

To simplify this division, we can multiply the numerator and the denominator by 100 to remove the decimal from the denominator:

$\frac{32}{0.05} = \frac{32 \times 100}{0.05 \times 100} = \frac{3200}{5}$

Now, perform the division:

$\frac{3200}{5} = 640$

Value of option (a) = 640.

Option (b): $\frac{0.320}{50}$

The trailing zero in the numerator does not change the value, so $\frac{0.32}{50}$.

We can perform the division directly or by writing it as a decimal division:

$0.32 \div 50$

Alternatively, convert to a fraction: $\frac{\frac{32}{100}}{50} = \frac{32}{100 \times 50} = \frac{32}{5000}$

To convert $\frac{32}{5000}$ to a decimal, divide 32 by 5000.

$\frac{32}{5000} = \frac{32 \div 50}{5000 \div 50} = \frac{0.64}{100} = 0.0064$

Value of option (b) = 0.0064.

Option (c): $\frac{3.2}{0.05}$

Multiply the numerator and the denominator by 100 to remove the decimal from the denominator:

$\frac{3.2}{0.05} = \frac{3.2 \times 100}{0.05 \times 100} = \frac{320}{5}$

Now, perform the division:

$\frac{320}{5} = 64$

Value of option (c) = 64.

Option (d): $\frac{3.2}{50}$

Perform the division:

$3.2 \div 50$

We can divide 3.2 by 5 and then divide the result by 10.

$3.2 \div 5 = 0.64$

$0.64 \div 10 = 0.064$

Value of option (d) = 0.064.

Comparing the values of the four options:

Value (a) = 640

Value (b) = 0.0064

Value (c) = 64

Value (d) = 0.064

Arranging the values in ascending order:

0.0064, 0.064, 64, 640

The largest value is 640, which corresponds to option (a).

Answer:

The correct option is (a).

Solution:

We need to evaluate the value of each option and then compare them to find the largest value.

Option (a):

The value is given directly as a decimal.

Value (a) = 0.0001

Option (b):

The value is given as a fraction. We convert it to a decimal by dividing 1 by 1000.

Value (b) = $\frac{1}{1000}$

Dividing by 1000 moves the decimal point 3 places to the left from the implied position after the digit 1.

Value (b) = 0.001

Option (c):

The value is given as a square of a decimal. Note that 0.100 is the same as 0.1.

Value (c) = $(0.100)^2 = (0.1)^2$

$(0.1)^2 = 0.1 \times 0.1 = 0.01$

Value (c) = 0.01

Option (d):

The value is given as a division. We can evaluate the fraction and the decimal first.

$\frac{1}{10} = 0.1$

So the expression becomes $0.1 \;÷\; 0.1$.

Any non-zero number divided by itself is 1.

Value (d) = $0.1 \div 0.1 = 1$

Now, let's list the values of all options in decimal form for comparison:

Value (a) = 0.0001

Value (b) = 0.001

Value (c) = 0.01

Value (d) = 1

Comparing these decimal values:

0.0001

0.001

0.01

1

The largest value among these is 1.

The largest value corresponds to option (d).

Answer:

The correct option is (d).

Solution:

A fraction often acts as an operator when it precedes the word 'of'. For example, $\frac{1}{2}$ of 10 means $\frac{1}{2} \times 10$. The word 'of' indicates the operation of multiplication.

In this context, the fraction acts as a multiplier or a scaling factor on the quantity that follows the word 'of'.

The blank can be filled with the word 'of'.

Answer:

A fraction acts as an operator of.

Solution:

The reciprocal of a fraction is obtained by interchanging its numerator and denominator.

If a fraction is $\frac{a}{b}$ (where $a \neq 0$ and $b \neq 0$), its reciprocal is $\frac{b}{a}$.

The given fraction is $\frac{2}{3}$.

Here, the numerator is 2 and the denominator is 3.

To find the reciprocal, we swap the numerator and the denominator.

Reciprocal of $\frac{2}{3}$ = $\frac{3}{2}$.

We can verify this by multiplying the fraction by its reciprocal:

$\frac{2}{3} \times \frac{3}{2} = \frac{2 \times 3}{3 \times 2} = \frac{6}{6} = 1$.

Since the product is 1, $\frac{3}{2}$ is indeed the reciprocal of $\frac{2}{3}$.

The blank should be filled with $\frac{3}{2}$.

Answer:

Fraction which is reciprocal of $\frac{2}{3}$ is $\frac{3}{2}$.

Solution:

A proper fraction is a fraction where the numerator is smaller than the denominator. Its value is always between 0 and 1 (excluding 0 and 1). For example, $\frac{1}{2}, \frac{3}{4}, \frac{5}{6}$.

An improper fraction is a fraction where the numerator is greater than or equal to the denominator. Its value is always greater than or equal to 1. For example, $\frac{3}{2}, \frac{5}{4}, \frac{6}{6}$.

Let $F_p$ be a proper fraction and $F_i$ be an improper fraction.

By definition of a proper fraction, $0 < F_p < 1$.

By definition of an improper fraction, $F_i \ge 1$. Since fractions are usually positive in this context, we can assume $F_i > 0$ (specifically $F_i \ge 1$).

We want to compare the product $P = F_p \times F_i$ with $F_i$.

Consider the inequality for the proper fraction: $F_p < 1$.

Multiply both sides of this inequality by $F_i$. Since $F_i > 0$, the inequality sign remains the same.

$F_p \times F_i < 1 \times F_i$

$P < F_i$

This shows that the product of a proper fraction and an improper fraction is always less than the improper fraction.

Let's check with an example:

Let proper fraction be $\frac{1}{3}$ and improper fraction be $\frac{5}{2}$.

Their product is $\frac{1}{3} \times \frac{5}{2} = \frac{5}{6}$.

Comparing the product $\frac{5}{6}$ with the improper fraction $\frac{5}{2}$:

$\frac{5}{6} = 0.833...$

$\frac{5}{2} = 2.5$

Clearly, $0.833... < 2.5$, so $\frac{5}{6} < \frac{5}{2}$.

The blank should be filled with "less than".

Answer:

Product of a proper and improper fraction is less than the improper fraction.

Solution:

Two numbers are called reciprocals of each other if their product is 1.

For example, the reciprocal of $\frac{a}{b}$ is $\frac{b}{a}$ because $\frac{a}{b} \times \frac{b}{a} = \frac{ab}{ba} = 1$ (assuming $a \neq 0$ and $b \neq 0$).

The statement describes the property of reciprocals.

The blank should be filled with "reciprocals".

Answer:

The two non-zero fractions whose product is 1, are called the reciprocals of each other.

Solution:

We need to express 5 rupees and 5 paise as a single value in rupees using decimal notation.

We know the conversion between rupees and paise:

1 Rupee = 100 Paise

To convert paise to rupees, we divide the number of paise by 100.

5 paise = $\frac{5}{100}$ Rupees

Now, we convert the fraction to a decimal:

$\frac{5}{100} = 0.05$ Rupees

The total amount is the sum of the rupees part and the converted paise part:

Total amount = 5 Rupees + 0.05 Rupees

= $5 + 0.05$ Rupees

= 5.05 Rupees

So, 5 rupees 5 paise is equal to $\textsf{₹}$ 5.05.

The blank should be filled with 5.05.

Answer:

5 rupees 5 paise = $\textsf{₹}$ 5.05.

Solution:

We are asked to convert a length given in millimeters (mm) to meters (m).

We need to know the relationship between millimeters and meters.

The standard relationship is:

1 meter = 1000 millimeters

To convert from millimeters to meters, we need to divide the number of millimeters by 1000.

So, 1 millimeter = $\frac{1}{1000}$ meter.

We are given 45 mm.

To convert 45 mm to meters, we multiply 45 by $\frac{1}{1000}$ or divide 45 by 1000.

$45 \text{ mm} = 45 \times \frac{1}{1000} \text{ m}$

$45 \text{ mm} = \frac{45}{1000} \text{ m}$

Now, we convert the fraction $\frac{45}{1000}$ to a decimal.

Dividing by 1000 means moving the decimal point three places to the left from its current position (which is after the digit 5 in 45).

$45 = 45.0$

Moving the decimal point 3 places to the left gives 0.045.

So, $\frac{45}{1000} = 0.045$.

Therefore, 45 mm = 0.045 m.

The blank should be filled with 0.045.

Answer:

45mm = 0.045 m.

Solution:

We need to calculate the product of 2.4 and 1000.

Multiplying a decimal number by a power of 10 (like 10, 100, 1000, etc.) involves shifting the decimal point to the right.

The number 1000 has three zeros.

When we multiply a number by 1000, we move the decimal point three places to the right.

Starting with the number 2.4, the decimal point is between 2 and 4.

To multiply by 1000, we move the decimal point 3 places to the right:

Original: 2.4

Move 1 place right: 24.

Move 2 places right: 240.

Move 3 places right: 2400.

We add zeros as placeholders in the empty places to the right of the digits.

Therefore, $2.4 \times 1000 = 2400$.

The blank should be filled with 2400.

Answer:

2.4 × 1000 = 2400.

Solution:

When dividing a decimal number by a power of 10, we shift the decimal point to the left.

The number of places we shift the decimal point is equal to the number of zeros in the power of 10.

We are dividing by 100.

The number 100 has two zeros.

Therefore, when dividing by 100, we shift the decimal point to the left by 2 places.

For example, consider the number 123.45. When divided by 100:

$123.45 \div 100 = 1.2345$

The decimal point moved from between 3 and 4 to between 1 and 2, which is a shift of 2 places to the left.

The statement should be filled with "left" and "two".

Answer:

To divide a decimal number by 100, we shift the decimal point in the number to the left by two places.

Solution:

An improper fraction is a fraction where the numerator is greater than or equal to the denominator. Examples: $\frac{3}{2}, \frac{5}{4}, \frac{7}{7}$. The value of an improper fraction is $\ge 1$.

The reciprocal of a fraction is obtained by interchanging the numerator and the denominator. If the fraction is $\frac{a}{b}$, its reciprocal is $\frac{b}{a}$.

Let's consider an improper fraction, for example, $\frac{3}{2}$.

Here, the numerator (3) is greater than the denominator (2), so it is an improper fraction. Its value is $1.5$, which is $\ge 1$.

The reciprocal of $\frac{3}{2}$ is obtained by swapping the numerator and denominator, which is $\frac{2}{3}$.

Now let's examine the reciprocal $\frac{2}{3}$. Here, the numerator (2) is less than the denominator (3).

A fraction where the numerator is less than the denominator is called a proper fraction.

The value of $\frac{2}{3}$ is approximately 0.667, which is less than 1.

So, the reciprocal of the improper fraction $\frac{3}{2}$ is the proper fraction $\frac{2}{3}$.

This one example where the reciprocal of an improper fraction is a proper fraction is sufficient to show that the given statement is not always true. The only case where the reciprocal of an improper fraction is also improper is when the original fraction is equal to 1 (e.g., $\frac{5}{5}$, reciprocal is $\frac{5}{5}$). But the statement claims it's true for "an" improper fraction, implying any improper fraction.

Therefore, the statement "Reciprocal of an improper fraction is an improper fraction" is false.

Answer:

False.

Solution:

We need to check if the given statement is true or false by evaluating the left-hand side of the equation and comparing it with the right-hand side.

The left-hand side (LHS) is $2\frac{2}{5} \;÷ \; 2\frac{1}{5}$.

First, convert the mixed numbers to improper fractions:

$2\frac{2}{5} = \frac{(2 \times 5) + 2}{5} = \frac{10 + 2}{5} = \frac{12}{5}$

$2\frac{1}{5} = \frac{(2 \times 5) + 1}{5} = \frac{10 + 1}{5} = \frac{11}{5}$

So, the expression becomes $\frac{12}{5} ÷ \frac{11}{5}$.

To divide a fraction by another fraction, we multiply the first fraction by the reciprocal of the second fraction.

The reciprocal of $\frac{11}{5}$ is $\frac{5}{11}$.

LHS = $\frac{12}{5} \times \frac{5}{11}$

Multiply the numerators and the denominators:

LHS = $\frac{12 \times 5}{5 \times 11}$

Cancel out the common factor of 5 in the numerator and denominator:

LHS = $\frac{12 \times \cancel{5}^{1}}{\cancel{5}_{1} \times 11}$

LHS = $\frac{12 \times 1}{1 \times 11} = \frac{12}{11}$

The right-hand side (RHS) of the equation is 2.

Now, compare the LHS and RHS:

LHS = $\frac{12}{11}$

RHS = 2

Since $\frac{12}{11}$ is not equal to 2 ($12/11 \approx 1.09$), the statement is false.

Answer:

False.

Solution:

We need to verify if the given equation is true by calculating the value of the left-hand side (LHS) and comparing it with the right-hand side (RHS).

The LHS is $0.04 \div 0.2$.

To perform this division, we can write it as a fraction:

$\frac{0.04}{0.2}$

To eliminate the decimals, we can multiply both the numerator and the denominator by a power of 10. Since the denominator has one decimal place and the numerator has two, we multiply by 100 to make both numbers integers.

$\frac{0.04 \times 100}{0.2 \times 100} = \frac{4}{20}$

Now, simplify the fraction:

$\frac{\cancel{4}^{1}}{\cancel{20}_{5}} = \frac{1}{5}$

Convert the fraction $\frac{1}{5}$ to a decimal:

$\frac{1}{5} = 0.2$

Alternatively, we can perform decimal division directly:

$0.04 \div 0.2$

Move the decimal point in the divisor (0.2) one place to the right to make it a whole number (2).

Move the decimal point in the dividend (0.04) the same number of places to the right (one place), resulting in 0.4.

The division becomes $0.4 \div 2$.

$0.4 \div 2 = 0.2$

The value of the LHS is 0.2.

The RHS is given as 0.2.

Since LHS = RHS ($0.2 = 0.2$), the statement is true.

Answer:

True.

Solution:

We need to determine if the given equation $0.2 \times 0.3 = 0.6$ is true or false.

Let's calculate the product on the left-hand side (LHS).

To multiply decimal numbers, we can multiply them as whole numbers and then place the decimal point in the product.

Multiply the numbers without considering the decimal points: $2 \times 3 = 6$.

Now, count the total number of decimal places in the numbers being multiplied. The number 0.2 has one decimal place, and the number 0.3 has one decimal place. The total number of decimal places is $1 + 1 = 2$.

So, the product must have two decimal places.

Starting from the right of the product of the whole numbers (6), we move the decimal point 2 places to the left.

6 becomes 0.6, then 0.06.

Thus, $0.2 \times 0.3 = 0.06$.

The left-hand side (LHS) is 0.06.

The right-hand side (RHS) is 0.6.

Comparing the LHS and RHS:

$0.06 \neq 0.6$

Therefore, the statement is false.

Answer:

False.

Solution:

We are asked to find $\frac{2}{3}$ of 6 using circles with shaded parts.

Finding $\frac{2}{3}$ of 6 is equivalent to calculating $\frac{2}{3} \times 6$.

We can represent the whole quantity, which is 6, using 6 circles.

$\bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc$

The denominator of the fraction $\frac{2}{3}$ is 3. This means we need to divide the total number of circles (6) into 3 equal groups.

Number of circles in each group = Total circles $\div$ Denominator = $6 \div 3 = 2$.

So, we divide the 6 circles into 3 groups, with 2 circles in each group.

Group 1: $\bigcirc \bigcirc$

Group 2: $\bigcirc \bigcirc$

Group 3: $\bigcirc \bigcirc$

The numerator of the fraction $\frac{2}{3}$ is 2. This means we need to take 2 out of these 3 equal groups.

We select 2 groups:

Selected Group 1: $\bigcirc \bigcirc$

Selected Group 2: $\bigcirc \bigcirc$

Now, we count the total number of circles in the selected groups.

Total circles in selected groups = (Circles in Group 1) + (Circles in Group 2)

= $2 + 2 = 4$

Thus, $\frac{2}{3}$ of 6 is 4.

Alternatively, by direct calculation: $\frac{2}{3} \times 6 = \frac{2 \times 6}{3} = \frac{12}{3} = 4$.

To show this with shading, we would draw 6 circles, group them into sets of 2, and shade 2 of these groups (a total of 4 circles).

Answer:

Using circles with shaded parts, $\frac{2}{3}$ of 6 is represented by taking 2 out of 3 equal parts of 6. Each part contains $6 \div 3 = 2$ circles. Taking 2 such parts gives $2 \times 2 = 4$ circles.

So, $\frac{2}{3}$ of 6 is 4.

Solution:

We are asked to find the value of the expression:

$\frac{1}{4\frac{2}{7}} + \frac{1}{3\frac{11}{13}} + \frac{1}{\left( \frac{5}{9} \right)}$

The expression is the sum of the reciprocals of the numbers in the denominators.

Recall that the reciprocal of a number $x$ is $\frac{1}{x}$.

Let's evaluate each term by finding the reciprocal of its denominator:

Term 1: $\frac{1}{4\frac{2}{7}}$

First, convert the mixed number $4\frac{2}{7}$ to an improper fraction:

$4\frac{2}{7} = \frac{(4 \times 7) + 2}{7} = \frac{28 + 2}{7} = \frac{30}{7}$

The first term is the reciprocal of $\frac{30}{7}$.

Reciprocal of $\frac{30}{7} = \frac{7}{30}$.

Term 2: $\frac{1}{3\frac{11}{13}}$

Convert the mixed number $3\frac{11}{13}$ to an improper fraction:

$3\frac{11}{13} = \frac{(3 \times 13) + 11}{13} = \frac{39 + 11}{13} = \frac{50}{13}$

The second term is the reciprocal of $\frac{50}{13}$.

Reciprocal of $\frac{50}{13} = \frac{13}{50}$.

Term 3: $\frac{1}{\left( \frac{5}{9} \right)}$

This term is the reciprocal of the fraction $\frac{5}{9}$.

Reciprocal of $\frac{5}{9} = \frac{9}{5}$.

Now, we need to find the sum of these reciprocals:

Sum = $\frac{7}{30} + \frac{13}{50} + \frac{9}{5}$

To add fractions with different denominators, we find the Least Common Multiple (LCM) of the denominators (30, 50, and 5).

Prime factorization of denominators:

$30 = 2 \times 3 \times 5$

$50 = 2 \times 5^2$

$5 = 5$

LCM(30, 50, 5) = $2 \times 3 \times 5^2 = 2 \times 3 \times 25 = 150$.

Convert each fraction to an equivalent fraction with the denominator 150:

$\frac{7}{30} = \frac{7 \times (150 \div 30)}{30 \times (150 \div 30)} = \frac{7 \times 5}{30 \times 5} = \frac{35}{150}$

$\frac{13}{50} = \frac{13 \times (150 \div 50)}{50 \times (150 \div 50)} = \frac{13 \times 3}{50 \times 3} = \frac{39}{150}$

$\frac{9}{5} = \frac{9 \times (150 \div 5)}{5 \times (150 \div 5)} = \frac{9 \times 30}{5 \times 30} = \frac{270}{150}$

Now, add the equivalent fractions:

Sum = $\frac{35}{150} + \frac{39}{150} + \frac{270}{150}$

Sum = $\frac{35 + 39 + 270}{150}$

Sum = $\frac{74 + 270}{150}$

Sum = $\frac{344}{150}$

Simplify the resulting fraction by dividing the numerator and the denominator by their greatest common divisor. Both are divisible by 2.

$\frac{344 \div 2}{150 \div 2} = \frac{172}{75}$

The fraction $\frac{172}{75}$ cannot be simplified further as the prime factors of 172 are $2^2 \times 43$ and the prime factors of 75 are $3 \times 5^2$.

The value of the expression is $\frac{172}{75}$. This can also be expressed as a mixed number: $172 \div 75 = 2$ with a remainder of $172 - (2 \times 75) = 172 - 150 = 22$.

So, $\frac{172}{75} = 2\frac{22}{75}$.

Answer:

The value of the expression is $\frac{172}{75}$ or $2\frac{22}{75}$.

Solution:

Given:

A 3 × 3 × 3 cube made up of twenty seven 1 × 1 × 1 cubes.

Cubes are removed to form a tunnel through the coloured squares shown in the figure.

To Find:

(i) Fraction of number of small cubes removed to the number of small cubes left.

(ii) Fraction of the number of small cubes removed to the total number of small cubes.

(iii) What part is (ii) of (i).

Solution:

The large cube is a 3 × 3 × 3 cube. The total number of small 1 × 1 × 1 cubes is the volume of the large cube in terms of small cubes.

Total number of small cubes = $3 \times 3 \times 3 = 27$.

The coloured squares on the faces indicate that a tunnel of cubes is removed passing through the center of the cube, parallel to the edges, starting from each coloured face.

Since the coloured squares are on the centers of three adjacent faces, this means tunnels are removed along three perpendicular directions (e.g., along the x, y, and z axes if we consider the cube aligned with axes).

Each tunnel is formed by removing a line of 3 small cubes passing through the 3 × 3 × 3 cube.

Number of cubes in one tunnel = 3.

There are 3 such tunnels.

If the tunnels did not intersect, the total number of removed cubes would be $3 \times 3 = 9$.

However, these tunnels intersect at the very center of the 3 × 3 × 3 cube. The central 1 × 1 × 1 cube is part of all three tunnels.

Let $N_x$ be the number of cubes in the tunnel along the x-direction, $N_y$ along the y-direction, and $N_z$ along the z-direction.

$N_x = 3$, $N_y = 3$, $N_z = 3$.

The intersection of any two tunnels is the single central cube. The intersection of all three tunnels is also the single central cube.

Using the principle of inclusion-exclusion to find the total number of unique cubes removed:

Number of removed cubes = $N_x + N_y + N_z - (\text{Intersections of pairs}) + (\text{Intersection of all three})$

Number of removed cubes = $3 + 3 + 3 - (1 + 1 + 1) + 1$

= $9 - 3 + 1 = 7$.

So, 7 small cubes are removed.

Number of small cubes left = Total number of small cubes - Number of small cubes removed

= $27 - 7 = 20$.

So, 20 small cubes are left.

(i) Fraction of number of small cubes removed to the number of small cubes left:

Fraction (i) = $\frac{\text{Number of small cubes removed}}{\text{Number of small cubes left}}$

Fraction (i) = $\frac{7}{20}$

(ii) Fraction of the number of small cubes removed to the total number of small cubes:

Fraction (ii) = $\frac{\text{Number of small cubes removed}}{\text{Total number of small cubes}}$

Fraction (ii) = $\frac{7}{27}$

(iii) What part is (ii) of (i)?

Let $x$ be the part such that (ii) $= x \times$ (i).

So, $\frac{7}{27} = x \times \frac{7}{20}$

To find $x$, divide $\frac{7}{27}$ by $\frac{7}{20}$.

$x = \frac{7}{27} \div \frac{7}{20}$

$x = \frac{7}{27} \times \frac{20}{7}$

Cancel out the common factor 7:

$x = \frac{\cancel{7}}{27} \times \frac{20}{\cancel{7}}$

$x = \frac{1}{27} \times \frac{20}{1} = \frac{20}{27}$

So, (ii) is $\frac{20}{27}$ part of (i).

Answer:

(i) The fraction of the number of small cubes removed to the number of small cubes left is $\frac{7}{20}$.

(ii) The fraction of the number of small cubes removed to the total number of small cubes is $\frac{7}{27}$.

(iii) Part (ii) is $\frac{20}{27}$ of part (i).

Given:

Ramu finishes $\frac{1}{3}$ part of a work in 1 hour.

Time spent = $2\frac{1}{5}$ hours.

To Find:

The part of the work finished in $2\frac{1}{5}$ hours.

Solution:

Ramu's work rate is the amount of work he finishes per hour.

Work finished in 1 hour = $\frac{1}{3}$ of the total work.

We need to find the work finished in $2\frac{1}{5}$ hours.

First, convert the mixed number $2\frac{1}{5}$ to an improper fraction:

$2\frac{1}{5} = \frac{(2 \times 5) + 1}{5} = \frac{10 + 1}{5} = \frac{11}{5}$ hours.

Assuming Ramu works at a constant rate, the amount of work finished is directly proportional to the time spent.

Work finished in $2\frac{1}{5}$ hours = (Work finished in 1 hour) $\times$ (Number of hours)

= $\frac{1}{3} \times \frac{11}{5}$

Multiply the numerators together and the denominators together:

= $\frac{1 \times 11}{3 \times 5}$

= $\frac{11}{15}$

So, Ramu will finish $\frac{11}{15}$ part of the work in $2\frac{1}{5}$ hours.

Answer:

The part of the work that will be finished in $2\frac{1}{5}$ hours is $\frac{11}{15}$.

Given:

Weight of the cake = 4 kg.

Weight of each piece to be cut = $\frac{2}{3}$ kg.

To Find:

The number of $\frac{2}{3}$ kg pieces that can be cut from a 4 kg cake.

Solution:

To find out how many pieces of a certain weight can be cut from a total weight, we need to divide the total weight by the weight of each piece.

Number of pieces = $\frac{\text{Total weight of the cake}}{\text{Weight of each piece}}$

Number of pieces = $\frac{4 \text{ kg}}{\frac{2}{3} \text{ kg}}$

Number of pieces = $4 \div \frac{2}{3}$

To divide a whole number by a fraction, we multiply the whole number by the reciprocal of the fraction.

The reciprocal of $\frac{2}{3}$ is $\frac{3}{2}$.

Number of pieces = $4 \times \frac{3}{2}$

Multiply the numbers:

Number of pieces = $\frac{4}{1} \times \frac{3}{2}$

= $\frac{4 \times 3}{1 \times 2}$

= $\frac{12}{2}$

= 6

Alternatively, using cancellation:

Number of pieces = $\cancel{4}^{2} \times \frac{3}{\cancel{2}_{1}}$

= $2 \times 3 = 6$

So, 6 pieces of $\frac{2}{3}$ kg can be cut from a 4 kg cake.

Answer:

From a cake of weight 4 kg, 6 pieces of $\frac{2}{3}$ kg can be cut.

Given:

Quantity of potatoes purchased = 3.5 kg.

Rate per kg of potatoes = $\textsf{₹}$ 13.75.

To Find:

Total amount to be paid in nearest rupees.

Solution:

The total cost is found by multiplying the quantity of potatoes by the rate per kg.

Total cost = Quantity $\times$ Rate

Total cost = $3.5 \text{ kg} \times \textsf{₹} 13.75/\text{kg}$

Total cost = $\textsf{₹} (3.5 \times 13.75)$

We calculate the product $3.5 \times 13.75$. We can multiply the numbers as if they were whole numbers first and then place the decimal point.

Multiply 35 by 1375:

$\begin{array}{cc}& & 1 & 3 & 7 & 5 \\ \times & & & & 3 & 5 \\ \hline && 6 & 8 & 7 & 5 \\ & 4 & 1 & 2 & 5 & \times \\ \hline 4 & 8 & 1 & 2 & 5 \\ \hline \end{array}$

The product of 35 and 1375 is 48125.

Now, we count the total number of decimal places in the numbers being multiplied. 3.5 has one decimal place and 13.75 has two decimal places.

Total number of decimal places = $1 + 2 = 3$.

So, we place the decimal point 3 places from the right in the product 48125.

$48.125$

Total cost = $\textsf{₹}$ 48.125.

We need to find the amount to be paid in nearest rupees. This means rounding the total cost to the nearest whole number.

The total cost is $\textsf{₹}$ 48.125.

To round to the nearest rupee, we look at the first digit after the decimal point, which is 1.

Since the digit 1 is less than 5, we round down. This means we keep the whole number part (48) as it is and drop the decimal part.

Amount in nearest rupees = $\textsf{₹}$ 48.

Answer:

Harmeet should pay $\textsf{₹}$ 48 in nearest rupees.

Given:

Total length of the rope = 9.5 m.

Length of each small piece required = 1.9 m.

To Find:

The number of small pieces of the required length that can be cut from the rope.

Solution:

To find the number of pieces of length 1.9 m that can be cut from a rope of length 9.5 m, we need to divide the total length by the length of each piece.

Number of pieces = $\frac{\text{Total length of rope}}{\text{Length of each piece}}$

Number of pieces = $\frac{9.5 \text{ m}}{1.9 \text{ m}}$

Number of pieces = $9.5 \div 1.9$

To divide decimal numbers, we can remove the decimal points by multiplying both the numerator and the denominator by a power of 10. In this case, both numbers have one decimal place, so we multiply by 10.

Number of pieces = $\frac{9.5 \times 10}{1.9 \times 10} = \frac{95}{19}$

Now, we perform the division of whole numbers:

$95 \div 19$.

We know that $19 \times 5 = 95$.

So, $\frac{95}{19} = 5$.

Thus, Kavita will get 5 pieces of rope of length 1.9 m each from the 9.5 m rope.

Answer:

Kavita will get 5 pieces of the required length from the rope.

Given:

Total amount earned by three boys = $\textsf{₹}$ 235.50.

Number of boys = 3.

To Find:

The average amount earned per boy.

Solution:

The average amount earned per boy is calculated by dividing the total amount earned by the number of boys.

Average amount earned = $\frac{\text{Total amount earned}}{\text{Number of boys}}$

Average amount earned = $\frac{\textsf{₹} 235.50}{3}$

Average amount earned = $\textsf{₹} (235.50 \div 3)$

Perform the division:

$235.50 \div 3$

Divide 235.50 by 3:

23 divided by 3 is 7 with a remainder of 2.

Bring down the next digit (5), making it 25. 25 divided by 3 is 8 with a remainder of 1.

Place the decimal point in the quotient as we cross the decimal point in the dividend.

Bring down the next digit (5), making it 15. 15 divided by 3 is 5 with a remainder of 0.

Bring down the last digit (0), making it 0. 0 divided by 3 is 0 with a remainder of 0.

So, $235.50 \div 3 = 78.50$.

The average amount earned per boy is $\textsf{₹}$ 78.50.

Answer:

The average amount earned per boy was $\textsf{₹}$ 78.50.

Solution:

To find the product of two fractions, we multiply the numerators together and the denominators together.

If the fractions are $\frac{a}{b}$ and $\frac{c}{d}$, their product is given by $\frac{a}{b} \times \frac{c}{d} = \frac{a \times c}{b \times d}$.

(i) Find the product of $\frac{1}{2}$ and $\frac{5}{8}$.

Product = $\frac{1}{2} \times \frac{5}{8}$

Multiply the numerators ($1 \times 5$) and the denominators ($2 \times 8$).

Product = $\frac{1 \times 5}{2 \times 8}$

Product = $\frac{5}{16}$

(ii) Find the product of $\frac{1}{3}$ and $\frac{7}{5}$.

Product = $\frac{1}{3} \times \frac{7}{5}$

Multiply the numerators ($1 \times 7$) and the denominators ($3 \times 5$).

Product = $\frac{1 \times 7}{3 \times 5}$

Product = $\frac{7}{15}$

(iii) Find the product of $\frac{4}{3}$ and $\frac{5}{2}$.

Product = $\frac{4}{3} \times \frac{5}{2}$

Multiply the numerators ($4 \times 5$) and the denominators ($3 \times 2$).

Product = $\frac{4 \times 5}{3 \times 2}$

Product = $\frac{20}{6}$

Simplify the resulting fraction by dividing the numerator and the denominator by their greatest common divisor, which is 2.

Product = $\frac{\cancel{20}^{10}}{\cancel{6}_{3}}$

Product = $\frac{10}{3}$

Answer:

(i) The product of $\frac{1}{2}$ and $\frac{5}{8}$ is $\frac{5}{16}$.

(ii) The product of $\frac{1}{3}$ and $\frac{7}{5}$ is $\frac{7}{15}$.

(iii) The product of $\frac{4}{3}$ and $\frac{5}{2}$ is $\frac{10}{3}$.

Solution:

We will use the results from Example 32 to answer the questions.

From Example 32, the products are:

(i) $\frac{1}{2} \times \frac{5}{8} = \frac{5}{16}$

(ii) $\frac{1}{3} \times \frac{7}{5} = \frac{7}{15}$

(iii) $\frac{4}{3} \times \frac{5}{2} = \frac{10}{3}$

(i) Does interchanging the fractions in the example, $\frac{1}{12}$ × $\frac{5}{8}$ affect the answer?

The multiplication of fractions (or any numbers) is commutative, which means that the order of the factors does not affect the product. That is, for any two fractions $\frac{a}{b}$ and $\frac{c}{d}$, we have $\frac{a}{b} \times \frac{c}{d} = \frac{c}{d} \times \frac{a}{b}$.

The question in the prompt mentions $\frac{1}{12} \times \frac{5}{8}$, which was not one of the products calculated in Example 32. Assuming there is a typo and the question refers to the first product in Example 32, which is $\frac{1}{2} \times \frac{5}{8}$, let's consider interchanging these fractions.

The original product from Example 32 (i) is $\frac{1}{2} \times \frac{5}{8} = \frac{5}{16}$.

If we interchange the fractions and multiply $\frac{5}{8}$ by $\frac{1}{2}$:

$\frac{5}{8} \times \frac{1}{2} = \frac{5 \times 1}{8 \times 2} = \frac{5}{16}$.

The product is still $\frac{5}{16}$.

Therefore, interchanging the fractions does not affect the answer.

(ii) Is the value of the fraction in the product greater or less than the value of either fraction?

Let's examine each case from Example 32:

Case (i): $\frac{1}{2} \times \frac{5}{8} = \frac{5}{16}$

The fractions are $\frac{1}{2}$ and $\frac{5}{8}$. Both are proper fractions (value less than 1).

The product is $\frac{5}{16}$.

Compare the product with the first fraction: $\frac{5}{16}$ versus $\frac{1}{2}$.

$\frac{1}{2} = \frac{1 \times 8}{2 \times 8} = \frac{8}{16}$.

Since $5 < 8$, $\frac{5}{16} < \frac{8}{16}$, so $\frac{5}{16} < \frac{1}{2}$.

Compare the product with the second fraction: $\frac{5}{16}$ versus $\frac{5}{8}$.

$\frac{5}{8} = \frac{5 \times 2}{8 \times 2} = \frac{10}{16}$.

Since $5 < 10$, $\frac{5}{16} < \frac{10}{16}$, so $\frac{5}{16} < \frac{5}{8}$.

In this case (multiplying two proper fractions), the product is less than either fraction.

Case (ii): $\frac{1}{3} \times \frac{7}{5} = \frac{7}{15}$

The fractions are $\frac{1}{3}$ (proper, value $<1$) and $\frac{7}{5}$ (improper, value $>1$).

The product is $\frac{7}{15}$ (proper, value $<1$).

Compare the product with the first fraction: $\frac{7}{15}$ versus $\frac{1}{3}$.

$\frac{1}{3} = \frac{1 \times 5}{3 \times 5} = \frac{5}{15}$.

Since $7 > 5$, $\frac{7}{15} > \frac{5}{15}$, so $\frac{7}{15} > \frac{1}{3}$.

Compare the product with the second fraction: $\frac{7}{15}$ versus $\frac{7}{5}$.

$\frac{7}{5} = \frac{7 \times 3}{5 \times 3} = \frac{21}{15}$.

Since $7 < 21$, $\frac{7}{15} < \frac{21}{15}$, so $\frac{7}{15} < \frac{7}{5}$.

In this case (multiplying a proper and an improper fraction), the product is greater than the proper fraction and less than the improper fraction.

Case (iii): $\frac{4}{3} \times \frac{5}{2} = \frac{10}{3}$

The fractions are $\frac{4}{3}$ and $\frac{5}{2}$. Both are improper fractions (value $>1$).

$\frac{4}{3} \approx 1.33$, $\frac{5}{2} = 2.5$.

The product is $\frac{10}{3} \approx 3.33$ (improper, value $>1$).

Compare the product with the first fraction: $\frac{10}{3}$ versus $\frac{4}{3}$.

Since $10 > 4$, $\frac{10}{3} > \frac{4}{3}$.

Compare the product with the second fraction: $\frac{10}{3}$ versus $\frac{5}{2}$.

To compare, find a common denominator (LCM of 3 and 2 is 6).

$\frac{10}{3} = \frac{10 \times 2}{3 \times 2} = \frac{20}{6}$.

$\frac{5}{2} = \frac{5 \times 3}{2 \times 3} = \frac{15}{6}$.

Since $20 > 15$, $\frac{20}{6} > \frac{15}{6}$, so $\frac{10}{3} > \frac{5}{2}$.

In this case (multiplying two improper fractions), the product is greater than either fraction.

Conclusion for (ii):

Based on the observations from the examples:

• When multiplying two proper fractions, the product is less than either fraction.
• When multiplying a proper fraction and an improper fraction, the product is greater than the proper fraction but less than the improper fraction.
• When multiplying two improper fractions, the product is greater than either fraction.

Thus, the value of the fraction in the product can be either greater than or less than the value of one or both of the original fractions, depending on whether the fractions being multiplied are proper or improper.

Answer:

(i) Interchanging the fractions does not affect the answer, due to the commutative property of multiplication.

(ii) The value of the fraction in the product is not always greater or always less than either fraction. It depends on whether the fractions being multiplied are proper or improper. As seen in the examples, the product can be less than both fractions, between the two fractions, or greater than both fractions.

Given:

Length of cloth needed to stitch one shirt = $\frac{3}{4}$ m.

Total length of cloth Reshma has = $2\frac{1}{4}$ m.

To Find:

The number of shirts Reshma can make with the total cloth.

Solution:

To find the number of shirts that can be made, we need to divide the total length of cloth by the length of cloth required for one shirt.

First, convert the total length of cloth from a mixed number to an improper fraction:

$2\frac{1}{4} = \frac{(2 \times 4) + 1}{4} = \frac{8 + 1}{4} = \frac{9}{4}$ m.

Number of shirts = $\frac{\text{Total length of cloth}}{\text{Length of cloth per shirt}}$

Number of shirts = $\frac{9}{4} \text{ m} \div \frac{3}{4} \text{ m}$

Number of shirts = $\frac{9}{4} \div \frac{3}{4}$

To divide by a fraction, we multiply by its reciprocal. The reciprocal of $\frac{3}{4}$ is $\frac{4}{3}$.

Number of shirts = $\frac{9}{4} \times \frac{4}{3}$

Multiply the numerators and the denominators:

Number of shirts = $\frac{9 \times 4}{4 \times 3}$

We can cancel out the common factors:

Number of shirts = $\frac{\cancel{9}^{3} \times \cancel{4}^{1}}{\cancel{4}_{1} \times \cancel{3}_{1}}$

Number of shirts = $\frac{3 \times 1}{1 \times 1}$

Number of shirts = 3

So, Reshma can make 3 shirts with $2\frac{1}{4}$ m of cloth.

Answer:

Reshma can make 3 shirts with $2\frac{1}{4}$ m cloth.

Given:

From the provided graphic:

Frequency of note D = 297 Hz

Frequency of note B = 495 Hz

To Find:

The fraction of the frequency of note D to the frequency of note B.

Solution:

The fraction of the frequency of note D to note B is given by:

Fraction = $\frac{\text{Frequency of D}}{\text{Frequency of B}}$

Substitute the given frequency values:

Fraction = $\frac{297}{495}$

Now, we need to simplify this fraction by finding the greatest common divisor (GCD) of the numerator (297) and the denominator (495) and dividing both by it.

Let's find the prime factorization of 297 and 495.

For 297:

So, $297 = 3 \times 3 \times 3 \times 11 = 3^3 \times 11$.

For 495:

So, $495 = 3 \times 3 \times 5 \times 11 = 3^2 \times 5 \times 11$.

The common prime factors are $3^2$ and 11.

GCD(297, 495) = $3^2 \times 11 = 9 \times 11 = 99$.

Now, divide the numerator and the denominator of the fraction by their GCD, 99:

Fraction = $\frac{297 \div 99}{495 \div 99}$

Fraction = $\frac{3}{5}$

The simplified fraction of the frequency of note D to note B is $\frac{3}{5}$.

Since the question mentions that if the fraction has a common factor, the notes are harmonious, let's check our finding. The original fraction $\frac{297}{495}$ has a common factor of 99 (or even 3, 9, 11, 33). Thus, according to the provided information, notes D and B are harmonious.

Answer:

The fraction of frequency of notes D and B is $\frac{3}{5}$.

Given:

Distance covered = 120 km.

The distance covered represents $\frac{2}{3}$ of the total distance to the campsite.

To Find:

The distance remaining to the campsite.

Solution:

Let the total distance to the campsite be $D$ km.

According to the problem, the distance covered (120 km) is equal to $\frac{2}{3}$ of the total distance $D$.

We can write this relationship as an equation:

Distance covered = $\frac{2}{3} \times \text{Total distance}$

So, $120 = \frac{2}{3} \times D$

To find the total distance $D$, we need to isolate $D$. We can do this by multiplying both sides of the equation by the reciprocal of $\frac{2}{3}$, which is $\frac{3}{2}$.

$120 \times \frac{3}{2} = \left( \frac{2}{3} \times D \right) \times \frac{3}{2}$

$120 \times \frac{3}{2} = D \times \left( \frac{2}{3} \times \frac{3}{2} \right)$

$120 \times \frac{3}{2} = D \times 1$

$D = \frac{120 \times 3}{2}$

$D = \frac{360}{2}$

$D = 180$

The total distance to the campsite is 180 km.

The distance remaining is the difference between the total distance and the distance already covered.

Distance remaining = Total distance - Distance covered

Distance remaining = $180 \text{ km} - 120 \text{ km}$

Distance remaining = $60 \text{ km}$

Alternatively, if $\frac{2}{3}$ of the way is covered, the remaining part of the way is $1 - \frac{2}{3}$.

Remaining fraction = $1 - \frac{2}{3} = \frac{3}{3} - \frac{2}{3} = \frac{3-2}{3} = \frac{1}{3}$.

So, the distance remaining is $\frac{1}{3}$ of the total distance.

We know that $\frac{2}{3}$ of the distance is 120 km.

If $\frac{2}{3}D = 120$, then $\frac{1}{3}D$ would be half of that amount (since $\frac{1}{3}$ is half of $\frac{2}{3}$).

$\frac{1}{3}D = \frac{1}{2} \times \left( \frac{2}{3}D \right) = \frac{1}{2} \times 120 \text{ km} = 60 \text{ km}$.

The distance remaining is 60 km.

Answer:

They have to go 60 km farther.

Solution:

We need to find the product of the fraction $\frac{2}{5}$ and the mixed number $5\frac{1}{5}$.

First, convert the mixed number to an improper fraction.

$5\frac{1}{5} = \frac{(5 \times 5) + 1}{5} = \frac{25 + 1}{5} = \frac{26}{5}$

Now, multiply the two fractions:

Product = $\frac{2}{5} \times \frac{26}{5}$

To multiply fractions, we multiply the numerators together and the denominators together:

Product = $\frac{2 \times 26}{5 \times 5}$

Product = $\frac{52}{25}$

The fraction $\frac{52}{25}$ is an improper fraction. It cannot be simplified further since 52 and 25 do not share any common factors other than 1.

Comparing our result with the given options:

(a) $\frac{26}{25}$

(b) $\frac{52}{25}$

(c) $\frac{2}{5}$

(d) 6

The calculated product $\frac{52}{25}$ matches option (b).

Answer:

The correct option is (b).

Solution:

We need to calculate the value of $3\frac{3}{4} \;÷ \; \frac{3}{4}$.

First, convert the mixed number $3\frac{3}{4}$ into an improper fraction:

$3\frac{3}{4} = \frac{(3 \times 4) + 3}{4} = \frac{12 + 3}{4} = \frac{15}{4}$

Now the expression is $\frac{15}{4} \;÷ \; \frac{3}{4}$.

To divide a fraction by another fraction, we multiply the first fraction by the reciprocal of the second fraction.

The reciprocal of $\frac{3}{4}$ is $\frac{4}{3}$.

So, $\frac{15}{4} \;÷ \; \frac{3}{4} = \frac{15}{4} \times \frac{4}{3}$

Multiply the numerators and the denominators:

= $\frac{15 \times 4}{4 \times 3}$

We can cancel out the common factors:

= $\frac{\cancel{15}^{5} \times \cancel{4}^{1}}{\cancel{4}_{1} \times \cancel{3}_{1}}$

= $\frac{5 \times 1}{1 \times 1}$

= $\frac{5}{1}$

= 5

The value of the expression is 5.

Comparing our result with the given options:

(a) 3

(b) 4

(c) 5

(d) $\frac{45}{16}$

The calculated value 5 matches option (c).

Answer:

The correct option is (c).

Solution:

Given:

Total length of the ribbon = $5\frac{1}{4}$ m.

Length of each small piece = $\frac{3}{4}$ m.

To Find:

The number of small pieces that can be cut from the ribbon.

Solution:

To find the number of pieces, we need to divide the total length of the ribbon by the length of each small piece.

First, convert the total length from a mixed number to an improper fraction:

$5\frac{1}{4} = \frac{(5 \times 4) + 1}{4} = \frac{20 + 1}{4} = \frac{21}{4}$ m.

Number of pieces = $\frac{\text{Total length of ribbon}}{\text{Length of each piece}}$

Number of pieces = $\frac{21}{4} \text{ m} \div \frac{3}{4} \text{ m}$

Number of pieces = $\frac{21}{4} \div \frac{3}{4}$

To divide a fraction by another fraction, we multiply the first fraction by the reciprocal of the second fraction.

The reciprocal of $\frac{3}{4}$ is $\frac{4}{3}$.

Number of pieces = $\frac{21}{4} \times \frac{4}{3}$

Multiply the numerators and the denominators:

Number of pieces = $\frac{21 \times 4}{4 \times 3}$

We can cancel out the common factors:

Number of pieces = $\frac{\cancel{21}^{7} \times \cancel{4}^{1}}{\cancel{4}_{1} \times \cancel{3}_{1}}$

Number of pieces = $\frac{7 \times 1}{1 \times 1}$

Number of pieces = 7

So, 7 pieces of length $\frac{3}{4}$ m can be cut from a ribbon of length $5\frac{1}{4}$ m.

Comparing our result with the given options:

(a) 5

(b) 6

(c) 7

(d) 8

The calculated number of pieces 7 matches option (c).

Answer:

The correct option is (c).

Solution:

We need to arrange the fractions $\frac{2}{3}$, $\frac{6}{7}$, and $\frac{13}{21}$ in ascending order.

To compare fractions, we need to express them with a common denominator.

The denominators are 3, 7, and 21.

We find the Least Common Multiple (LCM) of the denominators 3, 7, and 21.

Since $21 = 3 \times 7$, the LCM of 3, 7, and 21 is 21.

Now, we convert each fraction to an equivalent fraction with a denominator of 21.

For $\frac{2}{3}$: Multiply the numerator and denominator by $\frac{21}{3} = 7$.

$\frac{2}{3} = \frac{2 \times 7}{3 \times 7} = \frac{14}{21}$

For $\frac{6}{7}$: Multiply the numerator and denominator by $\frac{21}{7} = 3$.

$\frac{6}{7} = \frac{6 \times 3}{7 \times 3} = \frac{18}{21}$

For $\frac{13}{21}$: The denominator is already 21.

$\frac{13}{21} = \frac{13}{21}$

Now we compare the equivalent fractions: $\frac{14}{21}$, $\frac{18}{21}$, and $\frac{13}{21}$.

Since the denominators are the same, we compare the numerators: 14, 18, and 13.

In ascending order (smallest to largest), the numerators are 13, 14, 18.

Therefore, the fractions in ascending order are:

$\frac{13}{21}$ (corresponding to the original fraction $\frac{13}{21}$)

$\frac{14}{21}$ (corresponding to the original fraction $\frac{2}{3}$)

$\frac{18}{21}$ (corresponding to the original fraction $\frac{6}{7}$)

So, the ascending arrangement of the given fractions is $\frac{13}{21}$, $\frac{2}{3}$, $\frac{6}{7}$.

Comparing this order with the given options:

(a) $\frac{6}{7}$ , $\frac{2}{3}$ , $\frac{13}{21}$

(b) $\frac{13}{21}$ , $\frac{2}{3}$ , $\frac{6}{7}$

(c) $\frac{6}{7}$ , $\frac{13}{21}$ , $\frac{2}{3}$

(d) $\frac{2}{3}$ , $\frac{6}{7}$ , $\frac{13}{21}$

Option (b) matches our ascending arrangement.

Answer:

The correct option is (b).

Solution:

The reciprocal of a fraction is found by interchanging its numerator and its denominator.

If a fraction is $\frac{a}{b}$, its reciprocal is $\frac{b}{a}$, provided that $a \neq 0$ and $b \neq 0$.

The given fraction is $\frac{2}{3}$.

The numerator is 2 and the denominator is 3.

To find the reciprocal, we swap the numerator and the denominator.

The reciprocal of $\frac{2}{3}$ is $\frac{3}{2}$.

We can check this by multiplying the fraction by its reciprocal:

$\frac{2}{3} \times \frac{3}{2} = \frac{2 \times 3}{3 \times 2} = \frac{6}{6} = 1$.

Since their product is 1, $\frac{3}{2}$ is indeed the reciprocal of $\frac{2}{3}$.

Comparing our result with the given options:

(a) 2

(b) 3

(c) $\frac{2}{3}$

(d) $\frac{3}{2}$

The calculated reciprocal $\frac{3}{2}$ matches option (d).

Answer:

The correct option is (d).

Solution:

We need to find the product of the fraction $\frac{11}{13}$ and the whole number 4.

To multiply a fraction by a whole number, we can write the whole number as a fraction with a denominator of 1 and then multiply the fractions.

$4 = \frac{4}{1}$

Now, multiply the two fractions:

Product = $\frac{11}{13} \times \frac{4}{1}$

Multiply the numerators together and the denominators together:

Product = $\frac{11 \times 4}{13 \times 1}$

Product = $\frac{44}{13}$

The resulting fraction $\frac{44}{13}$ is an improper fraction, as the numerator (44) is greater than the denominator (13). The options are given in the form of mixed numbers.

To convert the improper fraction to a mixed number, we divide the numerator by the denominator.

Divide 44 by 13:

$44 \div 13 = 3$ with a remainder.

The remainder is $44 - (13 \times 3) = 44 - 39 = 5$.

The mixed number form is the quotient followed by the remainder over the original denominator.

$\frac{44}{13} = 3\frac{5}{13}$

The product of $\frac{11}{13}$ and 4 is $3\frac{5}{13}$.

Comparing our result with the given options:

(a) $3\frac{5}{13}$

(b) $5\frac{3}{13}$

(c) $13\frac{3}{5}$

(d) $13\frac{5}{3}$

The calculated mixed number $3\frac{5}{13}$ matches option (a).

Answer:

The correct option is (a).

Solution:

We need to find the product of the whole number 3 and the mixed number $4\frac{2}{5}$.

First, convert the mixed number $4\frac{2}{5}$ into an improper fraction:

$4\frac{2}{5} = \frac{(4 \times 5) + 2}{5} = \frac{20 + 2}{5} = \frac{22}{5}$

Now, we multiply the whole number 3 by the improper fraction $\frac{22}{5}$.

Product = $3 \times \frac{22}{5}$

We can write the whole number 3 as a fraction $\frac{3}{1}$:

Product = $\frac{3}{1} \times \frac{22}{5}$

To multiply fractions, multiply the numerators together and the denominators together:

Product = $\frac{3 \times 22}{1 \times 5}$

Product = $\frac{66}{5}$

The result is an improper fraction $\frac{66}{5}$. We should check if the options are in improper fraction or mixed number form. The options are in mixed number form or improper fraction form.

Let's convert our improper fraction $\frac{66}{5}$ to a mixed number by dividing the numerator by the denominator.

$66 \div 5$

$66 = (5 \times 13) + 1$

The quotient is 13 and the remainder is 1.

So, $\frac{66}{5} = 13\frac{1}{5}$.

Comparing our result with the given options:

(a) $17\frac{2}{5}$

(b) $\frac{24}{5}$

(c) $13\frac{1}{5}$

(d) $5\frac{1}{3}$

The calculated mixed number $13\frac{1}{5}$ matches option (c).

Answer:

The correct option is (c).

Solution:

We need to find the pictorial representation of $3 \times \frac{2}{3}$.

The expression $3 \times \frac{2}{3}$ means that the fraction $\frac{2}{3}$ is taken 3 times.

The fraction $\frac{2}{3}$ represents a whole quantity divided into 3 equal parts, and we are considering 2 of those parts.

Pictorially, this can be represented by a figure (like a circle, square, or rectangle) divided into 3 equal sections, with 2 of those sections shaded.

The multiplication by 3 ($3 \times \frac{2}{3}$) means we need to show this representation of $\frac{2}{3}$ three times.

So, the pictorial representation should consist of 3 identical figures, where each figure is divided into 3 equal parts and 2 parts are shaded.

Let's examine the typical structure of the options for such problems.

Option (a) usually shows the representation of the operation $3 \times \frac{2}{3}$ by depicting 3 figures, each having $\frac{2}{3}$ shaded.

The total shaded area in such a representation would be the sum of the shaded parts in each figure:

Total shaded parts = $\frac{2}{3} + \frac{2}{3} + \frac{2}{3} = 3 \times \frac{2}{3} = \frac{6}{3} = 2$ whole units.

So, the pictorial representation in the correct option should show three figures, each with 2 out of 3 parts shaded.

Based on standard textbook representations for this type of problem, option (a) is the pictorial representation that shows 3 units, each divided into 3 parts with 2 parts shaded.

Answer:

The correct pictorial representation of $3 \times \frac{2}{3}$ is the one showing 3 figures, each with 2 out of 3 equal parts shaded. This corresponds to option (a).

Solution:

We need to calculate the value of $\frac{1}{5} \;÷ \; \frac{4}{5}$.

To divide a fraction by another fraction, we multiply the first fraction by the reciprocal of the second fraction.

The reciprocal of the fraction $\frac{4}{5}$ is obtained by interchanging the numerator and denominator, which is $\frac{5}{4}$.

So, the division can be rewritten as:

$\frac{1}{5} \div \frac{4}{5} = \frac{1}{5} \times \frac{5}{4}$

Now, multiply the numerators together and the denominators together:

= $\frac{1 \times 5}{5 \times 4}$

Before multiplying, we can cancel out the common factor of 5 in the numerator and the denominator:

= $\frac{1 \times \cancel{5}^{1}}{\cancel{5}_{1} \times 4}$

= $\frac{1 \times 1}{1 \times 4}$

= $\frac{1}{4}$

The value of the expression is $\frac{1}{4}$.

Comparing our result with the given options:

(a) $\frac{4}{5}$

(b) $\frac{1}{5}$

(c) $\frac{5}{4}$

(d) $\frac{1}{4}$

The calculated value $\frac{1}{4}$ matches option (d).

Answer:

The correct option is (d).

Solution:

We need to find the product of the two decimal numbers 0.03 and 0.9.

To multiply decimal numbers, we first multiply them as if they were whole numbers, ignoring the decimal points for a moment.

Multiply 3 by 9:

$3 \times 9 = 27$

Next, we determine the position of the decimal point in the final product.

Count the total number of decimal places in the numbers being multiplied:

The number 0.03 has two decimal places (the digits 0 and 3 after the decimal point).

The number 0.9 has one decimal place (the digit 9 after the decimal point).

Total number of decimal places in the product = (Decimal places in 0.03) + (Decimal places in 0.9)

Total number of decimal places = $2 + 1 = 3$.

Starting from the rightmost digit of the product of the whole numbers (27), we move the decimal point 3 places to the left.

The whole number 27 has an implied decimal point after the 7 (27.0). We move the decimal 3 places left:

27.0 $\rightarrow$ 2.70 $\rightarrow$ 0.270 $\rightarrow$ 0.027

We add a zero as a placeholder in the empty position to the left of the digits.

So, the product is 0.027.

Therefore, $0.03 \times 0.9 = 0.027$.

Comparing our result with the given options:

(a) 2.7

(b) 0.27

(c) 0.027

(d) 0.0027

The calculated product 0.027 matches option (c).

Answer:

The correct option is (c).

Solution:

We need to calculate the value of $\frac{5}{7} \;÷ \; 6$.

Dividing by a whole number is the same as multiplying by its reciprocal.

The whole number is 6. We can write 6 as a fraction $\frac{6}{1}$.

The reciprocal of $\frac{6}{1}$ is $\frac{1}{6}$.

So, $\frac{5}{7} \div 6 = \frac{5}{7} \times \frac{1}{6}$.

To multiply fractions, multiply the numerators together and the denominators together:

= $\frac{5 \times 1}{7 \times 6}$

= $\frac{5}{42}$

The value of the expression is $\frac{5}{42}$. This fraction is in its simplest form because 5 and 42 have no common factors other than 1.

Comparing our result with the given options:

(a) $\frac{30}{7}$

(b) $\frac{5}{42}$

(c) $\frac{30}{42}$ (This simplifies to $\frac{5 \times 6}{7 \times 6} = \frac{5}{7}$, which is incorrect)

(d) $\frac{6}{7}$

The calculated value $\frac{5}{42}$ matches option (b).

Answer:

The correct option is (b).

Solution:

We need to calculate the value of $5\frac{1}{6} \;÷ \; \frac{9}{2}$.

First, convert the mixed number $5\frac{1}{6}$ into an improper fraction.

$5\frac{1}{6} = \frac{(5 \times 6) + 1}{6} = \frac{30 + 1}{6} = \frac{31}{6}$

Now, the expression becomes $\frac{31}{6} \;÷ \; \frac{9}{2}$.

To divide a fraction by another fraction, we multiply the first fraction by the reciprocal of the second fraction.

The reciprocal of $\frac{9}{2}$ is $\frac{2}{9}$.

So, $\frac{31}{6} \;÷ \; \frac{9}{2} = \frac{31}{6} \times \frac{2}{9}$

Multiply the numerators and the denominators:

= $\frac{31 \times 2}{6 \times 9}$

We can simplify by cancelling the common factor 2 from the numerator and denominator:

= $\frac{31 \times \cancel{2}^{1}}{\cancel{6}_{3} \times 9}$

= $\frac{31 \times 1}{3 \times 9}$

= $\frac{31}{27}$

The value of the expression is $\frac{31}{27}$. This is an improper fraction, but the options include improper fractions.

Comparing our result with the given options:

(a) $\frac{31}{6}$

(b) $\frac{1}{27}$

(c) $5\frac{1}{27}$ (which is $\frac{136}{27}$ as an improper fraction)

(d) $\frac{31}{27}$

The calculated value $\frac{31}{27}$ matches option (d).

Answer:

The correct option is (d).

Solution:

In mathematics, the word "of" between two numbers or quantities, especially when one of them is a fraction, indicates the operation of multiplication.

The phrase "$\frac{1}{3}$ of $\frac{1}{6}$" means taking a part of the quantity $\frac{1}{6}$. This is equivalent to multiplying $\frac{1}{3}$ by $\frac{1}{6}$.

Mathematically, this is represented as:

$\frac{1}{3} \times \frac{1}{6}$

Let's compare this expression with the given options:

(a) $\frac{1}{3}$ + $\frac{1}{6}$ (Addition)

(b) $\frac{1}{3}$ - $\frac{1}{6}$ (Subtraction)

(c) $\frac{1}{3}$ × $\frac{1}{6}$ (Multiplication)

(d) $\frac{1}{3}$ ÷ $\frac{1}{6}$ (Division)

Option (c) matches the mathematical representation of "$\frac{1}{3}$ of $\frac{1}{6}$".

The actual value of this product is $\frac{1 \times 1}{3 \times 6} = \frac{1}{18}$.

Answer:

The correct option representing $\frac{1}{3}$ of $\frac{1}{6}$ is (c).

Solution:

The phrase "$\frac{3}{7}$ of $\frac{2}{5}$" means we need to find the product of the two fractions $\frac{3}{7}$ and $\frac{2}{5}$.

"of" indicates multiplication.

So, we need to calculate $\frac{3}{7} \times \frac{2}{5}$.

To multiply fractions, we multiply the numerators together and the denominators together.

Product = $\frac{\text{Numerator 1} \times \text{Numerator 2}}{\text{Denominator 1} \times \text{Denominator 2}}$

Product = $\frac{3 \times 2}{7 \times 5}$

Product = $\frac{6}{35}$

The fraction $\frac{6}{35}$ is in its simplest form, as 6 and 35 do not have any common prime factors (prime factors of 6 are 2 and 3; prime factors of 35 are 5 and 7).

Comparing our result with the given options:

(a) $\frac{5}{12}$

(b) $\frac{5}{35}$

(c) $\frac{1}{35}$

(d) $\frac{6}{35}$

The calculated value $\frac{6}{35}$ matches option (d).

Answer:

The correct option is (d).

Solution:

Given:

Flour required for one packet = $2\frac{1}{2}$ cups.

Sugar required for one packet = $1\frac{2}{3}$ cups.

Number of packets = 10.

To Find:

Estimated total quantity of both ingredients used in 10 packets, choosing the correct range from the options.

Solution:

First, calculate the total quantity of both ingredients required for one packet.

Total ingredients per packet = Quantity of flour + Quantity of sugar

= $2\frac{1}{2} + 1\frac{2}{3}$ cups

Convert the mixed numbers to improper fractions:

$2\frac{1}{2} = \frac{(2 \times 2) + 1}{2} = \frac{4 + 1}{2} = \frac{5}{2}$ cups.

$1\frac{2}{3} = \frac{(1 \times 3) + 2}{3} = \frac{3 + 2}{3} = \frac{5}{3}$ cups.

Now, add the improper fractions:

Total ingredients per packet = $\frac{5}{2} + \frac{5}{3}$ cups.

To add fractions, find a common denominator. The LCM of 2 and 3 is 6.

$\frac{5}{2} = \frac{5 \times 3}{2 \times 3} = \frac{15}{6}$

$\frac{5}{3} = \frac{5 \times 2}{3 \times 2} = \frac{10}{6}$

Total ingredients per packet = $\frac{15}{6} + \frac{10}{6} = \frac{15 + 10}{6} = \frac{25}{6}$ cups.

Next, calculate the total quantity of ingredients needed for 10 packets. Multiply the quantity per packet by the number of packets.

Total ingredients for 10 packets = $10 \times \frac{25}{6}$ cups.

= $\frac{10 \times 25}{6}$ cups

= $\frac{250}{6}$ cups

Simplify the fraction $\frac{250}{6}$ by dividing both the numerator and the denominator by 2.

= $\frac{\cancel{250}^{125}}{\cancel{6}_{3}}$ cups

= $\frac{125}{3}$ cups.

Now, we estimate the value of $\frac{125}{3}$. Convert the improper fraction to a mixed number or decimal.

Divide 125 by 3:

$125 \div 3 = 41$ with a remainder of $125 - (3 \times 41) = 125 - 123 = 2$.

So, $\frac{125}{3} = 41\frac{2}{3}$ cups.

The total quantity is $41\frac{2}{3}$ cups. We need to estimate this value in relation to the given ranges.

$41\frac{2}{3}$ is greater than 41 but less than 42.

Let's check the options:

(a) less than 30 cups (False, $41\frac{2}{3}$ is not less than 30)

(b) between 30 cups and 40 cups (False, $41\frac{2}{3}$ is not between 30 and 40)

(c) between 40 cups and 50 cups (True, $41\frac{2}{3}$ is between 40 and 50)

(d) above 50 cups (False, $41\frac{2}{3}$ is not above 50)

The estimated total quantity is between 40 cups and 50 cups.

Answer:

The correct option is (c).

Solution:

We need to find the product of the whole number 7 and the mixed number $6\frac{3}{4}$.

First, convert the mixed number $6\frac{3}{4}$ into an improper fraction:

$6\frac{3}{4} = \frac{(6 \times 4) + 3}{4} = \frac{24 + 3}{4} = \frac{27}{4}$

Now, we multiply the whole number 7 by the improper fraction $\frac{27}{4}$.

Product = $7 \times \frac{27}{4}$

We can write the whole number 7 as a fraction $\frac{7}{1}$:

Product = $\frac{7}{1} \times \frac{27}{4}$

To multiply fractions, multiply the numerators together and the denominators together:

Product = $\frac{7 \times 27}{1 \times 4}$

Product = $\frac{189}{4}$

The result is an improper fraction $\frac{189}{4}$. The options are given in mixed number form.

To convert the improper fraction $\frac{189}{4}$ to a mixed number, we divide the numerator (189) by the denominator (4).

$189 \div 4$

Divide 18 by 4: $18 = 4 \times 4 + 2$. The quotient is 4, remainder is 2.

Bring down 9: 29. Divide 29 by 4: $29 = 4 \times 7 + 1$. The quotient is 7, remainder is 1.

So, 189 divided by 4 gives a quotient of 47 and a remainder of 1.

The mixed number form is $47\frac{1}{4}$.

The product of 7 and $6\frac{3}{4}$ is $47\frac{1}{4}$.

Comparing our result with the given options:

(a) $42\frac{1}{4}$

(b) $47\frac{1}{4}$

(c) $42\frac{3}{4}$

(d) $47\frac{3}{4}$

The calculated mixed number $47\frac{1}{4}$ matches option (b).

Answer:

The correct option is (b).

Solution:

We need to find the result of dividing the whole number 7 by the fraction $\frac{2}{5}$.

The division is $7 \div \frac{2}{5}$.

To divide a number by a fraction, we multiply the number by the reciprocal of the fraction.

The reciprocal of $\frac{2}{5}$ is $\frac{5}{2}$.

So, $7 \div \frac{2}{5} = 7 \times \frac{5}{2}$.

We can write the whole number 7 as a fraction $\frac{7}{1}$.

= $\frac{7}{1} \times \frac{5}{2}$

Multiply the numerators together and the denominators together:

= $\frac{7 \times 5}{1 \times 2}$

= $\frac{35}{2}$

The result of the division is $\frac{35}{2}$. This is an improper fraction. The options are given in fractional form.

Comparing our result with the given options:

(a) $\frac{14}{2}$ (This simplifies to 7)

(b) $\frac{35}{4}$

(c) $\frac{14}{5}$

(d) $\frac{35}{2}$

The calculated result $\frac{35}{2}$ matches option (d).

Answer:

The correct option is (d).

Solution:

We need to calculate the value of $2\frac{2}{3} \;÷ \; 5$.

First, convert the mixed number $2\frac{2}{3}$ into an improper fraction.

$2\frac{2}{3} = \frac{(2 \times 3) + 2}{3} = \frac{6 + 2}{3} = \frac{8}{3}$

Now, the expression becomes $\frac{8}{3} \;÷ \; 5$.

To divide a fraction by a whole number, we multiply the fraction by the reciprocal of the whole number.

The whole number is 5. We can write 5 as a fraction $\frac{5}{1}$.

The reciprocal of $\frac{5}{1}$ is $\frac{1}{5}$.

So, $\frac{8}{3} \;÷ \; 5 = \frac{8}{3} \times \frac{1}{5}$.

Multiply the numerators together and the denominators together:

= $\frac{8 \times 1}{3 \times 5}$

= $\frac{8}{15}$

The value of the expression is $\frac{8}{15}$. This fraction is in its simplest form because 8 and 15 have no common factors other than 1.

Comparing our result with the given options:

(a) $\frac{8}{15}$

(b) $\frac{40}{3}$

(c) $\frac{40}{5}$ (This simplifies to 8)

(d) $\frac{8}{3}$

The calculated value $\frac{8}{15}$ matches option (a).

Answer:

The correct option is (a).

Solution:

Given:

Total weight of apples = 5 kg.

Fraction of apples used on Monday = $\frac{4}{5}$.

Fraction of the remaining apples used on Tuesday = $\frac{1}{3}$.

To Find:

The weight (in kg) of apples left after Tuesday.

Solution:

First, calculate the weight of apples used on Monday.

Weight of apples used on Monday = $\frac{4}{5}$ of 5 kg

= $\frac{4}{5} \times 5$ kg

= $\frac{4 \times 5}{5}$ kg

= $\frac{20}{5}$ kg

= 4 kg.

Next, calculate the weight of apples left after Monday.

Apples left after Monday = Total weight of apples - Apples used on Monday

= $5 \text{ kg} - 4 \text{ kg}$

= 1 kg.

On Tuesday, $\frac{1}{3}$ of what was left was used. The amount left was 1 kg.

Weight of apples used on Tuesday = $\frac{1}{3}$ of 1 kg

= $\frac{1}{3} \times 1$ kg

= $\frac{1}{3}$ kg.

Finally, calculate the weight of apples left after Tuesday.

Apples left now = Apples left after Monday - Apples used on Tuesday

= $1 \text{ kg} - \frac{1}{3} \text{ kg}$

To subtract the fraction, write 1 as a fraction with denominator 3.

$1 = \frac{3}{3}$

Apples left now = $\frac{3}{3} - \frac{1}{3}$ kg

= $\frac{3-1}{3}$ kg

= $\frac{2}{3}$ kg.

The weight of apples left now is $\frac{2}{3}$ kg.

Comparing our result with the given options:

(a) $\frac{2}{7}$

(b) $\frac{1}{14}$

(c) $\frac{2}{3}$

(d) $\frac{4}{21}$

The calculated weight $\frac{2}{3}$ kg matches option (c).

Answer:

The correct option is (c).

Solution:

The given picture shows three identical figures (circles).

Each figure is divided into 4 equal parts.

In each figure, 1 out of the 4 equal parts is shaded.

Each individual figure therefore represents the fraction $\frac{1}{4}$.

The picture shows 3 such figures.

This represents taking the quantity $\frac{1}{4}$ three times.

Taking a quantity a certain number of times is represented mathematically by multiplication.

So, taking $\frac{1}{4}$ three times can be written as $3 \times \frac{1}{4}$.

Let's examine the given options:

(a) $\frac{1}{4}$ ÷ 3: This represents dividing the fraction $\frac{1}{4}$ into 3 equal parts.

(b) 3 × $\frac{1}{4}$: This represents taking the fraction $\frac{1}{4}$ three times.

(c) $\frac{3}{4}$ × 3: This represents taking the fraction $\frac{3}{4}$ three times.

(d) 3 ÷ $\frac{1}{4}$: This represents how many times the fraction $\frac{1}{4}$ is contained in the number 3.

The pictorial representation directly corresponds to taking the fraction $\frac{1}{4}$ three times, which is represented by the expression $3 \times \frac{1}{4}$.

The correct option is (b).

Answer:

The correct option is (b).

Solution:

Let the total cake be represented by 1 whole.

Part of the cake eaten by Rani = $\frac{2}{7}$

Part of the cake remaining after Rani ate:

Remaining part = Total cake - Part eaten by Rani

Remaining part = $1 - \frac{2}{7}$

Remaining part = $\frac{7}{7} - \frac{2}{7} = \frac{7-2}{7} = \frac{5}{7}$

Ravi ate $\frac{4}{5}$ of the remaining part.

Part eaten by Ravi = $\frac{4}{5} \times (\text{Remaining part})$

Part eaten by Ravi = $\frac{4}{5} \times \frac{5}{7}$

Part eaten by Ravi = $\frac{4 \times \cancel{5}}{\cancel{5} \times 7} = \frac{4}{7}$

The part of the cake left is the part that remained after Rani ate, minus the part that Ravi ate.

Part of cake left = (Remaining after Rani) - (Part eaten by Ravi)

Part of cake left = $\frac{5}{7} - \frac{4}{7} = \frac{5-4}{7} = \frac{1}{7}$

Alternate Solution:

Part of the cake remaining after Rani ate = $\frac{5}{7}$.

Ravi ate $\frac{4}{5}$ of this remaining part.

This means the part of the remaining cake that Ravi did NOT eat is $1 - \frac{4}{5}$.

Part of remaining cake not eaten by Ravi = $1 - \frac{4}{5} = \frac{5}{5} - \frac{4}{5} = \frac{1}{5}$

The part of the *whole* cake left is this fraction ($\frac{1}{5}$) of the remaining part ($\frac{5}{7}$).

Part of cake left = $\frac{1}{5} \times \frac{5}{7}$

Part of cake left = $\frac{1 \times \cancel{5}}{\cancel{5} \times 7} = \frac{1}{7}$

The part of the cake left is $\frac{1}{7}$.

So, the blank should be filled with $\frac{1}{7}$.

Solution:

The reciprocal of a non-zero fraction $\frac{a}{b}$ is obtained by swapping the numerator and the denominator, which results in $\frac{b}{a}$.

For the given fraction $\frac{3}{7}$, the numerator is 3 and the denominator is 7.

Swapping the numerator and the denominator, we get $\frac{7}{3}$.

The reciprocal of $\frac{3}{7}$ is $\frac{7}{3}$.

So, the blank should be filled with $\frac{7}{3}$.

Solution:

The phrase "$\frac{2}{3}$ of 27" means multiplying $\frac{2}{3}$ by 27.

We need to calculate $\frac{2}{3} \times 27$.

We can write 27 as $\frac{27}{1}$.

So, the calculation is $\frac{2}{3} \times \frac{27}{1}$.

Multiply the numerators and the denominators:

$\frac{2 \times 27}{3 \times 1} = \frac{54}{3}$

Now, simplify the fraction $\frac{54}{3}$ by dividing 54 by 3.

$\frac{54}{3} = 18$

Alternatively, we can simplify before multiplying:

$\frac{2}{\cancel{3}_{1}} \times \cancel{27}^{9} = 2 \times 9 = 18$

So, $\frac{2}{3}$ of 27 is 18.

The blank should be filled with 18.

Solution:

The phrase "$\frac{4}{5}$ of 45" means multiplying the fraction $\frac{4}{5}$ by the number 45.

We need to calculate the value of $\frac{4}{5} \times 45$.

We can write 45 as a fraction $\frac{45}{1}$.

So the expression becomes $\frac{4}{5} \times \frac{45}{1}$.

To multiply fractions, we multiply the numerators together and the denominators together:

$\frac{4 \times 45}{5 \times 1} = \frac{180}{5}$

Now, we simplify the fraction by dividing the numerator (180) by the denominator (5):

$\frac{180}{5} = 36$

Alternate Solution:

We can simplify before multiplying. Notice that 45 is divisible by 5.

$\frac{4}{\cancel{5}_{1}} \times \cancel{45}^{9}$

Now multiply the remaining numbers:

$4 \times 9 = 36$

Both methods give the same result. $\frac{4}{5}$ of 45 is 36.

The blank should be filled with 36.

Solution:

We are asked to find the value of the expression 4 × $6\frac{1}{3}$.

First, convert the mixed number $6\frac{1}{3}$ into an improper fraction.

$6\frac{1}{3} = 6 + \frac{1}{3}$

To add these, we find a common denominator, which is 3.

$6 = \frac{6 \times 3}{3} = \frac{18}{3}$

So, $6\frac{1}{3} = \frac{18}{3} + \frac{1}{3} = \frac{18+1}{3} = \frac{19}{3}$.

Now, we multiply 4 by the improper fraction $\frac{19}{3}$.

$4 \times 6\frac{1}{3} = 4 \times \frac{19}{3}$

To multiply a whole number by a fraction, we can write the whole number as a fraction with a denominator of 1:

$4 = \frac{4}{1}$

So, the multiplication becomes:

$\frac{4}{1} \times \frac{19}{3}$

Multiply the numerators together and the denominators together:

$\frac{4 \times 19}{1 \times 3} = \frac{76}{3}$

The result is the improper fraction $\frac{76}{3}$. If needed, we can convert this back to a mixed number by dividing 76 by 3.

$76 \div 3 = 25$ with a remainder of 1.

So, $\frac{76}{3} = 25\frac{1}{3}$.

The value of 4 × $6\frac{1}{3}$ is $\frac{76}{3}$ or $25\frac{1}{3}$.

The blank can be filled with either form, but $\frac{76}{3}$ is usually preferred unless a mixed number is specified.

So, the blank should be filled with $\frac{76}{3}$.

Solution:

We need to find the value of "$\frac{1}{2}$ of $4\frac{2}{7}$". The word "of" indicates multiplication.

First, convert the mixed number $4\frac{2}{7}$ into an improper fraction.

$4\frac{2}{7} = 4 + \frac{2}{7}$

To add 4 and $\frac{2}{7}$, express 4 as a fraction with denominator 7:

$4 = \frac{4 \times 7}{7} = \frac{28}{7}$

So, $4\frac{2}{7} = \frac{28}{7} + \frac{2}{7} = \frac{28+2}{7} = \frac{30}{7}$.

Now, multiply the fraction $\frac{1}{2}$ by the improper fraction $\frac{30}{7}$.

$\frac{1}{2} \times 4\frac{2}{7} = \frac{1}{2} \times \frac{30}{7}$

Multiply the numerators together and the denominators together:

$\frac{1 \times 30}{2 \times 7} = \frac{30}{14}$

Now, simplify the resulting fraction $\frac{30}{14}$ by dividing both the numerator and the denominator by their greatest common divisor, which is 2.

$\frac{\cancel{30}^{15}}{\cancel{14}_{7}} = \frac{15}{7}$

Alternate Solution:

Convert the mixed number $4\frac{2}{7}$ to $\frac{30}{7}$ as shown above.

$\frac{1}{2} \times \frac{30}{7}$

We can simplify before multiplying by dividing 30 in the numerator by 2 in the denominator.

$\frac{1}{\cancel{2}_{1}} \times \frac{\cancel{30}^{15}}{7} = \frac{1 \times 15}{1 \times 7} = \frac{15}{7}$

The result is the improper fraction $\frac{15}{7}$. If needed, we can convert this back to a mixed number by dividing 15 by 7.

$15 \div 7 = 2$ with a remainder of 1.

So, $\frac{15}{7} = 2\frac{1}{7}$.

The value of $\frac{1}{2}$ of $4\frac{2}{7}$ is $\frac{15}{7}$ or $2\frac{1}{7}$.

The blank can be filled with either form, but $\frac{15}{7}$ is usually preferred unless a mixed number is specified.

So, the blank should be filled with $\frac{15}{7}$.

Solution:

The phrase "$\frac{1}{9}$ of $\frac{6}{5}$" means we need to multiply the two fractions $\frac{1}{9}$ and $\frac{6}{5}$.

We need to calculate $\frac{1}{9} \times \frac{6}{5}$.

To multiply fractions, we multiply the numerators together and the denominators together.

Numerator: $1 \times 6 = 6$

Denominator: $9 \times 5 = 45$

So, the product is $\frac{6}{45}$.

Now, we simplify the fraction $\frac{6}{45}$ by dividing both the numerator and the denominator by their greatest common divisor, which is 3.

Divide the numerator by 3: $6 \div 3 = 2$

Divide the denominator by 3: $45 \div 3 = 15$

So, the simplified fraction is $\frac{2}{15}$.

Alternate Solution:

We can simplify before multiplying. Notice that the numerator 6 and the denominator 9 share a common factor of 3.

$\frac{1}{\cancel{9}_{3}} \times \frac{\cancel{6}^{2}}{5} = \frac{1 \times 2}{3 \times 5} = \frac{2}{15}$

Both methods yield the same result. $\frac{1}{9}$ of $\frac{6}{5}$ is $\frac{2}{15}$.

The blank should be filled with $\frac{2}{15}$.

Solution:

We are asked to find the product of $2\frac{3}{7}$ and $\frac{7}{9}$ and express it in its lowest form.

First, convert the mixed number $2\frac{3}{7}$ into an improper fraction.

$2\frac{3}{7} = 2 + \frac{3}{7}$

$2\frac{3}{7} = \frac{2 \times 7}{7} + \frac{3}{7} = \frac{14}{7} + \frac{3}{7} = \frac{14+3}{7} = \frac{17}{7}$

Now, multiply the improper fraction $\frac{17}{7}$ by the fraction $\frac{7}{9}$.

Product = $\frac{17}{7} \times \frac{7}{9}$

To multiply fractions, we multiply the numerators and the denominators.

Product = $\frac{17 \times 7}{7 \times 9} = \frac{119}{63}$

Now, we need to simplify the fraction $\frac{119}{63}$ to its lowest form.

We find the greatest common divisor (GCD) of 119 and 63.

Let's find the prime factors of 119 and 63.

Prime factors of 119: $119 = 7 \times 17$

Prime factors of 63: $63 = 3 \times 21 = 3 \times 3 \times 7 = 3^2 \times 7$

The common prime factor is 7.

GCD(119, 63) = 7.

Divide both the numerator and the denominator by the GCD (7).

$\frac{119 \div 7}{63 \div 7} = \frac{17}{9}$

Alternate Solution:

Convert the mixed number $2\frac{3}{7}$ to the improper fraction $\frac{17}{7}$.

The multiplication is $\frac{17}{7} \times \frac{7}{9}$.

We can cancel out the common factor of 7 in the numerator and the denominator before multiplying.

$\frac{17}{\cancel{7}_{1}} \times \frac{\cancel{7}^{1}}{9} = \frac{17 \times 1}{1 \times 9} = \frac{17}{9}$

The resulting fraction $\frac{17}{9}$ is in its lowest form because the numerator (17) and the denominator (9) have no common factors other than 1.

The lowest form of the product is $\frac{17}{9}$.

The blank should be filled with $\frac{17}{9}$.

Solution:

We need to find the value of the expression $\frac{4}{5} \div 4$.

Dividing by a number is the same as multiplying by its reciprocal.

The reciprocal of a whole number $n$ is $\frac{1}{n}$.

The reciprocal of 4 is $\frac{1}{4}$.

So, $\frac{4}{5} \div 4$ is equivalent to $\frac{4}{5} \times \frac{1}{4}$.

Now, we multiply the two fractions:

$\frac{4}{5} \times \frac{1}{4} = \frac{4 \times 1}{5 \times 4}$

$\frac{4}{20}$

Now, simplify the fraction $\frac{4}{20}$ to its lowest form by dividing both the numerator and the denominator by their greatest common divisor, which is 4.

$\frac{\cancel{4}^{1}}{\cancel{20}_{5}} = \frac{1}{5}$

Alternate Solution:

$\frac{4}{5} \div 4 = \frac{4}{5} \times \frac{1}{4}$

We can cancel the common factor of 4 in the numerator and the denominator before multiplying.

$\frac{\cancel{4}^{1}}{5} \times \frac{1}{\cancel{4}_{1}} = \frac{1 \times 1}{5 \times 1} = \frac{1}{5}$

Both methods give the same result. $\frac{4}{5} \div 4$ is equal to $\frac{1}{5}$.

The blank should be filled with $\frac{1}{5}$.

Solution:

The phrase "$\frac{2}{5}$ of 25" means we need to multiply the fraction $\frac{2}{5}$ by the number 25.

We need to calculate the value of $\frac{2}{5} \times 25$.

We can write 25 as a fraction $\frac{25}{1}$.

So the expression becomes $\frac{2}{5} \times \frac{25}{1}$.

To multiply fractions, we multiply the numerators together and the denominators together:

$\frac{2 \times 25}{5 \times 1} = \frac{50}{5}$

Now, we simplify the fraction by dividing the numerator (50) by the denominator (5):

$\frac{50}{5} = 10$

Alternate Solution:

We can simplify before multiplying. Notice that 25 is divisible by 5.

$\frac{2}{\cancel{5}_{1}} \times \cancel{25}^{5}$

Now multiply the remaining numbers:

$2 \times 5 = 10$

Both methods give the same result. $\frac{2}{5}$ of 25 is 10.

The blank should be filled with 10.

Solution:

We are asked to find the operation that relates $\frac{1}{5} \div \frac{5}{6}$ to $\frac{1}{5}$ and $\frac{6}{5}$.

When dividing a fraction by another fraction, we multiply the first fraction by the reciprocal of the second fraction.

The second fraction is $\frac{5}{6}$.

The reciprocal of $\frac{5}{6}$ is obtained by inverting the fraction, which is $\frac{6}{5}$.

According to the rule for division of fractions:

$\frac{1}{5} \div \frac{5}{6} = \frac{1}{5} \times (\text{reciprocal of } \frac{5}{6})$

$\frac{1}{5} \div \frac{5}{6} = \frac{1}{5} \times \frac{6}{5}$

Comparing this to the given expression $\frac{1}{5}$ ______ $\frac{6}{5}$, we see that the blank must be filled with the multiplication symbol ($\times$).

The blank should be filled with $\times$.

Solution:

We need to find the product of 3.2 and 10.

When multiplying a decimal number by 10, the decimal point is shifted one place to the right.

The given decimal number is 3.2.

The decimal point is currently between 3 and 2.

Shifting the decimal point one place to the right from its current position (between 3 and 2) moves it after the digit 2.

So, 3.2 becomes 32.

Thus, 3.2 × 10 = 32.

Alternatively, we can perform the multiplication directly:

$\begin{array}{cc} & 3.2 \\ \times & 10 \\ \hline & 00 \\ + & 320 \\ \hline & 32.0 \\ \hline \end{array}$

Since there is one digit after the decimal point in 3.2, the product will have one digit after the decimal point. The result is 32.0, which is equal to 32.

The result of 3.2 × 10 is 32.

The blank should be filled with 32.

Solution:

We need to find the product of 25.4 and 1000.

When multiplying a decimal number by a power of 10 (like 10, 100, 1000, etc.), we shift the decimal point to the right by the number of zeros in the power of 10.

In this case, we are multiplying by 1000, which has 3 zeros.

The given decimal number is 25.4.

We need to shift the decimal point 3 places to the right.

The decimal point is currently between 5 and 4.

Shifting one place right gives 254.

To shift two more places right, we add zeros after the digit 4.

Shifting two places right from 254 adds two zeros: 25400.

So, 25.4 × 1000 = 25400.

Alternate Solution:

We can also perform the multiplication directly using the method for multiplying decimals.

First, multiply the numbers ignoring the decimal point: $254 \times 1000 = 254000$.

Now, count the number of decimal places in the original decimal number (25.4). There is 1 decimal place.

The product should have the same number of decimal places. So, starting from the rightmost digit of 254000, move the decimal point 1 place to the left.

$25400\text{.}0$

This is equal to 25400.

$\begin{array}{cc}& & 25.4 \\ \times & 1000 \\ \hline && 00.0 \\ & 000\phantom{.0} \\ 25400\phantom{.00} \\ \hline 25400.0 \\ \hline \end{array}$

The result of 25.4 × 1000 is 25400.

The blank should be filled with 25400.

Solution:

We need to find the product of 93.5 and 100.

When multiplying a decimal number by a power of 10, such as 100, we shift the decimal point to the right by the number of zeros in the power of 10.

The number 100 has 2 zeros.

The given decimal number is 93.5.

The decimal point is located between 3 and 5.

We need to shift the decimal point 2 places to the right.

Shifting the decimal point one place to the right gives 935.

To shift it a second place to the right, we need to add a zero after the digit 5.

Shifting the decimal point two places right from 93.5 results in 9350.

So, 93.5 × 100 = 9350.

Alternate Solution:

Multiply the numbers ignoring the decimal point: $935 \times 100 = 93500$.

Count the number of decimal places in the original decimal number (93.5). There is 1 decimal place.

The product should have the same number of decimal places. Place the decimal point 1 place from the right in 93500.

$9350\text{.}0$

This is equal to 9350.

The result of 93.5 × 100 is 9350.

The blank should be filled with 9350.

Solution:

We need to find the result of dividing 4.7 by 10.

When dividing a decimal number by 10, we shift the decimal point one place to the left.

The given decimal number is 4.7.

The decimal point is located between 4 and 7.

Shifting the decimal point one place to the left moves it before the digit 4.

So, 4.7 becomes 0.47.

Thus, 4.7 ÷ 10 = 0.47.

Alternatively, we can perform the division:

$\frac{4.7}{10} = \frac{47/10}{10} = \frac{47}{10} \times \frac{1}{10} = \frac{47}{100} = 0.47$

The result of 4.7 ÷ 10 is 0.47.

The blank should be filled with 0.47.

Solution:

We need to find the result of dividing 4.7 by 100.

When dividing a decimal number by a power of 10, such as 100, we shift the decimal point to the left by the number of zeros in the power of 10.

The number 100 has 2 zeros.

The given decimal number is 4.7.

The decimal point is located between 4 and 7.

We need to shift the decimal point 2 places to the left.

Shifting the decimal point one place to the left moves it before the digit 4, giving 0.47.

To shift it a second place to the left, we need to add a zero before the digit 4.

Shifting the decimal point two places left from 4.7 results in 0.047.

Thus, 4.7 ÷ 100 = 0.047.

Alternatively, we can express the division as a fraction:

$\frac{4.7}{100}$

Convert the numerator to a fraction: $4.7 = \frac{47}{10}$.

So, $\frac{4.7}{100} = \frac{47/10}{100} = \frac{47}{10} \times \frac{1}{100} = \frac{47}{1000}$.

The fraction $\frac{47}{1000}$ corresponds to the decimal 0.047.

The result of 4.7 ÷ 100 is 0.047.

The blank should be filled with 0.047.

Solution:

We need to find the result of dividing 4.7 by 1000.

When dividing a decimal number by a power of 10, such as 1000, we shift the decimal point to the left by the number of zeros in the power of 10.

The number 1000 has 3 zeros.

The given decimal number is 4.7.

The decimal point is located between 4 and 7.

We need to shift the decimal point 3 places to the left.

Starting with 4.7:

Shift 1 place left: 0.47

Shift 2 places left: 0.047 (added a zero before 4)

Shift 3 places left: 0.0047 (added another zero before the first zero)

Thus, 4.7 ÷ 1000 = 0.0047.

Alternatively, we can express the division as a fraction:

$\frac{4.7}{1000}$

Convert the numerator to a fraction: $4.7 = \frac{47}{10}$.

So, $\frac{4.7}{1000} = \frac{47/10}{1000} = \frac{47}{10} \times \frac{1}{1000} = \frac{47}{10000}$.

The fraction $\frac{47}{10000}$ corresponds to the decimal 0.0047.

The result of 4.7 ÷ 1000 is 0.0047.

The blank should be filled with 0.0047.

Solution:

A proper fraction is a fraction where the numerator is less than the denominator. Its value is between 0 and 1 (exclusive of 0 and 1).

Let the two proper fractions be $\frac{a}{b}$ and $\frac{c}{d}$, where $0 < \frac{a}{b} < 1$ and $0 < \frac{c}{d} < 1$.

This means $0 < a < b$ and $0 < c < d$, where $a, b, c, d$ are positive integers.

The product of these two proper fractions is $\frac{a}{b} \times \frac{c}{d} = \frac{ac}{bd}$.

Let's compare the product $\frac{ac}{bd}$ with the first fraction $\frac{a}{b}$.

We are comparing $\frac{ac}{bd}$ and $\frac{a}{b}$.

We can write $\frac{a}{b}$ as $\frac{a}{b} \times 1$. Since $\frac{c}{d}$ is a proper fraction, $0 < \frac{c}{d} < 1$.

Multiplying $\frac{a}{b}$ by a number less than 1 will result in a product that is smaller than $\frac{a}{b}$.

$\frac{ac}{bd} = \frac{a}{b} \times \frac{c}{d}$

Since $0 < \frac{c}{d} < 1$, multiplying $\frac{a}{b}$ by $\frac{c}{d}$ results in a value less than $\frac{a}{b}$.

So, $\frac{ac}{bd} < \frac{a}{b}$.

Similarly, let's compare the product $\frac{ac}{bd}$ with the second fraction $\frac{c}{d}$.

We can write $\frac{c}{d}$ as $\frac{c}{d} \times 1$. Since $\frac{a}{b}$ is a proper fraction, $0 < \frac{a}{b} < 1$.

Multiplying $\frac{c}{d}$ by a number less than 1 will result in a product that is smaller than $\frac{c}{d}$.

$\frac{ac}{bd} = \frac{c}{d} \times \frac{a}{b}$

Since $0 < \frac{a}{b} < 1$, multiplying $\frac{c}{d}$ by $\frac{a}{b}$ results in a value less than $\frac{c}{d}$.

So, $\frac{ac}{bd} < \frac{c}{d}$.

Thus, the product of two proper fractions is less than each of the fractions that are multiplied.

Example:

Let the fractions be $\frac{1}{2}$ and $\frac{3}{4}$. Both are proper fractions.

Their product is $\frac{1}{2} \times \frac{3}{4} = \frac{3}{8}$.

Compare the product $\frac{3}{8}$ with the first fraction $\frac{1}{2}$.

$\frac{3}{8}$ vs $\frac{1}{2}$

Using a common denominator (8): $\frac{3}{8}$ vs $\frac{4}{8}$.

Since $3 < 4$, we have $\frac{3}{8} < \frac{4}{8}$, so $\frac{3}{8} < \frac{1}{2}$.

Compare the product $\frac{3}{8}$ with the second fraction $\frac{3}{4}$.

$\frac{3}{8}$ vs $\frac{3}{4}$

Using a common denominator (8): $\frac{3}{8}$ vs $\frac{6}{8}$.

Since $3 < 6$, we have $\frac{3}{8} < \frac{6}{8}$, so $\frac{3}{8} < \frac{3}{4}$.

In both comparisons, the product is less than the original fraction.

The blank should be filled with less.

Solution:

The rule for dividing fractions is a fundamental concept.

To divide a fraction $\frac{a}{b}$ by another fraction $\frac{c}{d}$ (where $\frac{c}{d} \ne 0$), the procedure is to multiply the first fraction $\frac{a}{b}$ by the reciprocal of the second fraction $\frac{c}{d}$.

The reciprocal of the second fraction $\frac{c}{d}$ is obtained by inverting it, which is $\frac{d}{c}$.

So, the division is performed as:

$\frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \times \frac{d}{c}$

Based on this rule, the first blank describes the operation performed on the first fraction, which is multiply.

The second blank describes what is multiplied by the first fraction, which is the reciprocal of the other fraction.

The statement should read: "While dividing a fraction by another fraction, we multiply the first fraction by the reciprocal of the other fraction."

The first blank should be filled with multiply.

The second blank should be filled with reciprocal.

Solution:

We are given the equation 8.4 ÷ $x$ = 2.1, where $x$ is the number that fills the blank.

We need to find the value of $x$.

The equation is $8.4 \div x = 2.1$.

We can rewrite the division as a fraction:

$\frac{8.4}{x} = 2.1$

To solve for $x$, we can rearrange the equation.

Multiply both sides of the equation by $x$:

$\frac{8.4}{x} \times x = 2.1 \times x$

$8.4 = 2.1 \times x$

Now, divide both sides by 2.1 to isolate $x$:

$\frac{8.4}{2.1} = \frac{2.1 \times x}{2.1}$

$x = \frac{8.4}{2.1}$

To calculate $\frac{8.4}{2.1}$, we can multiply both the numerator and the denominator by 10 to remove the decimal points:

$\frac{8.4 \times 10}{2.1 \times 10} = \frac{84}{21}$

Now, perform the division $84 \div 21$.

We know that $21 \times 4 = 84$.

So, $\frac{84}{21} = 4$.

Therefore, $x = 4$.

Check the answer:

8.4 ÷ 4 = 2.1, which is correct.

The blank should be filled with 4.

Solution:

We are given the equation 52.7 ÷ $x$ = 0.527, where $x$ is the number that fills the blank.

We need to find the value of $x$.

The equation is $52.7 \div x = 0.527$.

We can rewrite this equation to solve for $x$. If $a \div b = c$, then $b = a \div c$.

So, $x = 52.7 \div 0.527$.

To calculate the division $52.7 \div 0.527$, it's helpful to remove the decimal points.

We can multiply both the dividend (52.7) and the divisor (0.527) by a power of 10 such that the divisor becomes a whole number.

The divisor 0.527 has 3 decimal places. Multiplying by 1000 will make it a whole number (0.527 × 1000 = 527).

We must also multiply the dividend by 1000: $52.7 \times 1000$.

To multiply 52.7 by 1000, shift the decimal point 3 places to the right. $52.7 \to 527.0 \to 5270.0 \to 52700.0$. So, $52.7 \times 1000 = 52700$.

The division becomes $52700 \div 527$.

We can see that 52700 is 527 multiplied by 100 ($527 \times 100 = 52700$).

So, $52700 \div 527 = 100$.

Therefore, $x = 100$.

Check the answer:

52.7 ÷ 100 = 0.527 (by shifting the decimal point 2 places to the left, which is correct).

The blank should be filled with 100.

Solution:

We need to determine which arithmetic operation goes in the blank to make the statement 0.5 ______ 0.7 = 0.35 true.

Let's consider the common operations: addition (+), subtraction (-), multiplication (×), and division (÷).

Case 1: Addition ($0.5 + 0.7$)

$0.5 + 0.7 = 1.2$. This is not equal to 0.35.

Case 2: Subtraction ($0.5 - 0.7$)

$0.5 - 0.7 = -0.2$. This is not equal to 0.35.

Case 3: Multiplication ($0.5 \times 0.7$)

To multiply decimals, multiply the numbers as if they were whole numbers: $5 \times 7 = 35$.

Count the total number of decimal places in the original numbers (0.5 has 1 decimal place, 0.7 has 1 decimal place, so $1 + 1 = 2$ total decimal places).

Place the decimal point in the result so there are 2 decimal places. Starting from the right of 35, move the decimal point 2 places left: 0.35.

So, $0.5 \times 0.7 = 0.35$. This matches the right side of the equation.

Case 4: Division ($0.5 \div 0.7$)

$0.5 \div 0.7 = \frac{0.5}{0.7} = \frac{5}{7}$. As a decimal, $\frac{5}{7}$ is approximately 0.714, which is not equal to 0.35.

The operation that makes the statement true is multiplication ($\times$).

The blank should be filled with $\times$.

Solution:

We need to determine which arithmetic operation goes in the blank to make the statement 2 ______ $\frac{5}{3}$ = $\frac{10}{3}$ true.

Let's consider the common operations: addition (+), subtraction (-), multiplication (×), and division (÷).

We can write the whole number 2 as a fraction $\frac{2}{1}$.

Case 1: Addition ($2 + \frac{5}{3}$)

$2 + \frac{5}{3} = \frac{2 \times 3}{3} + \frac{5}{3} = \frac{6}{3} + \frac{5}{3} = \frac{6+5}{3} = \frac{11}{3}$. This is not equal to $\frac{10}{3}$.

Case 2: Subtraction ($2 - \frac{5}{3}$)

$2 - \frac{5}{3} = \frac{6}{3} - \frac{5}{3} = \frac{6-5}{3} = \frac{1}{3}$. This is not equal to $\frac{10}{3}$.

Case 3: Multiplication ($2 \times \frac{5}{3}$)

$2 \times \frac{5}{3} = \frac{2}{1} \times \frac{5}{3} = \frac{2 \times 5}{1 \times 3} = \frac{10}{3}$. This matches the right side of the equation.

Case 4: Division ($2 \div \frac{5}{3}$)

$2 \div \frac{5}{3} = 2 \times (\text{reciprocal of } \frac{5}{3}) = 2 \times \frac{3}{5} = \frac{2}{1} \times \frac{3}{5} = \frac{2 \times 3}{1 \times 5} = \frac{6}{5}$. This is not equal to $\frac{10}{3}$.

The operation that makes the statement true is multiplication ($\times$).

The blank should be filled with $\times$.

Solution:

We need to find the result of dividing 2.001 by 0.003.

When dividing by a decimal number, we first convert the divisor into a whole number by multiplying both the dividend and the divisor by a suitable power of 10.

The divisor is 0.003. It has 3 decimal places.

To make it a whole number, we multiply by $10^3 = 1000$.

Multiply the divisor by 1000:

$0.003 \times 1000 = 3$

Multiply the dividend by 1000:

$2.001 \times 1000 = 2001$

The division problem now becomes the equivalent whole number division:

$2001 \div 3$

Perform the division:

$2001 \div 3 = 667$

So, 2.001 ÷ 0.003 = 667.

The blank should be filled with 667.

Solution:

A proper fraction is a fraction where the numerator is less than the denominator.

Let's consider a proper fraction, for example, $\frac{2}{5}$. Here, the numerator (2) is less than the denominator (5).

The reciprocal of a fraction is obtained by inverting it (swapping the numerator and the denominator).

The reciprocal of $\frac{2}{5}$ is $\frac{5}{2}$.

Now let's examine the reciprocal $\frac{5}{2}$. In this fraction, the numerator (5) is greater than the denominator (2).

A fraction where the numerator is greater than or equal to the denominator is called an improper fraction (or a whole number if the numerator is a multiple of the denominator).

So, $\frac{5}{2}$ is an improper fraction.

In general, if $\frac{a}{b}$ is a proper fraction, then $a < b$ (assuming $a, b$ are positive).

The reciprocal is $\frac{b}{a}$.

Since $a < b$, it means $b > a$. Therefore, the numerator of the reciprocal ($b$) is greater than the denominator ($a$).

This means the reciprocal $\frac{b}{a}$ is an improper fraction (or a whole number if $b$ is a multiple of $a$).

The statement "The reciprocal of a proper fraction is a proper fraction" is incorrect.

The final answer is False.

Solution:

An improper fraction is a fraction where the numerator is greater than or equal to the denominator.

Let's consider an improper fraction, for example, $\frac{7}{4}$. Here, the numerator (7) is greater than the denominator (4).

The reciprocal of a fraction is found by inverting it (swapping the numerator and the denominator).

The reciprocal of $\frac{7}{4}$ is $\frac{4}{7}$.

Now let's examine the reciprocal $\frac{4}{7}$. In this fraction, the numerator (4) is less than the denominator (7).

A fraction where the numerator is less than the denominator is called a proper fraction.

So, $\frac{4}{7}$ is a proper fraction.

Let's consider another example, an improper fraction that is a whole number, like $\frac{6}{3} = 2$.

The reciprocal of 2 (or $\frac{6}{3}$) is $\frac{1}{2}$ (or $\frac{3}{6}$, which simplifies to $\frac{1}{2}$).

$\frac{1}{2}$ is a proper fraction.

In general, if $\frac{a}{b}$ is an improper fraction, then $a \ge b$ (assuming $a, b$ are positive and $b \ne 0$).

The reciprocal is $\frac{b}{a}$.

Since $a \ge b$, if $a > b$, then $b < a$, which means the numerator of the reciprocal ($b$) is less than the denominator ($a$), making it a proper fraction.

If $a = b$, the improper fraction is $\frac{a}{a}=1$, and its reciprocal is $\frac{a}{a}=1$, which can be considered an improper fraction (numerator equals denominator), but it's a special case.

However, the statement says "an improper fraction", implying it covers all cases where $a \ge b$. The reciprocal is not always an improper fraction.

The statement "The reciprocal of an improper fraction is an improper fraction" is generally false, as the reciprocal is usually a proper fraction (unless the original improper fraction was 1).

The final answer is False.

Solution:

Let the two fractions be $\frac{a}{b}$ and $\frac{c}{d}$, where $b \ne 0$ and $d \ne 0$.

The rule for multiplying two fractions is to multiply the numerators together and multiply the denominators together. The product is the fraction formed by these results.

The product of the two fractions $\frac{a}{b}$ and $\frac{c}{d}$ is:

Product = $\frac{a}{b} \times \frac{c}{d} = \frac{a \times c}{b \times d} = \frac{\text{Product of their numerators}}{\text{Product of their denominators}}$

The given statement is: Product of two fractions = $\frac{Product of their denominators}{Product of their numerators}$.

Comparing the rule with the given statement, we see that the numerator and denominator in the formula on the right side of the equality are swapped.

Let's use an example.

Let the fractions be $\frac{1}{2}$ and $\frac{3}{4}$.

Their product is $\frac{1}{2} \times \frac{3}{4} = \frac{1 \times 3}{2 \times 4} = \frac{3}{8}$.

According to the given statement, the product should be $\frac{\text{Product of their denominators}}{\text{Product of their numerators}} = \frac{2 \times 4}{1 \times 3} = \frac{8}{3}$.

Since $\frac{3}{8} \ne \frac{8}{3}$, the statement is false.

The correct statement is: Product of two fractions = $\frac{Product of their numerators}{Product of their denominators}$.

The final answer is False.

Solution:

An improper fraction is a fraction where the numerator is greater than or equal to the denominator. This means the value of an improper fraction is greater than or equal to 1.

Let the two improper fractions be $\frac{a}{b}$ and $\frac{c}{d}$, where $\frac{a}{b} \ge 1$ and $\frac{c}{d} \ge 1$.

This implies $a \ge b$ and $c \ge d$ (assuming $a, b, c, d$ are positive integers, $b \ne 0, d \ne 0$).

The product of these two improper fractions is $\frac{a}{b} \times \frac{c}{d} = \frac{ac}{bd}$.

Let's compare the product $\frac{ac}{bd}$ with the first fraction $\frac{a}{b}$.

We can write the product as $\frac{a}{b} \times \frac{c}{d}$.

Since $\frac{c}{d}$ is an improper fraction, its value is greater than or equal to 1 ($\frac{c}{d} \ge 1$).

When we multiply a number ($\frac{a}{b}$) by a number greater than or equal to 1 ($\frac{c}{d}$), the product will be greater than or equal to the original number ($\frac{a}{b}$).

So, $\frac{ac}{bd} = \frac{a}{b} \times \frac{c}{d} \ge \frac{a}{b}$ because $\frac{c}{d} \ge 1$.

Similarly, let's compare the product $\frac{ac}{bd}$ with the second fraction $\frac{c}{d}$.

We can write the product as $\frac{c}{d} \times \frac{a}{b}$.

Since $\frac{a}{b}$ is an improper fraction, its value is greater than or equal to 1 ($\frac{a}{b} \ge 1$).

When we multiply a number ($\frac{c}{d}$) by a number greater than or equal to 1 ($\frac{a}{b}$), the product will be greater than or equal to the original number ($\frac{c}{d}$).

So, $\frac{ac}{bd} = \frac{c}{d} \times \frac{a}{b} \ge \frac{c}{d}$ because $\frac{a}{b} \ge 1$.

The product of two improper fractions is greater than or equal to each of the fractions.

The statement says "The product of two improper fractions is less than both the fractions". This contradicts our finding.

Example:

Let the fractions be $\frac{3}{2}$ and $\frac{5}{4}$. Both are improper fractions ($1.5 \ge 1$ and $1.25 \ge 1$).

Their product is $\frac{3}{2} \times \frac{5}{4} = \frac{15}{8}$.

Convert to decimal for easy comparison: $\frac{15}{8} = 1.875$.

Compare the product (1.875) with the first fraction ($\frac{3}{2} = 1.5$). $1.875 > 1.5$.

Compare the product (1.875) with the second fraction ($\frac{5}{4} = 1.25$). $1.875 > 1.25$.

The product is greater than both fractions.

The final answer is False.

Solution:

Let the fraction be $\frac{a}{b}$, where $b \ne 0$.

The reciprocal of a fraction $\frac{a}{b}$ is defined as the fraction $\frac{c}{d}$ such that when you multiply the original fraction by its reciprocal, the result is 1.

$\frac{a}{b} \times \frac{c}{d} = 1$

If we choose $\frac{c}{d} = \frac{b}{a}$ (assuming $a \ne 0$, so the reciprocal exists), then:

$\frac{a}{b} \times \frac{b}{a} = \frac{a \times b}{b \times a} = \frac{ab}{ab} = 1$

The process of changing the numerator and denominator of a fraction is often referred to as "inverting" or "flipping" the fraction.

If the original fraction is $\frac{a}{b}$, its reciprocal is $\frac{b}{a}$. This is indeed obtained by swapping the positions of the numerator and the denominator, which is equivalent to inverting it upside down.

Example:

The fraction is $\frac{3}{4}$. Inverting it upside down gives $\frac{4}{3}$.

Let's check if this is the reciprocal: $\frac{3}{4} \times \frac{4}{3} = \frac{3 \times 4}{4 \times 3} = \frac{12}{12} = 1$. Yes, it is the reciprocal.

The statement accurately describes the method for finding the reciprocal of a fraction.

The final answer is True.

Solution:

Consider a decimal number, say 4.567.

We want to multiply this number by 1000.

$4.567 \times 1000$

We know that multiplying by a power of 10 involves shifting the decimal point.

The number 1000 is $10^3$, which means it has 3 zeros.

The rule for multiplying a decimal by a power of 10 is to shift the decimal point to the right by the number of zeros in the power of 10.

Since 1000 has 3 zeros, we should shift the decimal point 3 places to the right.

Let's apply this rule to 4.567:

The decimal point is between 4 and 5.

Shift 1 place right: 45.67

Shift 2 places right: 456.7

Shift 3 places right: 4567.

So, $4.567 \times 1000 = 4567$.

This confirms that moving the decimal point to the right by three places is the correct procedure when multiplying by 1000.

The statement is "To multiply a decimal number by 1000, we move the decimal point in the number to the right by three places."

This statement is consistent with the rule of multiplying decimals by powers of 10.

The final answer is True.

Solution:

Consider a decimal number, say 123.45.

We want to divide this number by 100.

$123.45 \div 100$

We know that dividing by a power of 10 involves shifting the decimal point.

The number 100 is $10^2$, which means it has 2 zeros.

The rule for dividing a decimal by a power of 10 is to shift the decimal point to the left by the number of zeros in the power of 10.

Since 100 has 2 zeros, we should shift the decimal point 2 places to the left.

Let's apply this rule to 123.45:

The decimal point is between 3 and 4.

Shift 1 place left: 12.345

Shift 2 places left: 1.2345

So, $123.45 \div 100 = 1.2345$.

This confirms that moving the decimal point to the left by two places is the correct procedure when dividing by 100.

The statement is "To divide a decimal number by 100, we move the decimal point in the number to the left by two places."

This statement is consistent with the rule of dividing decimals by powers of 10.

The final answer is True.

Solution:

The reciprocal of a number $x$ is the number $y$ such that $x \times y = 1$. The reciprocal of $x$ is usually denoted as $\frac{1}{x}$ or $x^{-1}$.

We are looking for a number $x$ such that its reciprocal is equal to itself. So, we want to find $x$ such that:

$x = \frac{1}{x}$

To solve this equation, multiply both sides by $x$ (assuming $x \ne 0$, because the reciprocal of 0 is undefined):

$x \times x = \frac{1}{x} \times x$

$x^2 = 1$

To find the value(s) of $x$, we take the square root of both sides:

$\sqrt{x^2} = \sqrt{1}$

$|x| = 1$

This means $x$ can be either 1 or -1.

$x = 1$ or $x = -1$

Let's check both values:

If $x = 1$, the reciprocal is $\frac{1}{1} = 1$. So, 1 is its own reciprocal.

If $x = -1$, the reciprocal is $\frac{1}{-1} = -1$. So, -1 is also its own reciprocal.

The statement is "1 is the only number which is its own reciprocal."

Based on our calculation, there are two such numbers: 1 and -1.

Therefore, the statement that 1 is the *only* such number is false.

The final answer is False.

Solution:

We need to compare the values of "$\frac{2}{3}$ of 8" and "$\frac{2}{3} \div 8$".

Calculate "$\frac{2}{3}$ of 8". The word "of" means multiplication.

$\frac{2}{3} \text{ of } 8 = \frac{2}{3} \times 8$

Multiply the fraction by the whole number:

$\frac{2}{3} \times 8 = \frac{2}{3} \times \frac{8}{1} = \frac{2 \times 8}{3 \times 1} = \frac{16}{3}$

Calculate "$\frac{2}{3} \div 8$". Dividing by a number is the same as multiplying by its reciprocal.

The reciprocal of 8 is $\frac{1}{8}$.

$\frac{2}{3} \div 8 = \frac{2}{3} \times \frac{1}{8}$

Multiply the two fractions:

$\frac{2}{3} \times \frac{1}{8} = \frac{2 \times 1}{3 \times 8} = \frac{2}{24}$

Simplify the fraction $\frac{2}{24}$ by dividing the numerator and denominator by their greatest common divisor, which is 2.

$\frac{\cancel{2}^{1}}{\cancel{24}_{12}} = \frac{1}{12}$

Now compare the results:

"$\frac{2}{3}$ of 8" is $\frac{16}{3}$.

"$\frac{2}{3} \div 8$" is $\frac{1}{12}$.

Since $\frac{16}{3} \ne \frac{1}{12}$, the two expressions are not the same.

The statement is "$\frac{2}{3}$ of 8 is same as $\frac{2}{3}$ ÷ 8."

This statement is false.

The final answer is False.

Solution:

The reciprocal of a fraction is obtained by inverting the fraction, which means swapping the numerator and the denominator.

The given fraction is $\frac{4}{7}$.

The numerator is 4 and the denominator is 7.

To find the reciprocal, we swap the numerator and the denominator.

The reciprocal of $\frac{4}{7}$ is $\frac{7}{4}$.

The statement says that the reciprocal of $\frac{4}{7}$ is $\frac{4}{7}$.

We found that the reciprocal is $\frac{7}{4}$.

Since $\frac{7}{4}$ is not equal to $\frac{4}{7}$, the statement is false.

The only numbers that are equal to their reciprocals are 1 and -1.

$\frac{4}{7}$ is not equal to 1 or -1.

The final answer is False.

Solution:

The original fraction is $\frac{5}{9}$.

5 is added to both the numerator and the denominator.

New numerator = Original numerator + 5 = 5 + 5 = 10

New denominator = Original denominator + 5 = 9 + 5 = 14

The new fraction is $\frac{10}{14}$.

First, we need to check if the value of the fraction is changed. We can do this by comparing the original fraction $\frac{5}{9}$ with the new fraction $\frac{10}{14}$.

We can compare the fractions by finding a common denominator or by cross-multiplication.

Using common denominator: The least common multiple (LCM) of 9 and 14 is $9 \times 14 = 126$ (since 9 and 14 have no common factors other than 1).

Original fraction: $\frac{5}{9} = \frac{5 \times 14}{9 \times 14} = \frac{70}{126}$

New fraction: $\frac{10}{14} = \frac{10 \times 9}{14 \times 9} = \frac{90}{126}$

Comparing $\frac{70}{126}$ and $\frac{90}{126}$, we see that $70 \ne 90$. Therefore, the value of the fraction is changed.

Yes, the value of the fraction is changed.

Next, we need to determine if the value increases or decreases.

We compare the original value $\frac{70}{126}$ with the new value $\frac{90}{126}$.

Since $70 < 90$, we have $\frac{70}{126} < \frac{90}{126}$.

Original value < New value.

This means the value of the fraction increased.

Alternatively, for a proper fraction $\frac{a}{b}$ where $a < b$, adding a positive number $k$ to both the numerator and the denominator results in the fraction $\frac{a+k}{b+k}$.

We can compare $\frac{a}{b}$ and $\frac{a+k}{b+k}$ by cross-multiplication:

$a \times (b+k)$ vs $b \times (a+k)$

$ab + ak$ vs $ba + bk$

$ab + ak$ vs $ab + bk$

Subtract $ab$ from both sides:

$ak$ vs $bk$

Since $k=5$ is positive ($k > 0$), the comparison depends on $a$ vs $b$.

For a proper fraction, $a < b$.

Multiplying both sides of $a < b$ by a positive number $k$, we get $ak < bk$.

So, $ak < bk$.

This implies $ab + ak < ab + bk$, which means $a(b+k) < b(a+k)$.

Dividing by the positive denominators $b(b+k)$, we get $\frac{a}{b} < \frac{a+k}{b+k}$.

This shows that adding a positive number to both the numerator and the denominator of a proper fraction increases its value.

Our original fraction $\frac{5}{9}$ is a proper fraction (5 < 9), and we added a positive number (5).

Therefore, the value of the fraction will increase.

The value of the fraction will be changed, and the value will increase.

Solution:

Let the original fraction be $\frac{n}{d}$, where $n$ is the numerator and $d$ is the denominator. We assume $n > 0$ and $d > 0$.

The numerator $n$ is kept unchanged.

The denominator $d$ is decreased to a new value, let's call it $d'$, such that $0 < d' < d$. The new fraction is $\frac{n}{d'}$.

We need to compare the value of the original fraction $\frac{n}{d}$ with the value of the new fraction $\frac{n}{d'}$.

Since $d' < d$, we know that $\frac{1}{d'}$ is greater than $\frac{1}{d}$. For example, if $d=5$ and $d'=2$, then $\frac{1}{2} > \frac{1}{5}$.

To get the value of the fraction $\frac{n}{d'}$, we multiply $\frac{1}{d'}$ by $n$. To get the value of the fraction $\frac{n}{d}$, we multiply $\frac{1}{d}$ by $n$.

Since $n$ is a positive number and $\frac{1}{d'} > \frac{1}{d}$, multiplying both sides by $n$ maintains the inequality:

$n \times \frac{1}{d'} > n \times \frac{1}{d}$

$\frac{n}{d'} > \frac{n}{d}$

This shows that the new fraction $\frac{n}{d'}$ is greater than the original fraction $\frac{n}{d}$.

Therefore, if the denominator of a fraction is decreased while the numerator is kept unchanged (and is positive), the value of the fraction increases.

Example:

Let the original fraction be $\frac{3}{4}$. Numerator = 3, Denominator = 4.

Decrease the denominator to 2 (while keeping the numerator as 3). The new fraction is $\frac{3}{2}$.

Compare $\frac{3}{4}$ and $\frac{3}{2}$.

$\frac{3}{4} = 0.75$

$\frac{3}{2} = 1.5$

Since $1.5 > 0.75$, the value increased.

The value of the fraction increases.

The letters among A and J are A, B, C, D, E, F, G, H, I, J.

There are 10 letters in this sequence.

To find the letter that is $\frac{2}{5}$ of the way among these 10 letters, we calculate the position.

Position = Total number of letters $\times$ Fraction of the way

$= 10 \times \frac{2}{5}$

$= \frac{10 \times 2}{5}$

$= \frac{20}{5}$

$= 4$

This means we are looking for the 4th letter in the sequence starting from A.

Counting from A:

1st letter is A

2nd letter is B

3rd letter is C

4th letter is D

Therefore, the letter that comes $\frac{2}{5}$ of the way among A and J is D.

Let the unknown number be $x$.

According to the problem statement, $\frac{2}{3}$ of the number is 10.

To find the value of $x$, we multiply both sides of equation (i) by $\frac{3}{2}$.

Now we need to find 1.75 times this number ($x$).

We can write 1.75 as a fraction:

$1.75 = 1 \frac{75}{100} = 1 \frac{3}{4} = \frac{1 \times 4 + 3}{4} = \frac{7}{4}$

So, we need to calculate $1.75 \times x = \frac{7}{4} \times 15$.

To express the result as a decimal, we divide 105 by 4.

$\frac{105}{4} = \frac{104 + 1}{4} = \frac{104}{4} + \frac{1}{4} = 26 + 0.25 = 26.25$

Alternatively, we can directly multiply 1.75 by 15.

$\begin{array}{cc}& & 1 & . & 7 & 5 \\ \times & & & 1 & 5 \\ \hline && 8 & . & 7 & 5 \\ & 1 & 7 & . & 5 & \times \\ \hline & 2 & 6 & . & 2 & 5 \\ \hline \end{array}$

Thus, 1.75 times the number is 26.25.

Given:

Fraction of students who like to eat rice only = $\frac{1}{5}$

Fraction of students who like to eat chapati only = $\frac{2}{5}$

The students who like to eat 'both' are the remaining students after accounting for those who like 'only rice' and 'only chapati'.

The total fraction representing the whole class is 1.

First, find the fraction of students who like to eat only rice or only chapati.

Now, subtract this fraction from the total fraction (1) to find the fraction of students who like to eat both.

Therefore, the fraction of the total number of students who like to eat both is $\frac{2}{5}$.

(Note: The actual number of students, 40, is not needed to find the fraction.)

Solution:

Given that Renu completed $\frac{2}{3}$ part of her homework in 2 hours.

We can find the rate at which she completes her homework (part of homework per hour).

Now, we need to find the part of homework completed in $1\frac{1}{4}$ hours.

First, convert the mixed number to an improper fraction:

$1\frac{1}{4} = \frac{1 \times 4 + 1}{4} = \frac{5}{4}$ hours.

Now, calculate the part of homework completed in $\frac{5}{4}$ hours using the rate.

Therefore, Renu completed $\frac{5}{12}$ part of her homework in $1\frac{1}{4}$ hours.

Given:

Initial fraction of book read = $\frac{1}{5}$

Additional pages read = 40

Total fraction of book read after reading 40 more pages = $\frac{7}{10}$

To Find:

Number of pages left to be read.

Solution:

Let the total number of pages in the book be $P$.

Initially, Reemu read $\frac{1}{5}$ of the book, which is $\frac{1}{5}P$ pages.

When she reads 40 more pages, the total number of pages read is $\frac{1}{5}P + 40$.

According to the problem, this total number of pages read is equal to $\frac{7}{10}$ of the book, which is $\frac{7}{10}P$.

We can set up the equation:

Now, we solve for $P$. Subtract $\frac{1}{5}P$ from both sides of the equation (i):

To subtract the fractions, we find a common denominator, which is 10.

Substitute this back into the equation:

Simplify the fraction $\frac{5}{10}$:

So the equation becomes:

Multiply both sides by 2 to solve for $P$:

The total number of pages in the book is 80.

Reemu has read a total of $\frac{7}{10}$ of the book.

So, Reemu has read 56 pages.

To find the number of pages left to be read, subtract the pages read from the total number of pages.

The number of pages left to be read is 24.

Let the number in the box be $x$. The given equation is:

To find the value of $x$, we need to isolate $x$. We can do this by multiplying both sides of equation (i) by the reciprocal of $\frac{3}{7}$, which is $\frac{7}{3}$.

Now, we can simplify the expression by cancelling common factors between the numerator and the denominator.

We can cancel 15 in the numerator with 3 in the denominator (since $15 = 5 \times 3$).

We can cancel 7 in the numerator with 98 in the denominator (since $98 = 14 \times 7$).

Multiplying the remaining terms:

So, the number in the box ⬜ is $\frac{5}{14}$.

The completed equation is:

$\frac{3}{7} \times \frac{5}{14} = \frac{15}{98}$

To determine if the quotient is greater than or less than 1.5, we first need to calculate the quotient.

Convert the mixed numbers to improper fractions:

Now, perform the division. Dividing by a fraction is the same as multiplying by its reciprocal.

Multiply the numerators and the denominators:

Simplify by cancelling the common factor of 3:

Now, we compare the quotient $\frac{43}{22}$ with 1.5.

We can convert 1.5 to a fraction or the quotient to a decimal.

Method 1: Convert the quotient to a decimal.

$\frac{43}{22} \approx 1.9545...$

Comparing 1.9545... with 1.5:

Method 2: Compare fractions by finding a common denominator or cross-multiplication.

Compare $\frac{43}{22}$ and $1.5 = \frac{3}{2}$.

Using cross-multiplication: Compare $43 \times 2$ and $22 \times 3$.

Since $86 > 66$, it means $\frac{43}{22} > \frac{3}{2}$.

In both methods, we find that the quotient $\frac{43}{22}$ is greater than 1.5.

Explanation:

The quotient $\frac{43}{22}$ is equivalent to $1 \frac{21}{22}$.

The value 1.5 is equivalent to $1 \frac{1}{2}$.

Comparing the fractional parts $\frac{21}{22}$ and $\frac{1}{2}$:

$\frac{1}{2} = \frac{11}{22}$.

Since $\frac{21}{22} > \frac{11}{22}$, it follows that $1 \frac{21}{22} > 1 \frac{1}{2}$.

Therefore, the quotient $\frac{43}{22}$ is greater than 1.5.

The quotient $7\frac{1}{6}$ ÷ $3\frac{2}{3}$ will be a fraction greater than 1.5.

We need to compare the fraction $\frac{13}{17}$ and the decimal 0.82.

Method 1: Convert the fraction to a decimal.

We convert the fraction $\frac{13}{17}$ into a decimal by dividing the numerator by the denominator.

Performing the division:

$\begin{array}{r} 0.\ 7 \ 6 \ 4 \ 7 \ ... \\ 17{\overline{\smash{\big)}\,13.0000\phantom{)}}} \\ \underline{-~\phantom{(}11\ 9\phantom{000)}} \\ 1\ 10\phantom{00} \\ \underline{-~\phantom{()}1\ 02\phantom{0}} \\ 80\phantom{0} \\ \underline{-~\phantom{()}68\phantom{}} \\ 120 \\ \underline{-~119} \\ 1 \end{array}$

So, $\frac{13}{17} \approx 0.7647$.

Now, we compare 0.7647... with 0.82.

Thus, $\frac{13}{17} < 0.82$.

Method 2: Convert the decimal to a fraction and compare the fractions.

Convert 0.82 to a fraction:

We can simplify this fraction by dividing the numerator and denominator by their greatest common divisor, which is 2.

Now, we compare $\frac{13}{17}$ and $\frac{41}{50}$.

To compare two fractions $\frac{a}{b}$ and $\frac{c}{d}$, we can compare the cross-products $a \times d$ and $b \times c$.

$\begin{array}{cc}& & 1 & 7 \\ \times & & 4 & 1 \\ \hline && 1 & 7 \\ & 6 & 8 & \times \\ \hline & 6 & 9 & 7 \\ \hline \end{array}$

Comparing the cross-products:

Since $13 \times 50 < 17 \times 41$, it follows that $\frac{13}{17} < \frac{41}{50}$.

Thus, $\frac{13}{17} < 0.82$.

Comparison of Methods:

Method 1 requires long division, which can be tedious and may result in a non-terminating decimal, requiring approximation. If the values were very close, rounding could potentially lead to an incorrect conclusion.

Method 2 involves converting the decimal to a fraction (which is straightforward for terminating decimals) and then comparing the two fractions, for example, using cross-multiplication. This involves integer multiplications and a simple comparison of the products. This method provides an exact comparison without approximation.

Conclusion on which is easier:

For most people, especially without a calculator, Method 2 (converting the decimal to a fraction and using cross-multiplication) is likely easier.

This is because converting 0.82 to $\frac{41}{50}$ is simple, and the subsequent integer multiplications ($13 \times 50$ and $17 \times 41$) are generally less prone to error and quicker than performing long division of 13 by 17 to several decimal places.

Given:

Weight of each tablet = $\frac{4}{25}$ gram.

Recommended dose for adult $\geq$ 60 kg = 4 tablets.

Recommended dose for adult between 40 kg and 50 kg = $2\frac{1}{2}$ tablets.

(a) Grams of pain reliever for a 72 kg adult taking 4 tablets.

A 72 kg adult is in the category "60 kg and over", so the recommended dose is 4 tablets.

To convert this to a decimal:

A 72 kg adult taking 4 tablets receives $\frac{16}{25}$ grams (or 0.64 grams) of pain reliever.

(b) Grams of pain reliever for a 46 kg adult.

A 46 kg adult is in the category "between 40 and 50 kg", so the recommended dose is $2\frac{1}{2}$ tablets.

Convert the mixed number dose to an improper fraction:

Calculate the total grams of pain reliever:

Multiply the numerators and denominators, then simplify:

To convert this to a decimal:

The recommended dose for an adult weighing 46 kg is $\frac{2}{5}$ grams (or 0.4 grams) of pain reliever.

Note: The dosage guidelines from the label are missing from the problem description. Without the specific recommended dosages for different animal types or weight ranges, it is not possible to calculate the exact dosage for each animal listed.

However, based on typical pet vitamin labels, we can infer the categories these animals would likely fall into if the guidelines were provided:

(a) a 18 kg adult dog

This dog falls into the category of an 18 kg adult dog. The dosage would depend on the specific weight range provided in the guidelines for adult dogs (e.g., a dose for 10-20 kg dogs, or a dose per kg).

(b) a 6 kg cat

This cat falls into the category of a 6 kg cat. The dosage would depend on the specific guidelines for cats, likely based on their weight (e.g., a dose for cats weighing up to 7 kg).

(c) a 18 kg pregnant dog

This dog falls into the category of an 18 kg pregnant dog. Dosage guidelines often have specific recommendations or adjustments for pregnant or lactating animals. This dosage would depend on the guidelines for pregnant dogs of this weight (e.g., a specific dose for pregnant dogs or a modified dose from the standard adult dog category).

To provide specific numerical answers for (a), (b), and (c), the dosage guidelines from the pet vitamin bottle label are required.

Given:

Total quantity of chocolates = $1\frac{1}{2}$ kg

Weight of each small box = $\frac{1}{16}$ kg

To Find:

Number of $\frac{1}{16}$ kg boxes that can be made.

Solution:

First, convert the total quantity of chocolates from a mixed number to an improper fraction.

To find the number of boxes, we need to divide the total quantity of chocolates by the weight of one box.

Dividing by a fraction is the same as multiplying by its reciprocal.

Now, multiply the fractions and simplify.

Cancel the common factor of 2 from the numerator and the denominator.

Therefore, 24 boxes of $\frac{1}{16}$ kg chocolates can be made with $1\frac{1}{2}$ kg chocolates.

Given:

Total length of the ribbon = 15 m

Assuming the length of one bookmark from Figure 2.6 is $\frac{3}{4}$ m (as indicated in the visual context typically associated with this type of problem).

To Find:

Number of bookmarks Anvi can make.

Solution:

To find the number of bookmarks that can be made, we need to divide the total length of the ribbon by the length of one bookmark.

To divide by a fraction, we multiply by its reciprocal.

The reciprocal of $\frac{3}{4}$ is $\frac{4}{3}$.

We can write 15 as $\frac{15}{1}$ and then multiply the fractions.

Now, multiply the numerators and the denominators.

Simplify the expression by cancelling the common factor of 3 from the numerator and the denominator.

Therefore, Anvi can make 20 bookmarks from a 15 m long ribbon.

Given Rule:

Approximate length of diagonal = Length of a side $\times$ 1.414

(a) When the length of a side is 8.3 cm.

Performing the multiplication:

$\begin{array}{cc}& & & 1 & . & 4 & 1 & 4 \\ \times & & & & & 8 & . & 3 \\ \hline & & & & 4 & 2 & 4 & 2 \\ & 1 & 1 & 3 & 1 & 2 & \times & \\ \hline & 1 & 1 & . & 7 & 3 & 6 & 2 \\ \hline \end{array}$

The approximate length of the diagonal is 11.7362 cm.

(b) When the length of a side is exactly 7.875 cm.

Performing the multiplication:

$\begin{array}{cc}& & & & 7 & . & 8 & 7 & 5 \\ \times & & & & 1 & . & 4 & 1 & 4 \\ \hline & & & & 3 & 1 & 5 & 0 & 0 \\ & & & 7 & 8 & 7 & 5 & \times & \\ & & 3 & 1 & 5 & 0 & \times & \times & \\ & 7 & 8 & 7 & 5 & \times & \times & \times & \\ \hline & 1 & 1 & . & 1 & 4 & 6 & 0 & 5 & 0 \\ \hline \end{array}$

The approximate length of the diagonal is 11.146050 cm.

Given Rule:

Length of side of the largest inscribed square = 0.707 $\times$ Diameter of the circle

(a) When the diameter of the circle is 14.35 cm.

Performing the multiplication:

$\begin{array}{cc}& & & & 1 & 4 & . & 3 & 5 \\ \times & & & & & 0 & . & 7 & 0 & 7 \\ \hline & & & & & 1 & 0 & 0 & 4 & 5 \\ & & & 0 & 0 & 0 & 0 & \times & & \\ & 1 & 0 & 0 & 4 & 5 & \times & \times & & \\ \hline & 1 & 0 & . & 1 & 5 & 5 & 0 & 4 & 5 \\ \hline \end{array}$

The approximate length of the side of the square is 10.155045 cm.

(b) When the diameter of the circle is 8.63 cm.

Performing the multiplication:

$\begin{array}{cc}& & & & & 8 & . & 6 & 3 \\ \times & & & & & 0 & . & 7 & 0 & 7 \\ \hline & & & & & 6 & 0 & 4 & 1 \\ & & & 0 & 0 & 0 & 0 & \times & \\ & & 6 & 0 & 4 & 1 & \times & \times & \\ \hline & & 6 & . & 0 & 9 & 7 & 5 & 4 & 1 \\ \hline \end{array}$

The approximate length of the side of the square is 6.097541 cm.

Given Rule:

Distance around a circular disc (Circumference) = Diameter $\times$ 3.14

(Using 3.14 as an approximation for $\pi$)

(a) When the diameter is 18.7 cm.

Performing the multiplication:

$\begin{array}{cc}& & & 1 & 8 & . & 7 \\ \times & & & 3 & . & 1 & 4 \\ \hline & & & & 7 & 4 & 8 \\ & & 1 & 8 & 7 & \times & \\ & 5 & 6 & 1 & \times & \times & \\ \hline & 5 & 8 & . & 7 & 1 & 8 \\ \hline \end{array}$

The distance around the disc is 58.718 cm.

(b) When the radius is 6.45 cm.

The diameter is twice the radius.

Performing the multiplication:

$\begin{array}{cc}& & 6 & . & 4 & 5 \\ \times & & & & 2 \\ \hline 1 & 2 & . & 9 & 0 \\ \hline \end{array}$

Now, apply the given rule using the calculated diameter.

Performing the multiplication:

$\begin{array}{cc}& & & 1 & 2 & . & 9 & 0 \\ \times & & & & 3 & . & 1 & 4 \\ \hline & & & & 5 & 1 & 6 & 0 \\ & & 1 & 2 & 9 & 0 & \times & \\ & 3 & 8 & 7 & 0 & \times & \times & \\ \hline & 4 & 0 & . & 5 & 0 & 6 & 0 \\ \hline \end{array}$

The distance around the disc is 40.506 cm.

Given:

Length of cloth = 27.5 m

Cost per metre = $\textsf{₹}$ 53.50

To Find:

The total cost of the cloth.

Solution:

To find the total cost, we multiply the length of the cloth by the cost per metre.

Performing the multiplication:

$\begin{array}{cc}& & 2 & 7 & . & 5 \\ \times & & 5 & 3 & . & 5 \\ \hline && 1 & 3 & 7 & 5 \\ & & 8 & 2 & 5 & \times \\ 1 & 3 & 7 & 5 & \times & \times \\ \hline 1 & 4 & 7 & 1 & . & 2 & 5 \\ \hline \end{array}$

The cost of 27.5 m of cloth at $\textsf{₹}$ 53.50 per metre is $\textsf{₹}$ 1471.25.

From the problem statement and the figure (Fig. 2.7), we understand that the race is divided into parts by the hurdles.

Nidhi is at hurdle B when she is $\frac{2}{6}$ of the way through the race.

This suggests that the hurdles A, B, C, D, E divide the race into 6 equal parts, with hurdles positioned at the end of certain parts.

Assuming the start is at 0 and hurdles are placed at equal fractional points:

Start: 0

Hurdle A: $\frac{1}{6}$ of the way

Hurdle B: $\frac{2}{6}$ of the way (Given)

Hurdle C: $\frac{3}{6}$ of the way

Hurdle D: $\frac{4}{6}$ of the way

Hurdle E: $\frac{5}{6}$ of the way

Finish: $\frac{6}{6}$ (or 1) of the way

(a) Where will Nidhi be, when she is $\frac{4}{6}$ of the way through the race?

Based on the positions derived above, $\frac{4}{6}$ of the way through the race corresponds to Hurdle D.

So, Nidhi will be over Hurdle D when she is $\frac{4}{6}$ of the way through the race.

(b) Where will Nidhi be when she is $\frac{5}{6}$ of the way through the race?

Based on the positions, $\frac{5}{6}$ of the way through the race corresponds to Hurdle E.

So, Nidhi will be over Hurdle E when she is $\frac{5}{6}$ of the way through the race.

(c) Give two fractions to tell what part of the race Nidhi has finished when she is over hurdle C.

Based on the positions, Hurdle C is at $\frac{3}{6}$ of the way through the race.

The fraction $\frac{3}{6}$ can be simplified by dividing the numerator and denominator by their greatest common divisor, which is 3.

So, when Nidhi is over Hurdle C, she has finished $\frac{3}{6}$ of the race, which is equivalent to $\frac{1}{2}$ of the race.

Two fractions that tell what part of the race Nidhi has finished when she is over hurdle C are $\frac{3}{6}$ and $\frac{1}{2}$.

Given:

Diameter of Earth = 12756000 m

Diameter of the new planet = $\frac{5}{86}$ of the diameter of Earth

To Find:

Diameter of the new planet in kilometers (km).

Solution:

First, calculate the diameter of the new planet in meters.

We need to calculate $12756000 \div 86$ first, then multiply by 5.

Let's perform the division $12756000 \div 86$:

$\begin{array}{r} 148325.58... \\ 86{\overline{\smash{\big)}\,12756000.00\phantom{)}}} \\ \underline{-~\phantom{()}(86)\phantom{00000.00)}} \\ 415\phantom{0000.00)} \\ \underline{-~\phantom{()}(344)\phantom{0000.00)}} \\ 716\phantom{000.00)} \\ \underline{-~\phantom{()}(688)\phantom{000.00)}} \\ 280\phantom{00.00)} \\ \underline{-~\phantom{()}(258)\phantom{00.00)}} \\ 220\phantom{0.00)} \\ \underline{-~\phantom{()}(172)\phantom{0.00)}} \\ 480\phantom{.00)} \\ \underline{-~\phantom{()}(430)\phantom{.00)}} \\ 500\phantom{0)} \\ \underline{-~\phantom{()}(430)\phantom{0)}} \\ 700\phantom{)} \\ \underline{-~\phantom{()}(688)} \\ 12 \end{array}$

Let's check if 12756000 is exactly divisible by 86. $12756000 \div 86 = 148325.58...$

Perhaps there is a typo in the numbers, or the intended calculation results in a neat number. Let's recheck the division: $12756 \div 86$.

$\begin{array}{r} 148.32... \\ 86{\overline{\smash{\big)}\,12756\phantom{)}}} \\ \underline{-~\phantom{(}(86)\phantom{0)}} \\ 415\phantom{0)} \\ \underline{-~\phantom{()}(344)\phantom{)}} \\ 716\phantom{)} \\ \underline{-~\phantom{()}(688)} \\ 28 \end{array}$

Ah, $12756$ is not divisible by 86. Let's check if $127560$ is divisible by 86. $127560 \div 86 = 1483.25...$

Let's assume the number is $1275600$ instead, $1275600 \div 86 = 14832.55...$

Let's assume the diameter is $12756$ km instead of meters. The diameter of Earth is approximately 12742 km. $12756 \div 86$ is not an integer. Let's assume the diameter of Earth in meters is $12756 \times 1000 = 12,756,000$ meters. Let's perform the division $12756000 \div 86$. We can divide 12756 by 86 first and then add the zeros.

Let's use a calculator to verify $12756000 / 86$. It is not an exact integer. Maybe there is a mistake in the problem statement or the number 86 is a factor of 12756000. Prime factors of 86 are 2 and 43. Let's check if 12756000 is divisible by 43. $12756000 \div 43 = 296651.16...$ Let's check if 12756 is divisible by 86. $12756 \div 86 = 148.32...$ Let's check if 127560 is divisible by 86. $127560 \div 86 = 1483.25...$ Let's check if 1275600 is divisible by 86. $1275600 \div 86 = 14832.55...$ Let's check if 12756000 is divisible by 86. $12756000 \div 86 = 148325.58...$

Let's assume there's a number in the question that makes the calculation simpler, for example, if the diameter of Earth was $12900000$ m, then $12900000 \div 86 = 150000$. Or if the fraction was $\frac{5}{85} = \frac{1}{17}$, then $12756000 \div 17 = 750352.9...$

Let's proceed with the given numbers and present the exact fraction or approximate decimal.

We can simplify the fraction $\frac{12756000}{86}$. Divide both by 2.

Let's perform the division $6378000 \div 43$.

$\begin{array}{r} 148325.58... \\ 43{\overline{\smash{\big)}\,6378000.00\phantom{)}}} \\ \underline{-~\phantom{(}(43)\phantom{0000.00)}} \\ 207\phantom{000.00)} \\ \underline{-~\phantom{()}(172)\phantom{000.00)}} \\ 358\phantom{00.00)} \\ \underline{-~\phantom{()}(344)\phantom{00.00)}} \\ 140\phantom{0.00)} \\ \underline{-~\phantom{()}(129)\phantom{0.00)}} \\ 110\phantom{.00)} \\ \underline{-~\phantom{()}(86)\phantom{.00)}} \\ 240\phantom{0)} \\ \underline{-~\phantom{()}(215)\phantom{0)}} \\ 250\phantom{)} \\ \underline{-~\phantom{()}(215)} \\ 35 \end{array}$

Since $6378000$ is not exactly divisible by $43$, the diameter will be a fraction or a repeating decimal.

Now, we need to convert this diameter from meters to kilometers.

1 km = 1000 m, so 1 m = $\frac{1}{1000}$ km.

Cancel the common factor of 1000:

Let's perform the division $31890 \div 43$.

$\begin{array}{r} 741.62... \\ 43{\overline{\smash{\big)}\,31890.00\phantom{)}}} \\ \underline{-~\phantom{(}(301)\phantom{0.00)}} \\ 179\phantom{0.00)} \\ \underline{-~\phantom{()}(172)\phantom{0.00)}} \\ 70\phantom{.00)} \\ \underline{-~\phantom{()}(43)\phantom{.00)}} \\ 270\phantom{0)} \\ \underline{-~\phantom{()}(258)\phantom{0)}} \\ 120\phantom{)} \\ \underline{-~\phantom{()}(86)} \\ 34 \end{array}$

The result is $741 \frac{27}{43}$ km, or approximately 741.63 km.

Given the structure of the previous problems, it's highly probable that the numbers were intended to result in a simpler fraction or integer. However, based strictly on the numbers provided:

If the question implies an approximate value using 3.14 or similar approximations as in previous questions, we might use an approximate diameter for Earth (e.g., 12800 km) or the fraction $\frac{5}{86}$ might be an approximation of a simpler fraction.

Using the exact numbers provided, the diameter of the new planet in km is $\frac{31890}{43}$ km.

As a decimal approximation: $\frac{31890}{43} \approx 741.6279...$ km. Rounding to two decimal places, it is approximately 741.63 km.

Given:

The fraction is $\frac{5}{129}$.

To Find:

The product of the fraction and its reciprocal.

Solution:

The reciprocal of a fraction $\frac{a}{b}$ is $\frac{b}{a}$.

So, the reciprocal of $\frac{5}{129}$ is $\frac{129}{5}$.

Now, we find the product of the fraction and its reciprocal.

Multiply the numerators together and the denominators together.

We can cancel the common factors in the numerator and the denominator.

The product of a non-zero number and its reciprocal is always 1.

The product of $\frac{5}{129}$ and its reciprocal is 1.

Given Expression:

$\frac{2\frac{1}{2}+\frac{1}{5}}{2\frac{1}{2}\div\frac{1}{5}}$

Solution:

First, convert the mixed number $2\frac{1}{2}$ to an improper fraction.

Substitute this into the expression:

Now, evaluate the numerator: $\frac{5}{2} + \frac{1}{5}$.

To add these fractions, find a common denominator. The least common multiple of 2 and 5 is 10.

Next, evaluate the denominator: $\frac{5}{2} \div \frac{1}{5}$.

To divide by a fraction, multiply by its reciprocal.

Now, substitute the values of the numerator and the denominator back into the main expression:

To divide by a fraction, multiply by the reciprocal of the denominator.

Multiply the numerators and the denominators, then simplify.

Cancel the common factor of 2 between the numerator (2) and the denominator (10).

The simplified value of the expression is $\frac{27}{125}$.

Given Expression:

$\frac{\frac{1}{4}+\frac{1}{5}}{1-\frac{3}{8}\times\frac{1}{5}}$

Solution:

We will simplify the numerator and the denominator separately.

Step 1: Simplify the Numerator

Numerator = $\frac{1}{4} + \frac{1}{5}$

To add these fractions, find a common denominator. The least common multiple of 4 and 5 is 20.

Step 2: Simplify the Denominator

Denominator = $1 - \frac{3}{8} \times \frac{1}{5}$

First, perform the multiplication in the denominator.

Now, perform the subtraction.

Write 1 as a fraction with a denominator of 40.

Step 3: Divide the Numerator by the Denominator

The original expression is the numerator divided by the denominator.

To divide by a fraction, multiply by its reciprocal.

Multiply the numerators and the denominators.

Simplify the expression by cancelling the common factor of 20 from the numerator and the denominator.

The simplified value of the expression is $\frac{18}{37}$.

Given:

We need to divide $\frac{3}{10}$ by the value of $\left( \frac{1}{4} \; of \; \frac{3}{5} \right)$.

To Find:

The result of the division: $\frac{3}{10} \div \left( \frac{1}{4} \; of \; \frac{3}{5} \right)$.

Solution:

First, we need to evaluate the term in the parenthesis: $\left( \frac{1}{4} \; of \; \frac{3}{5} \right)$.

In mathematics, the word "of" when used with fractions or percentages implies multiplication.

Multiply the numerators together and the denominators together:

Now, we need to divide $\frac{3}{10}$ by this result, $\frac{3}{20}$.

To divide by a fraction, we multiply by its reciprocal.

The reciprocal of $\frac{3}{20}$ is $\frac{20}{3}$.

Multiply the numerators and the denominators:

Simplify by cancelling common factors from the numerator and the denominator.

We can cancel 3 in the numerator with 3 in the denominator.

We can cancel 20 in the numerator with 10 in the denominator (since $20 = 2 \times 10$).

The result of dividing $\frac{3}{10}$ by $\left( \frac{1}{4} \; of \; \frac{3}{5} \right)$ is 2.

Given:

$\frac{1}{8}$ of an unknown number is equal to the result of the division $\frac{2}{5} \div \frac{1}{20}$.

To Find:

The unknown number.

Solution:

Let the unknown number be $N$.

The given condition can be written as an equation:

First, let's simplify the right side of the equation (i).

We need to evaluate $\frac{2}{5} \div \frac{1}{20}$.

To divide by a fraction, we multiply by its reciprocal.

The reciprocal of $\frac{1}{20}$ is $\frac{20}{1}$.

Multiply the numerators and the denominators:

Simplify by cancelling common factors. We can cancel 5 from the denominator (5) and the numerator (20, since $20 = 4 \times 5$).

Now, substitute this simplified value back into equation (i):

To find the value of $N$, we need to isolate $N$. Multiply both sides of the equation by the reciprocal of $\frac{1}{8}$, which is $\frac{8}{1}$ or just 8.

The unknown number is 64.

Given:

Amount of the electric bill = $\textsf{₹}$ 385.70

Amount paid by Heena's father = $\textsf{₹}$ 500

To Find:

The amount of change Heena's father should have received.

Solution:

To find the change received, we subtract the bill amount from the amount paid.

Performing the subtraction:

$\begin{array}{cc}& 5 & 0 & 0 & . & 0 & 0 \\ - & 3 & 8 & 5 & . & 7 & 0 \\ \hline & 1 & 1 & 4 & . & 3 & 0 \\ \hline \end{array}$

Heena's father should have received a change of $\textsf{₹}$ 114.30.

Given:

Normal body temperature = 98.6°F

Savitri's temperature = 103.1°F

To Find:

How many degrees above normal Savitri's temperature was.

Solution:

To find how many degrees above normal the temperature was, we subtract the normal temperature from Savitri's temperature.

Performing the subtraction:

$\begin{array}{cc} & 103 & . & 1 \\ - & \phantom{0}98 & . & 6 \\ \hline & \phantom{00}4 & . & 5 \\ \hline \end{array}$

Savitri's temperature was 4.5°F above normal.

Given Temperature Differences from Base (°C):

1958: 0.10

1964: -0.17

1965: -0.10

1978: $\frac{1}{50}$

2002: 0.54

Convert the fraction $\frac{1}{50}$ to a decimal:

So, the difference for 1978 is 0.02°C.

The temperature differences in decimal form are:

1958: 0.10

1964: -0.17

1965: -0.10

1978: 0.02

2002: 0.54

(a) Order the five years from coldest to warmest.

To order from coldest to warmest, we order the differences from the most negative to the most positive.

Ordering the differences: -0.17, -0.10, 0.02, 0.10, 0.54

The corresponding years are:

-0.17 corresponds to 1964

-0.10 corresponds to 1965

0.02 corresponds to 1978

0.10 corresponds to 1958

0.54 corresponds to 2002

Ordering the years from coldest to warmest: 1964, 1965, 1978, 1958, 2002.

(b) Where should 1946 fall in the ordered list?

The temperature variation for 1946 is -0.03°C.

Insert -0.03 into the ordered list of differences: -0.17, -0.10, -0.03, 0.02, 0.10, 0.54.

The value -0.03 falls between -0.10 (for 1965) and 0.02 (for 1978).

This corresponds to the years 1965 and 1978.

When the years are ordered from coldest to warmest, 1946 should fall between 1965 and 1978.

Given Atomic Weights:

Atomic weight of Helium = 4.0030

Atomic weight of Hydrogen = 1.0080

Atomic weight of Oxygen = 16.0000

(a) Difference between the atomic weights of Oxygen and Hydrogen.

Difference = Atomic weight of Oxygen - Atomic weight of Hydrogen

Performing the subtraction:

$\begin{array}{cc} & 16 & . & 0 & 0 & 0 & 0 \\ - & \phantom{0}1 & . & 0 & 0 & 8 & 0 \\ \hline & 14 & . & 9 & 9 & 2 & 0 \\ \hline \end{array}$

The difference between the atomic weights of Oxygen and Hydrogen is 14.9920.

(b) Difference between the atomic weights of Oxygen and Helium.

Difference = Atomic weight of Oxygen - Atomic weight of Helium

Performing the subtraction:

$\begin{array}{cc} & 16 & . & 0 & 0 & 0 & 0 \\ - & \phantom{0}4 & . & 0 & 0 & 3 & 0 \\ \hline & 11 & . & 9 & 9 & 7 & 0 \\ \hline \end{array}$

The difference between the atomic weights of Oxygen and Helium is 11.9970.

(c) Difference between the atomic weights of Helium and Hydrogen.

Difference = Atomic weight of Helium - Atomic weight of Hydrogen

Performing the subtraction:

$\begin{array}{cc} & 4 & . & 0 & 0 & 3 & 0 \\ - & 1 & . & 0 & 0 & 8 & 0 \\ \hline & 2 & . & 9 & 9 & 5 & 0 \\ \hline \end{array}$

The difference between the atomic weights of Helium and Hydrogen is 2.9950.

Given:

Correct length of the iron rod = 19.33 cm

Ravi's measurement = 19.34 cm

Kamal's measurement = 19.25 cm

Tabish's measurement = 19.27 cm

To Find:

The error made by each of the boys.

Solution:

The error in measurement is the difference between the measured value and the correct value. Error can be positive (if the measurement is higher) or negative (if the measurement is lower). We will calculate the signed error for each measurement.

Error made by Ravi:

Performing the subtraction:

$\begin{array}{cc} & 19 & . & 3 & 4 \\ - & 19 & . & 3 & 3 \\ \hline & \phantom{00}0 & . & 0 & 1 \\ \hline \end{array}$

Error made by Kamal:

Performing the subtraction (subtracting a larger number from a smaller number):

$\begin{array}{cc} & 19 & . & 3 & 3 \\ - & 19 & . & 2 & 5 \\ \hline & \phantom{00}0 & . & 0 & 8 \\ \hline \end{array}$

Since $19.25 < 19.33$, the difference is negative.

Error made by Tabish:

Performing the subtraction (subtracting a larger number from a smaller number):

$\begin{array}{cc} & 19 & . & 3 & 3 \\ - & 19 & . & 2 & 7 \\ \hline & \phantom{00}0 & . & 0 & 6 \\ \hline \end{array}$

Since $19.27 < 19.33$, the difference is negative.

The errors made by the boys are:

Ravi: 0.01 cm

Kamal: -0.08 cm

Tabish: -0.06 cm

(Note: The magnitude of the errors are 0.01 cm, 0.08 cm, and 0.06 cm respectively. The negative sign indicates the measurement was lower than the correct value.)

Given:

We need to find the quotient of $0.02964 \div 0.004$.

To Find:

The value of the quotient.

Solution:

We can write the division as a fraction:

To divide decimals, we can make the divisor a whole number by multiplying both the numerator and the denominator by the same power of 10.

The divisor 0.004 has three decimal places. So, we multiply both by $10^3 = 1000$.

Now, we perform the division of 29.64 by 4.

We can divide the whole number part first, then the decimal part.

Place the decimal point in the quotient and carry over the remainder 1 to the tenths place.

Putting it together, the quotient is 7.41.

Alternatively, using long division format:

$\begin{array}{r} 7.41\phantom{)} \\ 4{\overline{\smash{\big)}\,29.64\phantom{)}}} \\ \underline{-~\phantom{(}28\phantom{.00)}} \\ 1\,6\phantom{0)} \\ \underline{-~\phantom{()}(16)\phantom{0)}} \\ 04\phantom{)} \\ \underline{-~\phantom{()}(4)} \\ 0\phantom{)} \end{array}$

The quotient when 0.02964 is divided by 0.004 is 7.41.

Given:

Let the unknown number be $x$.

The quotient of $x$ divided by 520 is equal to the quotient of 85 divided by 0.625.

To Find:

The value of $x$.

Solution:

We can write the given information as an equation:

First, let's evaluate the quotient on the right side: $\frac{85}{0.625}$.

We can convert the decimal 0.625 to a fraction.

Simplify the fraction by dividing the numerator and denominator by their greatest common divisor (which is 125).

So, the right side of the equation is $85 \div \frac{5}{8}$.

To divide by a fraction, multiply by its reciprocal:

Multiply and simplify:

Now, substitute this value back into equation (i):

To find $x$, multiply both sides by 520:

Performing the multiplication:

$\begin{array}{cc}& & & 1 & 3 & 6 \\ \times & & & 5 & 2 & 0 \\ \hline & & & 0 & 0 & 0 \\ & & 2 & 7 & 2 & \times \\ & 6 & 8 & 0 & \times & \times \\ \hline & 7 & 0 & 7 & 2 & 0 \\ \hline \end{array}$

The number that divided by 520 gives the same quotient as 85 divided by 0.625 is 70720.

Given:

Length of floor = 4.5 m

Width of floor = 3.6 m

Side length of square tile = 6 cm

Cost per tile = $\textsf{₹}$ 23.25

To Find:

The total cost to cover the floor with these tiles.

Solution:

First, ensure all measurements are in the same unit. We will convert the floor dimensions from meters to centimeters.

1 meter = 100 centimeters.

Calculate the area of the floor.

$\begin{array}{cc}& & 4 & 5 & 0 \\ \times & & 3 & 6 & 0 \\ \hline & & 0 & 0 & 0 \\ & 2 & 7 & 0 & \times \\ 1 & 3 & 5 & \times & \times \\ \hline 1 & 6 & 2 & 0 & 0 & 0 \\ \hline \end{array}$

Calculate the area of one square tile.

Calculate the number of tiles needed to cover the floor.

Performing the division:

Calculate the total cost to cover the floor.

$\begin{array}{cc}& & & 2 & 3 & . & 2 & 5 \\ \times & & & 4 & 5 & 0 & 0 \\ \hline & & & 0 & 0 & 0 & 0 \\ & & 0 & 0 & 0 & 0 & \times \\ & 1 & 1 & 6 & 2 & 5 & \times & \times \\ 9 & 3 & 0 & 0 & \times & \times & \times \\ \hline 1 & 0 & 4 & 6 & 2 & 5 & . & 0 & 0 \\ \hline \end{array}$

The cost to cover the floor with these tiles will be $\textsf{₹}$ 104625.

Given:

Initial amount of cloth Sunita has = $\frac{3}{4}$ m

Fraction of the cloth Sunita gave to Rehana = $\frac{1}{3}$

To Find:

The amount of cloth Rehana received.

Solution:

The amount of cloth Sunita gave to Rehana is $\frac{1}{3}$ of the cloth Sunita had, which is $\frac{1}{3}$ of $\frac{3}{4}$ m.

In mathematics, "of" between fractions means multiplication.

Multiply the numerators together and the denominators together:

Simplify the fraction by cancelling common factors. We can cancel the '3' in the numerator with the '3' in the denominator.

Therefore, Rehana received $\frac{1}{4}$ m of cloth.

Given:

Length of the flower garden border = 22.50 m

Length of each brick = 0.25 m

To Find:

The number of bricks needed to make the border.

Solution:

To find the number of bricks needed, we divide the total length of the border by the length of one brick.

To divide decimals, we can make the divisor a whole number by multiplying both the numerator and the denominator by the same power of 10.

The divisor 0.25 has two decimal places. So, we multiply both by $10^2 = 100$.

Now, we perform the division of 2250 by 25.

We know that $25 \times 4 = 100$, so $25 \times 40 = 1000$.

We can also perform long division:

$\begin{array}{r} 90\phantom{)} \\ 25{\overline{\smash{\big)}\,2250\phantom{)}}} \\ \underline{-~\phantom{(}225\phantom{0)}} \\ 000\phantom{)} \\ \underline{-~\phantom{()}(000)} \\ 0\phantom{)} \end{array}$

Therefore, 90 bricks will be needed to make the border.

Given:

Number of shirts to be made = 6

Cloth required for each shirt (for the shirt itself) = $2\frac{1}{4}$ m

Cloth for waste in cutting and finishing for each shirt = $\frac{1}{8}$ m

To Find:

Total cloth used in making 6 shirts.

Solution:

First, calculate the total amount of cloth needed for one shirt, including the cloth for the shirt itself and the cloth for waste.

Convert the mixed number $2\frac{1}{4}$ to an improper fraction:

Total cloth needed for one shirt = (Cloth for shirt) + (Cloth for waste)

To add these fractions, find a common denominator. The least common multiple of 4 and 8 is 8.

Now, add the fractions:

Now, calculate the total cloth needed for 6 shirts by multiplying the cloth needed per shirt by the number of shirts.

Write 6 as $\frac{6}{1}$ and multiply the fractions:

Simplify the fraction by cancelling the common factor 2 from 6 and 8.

The answer can be left as an improper fraction, or converted to a mixed number or decimal.

As a mixed number: $\frac{57}{4} = \frac{56+1}{4} = \frac{56}{4} + \frac{1}{4} = 14 + \frac{1}{4} = 14\frac{1}{4}$ m.

As a decimal: $\frac{57}{4} = 14.25$ m.

Therefore, $\frac{57}{4}$ m (or $14\frac{1}{4}$ m or 14.25 m) of cloth will be used in making 6 shirts.

Given:

Total seating capacity of the picture hall = 820 persons

Actual number of tickets sold (number of persons) = 648

First usher's guess = $\frac{3}{4}$ full

Second usher's guess = $\frac{2}{3}$ full

To Find:

Which usher made the better guess.

Solution:

First, calculate the number of people corresponding to each usher's guess by multiplying the fraction by the total seating capacity.

First usher's guess:

Multiply and simplify:

Second usher's guess:

Multiply:

Performing the division $\frac{1640}{3}$: $1640 \div 3 \approx 546.67$ people.

Since the number of people must be a whole number, this fraction may indicate the guess was approximate, or there might be a slight inconsistency in the numbers provided for a perfectly exact match.

Let's express the fractions in terms of 820 and compare with 648. First usher: $\frac{3}{4} \times 820 = 615$. Second usher: $\frac{2}{3} \times 820 = \frac{1640}{3}$.

To compare whose guess is better, we calculate the difference (or the absolute difference) between each usher's guess and the actual number of people (648).

Difference for First usher:

Difference for Second usher:

To perform the subtraction, find a common denominator (3) for 648.

As a decimal, $\frac{304}{3} \approx 101.33$.

Comparing the absolute differences:

First usher's difference = 33 people

Second usher's difference = $\frac{304}{3} \approx 101.33$ people

Since 33 is less than $\frac{304}{3}$, the first usher's guess (615 people) was closer to the actual number of people (648).

The first usher made the better guess.

Given:

Cost of sweets = $\textsf{₹}$ 740.25

Cost of cold drink = $\textsf{₹}$ 70

Number of students who contributed = 35

Each student contributed equally.

To Find:

Amount contributed by each student.

Solution:

First, calculate the total cost of the sweets and cold drinks.

Performing the addition:

$\begin{array}{cc}& 7 & 4 & 0 & . & 2 & 5 \\ + & \phantom{00}7 & 0 & . & 0 & 0 \\ \hline & 8 & 1 & 0 & . & 2 & 5 \\ \hline \end{array}$

Since 35 students contributed equally to cover the total cost, the amount contributed by each student is the total cost divided by the number of students.

Performing the division $810.25 \div 35$:

$\begin{array}{r} 23.15\phantom{)} \\ 35{\overline{\smash{\big)}\,810.25\phantom{)}}} \\ \underline{-~\phantom{(}70\phantom{0.00)}} \\ 110\phantom{.00)} \\ \underline{-~\phantom{()}105\phantom{.00)}} \\ 52\phantom{0)} \\ \underline{-~\phantom{()}(35)\phantom{0)}} \\ 175\phantom{)} \\ \underline{-~\phantom{()}(175)} \\ 0\phantom{)} \end{array}$

Each student contributed $\textsf{₹}$ 23.15.

Given:

Rohan's times in five races (in minutes): 3.20, 3.37, 3.29, 3.17, 3.32

Number of races = 5

To Find:

The average time taken by Rohan in the races.

Solution:

The average time is calculated by summing all the times and then dividing by the number of races.

First, calculate the sum of the times:

Performing the addition:

$\begin{array}{cc} & 3 & . & 2 & 0 \\ & 3 & . & 3 & 7 \\ & 3 & . & 2 & 9 \\ & 3 & . & 1 & 7 \\ + & 3 & . & 3 & 2 \\ \hline & 16 & . & 3 & 5 \\ \hline \end{array}$

Now, divide the sum by the number of races (5).

Performing the division $16.35 \div 5$:

$\begin{array}{r} 3.27\phantom{)} \\ 5{\overline{\smash{\big)}\,16.35\phantom{)}}} \\ \underline{-~\phantom{(}15\phantom{.00)}} \\ 13\phantom{0)} \\ \underline{-~\phantom{()}10\phantom{0)}} \\ 35\phantom{)} \\ \underline{-~\phantom{()}35)} \\ 0\phantom{)} \end{array}$

The average time taken by Rohan in the races is 3.27 minutes.

Given:

Total length of the sewer line = $80\frac{1}{4}$ m

Length of sewer line completed in one day = 7.5 m

To Find:

The total time taken to complete the project (in days).

Solution:

First, convert the total length of the sewer line from a mixed number to an improper fraction or a decimal.

Alternatively, as a decimal: $\frac{1}{4} = 0.25$, so $80\frac{1}{4} = 80 + 0.25 = 80.25$ m.

The rate of work is 7.5 m per day. Convert this decimal to a fraction:

To find the total time taken, we divide the total length of the sewer line by the length completed in one day.

To divide by a fraction, multiply by its reciprocal.

Multiply the numerators and denominators:

Simplify by cancelling the common factor 2 from 2 and 4.

Simplify the fraction $\frac{321}{30}$ by cancelling the common factor 3.

To express this as a decimal:

The project will take 10.7 days to complete.

Given:

Weight on Moon = $\frac{1}{6} \times$ Weight on Earth

Weight of the object on Earth = $5\frac{3}{5}$ kg

To Find:

The weight of the object on the Moon.

Solution:

First, convert the weight on Earth from a mixed number to an improper fraction.

Now, use the given rule to find the weight on the Moon.

Multiply the numerators and the denominators:

Simplify the fraction by cancelling the common factor 2 from 28 and 6.

The weight of the object on the Moon would be $\frac{14}{15}$ kg.

Given:

Total number of students surveyed = 200

The circle graph shows the percentage of students influenced by different factors.

From the circle graph:

• Radio: 30%
• Music Video Channel: 25%
• Friend or Relative: 20%
• Heard in Shop: 15%
• Other: 10%

Check that the percentages sum to 100%: $30\% + 25\% + 20\% + 15\% + 10\% = 100\%$.

(a) How many students said radio influenced them most?

The percentage of students influenced by radio is 30%.

60 students said radio influenced them most.

(b) How many more students were influenced by radio than by a music video channel?

Number of students influenced by Radio (from part a) = 60.

Percentage of students influenced by Music Video Channel = 25%.

Difference in the number of students = Number (Radio) - Number (Music Video Channel)

10 more students were influenced by radio than by a music video channel.

(c) How many said a friend or relative influenced them or they heard the CD in a shop?

This question asks for the total number of students who were influenced by either a friend or relative OR heard the CD in a shop. We sum the percentages for these two categories and then find the number of students.

Percentage influenced by Friend or Relative = 20%

Percentage influenced by Heard in Shop = 15%

Total percentage for these two categories = $20\% + 15\% = 35\%$

70 students said a friend or relative influenced them or they heard the CD in a shop.

Given:

Initial amount of milk in the can = $5\frac{1}{2}$ L

Milk sold to Renu = $\frac{3}{4}$ L

Milk sold to Kamla = $\frac{3}{4}$ L

Milk sold to Renuka = $\frac{3}{4}$ L

Milk sold to Shadma = $\frac{7}{8}$ L

Milk given to Jassi = $1\frac{1}{2}$ L

To Find:

Amount of milk left in the can.

Solution:

First, calculate the total amount of milk sold/given away.

Convert all mixed numbers to improper fractions:

Total milk sold/given = (Milk to Renu) + (Milk to Kamla) + (Milk to Renuka) + (Milk to Shadma) + (Milk to Jassi)

Find a common denominator for the fractions (4, 4, 4, 8, 2). The least common multiple is 8.

Add the fractions with the common denominator:

Now, calculate the amount of milk left in the can by subtracting the total milk out from the initial amount.

Find a common denominator for 2 and 8, which is 8.

Subtract the fractions:

Therefore, $\frac{7}{8}$ L of milk is left in the can.

Given:

Total time taken by Anuradha to complete a piece of work = 6 hours.

To Find:

The part of the work done in 1 hour, 5 hours, and 6 hours.

Solution:

We assume the total piece of work is 1 unit.

If Anuradha completes the entire work (1 unit) in 6 hours, then her work rate (work done per hour) is:

Part of the work done in 1 hour:

Part of the work done in 5 hours:

Part of the work done in 6 hours:

Anuradha can do:

In 1 hour: $\frac{1}{6}$ part of the work.

In 5 hours: $\frac{5}{6}$ part of the work.

In 6 hours: 1 part (the entire work).

Given:

Total time required to paint the whole saree = 5 hours.

To Find:

The portion of the saree Rehana can paint in 1 hour, $4\frac{3}{5}$ hours, and $3\frac{1}{2}$ hours.

Solution:

We assume the whole saree represents 1 unit of work.

If Rehana paints the whole saree (1 unit) in 5 hours, her painting rate (portion painted per hour) is:

Portion of the saree painted in 1 hour:

Portion of the saree painted in $4\frac{3}{5}$ hours:

Convert the mixed number $4\frac{3}{5}$ to an improper fraction:

Now, calculate the portion painted:

Portion of the saree painted in $3\frac{1}{2}$ hours:

Convert the mixed number $3\frac{1}{2}$ to an improper fraction:

Now, calculate the portion painted:

Rehana can paint:

In 1 hour: $\frac{1}{5}$ portion of the saree.

In $4\frac{3}{5}$ hours: $\frac{23}{25}$ portion of the saree.

In $3\frac{1}{2}$ hours: $\frac{7}{10}$ portion of the saree.

Given:

Total quantity of cotton wool Rama has = $6\frac{1}{4}$ kg

Quantity of cotton wool needed for one pillow = $1\frac{1}{4}$ kg

To Find:

The number of pillows Rama can make.

Solution:

First, convert the mixed numbers to improper fractions.

Total cotton wool:

Cotton wool needed per pillow:

To find the number of pillows that can be made, we divide the total quantity of cotton wool by the quantity needed for one pillow.

To divide by a fraction, we multiply by its reciprocal.

The reciprocal of $\frac{5}{4}$ is $\frac{4}{5}$.

Multiply the numerators and the denominators:

Simplify the expression by cancelling common factors from the numerator and the denominator.

Therefore, Rama can make 5 pillows.

Given:

Length of cloth required to make one shirt = $2\frac{1}{3}$ m

Total length of cloth Radhika has = $9\frac{1}{3}$ m

To Find:

The number of shirts Radhika can make.

Solution:

First, convert the mixed numbers to improper fractions.

Length of cloth per shirt:

Total length of cloth:

To find the number of shirts that can be made, we divide the total length of cloth by the length of cloth needed for one shirt.

To divide by a fraction, we multiply by its reciprocal.

The reciprocal of $\frac{7}{3}$ is $\frac{3}{7}$.

Multiply the numerators and the denominators:

Simplify the expression by cancelling common factors from the numerator and the denominator.

Therefore, Radhika can make 4 shirts from the piece of cloth.

Given:

Distance Ravi can walk in one hour (Ravi's speed) = $3\frac{1}{3}$ km/hour

Distance from home to office = 10 km

To Find:

The time it will take Ravi to walk to his office.

Solution:

First, convert Ravi's speed from a mixed number to an improper fraction.

The relationship between distance, speed, and time is given by: Distance = Speed $\times$ Time.

We want to find the Time, so we rearrange the formula:

Substitute the given values into the formula:

To divide by a fraction, we multiply by its reciprocal.

The reciprocal of $\frac{10}{3}$ is $\frac{3}{10}$.

We can write 10 as $\frac{10}{1}$ and multiply the fractions:

Simplify by cancelling the common factor 10 from the numerator and the denominator.

It will take Ravi 3 hours to walk to his office.

Given:

Distance travelled on $\frac{3}{5}$ of the petrol tank = 360 km.

We assume the mileage (km per unit of petrol) remains constant.

To Find:

The distance Raj would travel with a full tank of petrol (which is $\frac{5}{5}$ or 1 whole tank).

Solution:

Let the total capacity of the petrol tank be $T$ (in some unit of volume). The amount of petrol used is $\frac{3}{5}T$. The distance travelled is 360 km.

Let the mileage of the car be $M$ km per unit of petrol.

We know that Distance = Mileage $\times$ Amount of petrol.

We want to find the distance travelled with a full tank, which is $T$.

From equation (i), we can find the value of $M \times T$.

To isolate $M \times T$, divide both sides by $\frac{3}{5}$ (or multiply by its reciprocal $\frac{5}{3}$).

Multiply and simplify:

So, $M \times T = 600$.

According to equation (ii), the distance travelled with a full tank is equal to $M \times T$.

Alternatively, we can reason as follows:

If $\frac{3}{5}$ of the tank gives a distance of 360 km, then $\frac{1}{5}$ of the tank will give $\frac{1}{3}$ of the distance.

A full tank is $\frac{5}{5}$, which is 5 times $\frac{1}{5}$.

Raj would travel 600 km with a full tank of petrol.

Given:

Amount Kajol has = $\textsf{₹}$ 75.

This amount ($\textsf{₹}$ 75) is $\frac{3}{8}$ of the total amount she earned.

To Find:

The total amount Kajol earned.

Solution:

Let the total amount Kajol earned be $E$ (in $\textsf{₹}$).

According to the problem statement, $\textsf{₹}$ 75 is $\frac{3}{8}$ of $E$.

We can write this as an equation:

To find the total amount earned ($E$), we need to isolate $E$. We can do this by multiplying both sides of equation (i) by the reciprocal of $\frac{3}{8}$, which is $\frac{8}{3}$.

We can write 75 as $\frac{75}{1}$ and multiply the fractions.

Simplify the expression by cancelling the common factor 3 from the numerator (75) and the denominator (3).

Alternatively, we can reason as follows:

If $\frac{3}{8}$ of the earning is $\textsf{₹}$ 75, then $\frac{1}{8}$ of the earning will be $\frac{1}{3}$ of $\textsf{₹}$ 75.

A full earning (the whole amount) is $\frac{8}{8}$, which is 8 times $\frac{1}{8}$.

Kajol earned $\textsf{₹}$ 200.

Given:

Number of trees needed to make 1 tonne of paper = 17

Total number of such trees in the forest = 221

(i) What fraction of the forest will be used to make:

The entire forest is represented by the total number of trees, which is 221.

The fraction of the forest used for a certain amount of paper is $\frac{\text{Number of trees used}}{\text{Total number of trees in the forest}}$.

Number of trees used to make 1 tonne of paper = 17.

Fraction of forest for 1 tonne of paper = $\frac{17}{221}$.

We can simplify this fraction. Let's find the prime factors of 17 and 221.

17 is a prime number.

For 221, try dividing by small primes: Not divisible by 2, 3, 5. Try 7: $221 \div 7 \approx 31.5$. Try 11: $221 \div 11 \approx 20.09$. Try 13: $221 \div 13 = 17$.

So, $221 = 13 \times 17$.

The fraction $\frac{17}{221} = \frac{\cancel{17}^{1}}{13 \times \cancel{17}_{1}} = \frac{1}{13}$.

So, $\frac{1}{13}$ of the forest is used to make 1 tonne of paper.

(a) 5 tonnes of paper.

To make 5 tonnes of paper, the number of trees needed is $5 \times 17 = 85$ trees.

Simplify the fraction. We know $221 = 13 \times 17$ and $85 = 5 \times 17$.

To make 5 tonnes of paper, $\frac{5}{13}$ part of the forest will be used.

(b) 10 tonnes of paper.

To make 10 tonnes of paper, the number of trees needed is $10 \times 17 = 170$ trees.

Simplify the fraction. We know $221 = 13 \times 17$ and $170 = 10 \times 17$.

To make 10 tonnes of paper, $\frac{10}{13}$ part of the forest will be used.

(ii) To save $\frac{7}{13}$ part of the forest, how much paper do we have to save?

Saving $\frac{7}{13}$ part of the forest means not using $\frac{7}{13}$ of the total trees.

Multiply and simplify (we know $221 = 13 \times 17$):

We know that 17 trees make 1 tonne of paper.

To find out how much paper 119 trees make, we divide the number of trees saved by the number of trees needed for 1 tonne.

Performing the division: $119 \div 17 = 7$.

To save $\frac{7}{13}$ part of the forest, we have to save 7 tonnes of paper.

Given Expression:

$\left(1 \div \frac{2}{9} \right) + \left(1 \div 3\frac{1}{5} \right) + \left(1 \div 2\frac{2}{3} \right)$

Solution:

We will simplify each term in the expression separately.

Remember that dividing by a fraction is the same as multiplying by its reciprocal.

Term 1: $1 \div \frac{2}{9}$

The reciprocal of $\frac{2}{9}$ is $\frac{9}{2}$.

Term 2: $1 \div 3\frac{1}{5}$

First, convert the mixed number to an improper fraction:

Now, find the reciprocal of $\frac{16}{5}$, which is $\frac{5}{16}$.

Term 3: $1 \div 2\frac{2}{3}$

First, convert the mixed number to an improper fraction:

Now, find the reciprocal of $\frac{8}{3}$, which is $\frac{3}{8}$.

Now, add the results of the three terms:

To add these fractions, find a common denominator. The least common multiple (LCM) of 2, 16, and 8 is 16.

Add the fractions:

Finally, write the result in decimal form by dividing the numerator by the denominator.

Performing the division:

$\begin{array}{r} 5.1875\phantom{)} \\ 16{\overline{\smash{\big)}\,83.0000\phantom{)}}} \\ \underline{-~\phantom{(}80\phantom{.000)}} \\ 3\,0\phantom{00)} \\ \underline{-~\phantom{()}1\,6\phantom{00)}} \\ 14\,0\phantom{0)} \\ \underline{-~\phantom{()}12\,8\phantom{0)}} \\ 1\,20\phantom{)} \\ \underline{-~\phantom{()}1\,12\phantom{)}} \\ 80\phantom{)} \\ \underline{-~\phantom{()}(80)} \\ 0\phantom{)} \end{array}$

The simplified result in decimal form is 5.1875.

We need to match each mathematical expression with the corresponding picture based on how multiplication of fractions is represented visually (typically as repeated addition).

Let's analyze each expression and the pictures:

(1) $2 \times \frac{1}{4}$: This represents adding $\frac{1}{4}$ two times. We look for a picture showing 2 groups, each with $\frac{1}{4}$ shaded.

Picture (d) shows 2 objects, each divided into 4 parts with 1 part shaded. This represents $2 \times \frac{1}{4}$.

(2) $2 \times \frac{3}{7}$: This represents adding $\frac{3}{7}$ two times. We look for a picture showing 2 groups, each with $\frac{3}{7}$ shaded.

Picture (e) shows 2 objects, each divided into 7 parts with 3 parts shaded. This represents $2 \times \frac{3}{7}$.

(3) $2 \times \frac{1}{3}$: This represents adding $\frac{1}{3}$ two times. We look for a picture showing 2 groups, each with $\frac{1}{3}$ shaded.

Picture (b) shows 2 objects, each divided into 3 parts with 1 part shaded. This represents $2 \times \frac{1}{3}$.

(4) $\frac{1}{4} \times 4$: This represents taking $\frac{1}{4}$ of 4 or adding $\frac{1}{4}$ four times. We look for a picture showing 4 groups, each with $\frac{1}{4}$ shaded.

Picture (f) shows 4 objects, each divided into 4 parts with 1 part shaded. This represents $4 \times \frac{1}{4}$ or $\frac{1}{4} \times 4$.

(5) $3 \times \frac{2}{9}$: This represents adding $\frac{2}{9}$ three times. We look for a picture showing 3 groups, each with $\frac{2}{9}$ shaded.

Picture (c) shows 3 objects, each divided into 9 parts with 2 parts shaded. This represents $3 \times \frac{2}{9}$.

(6) $\frac{1}{4} \times 3$: This represents taking $\frac{1}{4}$ of 3 or adding $\frac{1}{4}$ three times. We look for a picture showing 3 groups, each with $\frac{1}{4}$ shaded.

Picture (a) shows 3 objects, each divided into 4 parts with 1 part shaded. This represents $3 \times \frac{1}{4}$ or $\frac{1}{4} \times 3$.

Matching the expressions to the pictures:

(1) $2 \times \frac{1}{4}$ corresponds to Picture (d).

(2) $2 \times \frac{3}{7}$ corresponds to Picture (e).

(3) $2 \times \frac{1}{3}$ corresponds to Picture (b).

(4) $\frac{1}{4} \times 4$ corresponds to Picture (f).

(5) $3 \times \frac{2}{9}$ corresponds to Picture (c).

(6) $\frac{1}{4} \times 3$ corresponds to Picture (a).

Given Expression:

$(0.3) \times (0.3) – (0.2) \times (0.2)$

Solution:

We need to perform the multiplications first, and then the subtraction.

First term: $(0.3) \times (0.3)$

Second term: $(0.2) \times (0.2)$

Now, subtract the second term from the first term:

Performing the subtraction:

$\begin{array}{cc} & 0 & . & 0 & 9 \\ - & 0 & . & 0 & 4 \\ \hline & 0 & . & 0 & 5 \\ \hline \end{array}$

The value of the expression is 0.05.

Given Expression:

$\frac{0.6}{0.3} + \frac{0.16}{0.4}$

Solution:

We will evaluate each term of the expression separately and then add the results.

Term 1: $\frac{0.6}{0.3}$

To divide decimals, multiply the numerator and the denominator by a power of 10 such that the divisor becomes a whole number. In this case, multiply by 10.

Term 2: $\frac{0.16}{0.4}$

Multiply the numerator and the denominator by a power of 10 such that the divisor becomes a whole number. In this case, multiply by 10.

Now, divide 1.6 by 4.

Alternatively, we can multiply by 100 to remove decimals from both:

Simplify the fraction:

Convert the fraction to a decimal:

Add the results of Term 1 and Term 2:

The value of the expression is 2.4.

Given Expression:

$\frac{(0.2 \;×\; 0.14)\; + \;(0.5 \;×\; 0.91)}{(0.1 \;×\; 0.2)}$

Solution:

First, calculate the values of the terms in the numerator and the denominator.

Calculate the first product in the numerator: $0.2 \times 0.14$

Calculate the second product in the numerator: $0.5 \times 0.91$

Calculate the sum in the numerator: $(0.2 \times 0.14) + (0.5 \times 0.91)$

Performing the addition:

$\begin{array}{cc} & 0 & . & 0 & 2 & 8 \\ + & 0 & . & 4 & 5 & 5 \\ \hline & 0 & . & 4 & 8 & 3 \\ \hline \end{array}$

Calculate the product in the denominator: $0.1 \times 0.2$

Now, divide the numerator by the denominator:

To divide by a decimal, multiply the numerator and denominator by a power of 10 to make the denominator a whole number. Multiply by 100 (since 0.02 has 2 decimal places).

Performing the division $48.3 \div 2$:

$\begin{array}{r} 24.15\phantom{)} \\ 2{\overline{\smash{\big)}\,48.30\phantom{)}}} \\ \underline{-~\phantom{(}4\phantom{0.00)}} \\ 08\phantom{0)} \\ \underline{-~\phantom{()}8\phantom{0)}} \\ 03\phantom{)} \\ \underline{-~\phantom{()}2\phantom{)}} \\ 10\phantom{)} \\ \underline{-~\phantom{()}10)} \\ 0\phantom{)} \end{array}$

The value of the expression is 24.15.

Given:

An equilateral triangle and a square have a side in common.

The length of a side of the equilateral triangle = $\frac{4}{3}$ cm.

To Find:

The perimeter of the figure formed.

Solution:

Since the triangle is equilateral, all its sides are equal in length. Thus, each side of the triangle is $\frac{4}{3}$ cm.

The square shares one side with the equilateral triangle. Therefore, the side length of the square is also equal to the side length of the triangle.

A square has 4 equal sides.

The perimeter of the combined figure is the total length of its outer boundary.

The outer boundary consists of the sides of the square and the triangle that are NOT the common side.

Number of outer sides of the square = Total sides of square - Common side = $4 - 1 = 3$ sides.

Number of outer sides of the triangle = Total sides of triangle - Common side = $3 - 1 = 2$ sides.

Perimeter of the figure = (Length of 3 sides of the square) + (Length of 2 sides of the triangle)

Calculate the products:

Add the results:

To add a whole number and a fraction, write the whole number as a fraction with the same denominator as the other fraction (3).

The perimeter of the figure formed is $\frac{20}{3}$ cm.

(As a mixed number: $\frac{20}{3} = 6\frac{2}{3}$ cm)

Given:

Carpet size: Length = 4 m, Width = $6\frac{2}{3}$ m

Room size: Length = $3\frac{1}{3}$ m, Width = $5\frac{1}{3}$ m

To Find:

The fraction of the carpet's area that should be cut off.

Solution:

First, convert all mixed numbers to improper fractions.

Carpet width: $6\frac{2}{3} = \frac{(6 \times 3) + 2}{3} = \frac{18+2}{3} = \frac{20}{3}$ m

Room length: $3\frac{1}{3} = \frac{(3 \times 3) + 1}{3} = \frac{9+1}{3} = \frac{10}{3}$ m

Room width: $5\frac{1}{3} = \frac{(5 \times 3) + 1}{3} = \frac{15+1}{3} = \frac{16}{3}$ m

Calculate the area of the carpet.

Calculate the area of the room.

The area to be cut off is the difference between the carpet area and the room area.

To subtract the fractions, find a common denominator. The LCM of 3 and 9 is 9.

We need to find the fraction of the original carpet area that should be cut off.

To divide by a fraction, multiply by its reciprocal.

Multiply and simplify:

Therefore, $\frac{1}{3}$ of the carpet's area should be cut off to fit wall to wall into the room.

Given:

Photograph dimensions (excluding the border):

Length = $14\frac{2}{5}$ cm

Breadth = $10\frac{2}{5}$ cm

Width of the uniform border = $2\frac{3}{5}$ cm

To Find:

The area of the framed photograph (which includes the photograph and the border).

Solution:

First, convert all mixed numbers to improper fractions.

Photograph length: $14\frac{2}{5} = \frac{(14 \times 5) + 2}{5} = \frac{70+2}{5} = \frac{72}{5}$ cm

Photograph breadth: $10\frac{2}{5} = \frac{(10 \times 5) + 2}{5} = \frac{50+2}{5} = \frac{52}{5}$ cm

Border width: $2\frac{3}{5} = \frac{(2 \times 5) + 3}{5} = \frac{10+3}{5} = \frac{13}{5}$ cm

The framed photograph includes the border around the original photograph. Since the border is uniform, its width is added to both the length and the breadth of the photograph on both sides.

Now, calculate the area of the framed photograph.

Multiply the numerators and denominators:

The area of the framed photograph is $\frac{7644}{25}$ cm$^2$.

As a decimal: $\frac{7644}{25} = \frac{7644 \times 4}{25 \times 4} = \frac{30576}{100} = 305.76$ cm$^2$.

Given:

Cost of one burger = $\textsf{₹}$ $20\frac{3}{4}$

Cost of one Macpuff = $\textsf{₹}$ $15\frac{1}{2}$

To Find:

The total cost of 4 burgers and 14 Macpuffs.

Solution:

First, convert the mixed numbers to improper fractions.

Cost of one burger: $20\frac{3}{4} = \frac{(20 \times 4) + 3}{4} = \frac{80+3}{4} = \frac{83}{4}$ $\textsf{₹}$

Cost of one Macpuff: $15\frac{1}{2} = \frac{(15 \times 2) + 1}{2} = \frac{30+1}{2} = \frac{31}{2}$ $\textsf{₹}$

Calculate the cost of 4 burgers.

Multiply and simplify:

Calculate the cost of 14 Macpuffs.

Multiply and simplify:

Calculate the total cost.

The cost of 4 burgers and 14 Macpuffs is $\textsf{₹}$ 300.

Given:

Total height of the hill = $101\frac{1}{3}$ m

Fraction of the height that is under water = $\frac{1}{4}$

To Find:

The height of the hill visible above the water.

Solution:

First, convert the total height of the hill from a mixed number to an improper fraction.

Calculate the height of the hill that is under water. This is $\frac{1}{4}$ of the total height.

Multiply the fractions:

Simplify by cancelling the common factor 4 from 304 and 4.

Now, calculate the height of the hill visible above the water by subtracting the height under water from the total height.

Since the fractions have the same denominator, subtract the numerators.

Simplify the fraction by performing the division $228 \div 3$.

The height of the hill visible above the water is 76 m.

Alternate Solution:

If $\frac{1}{4}$ of the height is under water, then the remaining portion is visible above the water.

Now, calculate $\frac{3}{4}$ of the total height.

Multiply and simplify:

Both methods yield the same result.

Given:

Reaction time = 0.214 seconds

Total race time (including reaction time) = 11.03 seconds

To Find:

Time taken to run the actual distance.

Solution:

The total race time is the sum of the reaction time and the time taken to run the actual distance.

To find the time taken to run the actual distance, we subtract the reaction time from the total race time.

To perform the subtraction, align the decimal points and add a zero to 11.03 to match the number of decimal places in 0.214 (three decimal places).

Performing the subtraction:

$\begin{array}{cc} & 11 & . & 0 & 3 & 0 \\ - & \phantom{0}0 & . & 2 & 1 & 4 \\ \hline & 10 & . & 8 & 1 & 6 \\ \hline \end{array}$

It took her 10.816 seconds to run the actual distance.

To determine if the answer is greater than 1 or less than 1, we need to evaluate each expression.

1. $\frac{2}{3} \div \frac{1}{2} = \frac{2}{3} \times \frac{2}{1} = \frac{4}{3} = 1\frac{1}{3}$. This is greater than 1.

2. $\frac{2}{3} \div \frac{2}{1} = \frac{2}{3} \times \frac{1}{2} = \frac{2}{6} = \frac{1}{3}$. This is less than 1.

3. $6 \div \frac{1}{4} = 6 \times \frac{4}{1} = 24$. This is greater than 1.

4. $\frac{1}{5} \div \frac{1}{2} = \frac{1}{5} \times \frac{2}{1} = \frac{2}{5}$. This is less than 1.

5. $4 \frac{1}{3} \div 3 \frac{1}{2}$. Convert to improper fractions: $4\frac{1}{3} = \frac{13}{3}$, $3\frac{1}{2} = \frac{7}{2}$.

$\frac{13}{3} \div \frac{7}{2} = \frac{13}{3} \times \frac{2}{7} = \frac{26}{21} = 1\frac{5}{21}$. This is greater than 1.

6. $\frac{2}{3} \times 8 \frac{1}{2}$. Convert to improper fraction: $8\frac{1}{2} = \frac{17}{2}$.

$\frac{2}{3} \times \frac{17}{2} = \frac{\cancel{2}^{1} \times 17}{3 \times \cancel{2}_{1}} = \frac{17}{3} = 5\frac{2}{3}$. This is greater than 1.

Putting the results into the table:

Given:

The height of the smallest container is 10.5 cm.

The relationship between the height of the smallest container and $x$ is given by the equation $\frac{7}{25} x = 10.5$ cm.

From the figure, the heights of the four containers are $\frac{7}{25}x$, $\frac{2}{5}x$, $\frac{3}{4}x$, and $x$ in ascending order.

To Find:

The height of the largest container.

Solution:

From the figure, the largest container has a height equal to $x$.

We are given the equation for the height of the smallest container:

To find the value of $x$, we divide 10.5 by $\frac{7}{25}$. Dividing by a fraction is the same as multiplying by its reciprocal.

Convert 10.5 to a fraction: $10.5 = \frac{105}{10}$.

Simplify the expression by cancelling common factors.

Cancel 105 and 7 (since $105 = 15 \times 7$).

Cancel 10 and 25 (both are divisible by 5, $10 = 2 \times 5$, $25 = 5 \times 5$).

Convert the improper fraction to a decimal.

So, the value of $x$ is 37.5 cm.

The height of the largest container is $x$.

The height of the largest container is 37.5 cm.

The given figure shows a sequence of fractions in sectors of a circle.

The sequence of fractions is $\frac{1}{4}, \frac{1}{8}, \frac{1}{16}, \frac{1}{32}, ?$

Let's observe the pattern in the given sequence:

The second fraction is $\frac{1}{8}$.

$\frac{1}{8} = \frac{1}{2} \times \frac{1}{4}$

The third fraction is $\frac{1}{16}$.

$\frac{1}{16} = \frac{1}{2} \times \frac{1}{8}$

The fourth fraction is $\frac{1}{32}$.

$\frac{1}{32} = \frac{1}{2} \times \frac{1}{16}$

From the pattern, it is clear that each subsequent fraction is obtained by multiplying the previous fraction by $\frac{1}{2}$.

Therefore, to find the fraction represented by '?', we need to multiply the last given fraction $\frac{1}{32}$ by $\frac{1}{2}$.

$? = \frac{1}{2} \times \frac{1}{32}$

$? = \frac{1 \times 1}{2 \times 32}$

$? = \frac{1}{64}$

So, the appropriate fraction to replace '?' is $\frac{1}{64}$.

The given image shows a number line.

The number line is marked with fractions between $0$ and $1$.

The interval between $0$ and $1$ is divided into $8$ equal parts.

Each part represents $\frac{1}{8}$ of the unit length.

The marks on the number line are $\frac{1}{8}, \frac{2}{8}, \frac{3}{8}, \frac{4}{8}, \frac{5}{8}, \frac{6}{8}, \frac{7}{8}, \frac{8}{8}$ (which is equal to $1$).

The '?' is located at the fifth mark from $0$.

Therefore, the fraction represented by '?' is the fifth fraction in the sequence, which is $\frac{5}{8}$.

So, the appropriate fraction to replace '?' is $\frac{5}{8}$.

The given image shows a sequence of fractions represented by the shaded parts of different shapes.

Let's determine the fraction represented by the shaded area in each shape:

1. The first shape (circle) is divided into $2$ equal parts, and $1$ part is shaded. The fraction is $\frac{1}{2}$.

2. The second shape (rectangle) is divided into $4$ equal parts, and $1$ part is shaded. The fraction is $\frac{1}{4}$.

3. The third shape (rectangle) is divided into $8$ equal parts, and $1$ part is shaded. The fraction is $\frac{1}{8}$.

4. The fourth shape (circle) is divided into $16$ equal parts, and $1$ part is shaded. The fraction is $\frac{1}{16}$.

5. The fifth shape (rectangle) is divided into $32$ equal parts, and $1$ part is shaded. The fraction is $\frac{1}{32}$.

The sequence of fractions is $\frac{1}{2}, \frac{1}{4}, \frac{1}{8}, \frac{1}{16}, \frac{1}{32}, ?$

Let's observe the pattern in the denominators of the sequence: $2, 4, 8, 16, 32$.

We can see that each denominator is obtained by multiplying the previous denominator by $2$.

$4 = 2 \times 2$

$8 = 4 \times 2$

$16 = 8 \times 2$

$32 = 16 \times 2$

The numerators in the sequence are all $1$.

To find the fraction represented by '?', we follow the same pattern.

The next denominator will be $32 \times 2 = 64$.

The numerator will remain $1$.

Therefore, the fraction represented by '?' is $\frac{1}{64}$.

$? = \frac{1}{64}$

So, the appropriate fraction to replace '?' is $\frac{1}{64}$.

The given image shows a sequence of sums of fractions.

Each line shows the sum of two fractions resulting in $1$.

The sequence of sums is:

$\frac{1}{2} + \frac{1}{2} = 1$

$\frac{1}{4} + \frac{3}{4} = 1$

$\frac{1}{8} + \frac{7}{8} = 1$

$\frac{1}{16} + \frac{15}{16} = 1$

$\frac{1}{32} + \frac{31}{32} = 1$

$\frac{1}{64} + ? = 1$

We observe a pattern in the first fraction of each sum: $\frac{1}{2}, \frac{1}{4}, \frac{1}{8}, \frac{1}{16}, \frac{1}{32}, \frac{1}{64}$.

These are powers of $\frac{1}{2}$: $\frac{1}{2^1}, \frac{1}{2^2}, \frac{1}{2^3}, \frac{1}{2^4}, \frac{1}{2^5}, \frac{1}{2^6}$.

We also observe a pattern in the second fraction of each sum: $\frac{1}{2}, \frac{3}{4}, \frac{7}{8}, \frac{15}{16}, \frac{31}{32}$.

If the first fraction is $\frac{1}{2^n}$, the second fraction is $\frac{2^n-1}{2^n}$.

For $n=1$, first fraction is $\frac{1}{2^1} = \frac{1}{2}$. Second fraction is $\frac{2^1-1}{2^1} = \frac{2-1}{2} = \frac{1}{2}$. Sum: $\frac{1}{2} + \frac{1}{2} = 1$.

For $n=2$, first fraction is $\frac{1}{2^2} = \frac{1}{4}$. Second fraction is $\frac{2^2-1}{2^2} = \frac{4-1}{4} = \frac{3}{4}$. Sum: $\frac{1}{4} + \frac{3}{4} = 1$.

This pattern holds true for all the given sums.

In the last line, the first fraction is $\frac{1}{64}$. This corresponds to $n=6$ since $64 = 2^6$.

The equation is $\frac{1}{64} + ? = 1$.

To find the value of '?', we subtract $\frac{1}{64}$ from $1$.

$? = 1 - \frac{1}{64}$

$? = \frac{64}{64} - \frac{1}{64}$

$? = \frac{64 - 1}{64}$

$? = \frac{63}{64}$

Alternatively, using the pattern for the second fraction, when the first fraction is $\frac{1}{2^n}$, the second fraction is $\frac{2^n-1}{2^n}$.

Here, the first fraction is $\frac{1}{64} = \frac{1}{2^6}$, so $n=6$.

The second fraction (represented by '?') is $\frac{2^6-1}{2^6} = \frac{64-1}{64} = \frac{63}{64}$.

So, the appropriate fraction to replace '?' is $\frac{63}{64}$.

The student compared $-\frac{1}{4}$ and $-0.3$.

The student correctly converted $-\frac{1}{4}$ to its decimal form:

$-\frac{1}{4} = -0.25$

The student then compared the positive values $0.3$ and $0.25$ and correctly stated:

$\text{"Since } 0.3 \text{ is greater than } 0.25\text{"}$, which can be written as $0.3 > 0.25$.

Based on this, the student concluded:

$\text{"-0.3 is greater than -0.25"}$, which can be written as $-0.3 > -0.25$.

The student's error is in the final conclusion. The student assumed that if a positive number $a$ is greater than another positive number $b$ ($a > b$), then the negative of $a$ is greater than the negative of $b$ ($-a > -b$). This is incorrect.

When comparing negative numbers, the number with the greater absolute value is actually the smaller number.

The absolute value of $-0.3$ is $|-0.3| = 0.3$.

The absolute value of $-0.25$ is $|-0.25| = 0.25$.

Since $0.3 > 0.25$, it means that $-0.3$ is further away from zero on the negative side of the number line than $-0.25$.

Therefore, $-0.3$ is less than $-0.25$.

The correct comparison is $-0.3 < -0.25$.

The error: The student incorrectly concluded that because $0.3 > 0.25$, it follows that $-0.3 > -0.25$. The relationship between negative numbers is the opposite of the relationship between their corresponding positive values.

The student attempted to multiply the mixed fractions $2\frac{4}{7}$ and $3\frac{1}{4}$ as follows:

$2\frac{4}{7}$ × $3\frac{1}{4}$ = $6\frac{1}{7}$

Let's perform the correct multiplication.

To multiply mixed fractions, we must first convert them into improper fractions.

Convert $2\frac{4}{7}$ to an improper fraction:

$2\frac{4}{7} = \frac{(2 \times 7) + 4}{7} = \frac{14 + 4}{7} = \frac{18}{7}$

Convert $3\frac{1}{4}$ to an improper fraction:

$3\frac{1}{4} = \frac{(3 \times 4) + 1}{4} = \frac{12 + 1}{4} = \frac{13}{4}$

Now, multiply the improper fractions:

$\frac{18}{7} \times \frac{13}{4}$

We can cancel the common factor $2$ between $18$ and $4$:

$\frac{\cancel{18}^{9}}{7} \times \frac{13}{\cancel{4}_{2}} = \frac{9 \times 13}{7 \times 2} = \frac{117}{14}$

Now, convert the improper fraction $\frac{117}{14}$ back to a mixed fraction by dividing $117$ by $14$.

$117 \div 14 = 8$ with a remainder of $5$.

So, $\frac{117}{14} = 8\frac{5}{14}$.

The correct result of the multiplication is $8\frac{5}{14}$.

Comparing the student's result ($6\frac{1}{7}$) with the correct result ($8\frac{5}{14}$), we can see there is an error.

The student appears to have multiplied the whole number parts ($2 \times 3 = 6$) and the fractional parts ($\frac{4}{7} \times \frac{1}{4} = \frac{4}{28} = \frac{1}{7}$) separately and then combined them to get $6\frac{1}{7}$.

The error: The student multiplied the whole number parts and the fractional parts of the mixed numbers independently instead of converting the mixed numbers into improper fractions first and then multiplying.

The question asks to find which fraction, when added to the sum of the preceding fractions in the sequence $\frac{1}{3}, \frac{1}{4}, \frac{1}{5}, ...$, makes the total sum greater than $1$ for the first time.

We need to calculate the partial sums of the series starting from $\frac{1}{3}$.

Sum of the first fraction:

$S_1 = \frac{1}{3}$

$S_1 = \frac{1}{3} < 1$

Sum of the first two fractions:

$S_2 = \frac{1}{3} + \frac{1}{4}$

The least common multiple (LCM) of $3$ and $4$ is $12$.

$S_2 = \frac{1 \times 4}{3 \times 4} + \frac{1 \times 3}{4 \times 3} = \frac{4}{12} + \frac{3}{12} = \frac{4+3}{12} = \frac{7}{12}$

$S_2 = \frac{7}{12} < 1$

Sum of the first three fractions:

$S_3 = S_2 + \frac{1}{5} = \frac{7}{12} + \frac{1}{5}$

The LCM of $12$ and $5$ is $60$.

$S_3 = \frac{7 \times 5}{12 \times 5} + \frac{1 \times 12}{5 \times 12} = \frac{35}{60} + \frac{12}{60} = \frac{35+12}{60} = \frac{47}{60}$

$S_3 = \frac{47}{60} < 1$

Sum of the first four fractions:

$S_4 = S_3 + \frac{1}{6} = \frac{47}{60} + \frac{1}{6}$

The LCM of $60$ and $6$ is $60$.

$S_4 = \frac{47}{60} + \frac{1 \times 10}{6 \times 10} = \frac{47}{60} + \frac{10}{60} = \frac{47+10}{60} = \frac{57}{60}$

$S_4 = \frac{57}{60} < 1$

Sum of the first five fractions:

$S_5 = S_4 + \frac{1}{7} = \frac{57}{60} + \frac{1}{7}$

The LCM of $60$ and $7$ is $420$.

$S_5 = \frac{57 \times 7}{60 \times 7} + \frac{1 \times 60}{7 \times 60} = \frac{399}{420} + \frac{60}{420} = \frac{399+60}{420} = \frac{459}{420}$

Since $459 > 420$, $S_5 = \frac{459}{420} > 1$.

The sum first becomes greater than $1$ when the fraction $\frac{1}{7}$ is added.

The error:

The heading states "What is the Error in each of question 123 to 125?". Questions 123 and 124 describe specific errors made by a student in solving a problem.

Question 125, as presented, is a standard mathematical question asking for a specific value in a sequence. There is no mathematical error in the question itself.

The "error" referred to in the heading, in the context of this question, is likely not a mathematical flaw in the question's statement. Instead, it might refer to the fact that this question requires multiple steps of fraction addition with unlike denominators, which is a common source of calculation errors for students.

Alternatively, given the pattern of the preceding questions, the intended question 125 might have been to present a student's incorrect attempt to solve this problem (e.g., making an error in adding fractions or misinterpreting "greater than 1"), and ask for the analysis of that specific error. As presented, the question lacks the description of such a student error.