Chapter 12 Ratio and Proportion

Given:

Length of the rectangular field = $50$ m.

Breadth of the rectangular field = $15$ m.

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To Find:

The ratio of the length to the breadth of the field.

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Solution:

The ratio of the length to the breadth is given by:

Ratio = $\frac{\text{Length}}{\text{Breadth}}$

Substitute the given values for length and breadth:

Ratio = $\frac{50 \text{ m}}{15 \text{ m}}$

To simplify the ratio, we can divide both the numerator and the denominator by their greatest common divisor, which is 5.

Ratio = $\frac{\cancel{50}^{10}}{\cancel{15}_{3}}$

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

The ratio can also be written in the form 10 : 3.

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The ratio of the length to the breadth of the field is 10 : 3.

Given:

Quantity 1 = 90 cm.

Quantity 2 = 1.5 m.

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To Find:

The ratio of 90 cm to 1.5 m.

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Solution:

To find the ratio of two quantities, they must be in the same unit.

We need to convert meters to centimeters or centimeters to meters. It is usually easier to convert to the smaller unit, which is centimeters.

We know that $1$ meter = $100$ centimeters.

So, $1.5$ meters = $1.5 \times 100$ centimeters = $150$ centimeters.

Now we need to find the ratio of 90 cm to 150 cm.

Ratio = $\frac{90 \text{ cm}}{150 \text{ cm}}$

We can simplify this fraction by dividing both the numerator and the denominator by their greatest common divisor.

We can divide both numbers by 10:

$\frac{90}{150} = \frac{9}{15}$

Now, we can divide both 9 and 15 by 3:

$\frac{\cancel{9}^{3}}{\cancel{15}_{5}}$

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

The ratio can also be written as 3 : 5.

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The ratio of 90 cm to 1.5 m is 3 : 5.

Given:

Total number of persons in the office = 45.

Number of females = 25.

The remaining persons are males.

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To Find:

(a) The ratio of the number of females to the number of males.

(b) The ratio of the number of males to the number of females.

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Solution:

First, we need to find the number of males in the office.

Number of males = Total number of persons - Number of females

Number of males = $45 - 25 = 20$.

Now we can find the required ratios.

(a) Ratio of the number of females to the number of males:

Ratio = $\frac{\text{Number of females}}{\text{Number of males}}$

Ratio = $\frac{25}{20}$

To simplify the ratio, divide both the numerator and the denominator by their greatest common divisor, which is 5.

Ratio = $\frac{\cancel{25}^{5}}{\cancel{20}_{4}}$

Ratio = $\frac{5}{4}$

This ratio can also be written as 5 : 4.

(b) Ratio of the number of males to the number of females:

Ratio = $\frac{\text{Number of males}}{\text{Number of females}}$

Ratio = $\frac{20}{25}$

To simplify the ratio, divide both the numerator and the denominator by their greatest common divisor, which is 5.

Ratio = $\frac{\cancel{20}^{4}}{\cancel{25}_{5}}$

Ratio = $\frac{4}{5}$

This ratio can also be written as 4 : 5.

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The ratio of the number of females to the number of males is 5 : 4.

The ratio of the number of males to the number of females is 4 : 5.

Given:

The ratio is 6 : 4.

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To Find:

Two equivalent ratios of 6 : 4.

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Solution:

A ratio can be written as a fraction. The ratio 6 : 4 can be written as $\frac{6}{4}$.

Equivalent ratios are obtained by multiplying or dividing both the numerator and the denominator of the fraction by the same non-zero number.

Method 1: Multiplying the terms of the ratio

Multiply both terms of the ratio 6 : 4 by the same number, for example, 2.

New ratio = $(6 \times 2) : (4 \times 2) = 12 : 8$

The fraction form is $\frac{12}{8}$.

Multiply both terms of the ratio 6 : 4 by another number, for example, 3.

New ratio = $(6 \times 3) : (4 \times 3) = 18 : 12$

The fraction form is $\frac{18}{12}$.

So, two equivalent ratios are 12 : 8 and 18 : 12.

Method 2: Simplifying the ratio

We can also simplify the given ratio by dividing both terms by their greatest common divisor. The greatest common divisor of 6 and 4 is 2.

Simplified ratio = $(6 \div 2) : (4 \div 2) = 3 : 2$

The fraction form is $\frac{3}{2}$. This is also an equivalent ratio.

We can then multiply this simplified ratio by a number, say 4, to get another equivalent ratio.

New ratio = $(3 \times 4) : (2 \times 4) = 12 : 8$

The fraction form is $\frac{12}{8}$.

Using this method, two equivalent ratios are 3 : 2 and 12 : 8.

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Two equivalent ratios of 6 : 4 are 12 : 8 and 3 : 2.

Other possible equivalent ratios include 18 : 12, 24 : 16, etc.

Given:

The equivalent ratios: $\frac{14}{21} = \frac{\text{missing number 1}}{3} = \frac{6}{\text{missing number 2}}$.

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To Find:

The missing numbers.

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Solution:

We need to find the equivalent ratios by finding the relationship between the known parts of the fractions.

Part 1: Find the first missing number

We have the equality $\frac{14}{21} = \frac{\text{missing number 1}}{3}$.

Let the first missing number be $x$. So, $\frac{14}{21} = \frac{x}{3}$.

We can observe the relationship between the denominators: $21$ is divided by 7 to get 3 ($21 \div 7 = 3$).

To maintain the equality, we must perform the same operation on the numerator.

So, we divide the numerator 14 by 7.

$x = 14 \div 7 = 2$.

Alternatively, we can use cross-multiplication:

$14 \times 3 = 21 \times x$

$42 = 21x$

Divide both sides by 21:

$x = \frac{42}{21} = 2$.

So, the first missing number is 2.

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

Part 2: Find the second missing number

Now we use the equality $\frac{14}{21} = \frac{6}{\text{missing number 2}}$ or the simplified equivalent ratio $\frac{2}{3} = \frac{6}{\text{missing number 2}}$. Using the simplified ratio is easier.

Let the second missing number be $y$. So, $\frac{2}{3} = \frac{6}{y}$.

We can observe the relationship between the numerators: 2 is multiplied by 3 to get 6 ($2 \times 3 = 6$).

To maintain the equality, we must perform the same operation on the denominator.

So, we multiply the denominator 3 by 3.

$y = 3 \times 3 = 9$.

Alternatively, we can use cross-multiplication:

$2 \times y = 3 \times 6$

$2y = 18$

Divide both sides by 2:

$y = \frac{18}{2} = 9$.

So, the second missing number is 9.

$$\frac{2}{3} = \frac{6}{9}$$

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The completed equation with missing numbers filled in is:

$$\frac{14}{21} = \frac{2}{3} = \frac{6}{9}$$

The missing numbers are 2 and 9.

Given:

Ratio of distance of Mary’s home to school to the distance of John’s home to school is 2 : 1.

$$\frac{\text{Distance of Mary's home from school}}{\text{Distance of John's home from school}} = \frac{2}{1}$$

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To Find:

(a) Who lives nearer to the school.

(b) Complete the given table of distances.

(c) Who lives nearer if the ratio of Mary's distance to Kalam's distance is 1:2.

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Solution:

(a) Who lives nearer to the school?

The ratio of Mary's distance to John's distance is 2 : 1. This means that for every 2 units of distance Mary lives from the school, John lives 1 unit of distance from the school.

Since the ratio is 2:1, Mary's distance is twice John's distance ($2 \times \text{John's distance} = \text{Mary's distance}$).

Because John's distance is smaller than Mary's distance (in the ratio comparison), John lives nearer to the school.

(b) Complete the table:

The ratio of the distances is always $\frac{\text{Mary's distance}}{\text{John's distance}} = \frac{2}{1}$.

We can use this ratio to find the missing values in the table.

Let $D_M$ be the distance from Mary's home to school and $D_J$ be the distance from John's home to school.

$$\frac{D_M}{D_J} = \frac{2}{1}$$

From this, we can get $D_M = 2 \times D_J$ or $D_J = \frac{D_M}{2}$.

Distance from Mary’s home to school (in km) ($D_M$)	Distance from John’s home to school (in km) ($D_J$)
10	5
?	4
4	?
?	3
?	1

Filling the blanks:

• In the second column, $D_J = 4$ km. Since $D_M = 2 \times D_J$, $D_M = 2 \times 4 = 8$ km.
• In the third column, $D_M = 4$ km. Since $D_J = \frac{D_M}{2}$, $D_J = \frac{4}{2} = 2$ km.
• In the fourth column, $D_J = 3$ km. Since $D_M = 2 \times D_J$, $D_M = 2 \times 3 = 6$ km.
• In the fifth column, $D_J = 1$ km. Since $D_M = 2 \times D_J$, $D_M = 2 \times 1 = 2$ km.

The completed table is:

Distance from Mary’s home to school (in km) ($D_M$)	10	8	4	6	2
Distance from John’s home to school (in km) ($D_J$)	5	4	2	3	1

(c) Comparing distances of Mary and Kalam:

Given that the ratio of distance of Mary’s home to the distance of Kalam’s home from school is 1 : 2.

$$\frac{\text{Distance of Mary's home from school}}{\text{Distance of Kalam's home from school}} = \frac{1}{2}$$

Let $D_K$ be the distance of Kalam's home from school.

$$\frac{D_M}{D_K} = \frac{1}{2}$$

This means that for every 1 unit of distance Mary lives from the school, Kalam lives 2 units of distance from the school.

This implies $D_K = 2 \times D_M$ or $D_M = \frac{D_K}{2}$.

Since Mary's distance is half of Kalam's distance (or Kalam's distance is twice Mary's distance), Mary lives nearer to the school than Kalam.

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Summary:

(a) John lives nearer to the school.

(b) The completed table is provided above.

(c) Mary lives nearer to the school.

Given:

Total amount to be divided = $\textsf{₹}$ 60.

Ratio between Kriti and Kiran = 1 : 2.

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To Find:

The amount Kriti gets and the amount Kiran gets.

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Solution:

The given ratio is 1 : 2. This means that for every 1 part Kriti gets, Kiran gets 2 parts.

The total number of parts in the ratio is the sum of the individual parts:

Total parts = $1 + 2 = 3$.

The total amount $\textsf{₹}$ 60 is divided into 3 equal parts.

Kriti's share is 1 part out of the total 3 parts. So, Kriti gets $\frac{1}{3}$ of the total amount.

Kiran's share is 2 parts out of the total 3 parts. So, Kiran gets $\frac{2}{3}$ of the total amount.

Amount Kriti gets = $\frac{1}{3}$ of $\textsf{₹}$ 60

Amount Kriti gets = $\frac{1}{3} \times 60$

Amount Kriti gets = $\textsf{₹}$ $\frac{1 \times \cancel{60}^{20}}{\cancel{3}_{1}}$

Amount Kriti gets = $\textsf{₹}$ $20$.

Amount Kiran gets = $\frac{2}{3}$ of $\textsf{₹}$ 60

Amount Kiran gets = $\frac{2}{3} \times 60$

Amount Kiran gets = $\textsf{₹}$ $\frac{2 \times \cancel{60}^{20}}{\cancel{3}_{1}}$

Amount Kiran gets = $\textsf{₹}$ $2 \times 20$

Amount Kiran gets = $\textsf{₹}$ $40$.

We can check the total amount received: $\textsf{₹}$ 20 (Kriti) + $\textsf{₹}$ 40 (Kiran) = $\textsf{₹}$ 60, which is the total amount.

The ratio of the amounts received by Kriti and Kiran is 20 : 40, which simplifies to 1 : 2 ($\frac{20}{40} = \frac{\cancel{20}^{1}}{\cancel{40}_{2}} = \frac{1}{2}$). This matches the given ratio.

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Kriti gets $\textsf{₹}$ 20 and Kiran gets $\textsf{₹}$ 40.

Given:

Number of girls in the class = 20.

Number of boys in the class = 15.

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To Find:

(a) The ratio of the number of girls to the number of boys.

(b) The ratio of the number of girls to the total number of students in the class.

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Solution:

First, calculate the total number of students in the class.

Total number of students = Number of girls + Number of boys

Total number of students = $20 + 15 = 35$.

(a) Ratio of the number of girls to the number of boys:

Ratio = $\frac{\text{Number of girls}}{\text{Number of boys}}$

Ratio = $\frac{20}{15}$

To simplify the ratio, divide both the numerator and the denominator by their greatest common divisor, which is 5.

Ratio = $\frac{\cancel{20}^{4}}{\cancel{15}_{3}}$

Ratio = $\frac{4}{3}$

The ratio is 4 : 3.

(b) Ratio of the number of girls to the total number of students in the class:

Ratio = $\frac{\text{Number of girls}}{\text{Total number of students}}$

Ratio = $\frac{20}{35}$

To simplify the ratio, divide both the numerator and the denominator by their greatest common divisor, which is 5.

Ratio = $\frac{\cancel{20}^{4}}{\cancel{35}_{7}}$

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

The ratio is 4 : 7.

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The ratio of the number of girls to the number of boys is 4 : 3.

The ratio of the number of girls to the total number of students is 4 : 7.

Given:

Total number of students in a class = 30.

Number of students liking football = 6.

Number of students liking cricket = 12.

The remaining students like tennis.

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To Find:

(a) The ratio of the number of students liking football to the number of students liking tennis.

(b) The ratio of the number of students liking cricket to the total number of students.

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Solution:

First, calculate the number of students who like tennis.

Number of students liking tennis = Total number of students - (Number of students liking football + Number of students liking cricket)

Number of students liking tennis = $30 - (6 + 12)$

Number of students liking tennis = $30 - 18 = 12$.

Now we can find the required ratios.

(a) Ratio of number of students liking football to number of students liking tennis:

Ratio = $\frac{\text{Number of students liking football}}{\text{Number of students liking tennis}}$

Ratio = $\frac{6}{12}$

To simplify the ratio, divide both the numerator and the denominator by their greatest common divisor, which is 6.

Ratio = $\frac{\cancel{6}^{1}}{\cancel{12}_{2}}$

Ratio = $\frac{1}{2}$

The ratio is 1 : 2.

(b) Ratio of number of students liking cricket to total number of students:

Ratio = $\frac{\text{Number of students liking cricket}}{\text{Total number of students}}$

Ratio = $\frac{12}{30}$

To simplify the ratio, divide both the numerator and the denominator by their greatest common divisor, which is 6.

Ratio = $\frac{\cancel{12}^{2}}{\cancel{30}_{5}}$

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

The ratio is 2 : 5.

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The ratio of students liking football to students liking tennis is 1 : 2.

The ratio of students liking cricket to the total number of students is 2 : 5.

Given:

A figure containing triangles, squares, and circles inside a rectangle.

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To Find:

The specified ratios based on the figure.

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Solution:

First, count the number of each type of figure inside the rectangle from the given image.

Number of triangles = 3

Number of squares = 2

Number of circles = 2

Total number of figures inside the rectangle = Number of triangles + Number of squares + Number of circles = $3 + 2 + 2 = 7$.

(a) Ratio of the number of triangles to the number of circles inside the rectangle:

Ratio = $\frac{\text{Number of triangles}}{\text{Number of circles}}$

Ratio = $\frac{3}{2}$

The ratio is 3 : 2.

(b) Ratio of the number of squares to all the figures inside the rectangle:

Ratio = $\frac{\text{Number of squares}}{\text{Total number of figures}}$

Ratio = $\frac{2}{7}$

The ratio is 2 : 7.

(c) Ratio of the number of circles to all the figures inside the rectangle:

Ratio = $\frac{\text{Number of circles}}{\text{Total number of figures}}$

Ratio = $\frac{2}{7}$

The ratio is 2 : 7.

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The ratio of triangles to circles is 3 : 2.

The ratio of squares to all figures is 2 : 7.

The ratio of circles to all figures is 2 : 7.

Given:

The equation involving equivalent ratios: $\frac{15}{18} = \frac{\text{Blank 1}}{6} = \frac{10}{\text{Blank 2}} = \frac{\text{Blank 3}}{30}$.

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To Find:

The missing numbers in the blanks and whether these are equivalent ratios.

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Solution:

We start with the known ratio $\frac{15}{18}$ and find its simplest form by dividing the numerator and denominator by their greatest common divisor, which is 3.

$\frac{15}{18} = \frac{\cancel{15}^{5}}{\cancel{18}_{6}} = \frac{5}{6}$

Now we use this simplified ratio $\frac{5}{6}$ to find the missing numbers in the other fractions.

Finding Blank 1:

We have $\frac{15}{18} = \frac{\text{Blank 1}}{6}$, which simplifies to $\frac{5}{6} = \frac{\text{Blank 1}}{6}$.

Comparing the denominators, $18 \div 3 = 6$. So, we divide the numerator 15 by 3.

Blank 1 = $15 \div 3 = 5$.

Using the simplified form $\frac{5}{6}$: The denominator is the same (6), so the numerator must also be the same. Blank 1 = 5.

So, $\frac{15}{18} = \frac{5}{6}$.

Finding Blank 2:

We use the simplified ratio $\frac{5}{6} = \frac{10}{\text{Blank 2}}$.

Comparing the numerators, $5 \times 2 = 10$. So, we multiply the denominator 6 by 2.

Blank 2 = $6 \times 2 = 12$.

So, $\frac{5}{6} = \frac{10}{12}$.

Finding Blank 3:

We use the simplified ratio $\frac{5}{6} = \frac{\text{Blank 3}}{30}$.

Comparing the denominators, $6 \times 5 = 30$. So, we multiply the numerator 5 by 5.

Blank 3 = $5 \times 5 = 25$.

So, $\frac{5}{6} = \frac{25}{30}$.

Filling in the blanks, the equation becomes:

$$\frac{15}{18} = \frac{5}{6} = \frac{10}{12} = \frac{25}{30}$$

Are these equivalent ratios?

Yes, these are equivalent ratios. We found the missing numbers by maintaining the proportionality based on the known ratio $\frac{15}{18}$ (or its simplified form $\frac{5}{6}$).

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The missing numbers are 5, 12, and 25.

Yes, these are equivalent ratios.

Solution:

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We find the ratio of the first quantity to the second quantity by writing them as a fraction and simplifying the fraction to its lowest terms.

(a) Ratio of 81 to 108:

Ratio = $\frac{81}{108}$

We can find the greatest common divisor (GCD) of 81 and 108. Both are divisible by 9.

$\frac{81 \div 9}{108 \div 9} = \frac{9}{12}$

Now, 9 and 12 are both divisible by 3.

$\frac{9 \div 3}{12 \div 3} = \frac{3}{4}$

Alternatively, the GCD of 81 and 108 is 27 ($81 = 3 \times 27$, $108 = 4 \times 27$).

Ratio = $\frac{\cancel{81}^{3}}{\cancel{108}_{4}} = \frac{3}{4}$

The ratio is 3 : 4.

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(b) Ratio of 98 to 63:

Ratio = $\frac{98}{63}$

We can find the GCD of 98 and 63. Both are divisible by 7.

$\frac{98 \div 7}{63 \div 7} = \frac{14}{9}$

Alternatively, the GCD of 98 and 63 is 7.

Ratio = $\frac{\cancel{98}^{14}}{\cancel{63}_{9}} = \frac{14}{9}$

The ratio is 14 : 9.

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(c) Ratio of 33 km to 121 km:

The units are the same (km), so we can directly form the ratio.

Ratio = $\frac{33 \text{ km}}{121 \text{ km}} = \frac{33}{121}$

We can find the GCD of 33 and 121. Both are divisible by 11.

$\frac{33 \div 11}{121 \div 11} = \frac{3}{11}$

Alternatively, the GCD of 33 and 121 is 11.

Ratio = $\frac{\cancel{33}^{3}}{\cancel{121}_{11}} = \frac{3}{11}$

The ratio is 3 : 11.

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(d) Ratio of 30 minutes to 45 minutes:

The units are the same (minutes), so we can directly form the ratio.

Ratio = $\frac{30 \text{ minutes}}{45 \text{ minutes}} = \frac{30}{45}$

We can find the GCD of 30 and 45. Both are divisible by 15.

$\frac{30 \div 15}{45 \div 15} = \frac{2}{3}$

Alternatively, the GCD of 30 and 45 is 15.

Ratio = $\frac{\cancel{30}^{2}}{\cancel{45}_{3}} = \frac{2}{3}$

The ratio is 2 : 3.

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The ratios are:

(a) 3 : 4

(b) 14 : 9

(c) 3 : 11

(d) 2 : 3

Solution:

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To find the ratio of two quantities, they must be in the same unit. We convert the quantities to the same unit and then find the ratio in its simplest form.

(a) Ratio of 30 minutes to 1.5 hours:

We need to convert hours to minutes or minutes to hours. It's generally easier to convert to the smaller unit, which is minutes.

We know that 1 hour = 60 minutes.

So, 1.5 hours = $1.5 \times 60$ minutes = 90 minutes.

Now, we find the ratio of 30 minutes to 90 minutes.

Ratio = $\frac{30 \text{ minutes}}{90 \text{ minutes}} = \frac{30}{90}$

Simplify the fraction by dividing the numerator and denominator by their GCD, which is 30.

Ratio = $\frac{\cancel{30}^{1}}{\cancel{90}_{3}} = \frac{1}{3}$

The ratio is 1 : 3.

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(b) Ratio of 40 cm to 1.5 m:

We need to convert meters to centimeters. The smaller unit is centimeters.

We know that 1 meter = 100 cm.

So, 1.5 m = $1.5 \times 100$ cm = 150 cm.

Now, we find the ratio of 40 cm to 150 cm.

Ratio = $\frac{40 \text{ cm}}{150 \text{ cm}} = \frac{40}{150}$

Simplify the fraction by dividing the numerator and denominator by their GCD, which is 10.

Ratio = $\frac{\cancel{40}^{4}}{\cancel{150}_{15}} = \frac{4}{15}$

The ratio is 4 : 15.

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(c) Ratio of 55 paise to $\textsf{₹}$ 1:

We need to convert rupees to paise. The smaller unit is paise.

We know that $\textsf{₹}$ 1 = 100 paise.

Now, we find the ratio of 55 paise to 100 paise.

Ratio = $\frac{55 \text{ paise}}{100 \text{ paise}} = \frac{55}{100}$

Simplify the fraction by dividing the numerator and denominator by their GCD, which is 5.

Ratio = $\frac{\cancel{55}^{11}}{\cancel{100}_{20}} = \frac{11}{20}$

The ratio is 11 : 20.

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(d) Ratio of 500 mL to 2 litres:

We need to convert litres to milliliters. The smaller unit is milliliters (mL).

We know that 1 litre = 1000 mL.

So, 2 litres = $2 \times 1000$ mL = 2000 mL.

Now, we find the ratio of 500 mL to 2000 mL.

Ratio = $\frac{500 \text{ mL}}{2000 \text{ mL}} = \frac{500}{2000}$

Simplify the fraction by dividing the numerator and denominator by their GCD, which is 500.

Ratio = $\frac{\cancel{500}^{1}}{\cancel{2000}_{4}} = \frac{1}{4}$

The ratio is 1 : 4.

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The ratios are:

(a) 1 : 3

(b) 4 : 15

(c) 11 : 20

(d) 1 : 4

Given:

Seema's annual earning = $\textsf{₹}$ 1,50,000.

Seema's annual saving = $\textsf{₹}$ 50,000.

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To Find:

(a) The ratio of money earned to money saved.

(b) The ratio of money saved to money spent.

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Solution:

First, calculate the amount of money Seema spends in a year.

Money spent = Money earned - Money saved

Money spent = $\textsf{₹}$ 1,50,000 - $\textsf{₹}$ 50,000 = $\textsf{₹}$ 1,00,000.

Now we can find the required ratios.

(a) Ratio of money that Seema earns to the money she saves:

Ratio = $\frac{\text{Money earned}}{\text{Money saved}}$

Ratio = $\frac{1,50,000}{50,000}$

We can cancel out the common zeros.

Ratio = $\frac{150}{50}$

Divide the numerator and denominator by their GCD, which is 50.

Ratio = $\frac{\cancel{150}^{3}}{\cancel{50}_{1}} = \frac{3}{1}$

The ratio is 3 : 1.

(b) Ratio of money that she saves to the money she spends:

Ratio = $\frac{\text{Money saved}}{\text{Money spent}}$

Ratio = $\frac{50,000}{1,00,000}$

We can cancel out the common zeros.

Ratio = $\frac{50}{100}$

Divide the numerator and denominator by their GCD, which is 50.

Ratio = $\frac{\cancel{50}^{1}}{\cancel{100}_{2}} = \frac{1}{2}$

The ratio is 1 : 2.

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The ratio of money Seema earns to money she saves is 3 : 1.

The ratio of money she saves to money she spends is 1 : 2.

Given:

Number of teachers in the school = 102.

Number of students in the school = 3300.

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To Find:

The ratio of the number of teachers to the number of students.

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Solution:

The ratio of the number of teachers to the number of students is given by:

Ratio = $\frac{\text{Number of teachers}}{\text{Number of students}}$

Substitute the given numbers:

Ratio = $\frac{102}{3300}$

To simplify the ratio, we need to find the greatest common divisor (GCD) of 102 and 3300.

Let's divide both numbers by common factors. Both are even, so divide by 2:

$\frac{102 \div 2}{3300 \div 2} = \frac{51}{1650}$

Now, check for divisibility by 3 (sum of digits of 51 is $5+1=6$, which is divisible by 3; sum of digits of 1650 is $1+6+5+0=12$, which is divisible by 3).

$\frac{51 \div 3}{1650 \div 3} = \frac{17}{550}$

17 is a prime number. We check if 550 is divisible by 17. $550 \div 17 \approx 32.35$. 550 is not divisible by 17.

So, the simplified ratio is $\frac{17}{550}$.

Ratio = $\frac{17}{550}$

The ratio can also be written as 17 : 550.

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The ratio of the number of teachers to the number of students is 17 : 550.

Given:

Total number of students in the college = 4320.

Number of girls in the college = 2300.

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To Find:

(a) The ratio of the number of girls to the total number of students.

(b) The ratio of the number of boys to the number of girls.

(c) The ratio of the number of boys to the total number of students.

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Solution:

First, calculate the number of boys in the college.

Number of boys = Total number of students - Number of girls

Number of boys = $4320 - 2300 = 2020$.

Now we can find the required ratios.

(a) Ratio of number of girls to the total number of students:

Ratio = $\frac{\text{Number of girls}}{\text{Total number of students}}$

Ratio = $\frac{2300}{4320}$

Simplify the fraction by dividing the numerator and denominator by their GCD. Both numbers end in 0, so divide by 10.

$\frac{230}{432}$

Both numbers are even, so divide by 2.

$\frac{230 \div 2}{432 \div 2} = \frac{115}{216}$

115 is divisible by 5 ($115 = 5 \times 23$). 216 is not divisible by 5. Check for common factors of 115 and 216. The prime factors of 115 are 5 and 23. 216 is divisible by 2 and 3 (and powers of 2 and 3). There are no common prime factors other than 1.

The simplified ratio is $\frac{115}{216}$.

The ratio is 115 : 216.

(b) Ratio of number of boys to the number of girls:

Ratio = $\frac{\text{Number of boys}}{\text{Number of girls}}$

Ratio = $\frac{2020}{2300}$

Simplify the fraction by dividing by their GCD. Both numbers end in 0, so divide by 10.

$\frac{202}{230}$

Both numbers are even, so divide by 2.

$\frac{202 \div 2}{230 \div 2} = \frac{101}{115}$

101 is a prime number. 115 is $5 \times 23$. There are no common prime factors other than 1.

The simplified ratio is $\frac{101}{115}$.

The ratio is 101 : 115.

(c) Ratio of number of boys to the total number of students:

Ratio = $\frac{\text{Number of boys}}{\text{Total number of students}}$

Ratio = $\frac{2020}{4320}$

Simplify the fraction by dividing by their GCD. Both numbers end in 0, so divide by 10.

$\frac{202}{432}$

Both numbers are even, so divide by 2.

$\frac{202 \div 2}{432 \div 2} = \frac{101}{216}$

The simplified ratio is $\frac{101}{216}$.

The ratio is 101 : 216.

---

The ratios are:

(a) 115 : 216

(b) 101 : 115

(c) 101 : 216

Given:

Total number of students in the school = 1800.

Number of students who opted basketball = 750.

Number of students who opted cricket = 800.

The remaining students opted table tennis. Each student opted for only one game.

---

To Find:

(a) The ratio of the number of students who opted basketball to the number of students who opted table tennis.

(b) The ratio of the number of students who opted cricket to the number of students opting basketball.

(c) The ratio of the number of students who opted basketball to the total number of students.

---

Solution:

First, calculate the number of students who opted for table tennis.

Number of students who opted table tennis = Total number of students - (Number of students who opted basketball + Number of students who opted cricket)

Number of students who opted table tennis = $1800 - (750 + 800)$

Number of students who opted table tennis = $1800 - 1550 = 250$.

Now we can find the required ratios.

(a) Ratio of number of students who opted basketball to the number of students who opted table tennis:

Ratio = $\frac{\text{Number of students opted basketball}}{\text{Number of students opted table tennis}}$

Ratio = $\frac{750}{250}$

Simplify the fraction by dividing the numerator and denominator by their GCD, which is 250.

Ratio = $\frac{\cancel{750}^{3}}{\cancel{250}_{1}} = \frac{3}{1}$

The ratio is 3 : 1.

(b) Ratio of number of students who opted cricket to the number of students opting basketball:

Ratio = $\frac{\text{Number of students opted cricket}}{\text{Number of students opted basketball}}$

Ratio = $\frac{800}{750}$

Simplify the fraction by dividing the numerator and denominator by their GCD, which is 50.

Ratio = $\frac{\cancel{800}^{16}}{\cancel{750}_{15}} = \frac{16}{15}$

The ratio is 16 : 15.

(c) Ratio of number of students who opted basketball to the total number of students:

Ratio = $\frac{\text{Number of students opted basketball}}{\text{Total number of students}}$

Ratio = $\frac{750}{1800}$

Simplify the fraction by dividing the numerator and denominator by their GCD. Both end in 0, so divide by 10.

$\frac{75}{180}$

Both are divisible by 5.

$\frac{75 \div 5}{180 \div 5} = \frac{15}{36}$

Both are divisible by 3.

$\frac{15 \div 3}{36 \div 3} = \frac{5}{12}$

Alternatively, the GCD of 750 and 1800 is 150.

Ratio = $\frac{\cancel{750}^{5}}{\cancel{1800}_{12}} = \frac{5}{12}$

The ratio is 5 : 12.

---

The ratios are:

(a) 3 : 1

(b) 16 : 15

(c) 5 : 12

Given:

Cost of a dozen pens = $\textsf{₹}$ 180.

Cost of 8 ball pens = $\textsf{₹}$ 56.

---

To Find:

The ratio of the cost of a pen to the cost of a ball pen.

---

Solution:

First, we need to find the cost of a single pen and the cost of a single ball pen.

A dozen means 12.

Cost of 12 pens = $\textsf{₹}$ 180.

Cost of 1 pen = $\frac{\text{Total cost of pens}}{\text{Number of pens}}$

Cost of 1 pen = $\frac{\textsf{₹} \ 180}{12}$

Cost of 1 pen = $\textsf{₹}$ $\frac{\cancel{180}^{15}}{\cancel{12}_{1}}$

Cost of 1 pen = $\textsf{₹}$ 15.

Cost of 8 ball pens = $\textsf{₹}$ 56.

Cost of 1 ball pen = $\frac{\text{Total cost of ball pens}}{\text{Number of ball pens}}$

Cost of 1 ball pen = $\frac{\textsf{₹} \ 56}{8}$

Cost of 1 ball pen = $\textsf{₹}$ $\frac{\cancel{56}^{7}}{\cancel{8}_{1}}$

Cost of 1 ball pen = $\textsf{₹}$ 7.

Now, we find the ratio of the cost of a pen to the cost of a ball pen.

Ratio = $\frac{\text{Cost of 1 pen}}{\text{Cost of 1 ball pen}}$

Ratio = $\frac{\textsf{₹} \ 15}{\textsf{₹} \ 7}$

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

The ratio cannot be simplified further as 15 and 7 have no common factors other than 1.

The ratio is 15 : 7.

---

The ratio of the cost of a pen to the cost of a ball pen is 15 : 7.

Given:

The ratio of the breadth and length of a hall is 2 : 5.

---

To Find:

To complete the given table with the missing possible breadths and lengths of the hall.

---

Solution:

Let the breadth of the hall be B and the length of the hall be L.

According to the question, the ratio of breadth to length is given as:

This means that for any possible dimensions of the hall, the breadth and length must be in this proportion.

Case 1: When Length = 50 metres

We need to find the corresponding breadth (B). Using the ratio:

To find B, we multiply both sides by 50:

So, the missing breadth is 20 metres.

Case 2: When Breadth = 40 metres

We need to find the corresponding length (L). Using the ratio:

By cross-multiplication, we get:

So, the missing length is 100 metres.

Now, we can complete the table with the values we found.

Breadth of the hall (in metres)	10	20	40
Length of the hall (in metres)	25	50	100

---

Alternate Solution:

Since the ratio of breadth to length is 2 : 5, we can say that the breadth and length are multiples of a common number, say 'x'.

Breadth (B) = $2x$

Length (L) = $5x$

Case 1: When Length (L) = 50

Therefore, Breadth (B) = $2x = 2 \times 10 = 20$ metres.

Case 2: When Breadth (B) = 40

Therefore, Length (L) = $5x = 5 \times 20 = 100$ metres.

Given:

Total number of pens to be divided = 20.

Ratio of division between Sheela and Sangeeta = 3 : 2.

---

To Find:

The number of pens Sheela gets and the number of pens Sangeeta gets.

---

Solution:

The given ratio is 3 : 2. This means that the total number of pens is divided into parts, where Sheela gets 3 parts and Sangeeta gets 2 parts.

The total number of parts is the sum of the parts in the ratio:

Total parts = $3 + 2 = 5$.

The 20 pens are divided into 5 equal parts.

Value of one part = $\frac{\text{Total number of pens}}{\text{Total number of parts}}$

Value of one part = $\frac{20}{5} = 4$ pens.

Sheela's share is 3 parts of the total pens.

Number of pens Sheela gets = $3 \times \text{Value of one part}$

Number of pens Sheela gets = $3 \times 4 = 12$ pens.

Sangeeta's share is 2 parts of the total pens.

Number of pens Sangeeta gets = $2 \times \text{Value of one part}$

Number of pens Sangeeta gets = $2 \times 4 = 8$ pens.

To verify, the sum of the pens received by Sheela and Sangeeta should equal the total number of pens: $12 + 8 = 20$. This matches the given total.

The ratio of the pens received by Sheela and Sangeeta is 12 : 8, which simplifies to 3 : 2 ($\frac{12}{8} = \frac{\cancel{12}^{3}}{\cancel{8}_{2}} = \frac{3}{2}$). This matches the given ratio.

---

Sheela gets 12 pens and Sangeeta gets 8 pens.

Given:

Total amount to be divided = $\textsf{₹}$ 36.

Age of Shreya = 15 years.

Age of Bhoomika = 12 years.

The amount is to be divided in the ratio of their ages.

---

To Find:

The amount of money Shreya and Bhoomika will get.

---

Solution:

First, find the ratio of Shreya's age to Bhoomika's age.

Ratio of ages = Age of Shreya : Age of Bhoomika

Ratio of ages = 15 : 12

To simplify the ratio, divide both parts by their greatest common divisor, which is 3.

Ratio = $\frac{15}{12} = \frac{\cancel{15}^{5}}{\cancel{12}_{4}} = \frac{5}{4}$.

So, the money is to be divided in the ratio 5 : 4 between Shreya and Bhoomika.

The total number of parts in the ratio is the sum of the parts:

Total parts = $5 + 4 = 9$.

The total amount $\textsf{₹}$ 36 is divided into 9 equal parts.

Value of one part = $\frac{\text{Total amount}}{\text{Total number of parts}}$

Value of one part = $\frac{\textsf{₹} \ 36}{9} = \textsf{₹} \ 4$.

Shreya's share is 5 parts of the total amount.

Amount Shreya gets = $5 \times \text{Value of one part}$

Amount Shreya gets = $5 \times \textsf{₹} \ 4 = \textsf{₹} \ 20$.

Bhoomika's share is 4 parts of the total amount.

Amount Bhoomika gets = $4 \times \text{Value of one part}$

Amount Bhoomika gets = $4 \times \textsf{₹} \ 4 = \textsf{₹} \ 16$.

To verify, the sum of the amounts received by Shreya and Bhoomika should equal the total amount: $\textsf{₹}$ 20 + $\textsf{₹}$ 16 = $\textsf{₹}$ 36. This matches the given total.

The ratio of the amounts received is 20 : 16, which simplifies to 5 : 4 ($\frac{20}{16} = \frac{\cancel{20}^{5}}{\cancel{16}_{4}} = \frac{5}{4}$). This matches the age ratio.

---

Shreya will get $\textsf{₹}$ 20 and Bhoomika will get $\textsf{₹}$ 16.

Given:

Present age of father = 42 years.

Present age of son = 14 years.

---

To Find:

Various ratios of their ages at different times.

---

Solution:

(a) Ratio of present age of father to the present age of son:

Ratio = $\frac{\text{Present age of father}}{\text{Present age of son}}$

Ratio = $\frac{42}{14}$

Simplify the fraction by dividing the numerator and denominator by their GCD, which is 14.

Ratio = $\frac{\cancel{42}^{3}}{\cancel{14}_{1}} = \frac{3}{1}$

The ratio is 3 : 1.

(b) Age of the father to the age of son, when son was 12 years old:

The son's current age is 14 years. The son was 12 years old 2 years ago ($14 - 12 = 2$).

At that time (2 years ago), the father's age would also have been 2 years less than his present age.

Father's age when son was 12 = Present age of father - 2 years

Father's age = $42 - 2 = 40$ years.

Son's age at that time = 12 years.

Ratio = $\frac{\text{Father's age when son was 12}}{\text{Son's age at that time}}$

Ratio = $\frac{40}{12}$

Simplify the fraction by dividing the numerator and denominator by their GCD, which is 4.

Ratio = $\frac{\cancel{40}^{10}}{\cancel{12}_{3}} = \frac{10}{3}$

The ratio is 10 : 3.

(c) Age of father after 10 years to the age of son after 10 years:

Father's age after 10 years = Present age of father + 10 years = $42 + 10 = 52$ years.

Son's age after 10 years = Present age of son + 10 years = $14 + 10 = 24$ years.

Ratio = $\frac{\text{Father's age after 10 years}}{\text{Son's age after 10 years}}$

Ratio = $\frac{52}{24}$

Simplify the fraction by dividing the numerator and denominator by their GCD, which is 4.

Ratio = $\frac{\cancel{52}^{13}}{\cancel{24}_{6}} = \frac{13}{6}$

The ratio is 13 : 6.

(d) Age of father to the age of son when father was 30 years old:

The father's current age is 42 years. The father was 30 years old 12 years ago ($42 - 30 = 12$).

At that time (12 years ago), the son's age would also have been 12 years less than his present age.

Son's age when father was 30 = Present age of son - 12 years

Son's age = $14 - 12 = 2$ years.

Father's age at that time = 30 years.

Ratio = $\frac{\text{Father's age when father was 30}}{\text{Son's age at that time}}$

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

Simplify the fraction by dividing the numerator and denominator by their GCD, which is 2.

Ratio = $\frac{\cancel{30}^{15}}{\cancel{2}_{1}} = \frac{15}{1}$

The ratio is 15 : 1.

---

The ratios are:

(a) 3 : 1

(b) 10 : 3

(c) 13 : 6

(d) 15 : 1

Given:

Two ratios: 25g : 30g and 40 kg : 48 kg.

---

To Determine:

Whether the two given ratios are in proportion.

---

Solution:

To check if two ratios are in proportion, we need to compare their simplest forms. If the simplest forms of the two ratios are equal, then the ratios are in proportion.

Ratio 1: 25g : 30g

The first ratio is 25g : 30g. We can write this as a fraction $\frac{25}{30}$.

The units are the same (grams), so we can simplify the numbers.

We find the greatest common divisor (GCD) of 25 and 30. The factors of 25 are 1, 5, 25. The factors of 30 are 1, 2, 3, 5, 6, 10, 15, 30. The GCD is 5.

Divide both the numerator and the denominator by 5:

$$\frac{25 \div 5}{30 \div 5} = \frac{5}{6}$$

The simplified form of the first ratio is 5 : 6.

Ratio 2: 40 kg : 48 kg

The second ratio is 40 kg : 48 kg. We can write this as a fraction $\frac{40}{48}$.

The units are the same (kilograms), so we can simplify the numbers.

We find the greatest common divisor (GCD) of 40 and 48. We can divide by common factors:

Divide by 8:

$$\frac{40 \div 8}{48 \div 8} = \frac{5}{6}$$

Alternatively, using cancellation with GCD = 8:

$$\frac{\cancel{40}^{5}}{\cancel{48}_{6}} = \frac{5}{6}$$

The simplified form of the second ratio is 5 : 6.

Comparison:

The simplified form of the first ratio is 5 : 6.

The simplified form of the second ratio is 5 : 6.

Since the simplified forms of both ratios are the same (5 : 6), the ratios 25g : 30g and 40 kg : 48 kg are in proportion.

---

Yes, the ratios 25g : 30g and 40 kg : 48 kg are in proportion because their simplest forms are equal (5 : 6).

Given:

The four numbers: 30, 40, 45, and 60.

---

To Determine:

Whether the numbers 30, 40, 45, and 60 are in proportion.

---

Solution:

Four numbers $a, b, c, d$ are in proportion if the ratio $a:b$ is equal to the ratio $c:d$. This can be written as $a:b :: c:d$ or $\frac{a}{b} = \frac{c}{d}$.

In this case, the four numbers are 30, 40, 45, and 60. We need to check if the ratio of the first two numbers is equal to the ratio of the last two numbers.

Ratio of the first two numbers = 30 : 40 = $\frac{30}{40}$

Simplify the fraction by dividing the numerator and denominator by their GCD, which is 10.

$\frac{30}{40} = \frac{\cancel{30}^{3}}{\cancel{40}_{4}} = \frac{3}{4}$

Ratio of the last two numbers = 45 : 60 = $\frac{45}{60}$

Simplify the fraction by dividing the numerator and denominator by their GCD, which is 15.

$\frac{45}{60} = \frac{\cancel{45}^{3}}{\cancel{60}_{4}} = \frac{3}{4}$

Comparison:

The ratio of the first two numbers is $\frac{3}{4}$.

The ratio of the last two numbers is $\frac{3}{4}$.

Since the ratio of the first two numbers is equal to the ratio of the last two numbers ($\frac{3}{4} = \frac{3}{4}$), the numbers 30, 40, 45, and 60 are in proportion.

We can write this as 30 : 40 :: 45 : 60.

---

Yes, 30, 40, 45, and 60 are in proportion.

Given:

Two ratios: 15 cm to 2 m and 10 sec to 3 minutes.

---

To Determine:

Whether the two given ratios form a proportion.

---

Solution:

To compare ratios, the quantities in each ratio must be in the same units.

Ratio 1: 15 cm to 2 m

The units are centimeters (cm) and meters (m). We need to convert them to the same unit. Convert meters to centimeters.

We know that 1 m = 100 cm.

So, 2 m = $2 \times 100$ cm = 200 cm.

The first ratio is 15 cm : 200 cm.

Ratio 1 = $\frac{15 \text{ cm}}{200 \text{ cm}} = \frac{15}{200}$

Simplify the fraction by dividing the numerator and denominator by their GCD, which is 5.

$$\frac{15 \div 5}{200 \div 5} = \frac{3}{40}$$

The simplified form of the first ratio is 3 : 40.

Ratio 2: 10 sec to 3 minutes

The units are seconds (sec) and minutes. We need to convert them to the same unit. Convert minutes to seconds.

We know that 1 minute = 60 seconds.

So, 3 minutes = $3 \times 60$ seconds = 180 seconds.

The second ratio is 10 sec : 180 sec.

Ratio 2 = $\frac{10 \text{ sec}}{180 \text{ sec}} = \frac{10}{180}$

Simplify the fraction by dividing the numerator and denominator by their GCD, which is 10.

$$\frac{10 \div 10}{180 \div 10} = \frac{1}{18}$$

The simplified form of the second ratio is 1 : 18.

Comparison:

The simplified form of the first ratio is $\frac{3}{40}$.

The simplified form of the second ratio is $\frac{1}{18}$.

To check if these fractions are equal, we can compare them (e.g., cross-multiply or find a common denominator). $3 \times 18 = 54$ and $40 \times 1 = 40$. Since $54 \neq 40$, the fractions are not equal.

$$\frac{3}{40} \neq \frac{1}{18}$$

Since the simplified forms of the two ratios are not equal, the ratios 15 cm to 2 m and 10 sec to 3 minutes do not form a proportion.

---

No, the ratios 15 cm to 2 m and 10 sec to 3 minutes do not form a proportion.

Solution:

---

To determine if four numbers $a, b, c, d$ are in proportion, we check if the ratio $a:b$ is equal to the ratio $c:d$. This is equivalent to checking if $\frac{a}{b} = \frac{c}{d}$. We simplify each ratio and compare them.

(a) 15, 45, 40, 120

Ratio of the first two numbers: $\frac{15}{45}$.

$$\frac{15}{45} = \frac{\cancel{15}^{1}}{\cancel{45}_{3}} = \frac{1}{3}$$

Ratio of the last two numbers: $\frac{40}{120}$.

$$\frac{40}{120} = \frac{\cancel{40}^{1}}{\cancel{120}_{3}} = \frac{1}{3}$$

Since $\frac{1}{3} = \frac{1}{3}$, the ratios are equal.

Thus, 15, 45, 40, and 120 are in proportion.

---

(b) 33, 121, 9, 96

Ratio of the first two numbers: $\frac{33}{121}$.

$$\frac{33}{121} = \frac{\cancel{33}^{3}}{\cancel{121}_{11}} = \frac{3}{11}$$

Ratio of the last two numbers: $\frac{9}{96}$.

$$\frac{9}{96} = \frac{\cancel{9}^{3}}{\cancel{96}_{32}} = \frac{3}{32}$$

Since $\frac{3}{11} \neq \frac{3}{32}$, the ratios are not equal.

Thus, 33, 121, 9, and 96 are not in proportion.

---

(c) 24, 28, 36, 48

Ratio of the first two numbers: $\frac{24}{28}$.

$$\frac{24}{28} = \frac{\cancel{24}^{6}}{\cancel{28}_{7}} = \frac{6}{7}$$

Ratio of the last two numbers: $\frac{36}{48}$.

$$\frac{36}{48} = \frac{\cancel{36}^{3}}{\cancel{48}_{4}} = \frac{3}{4}$$

Since $\frac{6}{7} \neq \frac{3}{4}$, the ratios are not equal.

Thus, 24, 28, 36, and 48 are not in proportion.

---

(d) 32, 48, 70, 210

Ratio of the first two numbers: $\frac{32}{48}$.

$$\frac{32}{48} = \frac{\cancel{32}^{2}}{\cancel{48}_{3}} = \frac{2}{3}$$

Ratio of the last two numbers: $\frac{70}{210}$.

$$\frac{70}{210} = \frac{\cancel{70}^{1}}{\cancel{210}_{3}} = \frac{1}{3}$$

Since $\frac{2}{3} \neq \frac{1}{3}$, the ratios are not equal.

Thus, 32, 48, 70, and 210 are not in proportion.

---

(e) 4, 6, 8, 12

Ratio of the first two numbers: $\frac{4}{6}$.

$$\frac{4}{6} = \frac{\cancel{4}^{2}}{\cancel{6}_{3}} = \frac{2}{3}$$

Ratio of the last two numbers: $\frac{8}{12}$.

$$\frac{8}{12} = \frac{\cancel{8}^{2}}{\cancel{12}_{3}} = \frac{2}{3}$$

Since $\frac{2}{3} = \frac{2}{3}$, the ratios are equal.

Thus, 4, 6, 8, and 12 are in proportion.

---

(f) 33, 44, 75, 100

Ratio of the first two numbers: $\frac{33}{44}$.

$$\frac{33}{44} = \frac{\cancel{33}^{3}}{\cancel{44}_{4}} = \frac{3}{4}$$

Ratio of the last two numbers: $\frac{75}{100}$.

$$\frac{75}{100} = \frac{\cancel{75}^{3}}{\cancel{100}_{4}} = \frac{3}{4}$$

Since $\frac{3}{4} = \frac{3}{4}$, the ratios are equal.

Thus, 33, 44, 75, and 100 are in proportion.

---

Summary:

(a) 15, 45, 40, 120 are in proportion.

(b) 33, 121, 9, 96 are not in proportion.

(c) 24, 28, 36, 48 are not in proportion.

(d) 32, 48, 70, 210 are not in proportion.

(e) 4, 6, 8, 12 are in proportion.

(f) 33, 44, 75, 100 are in proportion.

Solution:

---

We check if the ratios on both sides of the '::' symbol are equivalent. '::' means 'is in proportion to', implying the ratios are equal.

(a) 16 : 24 :: 20 : 30

Ratio 1: $\frac{16}{24} = \frac{\cancel{16}^{2}}{\cancel{24}_{3}} = \frac{2}{3}$

Ratio 2: $\frac{20}{30} = \frac{\cancel{20}^{2}}{\cancel{30}_{3}} = \frac{2}{3}$

Since $\frac{2}{3} = \frac{2}{3}$, the statement is true.

Answer: T

---

(b) 21 : 6 :: 35 : 10

Ratio 1: $\frac{21}{6} = \frac{\cancel{21}^{7}}{\cancel{6}_{2}} = \frac{7}{2}$

Ratio 2: $\frac{35}{10} = \frac{\cancel{35}^{7}}{\cancel{10}_{2}} = \frac{7}{2}$

Since $\frac{7}{2} = \frac{7}{2}$, the statement is true.

Answer: T

---

(c) 12 : 18 :: 28 : 12

Ratio 1: $\frac{12}{18} = \frac{\cancel{12}^{2}}{\cancel{18}_{3}} = \frac{2}{3}$

Ratio 2: $\frac{28}{12} = \frac{\cancel{28}^{7}}{\cancel{12}_{3}} = \frac{7}{3}$

Since $\frac{2}{3} \neq \frac{7}{3}$, the statement is false.

Answer: F

---

(d) 8 : 9 :: 24 : 27

Ratio 1: $\frac{8}{9}$. This is already in simplest form.

Ratio 2: $\frac{24}{27} = \frac{\cancel{24}^{8}}{\cancel{27}_{9}} = \frac{8}{9}$

Since $\frac{8}{9} = \frac{8}{9}$, the statement is true.

Answer: T

---

(e) 5.2 : 3.9 :: 3 : 4

Ratio 1: $\frac{5.2}{3.9}$. To simplify, we can multiply the numerator and denominator by 10 to remove decimals.

$\frac{5.2 \times 10}{3.9 \times 10} = \frac{52}{39}$

The GCD of 52 and 39 is 13 ($52 = 4 \times 13$, $39 = 3 \times 13$).

$$\frac{52}{39} = \frac{\cancel{52}^{4}}{\cancel{39}_{3}} = \frac{4}{3}$$

Ratio 2: $\frac{3}{4}$. This is already in simplest form.

Since $\frac{4}{3} \neq \frac{3}{4}$, the statement is false.

Answer: F

---

(f) 0.9 : 0.36 :: 10 : 4

Ratio 1: $\frac{0.9}{0.36}$. To simplify, multiply the numerator and denominator by 100 to remove decimals.

$\frac{0.9 \times 100}{0.36 \times 100} = \frac{90}{36}$

The GCD of 90 and 36 is 18 ($90 = 5 \times 18$, $36 = 2 \times 18$).

$$\frac{90}{36} = \frac{\cancel{90}^{5}}{\cancel{36}_{2}} = \frac{5}{2}$$

Ratio 2: $\frac{10}{4} = \frac{\cancel{10}^{5}}{\cancel{4}_{2}} = \frac{5}{2}$

Since $\frac{5}{2} = \frac{5}{2}$, the statement is true.

Answer: T

---

Summary:

(a) T

(b) T

(c) F

(d) T

(e) F

(f) T

Solution:

---

To check if the given statements are true, we need to find the simplest form of the ratios on both sides of the equality sign and compare them.

(a) 40 persons : 200 persons = $\textsf{₹}$ 15 : $\textsf{₹}$ 75

Left side ratio: $\frac{40 \text{ persons}}{200 \text{ persons}} = \frac{40}{200} = \frac{\cancel{40}^{1}}{\cancel{200}_{5}} = \frac{1}{5}$.

Right side ratio: $\frac{\textsf{₹} \ 15}{\textsf{₹} \ 75} = \frac{15}{75} = \frac{\cancel{15}^{1}}{\cancel{75}_{5}} = \frac{1}{5}$.

Since $\frac{1}{5} = \frac{1}{5}$, the ratios are equal.

The statement is True (T).

---

(b) 7.5 litres : 15 litres = 5 kg : 10 kg

Left side ratio: $\frac{7.5 \text{ litres}}{15 \text{ litres}} = \frac{7.5}{15} = \frac{75}{150} = \frac{\cancel{75}^{1}}{\cancel{150}_{2}} = \frac{1}{2}$.

Right side ratio: $\frac{5 \text{ kg}}{10 \text{ kg}} = \frac{5}{10} = \frac{\cancel{5}^{1}}{\cancel{10}_{2}} = \frac{1}{2}$.

Since $\frac{1}{2} = \frac{1}{2}$, the ratios are equal.

The statement is True (T).

---

(c) 99 kg : 45 kg = $\textsf{₹}$ 44 : $\textsf{₹}$ 20

Left side ratio: $\frac{99 \text{ kg}}{45 \text{ kg}} = \frac{99}{45} = \frac{\cancel{99}^{11}}{\cancel{45}_{5}} = \frac{11}{5}$.

Right side ratio: $\frac{\textsf{₹} \ 44}{\textsf{₹} \ 20} = \frac{44}{20} = \frac{\cancel{44}^{11}}{\cancel{20}_{5}} = \frac{11}{5}$.

Since $\frac{11}{5} = \frac{11}{5}$, the ratios are equal.

The statement is True (T).

---

(d) 32 m : 64 m = 6 sec : 12 sec

Left side ratio: $\frac{32 \text{ m}}{64 \text{ m}} = \frac{32}{64} = \frac{\cancel{32}^{1}}{\cancel{64}_{2}} = \frac{1}{2}$.

Right side ratio: $\frac{6 \text{ sec}}{12 \text{ sec}} = \frac{6}{12} = \frac{\cancel{6}^{1}}{\cancel{12}_{2}} = \frac{1}{2}$.

Since $\frac{1}{2} = \frac{1}{2}$, the ratios are equal.

The statement is True (T).

---

(e) 45 km : 60 km = 12 hours : 15 hours

Left side ratio: $\frac{45 \text{ km}}{60 \text{ km}} = \frac{45}{60} = \frac{\cancel{45}^{3}}{\cancel{60}_{4}} = \frac{3}{4}$.

Right side ratio: $\frac{12 \text{ hours}}{15 \text{ hours}} = \frac{12}{15} = \frac{\cancel{12}^{4}}{\cancel{15}_{5}} = \frac{4}{5}$.

Since $\frac{3}{4} \neq \frac{4}{5}$, the ratios are not equal.

The statement is False (F).

---

Summary:

(a) T

(b) T

(c) T

(d) T

(e) F

Solution:

---

To determine if two ratios $a:b$ and $c:d$ form a proportion ($a:b :: c:d$), we check if $\frac{a}{b} = \frac{c}{d}$. If they form a proportion, the terms $a$ and $d$ are the extreme terms, and the terms $b$ and $c$ are the middle terms.

---

(a) 25 cm : 1 m and $\textsf{₹}$ 40 : $\textsf{₹}$ 160

Ratio 1: 25 cm : 1 m.

First, convert the units to be the same. Convert 1 meter to centimeters: 1 m = 100 cm.

The first ratio is 25 cm : 100 cm.

Ratio 1 = $\frac{25 \text{ cm}}{100 \text{ cm}} = \frac{25}{100}$.

Simplify the fraction by dividing the numerator and denominator by their greatest common divisor, which is 25.

$$\frac{25 \div 25}{100 \div 25} = \frac{1}{4}$$

The simplified form of the first ratio is 1 : 4.

Ratio 2: $\textsf{₹}$ 40 : $\textsf{₹}$ 160.

Ratio 2 = $\frac{\textsf{₹} \ 40}{\textsf{₹} \ 160} = \frac{40}{160}$.

Simplify the fraction by dividing the numerator and denominator by their greatest common divisor, which is 40.

$$\frac{40 \div 40}{160 \div 40} = \frac{1}{4}$$

The simplified form of the second ratio is 1 : 4.

Since the simplified forms of both ratios are equal ($\frac{1}{4} = \frac{1}{4}$), the ratios form a proportion.

The proportion is 25 cm : 1 m :: $\textsf{₹}$ 40 : $\textsf{₹}$ 160.

The four terms are 25 cm, 1 m, $\textsf{₹}$ 40, and $\textsf{₹}$ 160.

The middle terms are the second and third terms: 1 m and $\textsf{₹}$ 40.

The extreme terms are the first and fourth terms: 25 cm and $\textsf{₹}$ 160.

---

(b) 39 litres : 65 litres and 6 bottles : 10 bottles

Ratio 1: 39 litres : 65 litres.

Ratio 1 = $\frac{39 \text{ litres}}{65 \text{ litres}} = \frac{39}{65}$.

Simplify the fraction by dividing the numerator and denominator by their greatest common divisor, which is 13.

$$\frac{39 \div 13}{65 \div 13} = \frac{3}{5}$$

The simplified form of the first ratio is 3 : 5.

Ratio 2: 6 bottles : 10 bottles.

Ratio 2 = $\frac{6 \text{ bottles}}{10 \text{ bottles}} = \frac{6}{10}$.

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

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

The simplified form of the second ratio is 3 : 5.

Since the simplified forms of both ratios are equal ($\frac{3}{5} = \frac{3}{5}$), the ratios form a proportion.

The proportion is 39 litres : 65 litres :: 6 bottles : 10 bottles.

The four terms are 39 litres, 65 litres, 6 bottles, and 10 bottles.

The middle terms are the second and third terms: 65 litres and 6 bottles.

The extreme terms are the first and fourth terms: 39 litres and 10 bottles.

---

(c) 2 kg : 80 kg and 25 g : 625 g

Ratio 1: 2 kg : 80 kg.

Ratio 1 = $\frac{2 \text{ kg}}{80 \text{ kg}} = \frac{2}{80}$.

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

$$\frac{2 \div 2}{80 \div 2} = \frac{1}{40}$$

The simplified form of the first ratio is 1 : 40.

Ratio 2: 25 g : 625 g.

Ratio 2 = $\frac{25 \text{ g}}{625 \text{ g}} = \frac{25}{625}$.

Simplify the fraction by dividing the numerator and denominator by their greatest common divisor, which is 25.

$$\frac{25 \div 25}{625 \div 25} = \frac{1}{25}$$

The simplified form of the second ratio is 1 : 25.

Since the simplified forms of the two ratios are not equal ($\frac{1}{40} \neq \frac{1}{25}$), the ratios do not form a proportion.

---

(d) 200 mL : 2.5 litre and $\textsf{₹}$ 4 : $\textsf{₹}$ 50

Ratio 1: 200 mL : 2.5 litre.

First, convert the units to be the same. Convert 2.5 litres to milliliters: 1 litre = 1000 mL.

2.5 litre = $2.5 \times 1000$ mL = 2500 mL.

The first ratio is 200 mL : 2500 mL.

Ratio 1 = $\frac{200 \text{ mL}}{2500 \text{ mL}} = \frac{200}{2500}$.

Simplify the fraction by dividing the numerator and denominator by their greatest common divisor, which is 100.

$$\frac{200 \div 100}{2500 \div 100} = \frac{2}{25}$$

The simplified form of the first ratio is 2 : 25.

Ratio 2: $\textsf{₹}$ 4 : $\textsf{₹}$ 50.

Ratio 2 = $\frac{\textsf{₹} \ 4}{\textsf{₹} \ 50} = \frac{4}{50}$.

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

$$\frac{4 \div 2}{50 \div 2} = \frac{2}{25}$$

The simplified form of the second ratio is 2 : 25.

Since the simplified forms of both ratios are equal ($\frac{2}{25} = \frac{2}{25}$), the ratios form a proportion.

The proportion is 200 mL : 2.5 litre :: $\textsf{₹}$ 4 : $\textsf{₹}$ 50.

The four terms are 200 mL, 2.5 litre, $\textsf{₹}$ 4, and $\textsf{₹}$ 50.

The middle terms are the second and third terms: 2.5 litre and $\textsf{₹}$ 4.

The extreme terms are the first and fourth terms: 200 mL and $\textsf{₹}$ 50.

Given:

The cost of 6 cans of juice is $\textsf{₹}$ 210.

---

To Find:

The cost of 4 cans of juice.

---

Solution (Using Unitary Method):

In the unitary method, we first find the cost of one unit (one can of juice) and then multiply it by the required number of units (4 cans of juice).

Cost of 6 cans of juice = $\textsf{₹}$ 210.

Cost of 1 can of juice = $\frac{\text{Total cost}}{\text{Number of cans}}$

Cost of 1 can of juice = $\frac{\textsf{₹} \ 210}{6}$.

Performing the division:

$$\frac{210}{6} = \frac{\cancel{210}^{35}}{\cancel{6}_{1}} = 35$$

Cost of 1 can of juice = $\textsf{₹}$ 35.

Now, we need to find the cost of 4 cans of juice.

Cost of 4 cans of juice = Cost of 1 can $\times$ Number of cans

Cost of 4 cans of juice = $\textsf{₹} \ 35 \times 4$.

Performing the multiplication:

$\begin{array}{cc}& & 3 & 5 \\ \times & & & 4 \\ \hline & 1 & 4 & 0 \\ \hline \end{array}$

Cost of 4 cans of juice = $\textsf{₹}$ 140.

---

Alternate Solution (Using Proportion):

We can assume that the cost of juice is directly proportional to the number of cans, meaning the ratio of the number of cans is equal to the ratio of their costs.

Let the cost of 4 cans of juice be $\textsf{₹}$ $x$.

The ratio of the number of cans is 6 : 4.

The ratio of the costs is 210 : $x$.

Since they are in proportion:

6 cans : 4 cans :: $\textsf{₹}$ 210 : $\textsf{₹}$ $x$

$$\frac{6}{4} = \frac{210}{x}$$

To solve for $x$, we can cross-multiply:

$6 \times x = 4 \times 210$

$6x = 840$

Divide both sides by 6:

$x = \frac{840}{6}$

$x = 140$.

The cost of 4 cans of juice is $\textsf{₹}$ 140.

---

The cost of 4 cans of juice will be $\textsf{₹}$ 140.

Given:

Distance travelled by the motorbike in 5 litres of petrol = 220 km.

Amount of petrol = 1.5 litres.

---

To Find:

The distance the motorbike will cover in 1.5 litres of petrol.

---

Solution (Using Unitary Method):

First, find the distance covered by the motorbike in 1 litre of petrol (this is the mileage of the motorbike).

Distance covered in 5 litres = 220 km.

Distance covered in 1 litre = $\frac{\text{Total distance}}{\text{Total petrol consumed}}$

Distance covered in 1 litre = $\frac{220 \text{ km}}{5 \text{ litres}}$.

Performing the division:

$$\frac{220}{5} = \frac{\cancel{220}^{44}}{\cancel{5}_{1}} = 44$$

Distance covered in 1 litre = 44 km.

Now, we need to find the distance covered in 1.5 litres of petrol.

Distance covered in 1.5 litres = Distance covered in 1 litre $\times$ Amount of petrol

Distance covered in 1.5 litres = $44 \times 1.5$ km.

Performing the multiplication:

$\begin{array}{cc}& & 4 & 4 \\ \times & & 1 & 5 \\ \hline & 2 & 2 & 0 \\ + & 4 & 4 & \times \\ \hline & 6 & 6 & 0 \\ \hline \end{array}$

Since 1.5 has one decimal place, the result also has one decimal place.

$44 \times 1.5 = 66.0$

Distance covered in 1.5 litres = 66 km.

---

Alternate Solution (Using Proportion):

We can assume that the distance travelled is directly proportional to the amount of petrol consumed. The ratio of the amount of petrol is equal to the ratio of the distances travelled.

Let the distance covered in 1.5 litres be $x$ km.

The ratio of the amount of petrol is 5 litres : 1.5 litres.

The ratio of the distances is 220 km : $x$ km.

Since they are in proportion:

5 litres : 1.5 litres :: 220 km : $x$ km

$$\frac{5}{1.5} = \frac{220}{x}$$

To solve for $x$, we can cross-multiply:

$5 \times x = 1.5 \times 220$

$5x = 330$

Divide both sides by 5:

$x = \frac{330}{5}$

Performing the division:

$$\frac{330}{5} = \frac{\cancel{330}^{66}}{\cancel{5}_{1}} = 66$$

$x = 66$.

The distance covered in 1.5 litres of petrol is 66 km.

---

The motorbike will cover 66 km in 1.5 litres of petrol.

Given:

The cost of a dozen soaps is $\textsf{₹}$ 153.60.

A dozen means 12.

---

To Find:

The cost of 15 such soaps.

---

Solution (Using Unitary Method):

First, find the cost of one soap.

Cost of 12 soaps = $\textsf{₹}$ 153.60.

Cost of 1 soap = $\frac{\text{Total cost of 12 soaps}}{\text{Number of soaps}}$

Cost of 1 soap = $\frac{\textsf{₹} \ 153.60}{12}$.

Cost of 1 soap = $\textsf{₹}$ 12.80.

Now, we need to find the cost of 15 soaps.

Cost of 15 soaps = Cost of 1 soap $\times$ Number of soaps

Cost of 15 soaps = $\textsf{₹} \ 12.80 \times 15$.

$12.80 \times 15 = 192.00$.

Cost of 15 soaps = $\textsf{₹}$ 192.00.

---

Alternate Solution (Using Proportion):

We can assume that the cost of the soaps is directly proportional to the number of soaps. The ratio of the number of soaps is equal to the ratio of their costs.

Let the cost of 15 soaps be $\textsf{₹}$ $x$.

The number of soaps are 12 and 15.

The costs are $\textsf{₹}$ 153.60 and $\textsf{₹}$ $x$.

Since they are in proportion:

12 soaps : 15 soaps :: $\textsf{₹}$ 153.60 : $\textsf{₹}$ $x$

$$\frac{12}{15} = \frac{153.60}{x}$$

To solve for $x$, we can cross-multiply:

$12 \times x = 15 \times 153.60$

$12x = 2304.00$

$12x = 2304$

Divide both sides by 12:

$x = \frac{2304}{12}$

$x = 192$.

The cost of 15 soaps is $\textsf{₹}$ 192.

---

The cost of 15 such soaps will be $\textsf{₹}$ 192.00.

Given:

The cost of 105 envelopes is $\textsf{₹}$ 350.

The amount of money available is $\textsf{₹}$ 100.

---

To Find:

The number of envelopes that can be purchased for $\textsf{₹}$ 100.

---

Solution (Using Unitary Method):

In this case, it is easier to find the number of envelopes that can be purchased for $\textsf{₹}$ 1 (the unit cost per envelope is not a whole number, but the number of envelopes per rupee might be simpler). We find the number of envelopes per rupee and then multiply by the total amount of money.

For $\textsf{₹}$ 350, the number of envelopes is 105.

Number of envelopes for $\textsf{₹}$ 1 = $\frac{\text{Total number of envelopes}}{\text{Total cost}}$

Number of envelopes for $\textsf{₹}$ 1 = $\frac{105}{350}$.

Simplify the fraction by dividing the numerator and denominator by their GCD. Both end in 0 or 5, so divide by 5.

$\frac{105 \div 5}{350 \div 5} = \frac{21}{70}$

Both are divisible by 7.

$\frac{21 \div 7}{70 \div 7} = \frac{3}{10}$

Number of envelopes for $\textsf{₹}$ 1 = $\frac{3}{10}$ envelopes.

Now, we need to find the number of envelopes that can be purchased for $\textsf{₹}$ 100.

Number of envelopes for $\textsf{₹}$ 100 = (Number of envelopes for $\textsf{₹}$ 1) $\times$ Total amount

Number of envelopes for $\textsf{₹}$ 100 = $\frac{3}{10} \times 100$.

$$\frac{3}{10} \times 100 = \frac{3 \times \cancel{100}^{10}}{\cancel{10}_{1}} = 3 \times 10 = 30$$

Number of envelopes for $\textsf{₹}$ 100 = 30 envelopes.

---

Alternate Solution (Using Proportion):

We can assume that the number of envelopes is directly proportional to the cost. The ratio of the costs is equal to the ratio of the number of envelopes.

Let the number of envelopes that can be purchased for $\textsf{₹}$ 100 be $x$.

The costs are $\textsf{₹}$ 350 and $\textsf{₹}$ 100.

The number of envelopes are 105 and $x$.

Since they are in proportion:

$\textsf{₹}$ 350 : $\textsf{₹}$ 100 :: 105 envelopes : $x$ envelopes

$$\frac{350}{100} = \frac{105}{x}$$

To solve for $x$, we can cross-multiply:

$350 \times x = 100 \times 105$

$350x = 10500$

Divide both sides by 350:

$x = \frac{10500}{350}$

We can cancel out a zero from the numerator and denominator:

$x = \frac{1050}{35}$

Performing the division (or simplifying the fraction):

Both are divisible by 5:

$\frac{1050 \div 5}{35 \div 5} = \frac{210}{7}$

Divide by 7:

$\frac{210 \div 7}{7 \div 7} = \frac{30}{1} = 30$

$x = 30$.

The number of envelopes that can be purchased for $\textsf{₹}$ 100 is 30.

---

30 envelopes can be purchased for $\textsf{₹}$ 100.

Given:

Distance travelled by the car = 90 km.

Time taken to travel 90 km = $2\frac{1}{2}$ hours.

$2\frac{1}{2}$ hours = $2.5$ hours.

---

To Find:

(a) Time required to cover 30 km with the same speed.

(b) Distance covered in 2 hours with the same speed.

---

Solution:

Since the speed is the same in both parts, we can use the concept of proportionality or the unitary method. First, let's calculate the speed of the car.

Speed = $\frac{\text{Distance}}{\text{Time}}$

Speed = $\frac{90 \text{ km}}{2.5 \text{ hours}}$.

To calculate the speed, we can divide 90 by 2.5. Multiply numerator and denominator by 10 to remove the decimal:

Speed = $\frac{90 \times 10}{2.5 \times 10} = \frac{900}{25}$

Divide 900 by 25:

$$\frac{900}{25} = \frac{\cancel{900}^{36}}{\cancel{25}_{1}} = 36$$

Speed = 36 km/hour.

(a) Time required to cover 30 km:

We know the speed is 36 km/hour. We want to find the time taken to cover 30 km.

Time = $\frac{\text{Distance}}{\text{Speed}}$

Time = $\frac{30 \text{ km}}{36 \text{ km/hour}}$

Time = $\frac{30}{36}$ hours.

Simplify the fraction by dividing the numerator and denominator by their GCD, which is 6.

$$\frac{30 \div 6}{36 \div 6} = \frac{5}{6}$$

Time = $\frac{5}{6}$ hours.

We can also express this in minutes. 1 hour = 60 minutes.

Time in minutes = $\frac{5}{6} \times 60$ minutes = $\frac{5 \times \cancel{60}^{10}}{\cancel{6}_{1}} = 5 \times 10 = 50$ minutes.

Time required is $\frac{5}{6}$ hours or 50 minutes.

(b) Distance covered in 2 hours:

We know the speed is 36 km/hour. We want to find the distance covered in 2 hours.

Distance = Speed $\times$ Time

Distance = $36 \text{ km/hour} \times 2 \text{ hours}$

Distance = $36 \times 2 = 72$ km.

Distance covered is 72 km.

---

Summary:

(a) Time required to cover 30 km is $\frac{5}{6}$ hours (or 50 minutes).

(b) Distance covered in 2 hours is 72 km.

Given:

The cost of 7 m of cloth is $\textsf{₹}$ 1470.

---

To Find:

The cost of 5 m of cloth.

---

Solution (Using Unitary Method):

First, we find the cost of 1 meter of cloth.

Cost of 1 m of cloth = $\frac{\text{Cost of 7 m of cloth}}{\text{Length of cloth}}$

Cost of 1 m of cloth = $\frac{\textsf{₹} \ 1470}{7}$.

Performing the division:

$$ \textsf{₹} \ \frac{\cancel{1470}^{210}}{\cancel{7}_{1}} = \textsf{₹} \ 210 $$

So, the cost of 1 m of cloth is $\textsf{₹}$ 210.

Now, we need to find the cost of 5 meters of cloth.

Cost of 5 m of cloth = Cost of 1 m of cloth $\times$ 5

Cost of 5 m of cloth = $\textsf{₹} \ 210 \times 5$.

Performing the multiplication:

$$ 210 \times 5 = 1050 $$

So, the cost of 5 m of cloth is $\textsf{₹}$ 1050.

---

Alternate Solution (Using Proportion):

We can assume that the cost of the cloth is directly proportional to its length. This means that the ratio of the lengths is equal to the ratio of the costs.

Let the cost of 5 m of cloth be $\textsf{₹}$ $x$.

The length of cloth are 7 m and 5 m.

The corresponding costs are $\textsf{₹}$ 1470 and $\textsf{₹}$ $x$.

Forming the proportion:

$$7 \text{ m} : 5 \text{ m} :: \textsf{₹} \ 1470 : \textsf{₹} \ x$$

This can be written as a fraction equation:

$$\frac{7}{5} = \frac{1470}{x}$$

To solve for $x$, we can use cross-multiplication:

$$7 \times x = 5 \times 1470$$

$$7x = 7350$$

Now, divide both sides of the equation by 7:

$$x = \frac{7350}{7}$$

Performing the division:

$$ x = \frac{\cancel{7350}^{1050}}{\cancel{7}_{1}} = 1050 $$

So, the cost of 5 m of cloth is $\textsf{₹}$ 1050.

---

The cost of 5 m of cloth is $\textsf{₹}$ 1050.

Given:

Ekta earns $\textsf{₹}$ 3000 in 10 days.

---

To Find:

The amount Ekta will earn in 30 days.

---

Solution (Using Unitary Method):

First, we find Ekta's earning per day.

Earning in 10 days = $\textsf{₹}$ 3000.

Earning in 1 day = $\frac{\text{Total earning}}{\text{Number of days}}$

Earning in 1 day = $\frac{\textsf{₹} \ 3000}{10}$.

Performing the division:

$$ \textsf{₹} \ \frac{\cancel{3000}^{300}}{\cancel{10}_{1}} = \textsf{₹} \ 300 $$

So, Ekta earns $\textsf{₹}$ 300 per day.

Now, we need to find the amount she will earn in 30 days.

Earning in 30 days = Earning in 1 day $\times$ Number of days

Earning in 30 days = $\textsf{₹} \ 300 \times 30$.

Performing the multiplication:

$$ 300 \times 30 = 9000 $$

So, Ekta will earn $\textsf{₹}$ 9000 in 30 days.

---

Alternate Solution (Using Proportion):

Assuming Ekta earns at a constant rate, her earning is directly proportional to the number of days she works. The ratio of the number of days is equal to the ratio of the earnings.

Let the earning in 30 days be $\textsf{₹}$ $x$.

The number of days are 10 and 30.

The corresponding earnings are $\textsf{₹}$ 3000 and $\textsf{₹}$ $x$.

Forming the proportion:

$$10 \text{ days} : 30 \text{ days} :: \textsf{₹} \ 3000 : \textsf{₹} \ x$$

This can be written as a fraction equation:

$$\frac{10}{30} = \frac{3000}{x}$$

Simplify the left side:

$$\frac{\cancel{10}^{1}}{\cancel{30}_{3}} = \frac{1}{3}$$

So, the equation is:

$$\frac{1}{3} = \frac{3000}{x}$$

To solve for $x$, we can cross-multiply:

$$1 \times x = 3 \times 3000$$

$$x = 9000$$

So, Ekta will earn $\textsf{₹}$ 9000 in 30 days.

---

Ekta will earn $\textsf{₹}$ 9000 in 30 days.

Given:

Rainfall in the last 3 days = 276 mm.

The rain continues to fall at the same rate.

---

To Find:

The amount of rain that will fall in one full week (7 days) in centimeters (cm).

---

Solution (Using Unitary Method):

First, we find the amount of rain that falls per day.

Rainfall in 3 days = 276 mm.

Rainfall in 1 day = $\frac{\text{Total rainfall}}{\text{Number of days}}$

Rainfall in 1 day = $\frac{276 \text{ mm}}{3}$.

Performing the division:

$$ \frac{276}{3} = \frac{\cancel{276}^{92}}{\cancel{3}_{1}} = 92 $$

Rainfall in 1 day = 92 mm.

Now, we need to find the amount of rain that will fall in one full week (7 days).

Rainfall in 7 days = Rainfall in 1 day $\times$ 7

Rainfall in 7 days = $92 \text{ mm} \times 7$.

Performing the multiplication:

$$\begin{array}{cc}& & 9 & 2 \\ \times & & & 7 \\ \hline & 6 & 4 & 4 \\ \hline \end{array}$$

Rainfall in 7 days = 644 mm.

The question asks for the answer in centimeters (cm).

We know that 1 cm = 10 mm.

To convert millimeters to centimeters, we divide by 10.

Rainfall in cm = $\frac{644 \text{ mm}}{10 \text{ mm/cm}} = 64.4$ cm.

---

Alternate Solution (Using Proportion):

Assuming the rain falls at a constant rate, the amount of rainfall is directly proportional to the number of days. The ratio of the number of days is equal to the ratio of the rainfall amounts.

Let the rainfall in 7 days be $x$ mm.

The number of days are 3 and 7.

The corresponding rainfall amounts are 276 mm and $x$ mm.

Forming the proportion:

$$3 \text{ days} : 7 \text{ days} :: 276 \text{ mm} : x \text{ mm}$$

This can be written as a fraction equation:

$$\frac{3}{7} = \frac{276}{x}$$

To solve for $x$, we can cross-multiply:

$$3 \times x = 7 \times 276$$

$$3x = 1932$$

Now, divide both sides of the equation by 3:

$$x = \frac{1932}{3}$$

Performing the division:

$$ \frac{1932}{3} = \frac{\cancel{1932}^{644}}{\cancel{3}_{1}} = 644 $$

So, the rainfall in 7 days is 644 mm.

Convert this to centimeters:

Rainfall in cm = $\frac{644}{10}$ cm = 64.4 cm.

---

64.4 cm of rain will fall in one full week.

Given:

The cost of 5 kg of wheat is $\textsf{₹}$ 91.50.

---

To Find:

(a) The cost of 8 kg of wheat.

(b) The quantity of wheat that can be purchased for $\textsf{₹}$ 183.

---

Solution:

First, find the cost of 1 kg of wheat.

Cost of 5 kg of wheat = $\textsf{₹}$ 91.50.

Cost of 1 kg of wheat = $\frac{\text{Cost of 5 kg of wheat}}{\text{Quantity of wheat}}$

Cost of 1 kg of wheat = $\frac{\textsf{₹} \ 91.50}{5}$.

Cost of 1 kg of wheat = $\textsf{₹}$ 18.30.

(a) Cost of 8 kg of wheat:

Cost of 8 kg of wheat = Cost of 1 kg of wheat $\times$ 8

Cost of 8 kg of wheat = $\textsf{₹} \ 18.30 \times 8$.

Cost of 8 kg of wheat = $\textsf{₹}$ 146.40.

(b) Quantity of wheat that can be purchased for $\textsf{₹}$ 183:

We know the cost of 1 kg of wheat is $\textsf{₹}$ 18.30.

Quantity of wheat for $\textsf{₹}$ 1 = $\frac{1 \text{ kg}}{\textsf{₹} \ 18.30}$.

Quantity of wheat for $\textsf{₹}$ 183 = Quantity of wheat for $\textsf{₹}$ 1 $\times$ Total amount

Quantity of wheat for $\textsf{₹}$ 183 = $\frac{1}{18.30} \times 183$ kg.

Multiply the numerator and denominator of $\frac{1}{18.30}$ by 100 to remove the decimal:

$\frac{100}{1830}$

So, Quantity = $\frac{100}{1830} \times 183 = \frac{100 \times \cancel{183}}{\cancel{1830}_{10}} = \frac{100}{10} = 10$ kg.

Alternatively, perform the division $\frac{183}{18.30}$. Multiply numerator and denominator by 100:

$\frac{183 \times 100}{18.30 \times 100} = \frac{18300}{1830} = \frac{1830}{183} = 10$.

Quantity of wheat = 10 kg.

---

Summary:

(a) The cost of 8 kg of wheat will be $\textsf{₹}$ 146.40.

(b) 10 kg of wheat can be purchased for $\textsf{₹}$ 183.

Given:

Temperature dropped by 15$^\circ$C in the last 30 days.

The rate of temperature drop remains the same.

---

To Find:

The temperature drop in the next 10 days in degrees Celsius.

---

Solution (Using Unitary Method):

First, find the rate of temperature drop per day.

Temperature drop in 30 days = 15$^\circ$C.

Temperature drop in 1 day = $\frac{\text{Total temperature drop}}{\text{Number of days}}$

Temperature drop in 1 day = $\frac{15^\circ\text{C}}{30}$.

Simplify the fraction:

$$\frac{15}{30} = \frac{\cancel{15}^{1}}{\cancel{30}_{2}} = \frac{1}{2}$$

Temperature drop in 1 day = $\frac{1}{2}^\circ$C = 0.5$^\circ$C.

Now, we need to find the temperature drop in the next 10 days.

Temperature drop in 10 days = Temperature drop in 1 day $\times$ 10

Temperature drop in 10 days = $\frac{1}{2}^\circ\text{C} \times 10$.

$$\frac{1}{2} \times 10 = \frac{1 \times \cancel{10}^{5}}{\cancel{2}_{1}} = 5$$

Temperature drop in 10 days = 5$^\circ$C.

---

Alternate Solution (Using Proportion):

Assuming the rate of temperature drop is constant, the temperature drop is directly proportional to the number of days. The ratio of the number of days is equal to the ratio of the temperature drops.

Let the temperature drop in 10 days be $x^\circ$C.

The number of days are 30 and 10.

The corresponding temperature drops are 15$^\circ$C and $x^\circ$C.

Forming the proportion:

$$30 \text{ days} : 10 \text{ days} :: 15^\circ\text{C} : x^\circ\text{C}$$

This can be written as a fraction equation:

$$\frac{30}{10} = \frac{15}{x}$$

Simplify the left side:

$$\frac{30}{10} = 3$$

So, the equation is:

$$3 = \frac{15}{x}$$

To solve for $x$, multiply both sides by $x$ and then divide by 3:

$3x = 15$

$x = \frac{15}{3}$

$x = 5$

So, the temperature drop in the next 10 days is 5$^\circ$C.

---

The temperature will drop by 5 degrees Celsius in the next ten days.

Given:

Shaina pays $\textsf{₹}$ 15000 as rent for 3 months.

The rent per month remains the same.

---

To Find:

The total rent Shaina has to pay for a whole year.

---

Solution (Using Unitary Method):

First, find the rent paid per month.

Rent for 3 months = $\textsf{₹}$ 15000.

Rent for 1 month = $\frac{\text{Total rent for 3 months}}{\text{Number of months}}$

Rent for 1 month = $\frac{\textsf{₹} \ 15000}{3}$.

Performing the division:

$$ \textsf{₹} \ \frac{\cancel{15000}^{5000}}{\cancel{3}_{1}} = \textsf{₹} \ 5000 $$

So, the rent per month is $\textsf{₹}$ 5000.

Now, we need to find the total rent for a whole year. A whole year has 12 months.

Rent for 12 months = Rent for 1 month $\times$ 12

Rent for 12 months = $\textsf{₹} \ 5000 \times 12$.

Performing the multiplication:

$$ 5000 \times 12 = 60000 $$

So, Shaina has to pay $\textsf{₹}$ 60000 for a whole year.

---

Alternate Solution (Using Proportion):

Assuming the rent per month is constant, the total rent is directly proportional to the number of months. The ratio of the number of months is equal to the ratio of the total rent.

Let the rent for 12 months be $\textsf{₹}$ $x$.

The number of months are 3 and 12.

The corresponding rents are $\textsf{₹}$ 15000 and $\textsf{₹}$ $x$.

Forming the proportion:

$$3 \text{ months} : 12 \text{ months} :: \textsf{₹} \ 15000 : \textsf{₹} \ x$$

This can be written as a fraction equation:

$$\frac{3}{12} = \frac{15000}{x}$$

Simplify the left side:

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

So, the equation is:

$$\frac{1}{4} = \frac{15000}{x}$$

To solve for $x$, we can cross-multiply:

$$1 \times x = 4 \times 15000$$

$$x = 60000$$

So, Shaina has to pay $\textsf{₹}$ 60000 for a whole year.

---

Shaina has to pay $\textsf{₹}$ 60000 for a whole year.

Given:

The cost of 4 dozen bananas is $\textsf{₹}$ 180.

1 dozen = 12.

So, 4 dozen bananas = $4 \times 12 = 48$ bananas.

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To Find:

The number of bananas that can be purchased for $\textsf{₹}$ 90.

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Solution (Using Unitary Method):

We are given that 48 bananas cost $\textsf{₹}$ 180.

We need to find how many bananas can be purchased for $\textsf{₹}$ 90. It is easier to find the number of bananas per rupee.

Number of bananas for $\textsf{₹}$ 180 = 48 bananas.

Number of bananas for $\textsf{₹}$ 1 = $\frac{\text{Total number of bananas}}{\text{Total cost}}$

Number of bananas for $\textsf{₹}$ 1 = $\frac{48 \text{ bananas}}{\textsf{₹} \ 180}$.

Simplify the fraction by dividing the numerator and denominator by their GCD, which is 12.

$$\frac{48 \div 12}{180 \div 12} = \frac{4}{15}$$

Number of bananas for $\textsf{₹}$ 1 = $\frac{4}{15}$ bananas.

Now, we need to find the number of bananas that can be purchased for $\textsf{₹}$ 90.

Number of bananas for $\textsf{₹}$ 90 = (Number of bananas for $\textsf{₹}$ 1) $\times$ Total amount

Number of bananas for $\textsf{₹}$ 90 = $\frac{4}{15} \times 90$ bananas.

$$\frac{4}{15} \times 90 = \frac{4 \times \cancel{90}^{6}}{\cancel{15}_{1}} = 4 \times 6 = 24$$

Number of bananas for $\textsf{₹}$ 90 = 24 bananas.

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Alternate Solution (Using Proportion):

We can assume that the number of bananas is directly proportional to the cost. The ratio of the costs is equal to the ratio of the number of bananas.

Let the number of bananas that can be purchased for $\textsf{₹}$ 90 be $x$.

The costs are $\textsf{₹}$ 180 and $\textsf{₹}$ 90.

The number of bananas are 48 and $x$.

Forming the proportion:

$$\textsf{₹} \ 180 : \textsf{₹} \ 90 :: 48 \text{ bananas} : x \text{ bananas}$$

This can be written as a fraction equation:

$$\frac{180}{90} = \frac{48}{x}$$Simplify the left side:

$$\frac{180}{90} = \frac{\cancel{180}^{2}}{\cancel{90}_{1}} = 2$$

So, the equation is:

$$2 = \frac{48}{x}$$

To solve for $x$, multiply both sides by $x$ and then divide by 2:

$2x = 48$

$x = \frac{48}{2}$

$x = 24$

So, 24 bananas can be purchased for $\textsf{₹}$ 90.

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24 bananas can be purchased for $\textsf{₹}$ 90.

Given:

The weight of 72 books is 9 kg.

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To Find:

The weight of 40 such books in kilograms.

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Solution (Using Unitary Method):

First, find the weight of 1 book.

Weight of 72 books = 9 kg.

Weight of 1 book = $\frac{\text{Total weight}}{\text{Number of books}}$

Weight of 1 book = $\frac{9 \text{ kg}}{72}$.

Simplify the fraction by dividing the numerator and denominator by their GCD, which is 9.

$$ \frac{9}{72} = \frac{\cancel{9}^{1}}{\cancel{72}_{8}} = \frac{1}{8} $$

Weight of 1 book = $\frac{1}{8}$ kg.

Now, we need to find the weight of 40 such books.

Weight of 40 books = Weight of 1 book $\times$ 40

Weight of 40 books = $\frac{1}{8} \text{ kg} \times 40$.

$$ \frac{1}{8} \times 40 = \frac{1 \times \cancel{40}^{5}}{\cancel{8}_{1}} = 1 \times 5 = 5 $$

Weight of 40 books = 5 kg.

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Alternate Solution (Using Proportion):

Assuming the weight of the books is directly proportional to the number of books. The ratio of the number of books is equal to the ratio of their weights.

Let the weight of 40 books be $x$ kg.

The number of books are 72 and 40.

The corresponding weights are 9 kg and $x$ kg.

Forming the proportion:

$$72 \text{ books} : 40 \text{ books} :: 9 \text{ kg} : x \text{ kg}$$

This can be written as a fraction equation:

$$\frac{72}{40} = \frac{9}{x}$$

Simplify the left side by dividing numerator and denominator by their GCD, which is 8.

$$\frac{72 \div 8}{40 \div 8} = \frac{9}{5}$$

So, the equation is:

$$\frac{9}{5} = \frac{9}{x}$$

Comparing the numerators, they are the same (9). For the equality to hold, the denominators must also be the same.

So, $x = 5$.

Alternatively, cross-multiply:

$$9 \times x = 5 \times 9$$

$$9x = 45$$

Divide both sides by 9:

$$x = \frac{45}{9}$$$$x = 5$$

So, the weight of 40 such books is 5 kg.

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The weight of 40 such books is 5 kg.

Given:

Distance covered by the truck with 108 litres of diesel = 594 km.

New distance to be covered = 1650 km.

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To Find:

The amount of diesel required to cover a distance of 1650 km, assuming the truck consumes diesel at a constant rate.

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Solution (Using Unitary Method):

First, find the amount of diesel required to cover 1 km.

Diesel required for 594 km = 108 litres.

Diesel required for 1 km = $\frac{\text{Diesel required for 594 km}}{\text{Distance covered}}$

Diesel required for 1 km = $\frac{108 \text{ litres}}{594 \text{ km}}$.

Simplify the fraction $\frac{108}{594}$. We can divide both numerator and denominator by their greatest common divisor.

Divide by 54:

$$ \frac{108 \div 54}{594 \div 54} = \frac{2}{11} $$

Alternatively, using cancellation:

$$ \frac{\cancel{108}^{2}}{\cancel{594}_{11}} = \frac{2}{11} $$

So, the diesel required for 1 km is $\frac{2}{11}$ litres.

Now, we need to find the amount of diesel required for 1650 km.

Diesel required for 1650 km = (Diesel required for 1 km) $\times$ 1650

Diesel required for 1650 km = $\frac{2}{11} \text{ litres/km} \times 1650 \text{ km}$.

$$ \frac{2}{11} \times 1650 = \frac{2 \times \cancel{1650}^{150}}{\cancel{11}_{1}} = 2 \times 150 = 300 $$

Diesel required for 1650 km = 300 litres.

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Alternate Solution (Using Proportion):

Assuming the rate of diesel consumption is constant, the amount of diesel required is directly proportional to the distance covered. The ratio of the distances is equal to the ratio of the amounts of diesel.

Let the amount of diesel required for 1650 km be $x$ litres.

The distances are 594 km and 1650 km.

The corresponding diesel amounts are 108 litres and $x$ litres.

Forming the proportion:

$$594 \text{ km} : 1650 \text{ km} :: 108 \text{ litres} : x \text{ litres}$$

This can be written as a fraction equation:

$$\frac{594}{1650} = \frac{108}{x}$$

To solve for $x$, we can cross-multiply:

$$594 \times x = 1650 \times 108$$

$$594x = 178200$$

Now, divide both sides of the equation by 594:

$$x = \frac{178200}{594}$$

Performing the division:

$$ \frac{178200}{594} = \frac{\cancel{178200}^{300}}{\cancel{594}_{1}} = 300 $$

So, the diesel required for 1650 km is 300 litres.

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The truck will require 300 litres of diesel to cover a distance of 1650 km.

Given:

Raju purchases 10 pens for $\textsf{₹}$ 150.

Manish buys 7 pens for $\textsf{₹}$ 84.

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To Determine:

Who got the pens cheaper.

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Solution:

To find out who got the pens cheaper, we need to compare the cost per pen for both Raju and Manish. We will calculate the unit cost (cost of one pen) in each case.

Cost per pen for Raju:

Cost of 10 pens = $\textsf{₹}$ 150.

Cost of 1 pen for Raju = $\frac{\text{Total cost}}{\text{Number of pens}}$

Cost of 1 pen for Raju = $\frac{\textsf{₹} \ 150}{10}$.

Performing the division:

$$ \textsf{₹} \ \frac{\cancel{150}^{15}}{\cancel{10}_{1}} = \textsf{₹} \ 15 $$

So, Raju paid $\textsf{₹}$ 15 per pen.

Cost per pen for Manish:

Cost of 7 pens = $\textsf{₹}$ 84.

Cost of 1 pen for Manish = $\frac{\text{Total cost}}{\text{Number of pens}}$

Cost of 1 pen for Manish = $\frac{\textsf{₹} \ 84}{7}$.

Performing the division:

$$ \textsf{₹} \ \frac{\cancel{84}^{12}}{\cancel{7}_{1}} = \textsf{₹} \ 12 $$

So, Manish paid $\textsf{₹}$ 12 per pen.

Comparison of costs:

Cost per pen for Raju = $\textsf{₹}$ 15.

Cost per pen for Manish = $\textsf{₹}$ 12.

Comparing the unit costs, $\textsf{₹}$ 12 is less than $\textsf{₹}$ 15 ($\textsf{₹} \ 12 < \textsf{₹} \ 15$).

Therefore, Manish got the pens cheaper.

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Manish got the pens cheaper as the cost per pen for Manish is $\textsf{₹}$ 12, which is less than $\textsf{₹}$ 15 (cost per pen for Raju).

Given:

Anish made 42 runs in 6 overs.

Anup made 63 runs in 7 overs.

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To Determine:

Who made more runs per over.

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Solution:

To find out who made more runs per over, we need to calculate the runs scored per over for both Anish and Anup. Runs per over is calculated by dividing the total runs by the number of overs.

Runs per over for Anish:

Runs made by Anish = 42.

Overs bowled for Anish = 6.

Runs per over for Anish = $\frac{\text{Total runs}}{\text{Number of overs}}$

Runs per over for Anish = $\frac{42}{6}$.

Performing the division:

$$ \frac{42}{6} = 7 $$

Anish made 7 runs per over.

Runs per over for Anup:

Runs made by Anup = 63.

Overs bowled for Anup = 7.

Runs per over for Anup = $\frac{\text{Total runs}}{\text{Number of overs}}$

Runs per over for Anup = $\frac{63}{7}$.

Performing the division:

$$ \frac{63}{7} = 9 $$

Anup made 9 runs per over.

Comparison of runs per over:

Runs per over for Anish = 7.

Runs per over for Anup = 9.

Comparing the runs per over, 9 is greater than 7 ($9 > 7$).

Therefore, Anup made more runs per over.

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Anup made more runs per over (9 runs per over) compared to Anish (7 runs per over).