Chapter 1 Rational Numbers (Additional Questions)

A rational number is a number that can be expressed in the form $\frac{p}{q}$, where $p$ and $q$ are integers and $q \neq 0$.

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(A) $\sqrt{2}$: The square root of 2 is a non-terminating, non-repeating decimal. It cannot be expressed in the form $\frac{p}{q}$. Thus, it is an irrational number.

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(B) $\pi$: Pi ($\pi$) is a transcendental number, which means it is an irrational number. It is a non-terminating, non-repeating decimal. Thus, it is an irrational number.

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(C) $0.101101110...$: This decimal is non-terminating and non-repeating because the pattern of 1s between the 0s increases ($10$, $110$, $1110$, etc.). It cannot be expressed in the form $\frac{p}{q}$. Thus, it is an irrational number.

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(D) $-\frac{5}{7}$: This number is in the form $\frac{p}{q}$, where $p = -5$ and $q = 7$. Both $-5$ and $7$ are integers, and the denominator $7$ is not equal to zero. Thus, it is a rational number.

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Therefore, the only rational number among the given options is $-\frac{5}{7}$.

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The correct option is (D) $-\frac{5}{7}$.

The given equation is $\frac{3}{4} + (-\frac{3}{4}) = 0$.

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Let's examine the given options:

(A) Commutative property of addition states that for any two numbers $a$ and $b$, $a + b = b + a$. This property involves changing the order of the operands, which is not the primary illustration in the given equation.

(B) Additive identity states that for any number $a$, $a + 0 = a$ or $0 + a = a$. The number $0$ is the additive identity. The given equation results in $0$, but it shows the sum of a number and another number equaling $0$, not the sum of a number and $0$ equaling the number itself.

(C) Additive inverse states that for any number $a$, there exists a number $-a$ such that $a + (-a) = 0$ or $(-a) + a = 0$. The number $-a$ is called the additive inverse (or opposite) of $a$. In the given equation, $a = \frac{3}{4}$ and $-a = -\frac{3}{4}$. The equation shows that the sum of a number ($\frac{3}{4}$) and its additive inverse ($-\frac{3}{4}$) is equal to the additive identity ($0$).

(D) Associative property of addition states that for any three numbers $a$, $b$, and $c$, $(a + b) + c = a + (b + c)$. This property involves grouping of three or more operands, which is not applicable to the given equation involving only two numbers.

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The equation $\frac{3}{4} + (-\frac{3}{4}) = 0$ directly illustrates the property that the sum of a number and its opposite (additive inverse) is zero.

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Therefore, the property illustrated is the Additive inverse property.

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The correct option is (C) Additive inverse.

The multiplicative inverse (or reciprocal) of a non-zero rational number $\frac{a}{b}$ is $\frac{b}{a}$.

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The product of a non-zero rational number and its multiplicative inverse is always the multiplicative identity, which is $1$.

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

The multiplicative inverse of $\frac{2}{3}$ is $\frac{3}{2}$.

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Now, we find the product of the number and its multiplicative inverse:

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

Product $= \frac{\cancel{2}}{\cancel{3}} \times \frac{\cancel{3}}{\cancel{2}}$

Product $= 1$

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The product of $\frac{2}{3}$ and its multiplicative inverse is $1$.

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The correct option is (B) $1$.

Let's analyze each statement:

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(A) Every integer is a rational number.

A rational number is a number that can be expressed in the form $\frac{p}{q}$, where $p$ and $q$ are integers and $q \neq 0$.

An integer $n$ can always be written as $\frac{n}{1}$. Since $n$ and $1$ are integers and $1 \neq 0$, every integer fits the definition of a rational number.

For example, the integer $5$ can be written as $\frac{5}{1}$, which is a rational number. The integer $-3$ can be written as $\frac{-3}{1}$, which is a rational number.

This statement is TRUE.

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(B) Every fraction is a rational number.

In the context of rational numbers, a fraction is generally considered to be of the form $\frac{p}{q}$ where $p$ and $q$ are integers and $q \neq 0$. This is exactly the definition of a rational number.

For example, $\frac{2}{3}$, $-\frac{1}{4}$, $\frac{7}{5}$ are fractions, and they are all rational numbers.

This statement is TRUE.

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(C) Every rational number is an integer.

A rational number is of the form $\frac{p}{q}$, where $p$ and $q$ are integers and $q \neq 0$. An integer is a whole number ($\dots, -2, -1, 0, 1, 2, \dots$).

Consider the rational number $\frac{1}{2}$. Here, $p=1$ and $q=2$. It is a rational number because $1$ and $2$ are integers and $2 \neq 0$. However, $\frac{1}{2} = 0.5$, which is not an integer.

Therefore, there exist rational numbers that are not integers. This statement is FALSE.

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(D) $0$ is a rational number.

$0$ can be expressed as $\frac{0}{1}$. Here, $p=0$ and $q=1$. Both $0$ and $1$ are integers, and the denominator $1$ is not equal to $0$. Thus, $0$ fits the definition of a rational number.

This statement is TRUE.

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The statement which is FALSE is "Every rational number is an integer."

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The correct option is (C) Every rational number is an integer.

A rational number $\frac{p}{q}$ is positive if $p$ and $q$ have the same sign.

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Let's examine each option:

(A) $p > 0, q > 0$: If both $p$ and $q$ are positive, their quotient $\frac{p}{q}$ is positive.

Example: $\frac{3}{4}$ is positive, where $3 > 0$ and $4 > 0$.

This condition makes the rational number positive.

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(B) $p < 0, q < 0$: If both $p$ and $q$ are negative, their quotient $\frac{p}{q}$ is positive (negative divided by negative is positive).

Example: $\frac{-3}{-4} = \frac{3}{4}$ is positive, where $-3 < 0$ and $-4 < 0$.

This condition also makes the rational number positive.

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(C) $p > 0, q < 0$: If $p$ is positive and $q$ is negative, their quotient $\frac{p}{q}$ is negative (positive divided by negative is negative).

Example: $\frac{3}{-4} = -\frac{3}{4}$ is negative, where $3 > 0$ and $-4 < 0$.

This condition makes the rational number negative.

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(D) Both (A) and (B): Since the rational number $\frac{p}{q}$ is positive when condition (A) is met and also when condition (B) is met, the correct answer is that it is positive if both (A) and (B) conditions are valid possibilities for it to be positive.

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The rational number $\frac{p}{q}$ is positive if $p$ and $q$ are both positive OR if $p$ and $q$ are both negative.

Therefore, both conditions (A) and (B) result in a positive rational number.

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The correct option is (D) Both (A) and (B).

We need to evaluate the expression $\frac{5}{8} + (-\frac{3}{4})$.

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To add fractions, we need to find a common denominator. The denominators are $8$ and $4$.

The least common multiple (LCM) of $8$ and $4$ is $8$.

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We rewrite the second fraction with a denominator of $8$:

$-\frac{3}{4} = -\frac{3 \times 2}{4 \times 2} = -\frac{6}{8}$

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Now, substitute this back into the original expression:

$\frac{5}{8} + (-\frac{3}{4}) = \frac{5}{8} + (-\frac{6}{8})$

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To add fractions with the same denominator, we add the numerators and keep the common denominator:

$\frac{5 + (-6)}{8} = \frac{5 - 6}{8} = \frac{-1}{8}$

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So, the result is $-\frac{1}{8}$.

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Comparing this result with the given options, we see that it matches option (B).

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The correct option is (B) $-\frac{1}{8}$.

We need to evaluate $-\frac{7}{12} \div (-\frac{2}{3})$.

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Dividing by a fraction is the same as multiplying by its reciprocal.

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

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So, the expression becomes:

$-\frac{7}{12} \div (-\frac{2}{3}) = -\frac{7}{12} \times (-\frac{3}{2})$

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When multiplying two negative numbers, the result is positive.

$-\frac{7}{12} \times (-\frac{3}{2}) = \frac{7}{12} \times \frac{3}{2}$

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Now, multiply the numerators and the denominators:

$\frac{7}{12} \times \frac{3}{2} = \frac{7 \times 3}{12 \times 2}$

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We can simplify the expression by cancelling out common factors. Both $3$ and $12$ are divisible by $3$.

$\frac{7 \times \cancel{3}^{1}}{\cancel{12}_{4} \times 2} = \frac{7 \times 1}{4 \times 2} = \frac{7}{8}$

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So, $-\frac{7}{12} \div (-\frac{2}{3}) = \frac{7}{8}$.

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Comparing this result with the given options, we see that it matches option (B).

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The correct option is (B) $\frac{7}{8}$.

We are asked to subtract $-\frac{4}{9}$ from $\frac{2}{9}$.

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This can be written as:

$\frac{2}{9} - (-\frac{4}{9})$

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Subtracting a negative number is equivalent to adding the corresponding positive number:

$\frac{2}{9} - (-\frac{4}{9}) = \frac{2}{9} + \frac{4}{9}$

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Since the fractions have the same denominator ($9$), we can add the numerators directly:

$\frac{2 + 4}{9} = \frac{6}{9}$

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The result is $\frac{6}{9}$. This fraction can be simplified by dividing both the numerator and the denominator by their greatest common divisor, which is $3$:

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

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However, looking at the options, $\frac{6}{9}$ is given as option (A). While $\frac{2}{3}$ is the simplified form, $\frac{6}{9}$ is also a correct representation of the result.

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Comparing $\frac{6}{9}$ with the given options, we find that it matches option (A).

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The correct option is (A) $\frac{6}{9}$.

The additive inverse of a number $a$ is the number $-a$ such that when you add $a$ and $-a$, the result is the additive identity, $0$. That is, $a + (-a) = 0$.

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We want to find the additive inverse of $0$. Let the additive inverse of $0$ be $x$.

According to the definition of additive inverse, we must have:

$0 + x = 0$

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From the equation $0 + x = 0$, we can see that $x$ must be equal to $0$.

Also, the additive inverse of $0$ is $-0$, which is equal to $0$.

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Therefore, the additive inverse of $0$ is $0$.

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Comparing this result with the given options, we find that it matches option (C).

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The correct option is (C) $0$.

A rational number is a number that can be expressed in the form $\frac{p}{q}$, where $p$ and $q$ are integers and $q \neq 0$. Rational numbers include terminating decimals and repeating decimals.

An irrational number is a number that cannot be expressed in the form $\frac{p}{q}$. Irrational numbers are non-terminating and non-repeating decimals.

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Let's analyze each option:

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(A) $\sqrt{4}$: The square root of $4$ is $2$. $2$ is an integer, and any integer can be written in the form $\frac{p}{q}$ (e.g., $2 = \frac{2}{1}$). Since $2$ and $1$ are integers and $1 \neq 0$, $\sqrt{4}$ is a rational number.

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(B) $0$: $0$ is an integer, and it can be written in the form $\frac{p}{q}$ (e.g., $0 = \frac{0}{1}$). Since $0$ and $1$ are integers and $1 \neq 0$, $0$ is a rational number.

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(C) $1.23456...$ (non-terminating, non-repeating): This decimal representation goes on infinitely without any repeating block of digits. This is the definition of an irrational number.

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(D) $0.333...$: This is a repeating decimal, where the digit $3$ repeats infinitely. Repeating decimals can always be expressed as a fraction in the form $\frac{p}{q}$. For example, $0.333... = \frac{1}{3}$. Since $1$ and $3$ are integers and $3 \neq 0$, $0.333...$ is a rational number.

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We are looking for the number that is NOT a rational number. Based on our analysis, $\sqrt{4}$, $0$, and $0.333...$ are all rational numbers, while $1.23456...$ is an irrational number.

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Therefore, the number which is NOT a rational number is $1.23456...$ (non-terminating, non-repeating).

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The correct option is (C) $1.23456...$ (non-terminating, non-repeating).

Let's analyze the given Assertion (A) and Reason (R).

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Assertion (A): Between any two distinct rational numbers, there are infinitely many rational numbers.

Consider two distinct rational numbers $x$ and $y$, with $x < y$. We can find a rational number between them, for example, the average $\frac{x+y}{2}$. Since $x$ and $y$ are rational, $(x+y)$ is rational, and dividing by $2$ (a non-zero rational number) gives a rational number. Also, $x < \frac{x+y}{2} < y$.

We can repeat this process. Between $x$ and $\frac{x+y}{2}$, we can find another rational number. Between $\frac{x+y}{2}$ and $y$, we can find another rational number. This process can be continued indefinitely, generating an infinite number of distinct rational numbers between $x$ and $y$.

Thus, Assertion (A) is TRUE.

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Reason (R): The property that states there is at least one rational number between any two distinct rational numbers is called the density property.

The property described in Reason (R) is indeed known as the density property of rational numbers. It formally states that for any two distinct rational numbers $a$ and $b$, there exists a rational number $c$ such that $a < c < b$.

Thus, Reason (R) is TRUE.

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Now let's examine if Reason (R) is the correct explanation for Assertion (A).

Assertion (A) states that there are infinitely many rational numbers between any two distinct ones.

Reason (R) states that there is at least one rational number between any two distinct ones (this is the definition of the density property).

The density property (R) is the fundamental reason why Assertion (A) is true. Because we can always find *at least one* rational number between any two, we can apply this property repeatedly to generate an infinite sequence of distinct rational numbers within the given interval. For example, given $r_1 < r_2$, the density property guarantees a rational $r_3$ such that $r_1 < r_3 < r_2$. Applying the property again, there exists a rational $r_4$ such that $r_1 < r_4 < r_3$, and so on. This iterative process confirms the existence of infinitely many rational numbers between $r_1$ and $r_2$.

Therefore, Reason (R) correctly explains why Assertion (A) is true.

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Both the Assertion and the Reason are true, and the Reason is the correct explanation for the Assertion.

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The correct option is (A) Both A and R are true, and R is the correct explanation of A.

Let's examine each statement regarding the closure property of rational numbers.

A set is said to be closed under an operation if performing the operation on any two elements of the set always results in an element that is also in the same set.

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(A) Rational numbers are closed under addition.

Let $\frac{p}{q}$ and $\frac{r}{s}$ be two rational numbers, where $p, q, r, s$ are integers and $q \neq 0$, $s \neq 0$.

Their sum is $\frac{p}{q} + \frac{r}{s} = \frac{ps + qr}{qs}$.

Since $p, q, r, s$ are integers, $ps$, $qr$, and $qs$ are also integers. Also, since $q \neq 0$ and $s \neq 0$, $qs \neq 0$.

Thus, $\frac{ps + qr}{qs}$ is a rational number.

Statement (A) is TRUE.

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(B) Rational numbers are closed under subtraction.

Let $\frac{p}{q}$ and $\frac{r}{s}$ be two rational numbers.

Their difference is $\frac{p}{q} - \frac{r}{s} = \frac{ps - qr}{qs}$.

Since $p, q, r, s$ are integers, $ps$, $qr$, $ps - qr$, and $qs$ are also integers. Also, $qs \neq 0$.

Thus, $\frac{ps - qr}{qs}$ is a rational number.

Statement (B) is TRUE.

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(C) Rational numbers are closed under multiplication.

Let $\frac{p}{q}$ and $\frac{r}{s}$ be two rational numbers.

Their product is $\frac{p}{q} \times \frac{r}{s} = \frac{pr}{qs}$.

Since $p, q, r, s$ are integers, $pr$ and $qs$ are also integers. Also, $qs \neq 0$.

Thus, $\frac{pr}{qs}$ is a rational number.

Statement (C) is TRUE.

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(D) Rational numbers are closed under division.

Let $\frac{p}{q}$ and $\frac{r}{s}$ be two rational numbers. We need to consider division by a non-zero rational number. So, let $\frac{r}{s} \neq 0$, which means $r \neq 0$.

Their division is $\frac{p}{q} \div \frac{r}{s} = \frac{p}{q} \times \frac{s}{r} = \frac{ps}{qr}$.

Since $p, q, r, s$ are integers, $ps$ and $qr$ are also integers. Since $q \neq 0$ and we assumed $r \neq 0$, their product $qr \neq 0$.

Thus, $\frac{ps}{qr}$ is a rational number (provided the divisor is non-zero).

Under the standard definition of closure under division for number sets (excluding division by zero), this statement is TRUE.

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Based on the properties of rational numbers, they are closed under addition, subtraction, multiplication, and division (by non-zero rational numbers).

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All the given statements are TRUE.

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The statements that are TRUE are (A), (B), (C), and (D).

Let's match each expression with its corresponding property:

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(i) $a+b = b+a$

This expression shows that the order of the operands in addition does not affect the result. This is the definition of the Commutative property of addition.

So, (i) matches with (b).

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(ii) $a \times (b \times c) = (a \times b) \times c$

This expression shows that the grouping of the operands in multiplication does not affect the result. This is the definition of the Associative property of multiplication.

So, (ii) matches with (c).

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(iii) $a+0 = 0+a = a$

This expression shows that adding $0$ to any number results in the number itself. $0$ is the Additive identity.

So, (iii) matches with (d).

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(iv) $a \times 1 = 1 \times a = a$

This expression shows that multiplying any number by $1$ results in the number itself. $1$ is the Multiplicative identity.

So, (iv) matches with (a).

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The correct matching is: (i)-(b), (ii)-(c), (iii)-(d), (iv)-(a).

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Comparing this matching with the given options, we find that it corresponds to option (B).

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The correct option is (B) (i)-(b), (ii)-(c), (iii)-(d), (iv)-(a).

The multiplicative inverse (or reciprocal) of a non-zero rational number $\frac{a}{b}$ is the rational number $\frac{b}{a}$.

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The property of multiplicative inverse states that the product of a non-zero number and its multiplicative inverse is equal to the multiplicative identity.

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For rational numbers, the multiplicative identity is $1$, because for any rational number $x$, $x \times 1 = 1 \times x = x$.

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Let $\frac{p}{q}$ be a non-zero rational number (so $p \neq 0$ and $q \neq 0$). Its multiplicative inverse is $\frac{q}{p}$.

The product of the number and its multiplicative inverse is:

$\frac{p}{q} \times \frac{q}{p}$

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Multiplying the numerators and the denominators, we get:

$\frac{p \times q}{q \times p} = \frac{pq}{pq}$

Since $pq \neq 0$, the fraction simplifies to $1$.

$\frac{\cancel{pq}}{\cancel{pq}} = 1$

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So, the product of a rational number and its multiplicative inverse is $1$.

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The correct option is (B) $1$.

We need to determine the position of the rational number $-\frac{5}{3}$ on a number line.

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First, let's convert the fraction $\frac{5}{3}$ into a mixed number or a decimal.

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

$\frac{5}{3} = 1 \frac{2}{3}$

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Therefore, $-\frac{5}{3} = -(1 \frac{2}{3})$.

This means the number is $1$ whole unit less than $0$, plus an additional $\frac{2}{3}$ of a unit further to the left.

So, $-\frac{5}{3}$ is located between $-1$ and $-2$ on the number line.

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To determine whether it is closer to $-1$ or $-2$, we can compare the fractional part $\frac{2}{3}$ to $\frac{1}{2}$.

Since $\frac{2}{3} > \frac{1}{2}$ (because $2 \times 2 = 4$ and $3 \times 1 = 3$, and $4 > 3$), the point is more than halfway from $-1$ towards $-2$.

Alternatively, we can find the distance from $-1$ and $-2$:

Distance from $-1$ is $|-\frac{5}{3} - (-1)| = |-\frac{5}{3} + \frac{3}{3}| = |-\frac{2}{3}| = \frac{2}{3}$.

Distance from $-2$ is $|-\frac{5}{3} - (-2)| = |-\frac{5}{3} + \frac{6}{3}| = |\frac{1}{3}| = \frac{1}{3}$.

Since $\frac{1}{3} < \frac{2}{3}$, the point $-\frac{5}{3}$ is closer to $-2$ than to $-1$.

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Thus, $-\frac{5}{3}$ is a point between $-1$ and $-2$ closer to $-2$.

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Let's check the options:

(A) A point between $0$ and $1$: This is incorrect as $-\frac{5}{3}$ is negative.

(B) A point between $-1$ and $-2$ closer to $-2$: This matches our finding.

(C) A point between $-1$ and $-2$ closer to $-1$: This is incorrect as $-\frac{5}{3}$ is closer to $-2$.

(D) A point to the left of $-2$: This would mean the number is less than $-2$, but $-\frac{5}{3} = -1\frac{2}{3}$ is greater than $-2$. This is incorrect.

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The correct option is (B) A point between $-1$ and $-2$ closer to $-2$.

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Given:

Total quantity of sugar = $20 \text{ kg}$.

Fraction of sugar sold = $\frac{3}{4}$.

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To Find:

Quantity of sugar left with the shopkeeper.

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Solution:

We need to find the amount of sugar that was sold. The amount sold is $\frac{3}{4}$ of the total quantity of sugar.

Quantity of sugar sold = $\frac{3}{4} \times 20 \text{ kg}$.

Now, we calculate the value:

Quantity of sugar sold = $\frac{3}{\cancel{4}_{1}} \times \cancel{20}^{5} \text{ kg}$.

Quantity of sugar sold = $3 \times 5 \text{ kg}$.

Quantity of sugar sold = $15 \text{ kg}$.

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To find the quantity of sugar left, we subtract the quantity sold from the total quantity.

Quantity of sugar left = Total quantity of sugar - Quantity of sugar sold.

Quantity of sugar left = $20 \text{ kg} - 15 \text{ kg}$.

Quantity of sugar left = $5 \text{ kg}$.

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Alternate Solution:

We can first find the fraction of sugar that is left.

The total quantity of sugar can be represented as 1 whole.

Fraction of sugar left = $1 - \text{Fraction of sugar sold}$.

Fraction of sugar left = $1 - \frac{3}{4}$.

To subtract, we find a common denominator, which is 4.

Fraction of sugar left = $\frac{4}{4} - \frac{3}{4} = \frac{4-3}{4} = \frac{1}{4}$.

Now, we calculate the quantity of sugar corresponding to this fraction.

Quantity of sugar left = Fraction of sugar left $\times$ Total quantity of sugar.

Quantity of sugar left = $\frac{1}{4} \times 20 \text{ kg}$.

Quantity of sugar left = $\frac{1}{\cancel{4}_{1}} \times \cancel{20}^{5} \text{ kg}$.

Quantity of sugar left = $1 \times 5 \text{ kg}$.

Quantity of sugar left = $5 \text{ kg}$.

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Both methods show that the quantity of sugar left with the shopkeeper is $5 \text{ kg}$.

Comparing this result with the given options:

(A) $5 \text{ kg}$

(B) $15 \text{ kg}$

(C) $10 \text{ kg}$

(D) $4 \text{ kg}$

The result matches option (A).

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The final answer is $5 \text{ kg}$.

The correct option is (A).

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Given:

The number is $-\frac{8}{7}$.

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To Find:

The reciprocal of $-\frac{8}{7}$.

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Solution:

The reciprocal of a non-zero number is 1 divided by the number. If the number is a fraction $\frac{a}{b}$ (where $a \neq 0$ and $b \neq 0$), its reciprocal is $\frac{b}{a}$.

In this case, the given number is $-\frac{8}{7}$. We can write this as $\frac{-8}{7}$.

Using the definition of reciprocal for a fraction, we swap the numerator and the denominator.

The reciprocal of $\frac{-8}{7}$ is $\frac{7}{-8}$.

The fraction $\frac{7}{-8}$ is equivalent to $-\frac{7}{8}$.

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To verify, we multiply the original number by its reciprocal. The product should be 1.

$-\frac{8}{7} \times \left(-\frac{7}{8}\right) = \frac{-8}{7} \times \frac{-7}{8} = \frac{(-8) \times (-7)}{7 \times 8} = \frac{56}{56} = 1$.

Since the product is 1, our reciprocal is correct.

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The reciprocal of $-\frac{8}{7}$ is $-\frac{7}{8}$.

Comparing this result with the given options:

(A) $\frac{8}{7}$

(B) $-\frac{7}{8}$

(C) $\frac{7}{8}$

(D) $-\frac{8}{7}$

The result matches option (B).

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The final answer is $-\frac{7}{8}$.

The correct option is (B).

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Given:

The fraction is $\frac{4}{5}$.

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To Find:

Which of the given options is equivalent to $\frac{4}{5}$.

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Solution:

Two fractions are equivalent if one can be obtained from the other by multiplying or dividing both the numerator and the denominator by the same non-zero number.

We need to check each option to see which fraction is equivalent to $\frac{4}{5}$.

Option (A) $\frac{16}{25}$:

To get 16 from 4, we multiply by 4 ($4 \times 4 = 16$).

To get 25 from 5, we multiply by 5 ($5 \times 5 = 25$).

Since the multiplier for the numerator (4) is different from the multiplier for the denominator (5), $\frac{16}{25}$ is not equivalent to $\frac{4}{5}$.

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Option (B) $\frac{8}{10}$:

To get 8 from 4, we multiply by 2 ($4 \times 2 = 8$).

To get 10 from 5, we multiply by 2 ($5 \times 2 = 10$).

Since we multiplied both the numerator and the denominator by the same number (2), the fraction $\frac{8}{10}$ is equivalent to $\frac{4}{5}$.

We can write $\frac{4}{5} = \frac{4 \times 2}{5 \times 2} = \frac{8}{10}$.

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Option (C) $\frac{20}{30}$:

Let's simplify $\frac{20}{30}$ by dividing the numerator and denominator by their greatest common divisor, which is 10.

$\frac{20}{30} = \frac{20 \div 10}{30 \div 10} = \frac{2}{3}$.

Since $\frac{2}{3} \neq \frac{4}{5}$, $\frac{20}{30}$ is not equivalent to $\frac{4}{5}$.

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Option (D) $\frac{12}{20}$:

Let's simplify $\frac{12}{20}$ by dividing the numerator and denominator by their greatest common divisor, which is 4.

$\frac{12}{20} = \frac{12 \div 4}{20 \div 4} = \frac{3}{5}$.

Since $\frac{3}{5} \neq \frac{4}{5}$, $\frac{12}{20}$ is not equivalent to $\frac{4}{5}$.

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Based on the comparison of the options, the fraction equivalent to $\frac{4}{5}$ is $\frac{8}{10}$.

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The final answer is $\frac{8}{10}$.

The correct option is (B).

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Given:

The expression is $\frac{2}{7} \times (-\frac{1}{5}) + \frac{3}{7} \times (-\frac{1}{5})$.

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To Find:

The value of the given expression.

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Solution:

We need to evaluate the expression $\frac{2}{7} \times (-\frac{1}{5}) + \frac{3}{7} \times (-\frac{1}{5})$.

First, let's calculate the product of the first term:

$\frac{2}{7} \times \left(-\frac{1}{5}\right) = \frac{2 \times (-1)}{7 \times 5} = \frac{-2}{35}$.

Next, let's calculate the product of the second term:

$\frac{3}{7} \times \left(-\frac{1}{5}\right) = \frac{3 \times (-1)}{7 \times 5} = \frac{-3}{35}$.

Now, we add the results of the two products:

Value $= \frac{-2}{35} + \frac{-3}{35}$.

Since the denominators are the same, we can add the numerators directly:

Value $= \frac{-2 + (-3)}{35} = \frac{-2 - 3}{35} = \frac{-5}{35}$.

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Alternate Solution (Using Distributive Property):

The given expression is $\frac{2}{7} \times (-\frac{1}{5}) + \frac{3}{7} \times (-\frac{1}{5})$.

We observe that $(-\frac{1}{5})$ is a common factor in both terms.

Using the distributive property, which states that $a \times c + b \times c = (a+b) \times c$, we can rewrite the expression as:

Value $= \left(\frac{2}{7} + \frac{3}{7}\right) \times \left(-\frac{1}{5}\right)$.

First, perform the addition inside the parenthesis:

$\frac{2}{7} + \frac{3}{7} = \frac{2+3}{7} = \frac{5}{7}$.

Now, multiply the result by $(-\frac{1}{5})$:

Value $= \frac{5}{7} \times \left(-\frac{1}{5}\right) = \frac{5 \times (-1)}{7 \times 5} = \frac{-5}{35}$.

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Both methods yield the same result, which is $\frac{-5}{35}$ or $-\frac{5}{35}$.

Comparing this result with the given options:

(A) $-\frac{1}{5}$

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

(C) $-\frac{5}{35}$

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

The result $-\frac{5}{35}$ matches option (C).

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The final answer is $-\frac{5}{35}$.

The correct option is (C).

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Given:

The starting number is $-\frac{3}{4}$.

The desired result is $\frac{5}{6}$.

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To Find:

The number that should be added to $-\frac{3}{4}$ to get $\frac{5}{6}$.

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Solution:

Let the number that should be added be $x$.

According to the problem statement, we can write the equation:

To find the value of $x$, we need to isolate $x$ on one side of the equation. We can do this by adding $\frac{3}{4}$ to both sides of the equation.

Subtracting a negative number is the same as adding the positive number:

To add these fractions, we need to find a common denominator. The least common multiple (LCM) of 6 and 4 is 12.

We rewrite each fraction with the denominator 12:

$\frac{5}{6} = \frac{5 \times 2}{6 \times 2} = \frac{10}{12}$.

$\frac{3}{4} = \frac{3 \times 3}{4 \times 3} = \frac{9}{12}$.

Now substitute these equivalent fractions back into the equation for $x$:

Since the denominators are the same, we add the numerators:

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Thus, the number that should be added to $-\frac{3}{4}$ to get $\frac{5}{6}$ is $\frac{19}{12}$.

Comparing this result with the given options:

(A) $\frac{19}{12}$

(B) $-\frac{1}{12}$

(C) $\frac{1}{12}$

(D) $-\frac{19}{12}$

The result matches option (A).

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The final answer is $\frac{19}{12}$.

The correct option is (A).

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Given:

Cost of $1 \text{ meter}$ of cloth = $\textsf{₹} \frac{150}{2}$.

Quantity of cloth = $\frac{3}{5} \text{ meter}$.

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To Find:

The cost of $\frac{3}{5} \text{ meter}$ of cloth.

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Solution:

First, we can simplify the cost of $1 \text{ meter}$ of cloth:

Cost of $1 \text{ meter}$ = $\textsf{₹} \frac{150}{2} = \textsf{₹} 75$.

Now, to find the cost of $\frac{3}{5} \text{ meter}$ of cloth, we multiply the cost per meter by the quantity of cloth.

Cost of $\frac{3}{5} \text{ meter}$ = (Cost of $1 \text{ meter}$) $\times$ (Quantity in meters).

Cost = $\textsf{₹} 75 \times \frac{3}{5}$.

We perform the multiplication:

Cost = $\textsf{₹} \frac{75}{1} \times \frac{3}{5}$.

We can cancel out the common factor of 5 from the numerator (75) and the denominator (5).

$\frac{\cancel{75}^{15}}{1} \times \frac{3}{\cancel{5}_{1}} = 15 \times 3 = 45$.

So, the cost is $\textsf{₹} 45$.

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The cost of $\frac{3}{5} \text{ meter}$ of cloth is $\textsf{₹} 45$.

Comparing this result with the given options:

(A) $\textsf{₹} 75$

(B) $\textsf{₹} 45$

(C) $\textsf{₹} 90$

(D) $\textsf{₹} 120$

The result matches option (B).

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The final answer is $\textsf{₹} 45$.

The correct option is (B).

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Given:

The two rational numbers are $\frac{1}{3}$ and $\frac{1}{2}$.

The options are (A) $\frac{2}{5}$, (B) $\frac{3}{7}$, (C) $\frac{4}{9}$, (D) All of the above.

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To Find:

Which of the given rational numbers lies between $\frac{1}{3}$ and $\frac{1}{2}$.

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Solution:

To find a rational number between two fractions $\frac{a}{b}$ and $\frac{c}{d}$, we need to check which of the given options, say $\frac{x}{y}$, satisfies the condition $\frac{1}{3} < \frac{x}{y} < \frac{1}{2}$.

We can compare fractions by finding a common denominator or by comparing cross-products. To compare $\frac{a}{b}$ and $\frac{c}{d}$, we compare $a \times d$ and $b \times c$.

If $a \times d < b \times c$, then $\frac{a}{b} < \frac{c}{d}$.

If $a \times d > b \times c$, then $\frac{a}{b} > \frac{c}{d}$.

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Let's check each option:

Option (A) $\frac{2}{5}$:

We check if $\frac{1}{3} < \frac{2}{5}$. Comparing cross-products: $1 \times 5 = 5$ and $3 \times 2 = 6$. Since $5 < 6$, we have $\frac{1}{3} < \frac{2}{5}$.

We check if $\frac{2}{5} < \frac{1}{2}$. Comparing cross-products: $2 \times 2 = 4$ and $5 \times 1 = 5$. Since $4 < 5$, we have $\frac{2}{5} < \frac{1}{2}$.

Since $\frac{1}{3} < \frac{2}{5}$ and $\frac{2}{5} < \frac{1}{2}$, the inequality $\frac{1}{3} < \frac{2}{5} < \frac{1}{2}$ holds. So, $\frac{2}{5}$ lies between $\frac{1}{3}$ and $\frac{1}{2}$. Option (A) is a correct answer.

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Option (B) $\frac{3}{7}$:

We check if $\frac{1}{3} < \frac{3}{7}$. Comparing cross-products: $1 \times 7 = 7$ and $3 \times 3 = 9$. Since $7 < 9$, we have $\frac{1}{3} < \frac{3}{7}$.

We check if $\frac{3}{7} < \frac{1}{2}$. Comparing cross-products: $3 \times 2 = 6$ and $7 \times 1 = 7$. Since $6 < 7$, we have $\frac{3}{7} < \frac{1}{2}$.

Since $\frac{1}{3} < \frac{3}{7}$ and $\frac{3}{7} < \frac{1}{2}$, the inequality $\frac{1}{3} < \frac{3}{7} < \frac{1}{2}$ holds. So, $\frac{3}{7}$ lies between $\frac{1}{3}$ and $\frac{1}{2}$. Option (B) is a correct answer.

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Option (C) $\frac{4}{9}$:

We check if $\frac{1}{3} < \frac{4}{9}$. Comparing cross-products: $1 \times 9 = 9$ and $3 \times 4 = 12$. Since $9 < 12$, we have $\frac{1}{3} < \frac{4}{9}$.

We check if $\frac{4}{9} < \frac{1}{2}$. Comparing cross-products: $4 \times 2 = 8$ and $9 \times 1 = 9$. Since $8 < 9$, we have $\frac{4}{9} < \frac{1}{2}$.

Since $\frac{1}{3} < \frac{4}{9}$ and $\frac{4}{9} < \frac{1}{2}$, the inequality $\frac{1}{3} < \frac{4}{9} < \frac{1}{2}$ holds. So, $\frac{4}{9}$ lies between $\frac{1}{3}$ and $\frac{1}{2}$. Option (C) is a correct answer.

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Since options (A), (B), and (C) are all rational numbers that lie between $\frac{1}{3}$ and $\frac{1}{2}$, the correct answer is (D) All of the above.

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The final answer is All of the above.

The correct option is (D).

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Given:

Distance covered by the train = $360 \text{ km}$.

Time taken = $4 \frac{1}{2}$ hours.

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To Find:

The speed of the train in $\text{km/hr}$.

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Solution:

First, convert the time taken from a mixed number to an improper fraction.

Time taken $= 4 \frac{1}{2}$ hours $= 4 + \frac{1}{2}$ hours.

To add these, we find a common denominator:

$4 + \frac{1}{2} = \frac{4 \times 2}{2} + \frac{1}{2} = \frac{8}{2} + \frac{1}{2} = \frac{8+1}{2} = \frac{9}{2}$ hours.

So, the time taken is $\frac{9}{2}$ hours.

The formula for speed is:

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

Substitute the given values into the formula:

Speed $= \frac{360 \text{ km}}{\frac{9}{2} \text{ hours}}$.

Dividing by a fraction is the same as multiplying by its reciprocal. The reciprocal of $\frac{9}{2}$ is $\frac{2}{9}$.

Speed $= 360 \times \frac{2}{9} \text{ km/hr}$.

Now, we perform the multiplication. We can simplify by dividing 360 by 9:

Speed $= \frac{\cancel{360}^{40}}{1} \times \frac{2}{\cancel{9}_{1}} \text{ km/hr}$.

Speed $= 40 \times 2 \text{ km/hr}$.

Speed $= 80 \text{ km/hr}$.

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The speed of the train is $80 \text{ km/hr}$.

Comparing this result with the given options:

(A) $80 \text{ km/hr}$

(B) $90 \text{ km/hr}$

(C) $75 \text{ km/hr}$

(D) $60 \text{ km/hr}$

The calculated speed matches option (A).

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The final answer is $80 \text{ km/hr}$.

The correct option is (A).

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Given:

The sum of two rational numbers is $-\frac{2}{5}$.

One of the rational numbers is $\frac{3}{10}$.

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To Find:

The other rational number.

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Solution:

Let the other rational number be $x$.

According to the problem statement, the sum of the two numbers is $-\frac{2}{5}$. We can write this as an equation:

One number + Other number = Sum

To find the value of $x$, we need to isolate it on one side of the equation. We can do this by subtracting $\frac{3}{10}$ from both sides of the equation.

To subtract these fractions, we need to find a common denominator for 5 and 10. The least common multiple (LCM) of 5 and 10 is 10.

We need to rewrite the fraction $-\frac{2}{5}$ with a denominator of 10. To get 10 from 5, we multiply by 2. So, we multiply both the numerator and the denominator by 2:

$-\frac{2}{5} = -\frac{2 \times 2}{5 \times 2} = -\frac{4}{10}$.

Now, substitute this equivalent fraction back into the equation for $x$:

Since the denominators are now the same, we can subtract the numerators directly:

So, the other rational number is $-\frac{7}{10}$.

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We can verify the answer by adding the two numbers:

$\frac{3}{10} + \left(-\frac{7}{10}\right) = \frac{3 - 7}{10} = \frac{-4}{10}$.

Simplifying $\frac{-4}{10}$ by dividing the numerator and denominator by their greatest common divisor (2), we get:

$\frac{-4 \div 2}{10 \div 2} = \frac{-2}{5}$.

This matches the given sum, so our answer is correct.

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The other rational number is $-\frac{7}{10}$.

Comparing this result with the given options:

(A) $-\frac{7}{10}$

(B) $-\frac{1}{10}$

(C) $\frac{1}{10}$

(D) $\frac{7}{10}$

The calculated number matches option (A).

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The final answer is $-\frac{7}{10}$.

The correct option is (A).

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Given:

The mathematical statement $(\frac{2}{3} \times \frac{5}{7}) \times (-\frac{1}{4}) = \frac{2}{3} \times (\frac{5}{7} \times -\frac{1}{4})$.

---

To Find:

The property that justifies the given statement.

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Solution:

Let the three rational numbers be $a$, $b$, and $c$.

In the given statement, we have:

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

$b = \frac{5}{7}$

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

The statement is in the form $(a \times b) \times c = a \times (b \times c)$.

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Let's recall the properties of rational numbers under multiplication:

1. Closure Property: For any two rational numbers $a$ and $b$, the product $a \times b$ is also a rational number. This property states that the set of rational numbers is closed under multiplication.

2. Commutative Property: For any two rational numbers $a$ and $b$, $a \times b = b \times a$. This property states that the order of multiplication does not affect the result.

3. Associative Property: For any three rational numbers $a$, $b$, and $c$, $(a \times b) \times c = a \times (b \times c)$. This property states that the grouping of rational numbers in multiplication does not affect the result.

4. Distributive Property: This property relates multiplication and addition (or subtraction). For rational numbers $a$, $b$, and $c$, $a \times (b + c) = a \times b + a \times c$ or $(a + b) \times c = a \times c + b \times c$.

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Comparing the given statement $(\frac{2}{3} \times \frac{5}{7}) \times (-\frac{1}{4}) = \frac{2}{3} \times (\frac{5}{7} \times -\frac{1}{4})$ with the properties listed above, we see that it matches the form of the Associative Property of multiplication, which is $(a \times b) \times c = a \times (b \times c)$.

The statement shows that the product of the three rational numbers remains the same regardless of how the numbers are grouped (the first two numbers multiplied first, or the last two numbers multiplied first).

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Let's consider why the other options are incorrect for this specific statement:

(A) Commutative property of multiplication would involve changing the order of the numbers, like $\frac{2}{3} \times \frac{5}{7} = \frac{5}{7} \times \frac{2}{3}$. The given statement involves changing the grouping, not the order.

(C) Distributive property relates multiplication to addition or subtraction, which is not present in the given statement.

(D) Closure property of multiplication states that the result of the multiplication is a rational number, but it does not describe the equality shown in the statement.

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Therefore, the property that justifies the given statement is the Associative property of multiplication.

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The final answer is the Associative property of multiplication.

The correct option is (B).

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Given:

The expression is $(\frac{1}{2} - \frac{1}{3}) \div (\frac{1}{4} + \frac{1}{5})$.

---

To Find:

The value of the given expression.

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Solution:

We need to evaluate the expression in two parts: first the subtraction within the first parenthesis, and then the addition within the second parenthesis, followed by the division of the results.

Part 1: Calculate the value of the first parenthesis $(\frac{1}{2} - \frac{1}{3})$.

To subtract the fractions, we need to find a common denominator for 2 and 3. The least common multiple (LCM) of 2 and 3 is 6.

Rewrite the fractions with the denominator 6:

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

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

Now perform the subtraction:

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

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Part 2: Calculate the value of the second parenthesis $(\frac{1}{4} + \frac{1}{5})$.

To add the fractions, we need to find a common denominator for 4 and 5. The least common multiple (LCM) of 4 and 5 is 20.

Rewrite the fractions with the denominator 20:

$\frac{1}{4} = \frac{1 \times 5}{4 \times 5} = \frac{5}{20}$.

$\frac{1}{5} = \frac{1 \times 4}{5 \times 4} = \frac{4}{20}$.

Now perform the addition:

$\frac{1}{4} + \frac{1}{5} = \frac{5}{20} + \frac{4}{20} = \frac{5+4}{20} = \frac{9}{20}$.

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Part 3: Perform the division of the results from Part 1 and Part 2.

The original expression is equivalent to the result of Part 1 divided by the result of Part 2:

$(\frac{1}{2} - \frac{1}{3}) \div (\frac{1}{4} + \frac{1}{5}) = \frac{1}{6} \div \frac{9}{20}$.

To divide by a fraction, we multiply by its reciprocal. The reciprocal of $\frac{9}{20}$ is $\frac{20}{9}$.

Value $= \frac{1}{6} \times \frac{20}{9}$.

Multiply the numerators together and the denominators together:

Value $= \frac{1 \times 20}{6 \times 9} = \frac{20}{54}$.

The fraction $\frac{20}{54}$ can be simplified by dividing the numerator and the denominator by their greatest common divisor, which is 2.

Value $= \frac{20 \div 2}{54 \div 2} = \frac{10}{27}$.

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The value of the expression is $\frac{10}{27}$.

Comparing this result with the given options:

(A) $\frac{1}{6} \div \frac{9}{20} = \frac{1}{6} \times \frac{20}{9} = \frac{20}{54} = \frac{10}{27}$. This matches our calculation.

(B) $\frac{1}{6} \div \frac{5}{20} = \frac{1}{6} \times \frac{20}{5} = \frac{1}{6} \times 4 = \frac{4}{6} = \frac{2}{3}$. Incorrect.

(C) $\frac{5}{6} \div \frac{9}{20} = \frac{5}{6} \times \frac{20}{9} = \frac{100}{54} = \frac{50}{27}$. Incorrect (the first part of the calculation in the option is wrong).

(D) $\frac{1}{6} \div \frac{9}{10} = \frac{1}{6} \times \frac{10}{9} = \frac{10}{54} = \frac{5}{27}$. Incorrect (the divisor fraction is wrong).

Option (A) correctly represents the calculation steps and the final answer.

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The final answer is $\frac{10}{27}$.

The correct option is (A).

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Given:

Initial amount of water in the tank = $\frac{2}{3}$ of the full capacity.

Amount of water removed = $5 \text{ litres}$.

Amount of water left in the tank = $\frac{1}{2}$ of the full capacity.

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To Find:

The full capacity of the tank.

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Solution:

Let the full capacity of the tank be $C$ litres.

Initially, the amount of water in the tank is $\frac{2}{3}$ of the capacity, which is $\frac{2}{3}C$ litres.

When $5 \text{ litres}$ of water are removed, the amount of water left is the initial amount minus $5 \text{ litres}$.

Amount of water left = $\frac{2}{3}C - 5$ litres.

According to the problem, after removing $5 \text{ litres}$, the tank is $\frac{1}{2}$ full, which means the amount of water left is $\frac{1}{2}C$ litres.

So, we can set up the equation:

Now, we need to solve this equation for $C$.

Move the terms with $C$ to one side and the constant term to the other side:

To subtract the fractions $\frac{2}{3}$ and $\frac{1}{2}$, find a common denominator. The least common multiple (LCM) of 3 and 2 is 6.

Rewrite the fractions with the denominator 6:

$\frac{2}{3} = \frac{2 \times 2}{3 \times 2} = \frac{4}{6}$

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

Substitute these back into the equation:

Combine the terms on the left side:

To find $C$, multiply both sides of the equation by 6:

The full capacity of the tank is $30 \text{ litres}$.

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We can verify the answer:

Initial amount of water = $\frac{2}{3} \times 30 = \frac{2}{\cancel{3}_{1}} \times \cancel{30}^{10} = 2 \times 10 = 20 \text{ litres}$.

After removing $5 \text{ litres}$, the amount left is $20 - 5 = 15 \text{ litres}$.

The problem states the tank is then $\frac{1}{2}$ full. $\frac{1}{2}$ of the full capacity ($30 \text{ litres}$) is $\frac{1}{2} \times 30 = \frac{1}{\cancel{2}_{1}} \times \cancel{30}^{15} = 1 \times 15 = 15 \text{ litres}$.

Since the amount left ($15 \text{ litres}$) matches $\frac{1}{2}$ of the capacity ($15 \text{ litres}$), our calculation for the full capacity is correct.

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Comparing the result with the given options:

(A) $10 \text{ litres}$

(B) $20 \text{ litres}$

(C) $30 \text{ litres}$

(D) $40 \text{ litres}$

The calculated capacity matches option (C).

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The final answer is $30 \text{ litres}$.

The correct option is (C).

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Given:

Four pairs of rational numbers.

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To Find:

The pair of rational numbers which are additive inverses of each other.

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Solution:

Two rational numbers are called additive inverses of each other if their sum is $0$.

If $a$ is a rational number, its additive inverse is $-a$, because $a + (-a) = 0$.

We need to check the sum of the numbers in each given pair.

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Option (A) $\frac{2}{3}, \frac{3}{2}$:

Sum $= \frac{2}{3} + \frac{3}{2}$.

The LCM of the denominators 3 and 2 is 6.

Sum $= \frac{2 \times 2}{3 \times 2} + \frac{3 \times 3}{2 \times 3} = \frac{4}{6} + \frac{9}{6} = \frac{4+9}{6} = \frac{13}{6}$.

Since $\frac{13}{6} \neq 0$, this pair is not of additive inverses.

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Option (B) $-\frac{5}{8}, \frac{5}{8}$:

Sum $= -\frac{5}{8} + \frac{5}{8}$.

The denominators are the same.

Sum $= \frac{-5 + 5}{8} = \frac{0}{8} = 0$.

Since the sum is $0$, this pair are additive inverses of each other.

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Option (C) $\frac{1}{4}, -\frac{4}{1}$:

Sum $= \frac{1}{4} + \left(-\frac{4}{1}\right) = \frac{1}{4} - 4$.

To subtract, find a common denominator, which is 4.

Sum $= \frac{1}{4} - \frac{4 \times 4}{1 \times 4} = \frac{1}{4} - \frac{16}{4} = \frac{1-16}{4} = \frac{-15}{4}$.

Since $\frac{-15}{4} \neq 0$, this pair is not of additive inverses. Note that $-\frac{4}{1}$ is the reciprocal of $-\frac{1}{4}$, not the additive inverse of $\frac{1}{4}$.

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Option (D) $-\frac{6}{5}, -\frac{5}{6}$:

Sum $= -\frac{6}{5} + \left(-\frac{5}{6}\right) = -\frac{6}{5} - \frac{5}{6}$.

The LCM of the denominators 5 and 6 is 30.

Sum $= -\frac{6 \times 6}{5 \times 6} - \frac{5 \times 5}{6 \times 5} = -\frac{36}{30} - \frac{25}{30} = \frac{-36 - 25}{30} = \frac{-61}{30}$.

Since $\frac{-61}{30} \neq 0$, this pair is not of additive inverses.

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Only the pair in option (B) has a sum of $0$. Therefore, $-\frac{5}{8}$ and $\frac{5}{8}$ are additive inverses of each other.

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The final answer is $-\frac{5}{8}, \frac{5}{8}$.

The correct option is (B).

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Given:

The two rational numbers are $-\frac{1}{4}$ and $\frac{1}{2}$.

The options are (A) $-\frac{3}{8}$, (B) $\frac{3}{8}$, (C) $\frac{5}{8}$, (D) $-\frac{5}{8}$.

---

To Find:

Which of the given rational numbers lies between $-\frac{1}{4}$ and $\frac{1}{2}$.

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Solution:

To find a rational number between two given rational numbers, we can express all the numbers with a common denominator and then compare their numerators.

The denominators of the given numbers are 4 and 2. The denominators in the options are 8.

Let's find a common denominator for all these fractions. The least common multiple (LCM) of 4, 2, and 8 is 8.

Rewrite the given fractions with a denominator of 8:

$-\frac{1}{4} = -\frac{1 \times 2}{4 \times 2} = -\frac{2}{8}$.

$\frac{1}{2} = \frac{1 \times 4}{2 \times 4} = \frac{4}{8}$.

So we are looking for a rational number that lies between $-\frac{2}{8}$ and $\frac{4}{8}$. This means we are looking for a fraction $\frac{N}{8}$ such that $-\frac{2}{8} < \frac{N}{8} < \frac{4}{8}$. This inequality is equivalent to $-2 < N < 4$.

Now, let's look at the numerators of the given options when their denominator is 8:

(A) $-\frac{3}{8}$: The numerator is -3. Is $-2 < -3 < 4$? No, $-2 > -3$. So $-\frac{3}{8}$ is not between $-\frac{2}{8}$ and $\frac{4}{8}$.

(B) $\frac{3}{8}$: The numerator is 3. Is $-2 < 3 < 4$? Yes, $-2$ is less than $3$, and $3$ is less than $4$. So $\frac{3}{8}$ is between $-\frac{2}{8}$ and $\frac{4}{8}$.

(C) $\frac{5}{8}$: The numerator is 5. Is $-2 < 5 < 4$? No, $5$ is not less than $4$. So $\frac{5}{8}$ is not between $-\frac{2}{8}$ and $\frac{4}{8}$. In fact, $\frac{5}{8}$ is greater than $\frac{4}{8}$.

(D) $-\frac{5}{8}$: The numerator is -5. Is $-2 < -5 < 4$? No, $-2 > -5$. So $-\frac{5}{8}$ is not between $-\frac{2}{8}$ and $\frac{4}{8}$. In fact, $-\frac{5}{8}$ is less than $-\frac{2}{8}$.

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The only option whose numerator lies between -2 and 4 is option (B) with numerator 3.

Thus, $\frac{3}{8}$ lies between $-\frac{1}{4}$ and $\frac{1}{2}$.

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The final answer is $\frac{3}{8}$.

The correct option is (B).

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Given:

Amount of sugar required for one recipe = $\frac{1}{2}$ cup.

The recipe is to be made $1 \frac{1}{2}$ times the original amount.

---

To Find:

The total amount of sugar needed for $1 \frac{1}{2}$ times the recipe.

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Solution:

To find the total amount of sugar needed, we need to multiply the amount of sugar for one recipe by the scaling factor ($1 \frac{1}{2}$).

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

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

So, the scaling factor is $\frac{3}{2}$.

Now, multiply the amount of sugar per recipe by the scaling factor:

Total sugar needed = (Sugar per recipe) $\times$ (Scaling factor).

Total sugar needed = $\frac{1}{2} \text{ cup} \times \frac{3}{2}$.

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

Total sugar needed = $\frac{1 \times 3}{2 \times 2} \text{ cup}$.

Total sugar needed = $\frac{3}{4} \text{ cup}$.

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The amount of sugar needed is $\frac{3}{4}$ cup.

Comparing this result with the given options:

(A) $\frac{3}{4}$ cup

(B) $1$ cup

(C) $\frac{1}{4}$ cup

(D) $1 \frac{1}{4}$ cups

The calculated amount matches option (A).

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The final answer is $\frac{3}{4}$ cup.

The correct option is (A).

---

Given:

Four statements about properties of rational numbers.

---

To Identify:

The correct statement(s) among the given options.

---

Solution:

Let's examine each statement:

Statement (A): Division by zero is undefined for rational numbers.

For any rational number $a$, if we attempt to divide $a$ by $0$, the operation $\frac{a}{0}$ is undefined in the set of rational numbers (and indeed in the set of real numbers). Division is the inverse operation of multiplication. If $\frac{a}{0} = x$, it would mean $a = 0 \times x$. If $a \neq 0$, there is no value of $x$ that satisfies this equation because $0 \times x$ is always $0$. If $a = 0$, then $0 = 0 \times x$ is true for any value of $x$, meaning the result is not unique. Therefore, division by zero is undefined.

This statement is TRUE.

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Statement (B): Subtraction of rational numbers is commutative.

A binary operation is commutative if changing the order of the operands does not change the result. For subtraction, this would mean $a - b = b - a$ for any rational numbers $a$ and $b$. Let's take an example with rational numbers $\frac{1}{2}$ and $\frac{1}{4}$.

$\frac{1}{2} - \frac{1}{4} = \frac{2}{4} - \frac{1}{4} = \frac{1}{4}$.

$\frac{1}{4} - \frac{1}{2} = \frac{1}{4} - \frac{2}{4} = -\frac{1}{4}$.

Since $\frac{1}{4} \neq -\frac{1}{4}$, the commutative property does not hold for subtraction of rational numbers.

This statement is FALSE.

---

Statement (C): Multiplication of rational numbers is associative.

A binary operation is associative if the way in which operands are grouped does not affect the result. For multiplication, this means $(a \times b) \times c = a \times (b \times c)$ for any rational numbers $a$, $b$, and $c$. This property holds for multiplication of rational numbers.

Let's take an example: $a=\frac{2}{3}$, $b=\frac{5}{7}$, $c=-\frac{1}{4}$.

$(\frac{2}{3} \times \frac{5}{7}) \times (-\frac{1}{4}) = (\frac{2 \times 5}{3 \times 7}) \times (-\frac{1}{4}) = \frac{10}{21} \times (-\frac{1}{4}) = \frac{10 \times (-1)}{21 \times 4} = \frac{-10}{84}$.

$\frac{2}{3} \times (\frac{5}{7} \times (-\frac{1}{4})) = \frac{2}{3} \times (\frac{5 \times (-1)}{7 \times 4}) = \frac{2}{3} \times (-\frac{5}{28}) = \frac{2 \times (-5)}{3 \times 28} = \frac{-10}{84}$.

Since $\frac{-10}{84} = \frac{-10}{84}$, the associative property holds for multiplication.

This statement is TRUE.

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Statement (D): The reciprocal of $0$ is $0$.

The reciprocal of a non-zero number $a$ is $\frac{1}{a}$, because $a \times \frac{1}{a} = 1$. For the number $0$, its reciprocal would be $\frac{1}{0}$. As established in statement (A), division by zero is undefined. Therefore, the reciprocal of $0$ is undefined; it is not $0$.

This statement is FALSE.

---

Statements (A) and (C) are correct.

The options given are the statements themselves. Therefore, both option (A) and option (C) correspond to correct statements.

Given that this is likely intended as a single-choice question, there might be an issue with the question design as multiple options are correct statements. However, based on the evaluation:

Option (A) is a correct statement.

Option (B) is an incorrect statement.

Option (C) is a correct statement.

Option (D) is an incorrect statement.

---

Both Option (A) and Option (C) are correct based on the provided statements.

Assuming a single correct answer is expected, there might be a context where one property is specifically being emphasized or the question is flawed. However, based purely on the truth value of the statements, both (A) and (C) are correct.

If we must select only one option, and the question is from a source expecting a single answer, we would typically choose one of the correct statements listed as an option. Option (A) is listed first among the correct ones.

---

The correct statements are (A) and (C).

Assuming a single option answer is required, we select (A).

---

Given:

The rational number is $-\frac{7}{8}$.

---

To Find:

The decimal representation of $-\frac{7}{8}$.

---

Solution:

To find the decimal representation of a fraction, we divide the numerator by the denominator.

We need to find the decimal representation of $\frac{7}{8}$ and then apply the negative sign.

We perform the long division of 7 by 8.

Since 7 is less than 8, we place a decimal point after 7 and add zeros. We consider 70 first.

$70 \div 8$: $8 \times 8 = 64$. The quotient digit is 8, and the remainder is $70 - 64 = 6$.

Bring down the next zero to get 60.

$60 \div 8$: $8 \times 7 = 56$. The quotient digit is 7, and the remainder is $60 - 56 = 4$.

Bring down the next zero to get 40.

$40 \div 8$: $8 \times 5 = 40$. The quotient digit is 5, and the remainder is $40 - 40 = 0$.

The long division terminates when the remainder is 0.

The decimal representation of $\frac{7}{8}$ is $0.875$.

Since the original number is $-\frac{7}{8}$, its decimal representation is $-0.875$.

---

Comparing this result with the given options:

(A) $-0.875$

(B) $-0.78$

(C) $-0.8$

(D) $-0.75$

The calculated decimal representation matches option (A).

---

The final answer is $-0.875$.

The correct option is (A).

---

Given:

Daily rainfall amounts in inches for a week: $0, \frac{1}{4}, \frac{1}{2}, 0, \frac{3}{4}, \frac{1}{8}, \frac{1}{2}$.

---

To Find:

The total rainfall for the week.

---

Solution:

To find the total rainfall for the week, we need to sum the rainfall amounts for each day.

Total rainfall $= 0 + \frac{1}{4} + \frac{1}{2} + 0 + \frac{3}{4} + \frac{1}{8} + \frac{1}{2}$ inches.

We can rewrite this sum by grouping the fractions and ignoring the zeros:

Total rainfall $= \frac{1}{4} + \frac{1}{2} + \frac{3}{4} + \frac{1}{8} + \frac{1}{2}$ inches.

To add these fractions, we need to express them with a common denominator. The denominators are 4, 2, and 8.

The least common multiple (LCM) of 2, 4, and 8 is 8.

Now, we convert each fraction to an equivalent fraction with a denominator of 8:

$\frac{1}{4} = \frac{1 \times 2}{4 \times 2} = \frac{2}{8}$

$\frac{1}{2} = \frac{1 \times 4}{2 \times 4} = \frac{4}{8}$

$\frac{3}{4} = \frac{3 \times 2}{4 \times 2} = \frac{6}{8}$

$\frac{1}{8}$ is already in terms of eighths.

Substitute these equivalent fractions back into the sum:

Total rainfall $= \frac{2}{8} + \frac{4}{8} + \frac{6}{8} + \frac{1}{8} + \frac{4}{8}$ inches.

Now, add the numerators while keeping the common denominator:

Total rainfall $= \frac{2 + 4 + 6 + 1 + 4}{8}$ inches.

Calculate the sum of the numerators:

$2 + 4 = 6$

$6 + 6 = 12$

$12 + 1 = 13$

$13 + 4 = 17$

So, the sum of the numerators is 17.

Total rainfall $= \frac{17}{8}$ inches.

The answer options are given in mixed numbers. We convert the improper fraction $\frac{17}{8}$ to a mixed number.

Divide 17 by 8:

$17 \div 8 = 2$ with a remainder of 1.

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

$\frac{17}{8} = 2 \frac{1}{8}$.

The total rainfall for the week is $2 \frac{1}{8}$ inches.

---

Comparing this result with the given options:

(A) $2 \frac{1}{8}$ inches

(B) $2 \frac{3}{8}$ inches

(C) $1 \frac{7}{8}$ inches

(D) $2 \frac{5}{8}$ inches

Our calculated total rainfall, $2 \frac{1}{8}$ inches, matches option (A).

---

The final answer is $2 \frac{1}{8}$ inches.

The correct option is (A).

---

Given:

The first number is $\frac{1}{5}$.

The second number is $-\frac{1}{2}$.

---

To Find:

The result of multiplying the additive inverse of $\frac{1}{5}$ by the multiplicative inverse of $-\frac{1}{2}$.

---

Solution:

First, let's find the additive inverse of $\frac{1}{5}$.

The additive inverse of a rational number $a$ is $-a$. When added together, a number and its additive inverse sum to $0$.

Additive inverse of $\frac{1}{5} = -\frac{1}{5}$.

---

Next, let's find the multiplicative inverse of $-\frac{1}{2}$.

The multiplicative inverse (or reciprocal) of a non-zero rational number $a$ is $\frac{1}{a}$. When multiplied together, a number and its multiplicative inverse have a product of $1$.

The multiplicative inverse of $-\frac{1}{2}$ is the reciprocal of $-\frac{1}{2}$.

Multiplicative inverse of $-\frac{1}{2} = \frac{1}{-\frac{1}{2}} = 1 \div \left(-\frac{1}{2}\right)$.

To divide by a fraction, we multiply by its reciprocal. The reciprocal of $-\frac{1}{2}$ is $-\frac{2}{1}$, which is equal to $-2$.

So, the multiplicative inverse of $-\frac{1}{2}$ is $-2$.

We can check this: $-\frac{1}{2} \times (-2) = (-\frac{1}{2}) \times (-\frac{2}{1}) = \frac{(-1) \times (-2)}{2 \times 1} = \frac{2}{2} = 1$.

---

Now, we need to multiply the additive inverse of $\frac{1}{5}$ by the multiplicative inverse of $-\frac{1}{2}$.

Result $= (\text{Additive inverse of } \frac{1}{5}) \times (\text{Multiplicative inverse of } -\frac{1}{2})$

Result $= (-\frac{1}{5}) \times (-2)$.

To multiply $-\frac{1}{5}$ by $-2$, we can write $-2$ as $\frac{-2}{1}$.

Result $= \frac{-1}{5} \times \frac{-2}{1}$.

Multiply the numerators together and the denominators together:

Result $= \frac{(-1) \times (-2)}{5 \times 1} = \frac{2}{5}$.

---

The result of the multiplication is $\frac{2}{5}$.

Comparing this result with the given options:

(A) $(-\frac{1}{5}) \times (-2) = \frac{2}{5}$. This option shows the correct multiplication of the inverses and yields the correct result.

(B) $(\frac{1}{5}) \times (2) = \frac{2}{5}$. This option yields the correct result but does not multiply the required inverses.

(C) $(-\frac{1}{5}) \times (2) = -\frac{2}{5}$. Incorrect result.

(D) $(\frac{1}{5}) \times (-2) = -\frac{2}{5}$. Incorrect result.

Option (A) correctly represents the calculation requested in the question.

---

The final answer is $\frac{2}{5}$.

The correct option is (A).

---

Given:

The rational number is $\frac{7}{4}$.

---

To Represent:

The number $\frac{7}{4}$ on the number line.

---

To Find:

The two integers between which $\frac{7}{4}$ lies.

---

Solution:

To represent the fraction $\frac{7}{4}$ on the number line, we first convert it to a mixed number or a decimal to understand its value.

Converting $\frac{7}{4}$ to a mixed number: Divide 7 by 4.

$7 \div 4 = 1$ with a remainder of $7 - (4 \times 1) = 3$.

So, $\frac{7}{4} = 1 \frac{3}{4}$.

In decimal form, $\frac{7}{4} = 1.75$.

---

Now, we can represent $1 \frac{3}{4}$ on the number line. $1 \frac{3}{4}$ means 1 whole unit plus $\frac{3}{4}$ of the next unit.

On the number line, $1 \frac{3}{4}$ is greater than 1 but less than 2.

To locate it precisely, we consider the interval between the integers 1 and 2. We divide this interval into 4 equal parts, as indicated by the denominator of the fractional part $\frac{3}{4}$. The number $1 \frac{3}{4}$ is located at the third mark after 1.

Number line representation:

We have integers 0, 1, 2, 3, etc. Divide the segments between integers into four parts. The point representing $\frac{7}{4}$ or $1 \frac{3}{4}$ will be 3 segments away from 1 towards 2.

$\quad\qquad\qquad | \text{---} | \text{---} | \text{---} | \text{---} | \text{---} | \text{---} |$

$\quad\qquad\qquad 0 \quad \quad \frac{1}{4} \quad \frac{2}{4} \quad \frac{3}{4} \quad 1 \quad \frac{5}{4} \quad \frac{6}{4} \quad \textbf{\frac{7}{4}} \quad 2$

Or using the mixed number form:

$\quad\qquad\qquad | \text{---} | \text{---} | \text{---} | \text{---} | \text{---} | \text{---} |$

$\quad\qquad\qquad 0 \quad \quad \frac{1}{4} \quad \frac{2}{4} \quad \frac{3}{4} \quad 1 \quad 1\frac{1}{4} \quad 1\frac{2}{4} \quad \textbf{1\frac{3}{4}} \quad 2$

From the number line representation, it is clear that $\frac{7}{4}$ lies between the integers 1 and 2.

---

Comparing this with the given options:

(A) $0$ and $1$: Incorrect, since $\frac{7}{4} > 1$.

(B) $1$ and $2$: Correct, since $1 < \frac{7}{4} < 2$ (or $1 < 1.75 < 2$).

(C) $2$ and $3$: Incorrect, since $\frac{7}{4} < 2$.

(D) $-1$ and $0$: Incorrect, since $\frac{7}{4}$ is a positive number.

---

The number $\frac{7}{4}$ lies between the integers 1 and 2.

---

The final answer is that $\frac{7}{4}$ lies between 1 and 2.

The correct option is (B).

$p/q$, where $p$ and $q$ are integers and $q \neq 0$.

---

Yes, the number $0$ is a rational number.

---

To justify that $0$ is a rational number, we need to show that it can be written in the form $p/q$, where $p$ and $q$ are integers and $q$ is not zero.

The number $0$ can be written as $0/1$. In this representation, $p=0$ and $q=1$. Both $0$ and $1$ are integers, and the denominator $q=1$ is not equal to zero.

Similarly, $0$ can also be represented as $0/2$, $0/3$, $0/-5$, etc.

For instance, consider the representation $0/2$. Here, $p=0$ and $q=2$. Both are integers, and $q=2 \neq 0$.

Since $0$ satisfies the definition of a rational number (it can be expressed as a fraction of two integers where the denominator is non-zero), $0$ is a rational number.

To find rational numbers equivalent to a given rational number, we can multiply both the numerator and the denominator by the same non-zero integer.

---

The given rational number is $\frac{3}{7}$.

We can find four equivalent rational numbers by multiplying the numerator ($3$) and the denominator ($7$) by different non-zero integers, for example, $2, 3, 4,$ and $5$.

---

First equivalent rational number:

Multiply by $2$:

$\frac{3 \times 2}{7 \times 2} = \frac{6}{14}$

---

Second equivalent rational number:

Multiply by $3$:

$\frac{3 \times 3}{7 \times 3} = \frac{9}{21}$

---

Third equivalent rational number:

Multiply by $4$:

$\frac{3 \times 4}{7 \times 4} = \frac{12}{28}$

---

Fourth equivalent rational number:

Multiply by $5$:

$\frac{3 \times 5}{7 \times 5} = \frac{15}{35}$

---

Therefore, four rational numbers equivalent to $\frac{3}{7}$ are $\frac{6}{14}$, $\frac{9}{21}$, $\frac{12}{28}$, and $\frac{15}{35}$.

The standard form of a rational number $\frac{p}{q}$ is achieved when the denominator $q$ is positive and the greatest common divisor (GCD) of $p$ and $q$ is $1$.

---

(a) $\frac{-21}{35}$

The denominator $35$ is already positive.

We find the GCD of the absolute values of the numerator and the denominator, i.e., $\text{gcd}(|-21|, |35|) = \text{gcd}(21, 35)$.

Prime factorization of $21$ is $3 \times 7$.

Prime factorization of $35$ is $5 \times 7$.

The common factor is $7$. So, $\text{gcd}(21, 35) = 7$.

Now, divide both the numerator and the denominator by their GCD:

$\frac{-21 \div 7}{35 \div 7} = \frac{-3}{5}$

The denominator is $5$ (positive), and the $\text{gcd}(|-3|, |5|) = \text{gcd}(3, 5) = 1$.

Thus, the standard form of $\frac{-21}{35}$ is $\frac{-3}{5}$.

---

(b) $\frac{16}{-48}$

The denominator $-48$ is negative. First, we make the denominator positive by multiplying both the numerator and the denominator by $-1$.

$\frac{16 \times (-1)}{-48 \times (-1)} = \frac{-16}{48}$

Now, the denominator is $48$ (positive).

We find the GCD of the absolute values of the numerator and the denominator, i.e., $\text{gcd}(|-16|, |48|) = \text{gcd}(16, 48)$.

Prime factorization of $16$ is $2 \times 2 \times 2 \times 2 = 2^4$.

Prime factorization of $48$ is $2 \times 2 \times 2 \times 2 \times 3 = 2^4 \times 3$.

The common factors are $2^4$. So, $\text{gcd}(16, 48) = 16$.

Now, divide both the numerator and the denominator by their GCD:

$\frac{-16 \div 16}{48 \div 16} = \frac{-1}{3}$

The denominator is $3$ (positive), and the $\text{gcd}(|-1|, |3|) = \text{gcd}(1, 3) = 1$.

Thus, the standard form of $\frac{16}{-48}$ is $\frac{-1}{3}$.

To find the sum of $\frac{2}{9}$ and $\frac{-5}{6}$, we need to find a common denominator for the two fractions. The least common multiple (LCM) of the denominators $9$ and $6$ is the most suitable common denominator.

---

First, let's find the LCM of $9$ and $6$.

We can use the prime factorization method or the division method for LCM.

Using the division method:

$\begin{array}{c|cc} 2 & 9 \;, & 6 \\ \hline 3 & 9 \; , & 3 \\ \hline 3 & 3 \; , & 1 \\ \hline & 1 \; , & 1 \end{array}$

The LCM of $9$ and $6$ is $2 \times 3 \times 3 = 18$.

---

Now, we rewrite each fraction with the common denominator $18$.

For the fraction $\frac{2}{9}$, multiply the numerator and denominator by $18 \div 9 = 2$:

$\frac{2}{9} = \frac{2 \times 2}{9 \times 2} = \frac{4}{18}$

---

For the fraction $\frac{-5}{6}$, multiply the numerator and denominator by $18 \div 6 = 3$:

$\frac{-5}{6} = \frac{-5 \times 3}{6 \times 3} = \frac{-15}{18}$

---

Now, add the rewritten fractions:

$\frac{4}{18} + \frac{-15}{18}$

Add the numerators and keep the common denominator:

Sum $= \frac{4 + (-15)}{18} = \frac{4 - 15}{18} = \frac{-11}{18}$

---

The resulting fraction is $\frac{-11}{18}$. This fraction is in its standard form because the denominator ($18$) is positive and the greatest common divisor of $|-11|$ and $|18|$ is $1$.

---

Therefore, the sum of $\frac{2}{9}$ and $\frac{-5}{6}$ is $\frac{-11}{18}$.

We need to calculate $\frac{5}{12} - (\frac{-3}{8})$.

---

Subtracting a negative number is the same as adding its positive counterpart. So, $\frac{5}{12} - (\frac{-3}{8}) = \frac{5}{12} + \frac{3}{8}$.

---

To add these fractions, we need to find a common denominator. The least common multiple (LCM) of the denominators $12$ and $8$ is the most suitable common denominator.

---

Let's find the LCM of $12$ and $8$.

Using the prime factorization method:

$12 = 2 \times 2 \times 3 = 2^2 \times 3$

$8 = 2 \times 2 \times 2 = 2^3$

LCM$(12, 8) = 2^3 \times 3 = 8 \times 3 = 24$.

The common denominator is $24$.

---

Now, rewrite each fraction with the common denominator $24$.

For the fraction $\frac{5}{12}$, multiply the numerator and denominator by $24 \div 12 = 2$:

$\frac{5}{12} = \frac{5 \times 2}{12 \times 2} = \frac{10}{24}$

---

For the fraction $\frac{3}{8}$, multiply the numerator and denominator by $24 \div 8 = 3$:

$\frac{3}{8} = \frac{3 \times 3}{8 \times 3} = \frac{9}{24}$

---

Now, add the rewritten fractions:

$\frac{10}{24} + \frac{9}{24} = \frac{10 + 9}{24} = \frac{19}{24}$

---

The resulting fraction is $\frac{19}{24}$. The denominator is positive, and the greatest common divisor of $19$ and $24$ is $1$. Thus, it is in its standard form.

---

Therefore, $\frac{5}{12} - (\frac{-3}{8}) = \frac{19}{24}$.

To evaluate the product of two rational numbers, we multiply the numerators together and the denominators together.

---

The expression is $\frac{-4}{5} \times \frac{15}{16}$.

We can simplify the calculation by cancelling common factors between the numerator of one fraction and the denominator of the other fraction before multiplying.

---

We can cancel $4$ from the numerator $-4$ and the denominator $16$ (since $16 = 4 \times 4$).

We can cancel $5$ from the denominator $5$ and the numerator $15$ (since $15 = 5 \times 3$).

Performing the cancellations:

$\frac{\cancel{-4}^{-1}}{\cancel{5}^{1}} \times \frac{\cancel{15}^{3}}{\cancel{16}^{4}}$

---

Now, multiply the simplified numerators and denominators:

Numerator = $(-1) \times 3 = -3$

Denominator = $1 \times 4 = 4$

---

The product is $\frac{-3}{4}$.

---

The resulting fraction is $\frac{-3}{4}$. This is in standard form as the denominator is positive and the greatest common divisor of the numerator and the denominator is $1$.

---

Thus, $\frac{-4}{5} \times \frac{15}{16} = \frac{-3}{4}$.

To divide a rational number by another non-zero rational number, we multiply the first rational number by the reciprocal of the second rational number.

---

The expression is $\frac{-7}{12} \div \frac{21}{24}$.

The reciprocal of $\frac{21}{24}$ is $\frac{24}{21}$.

---

So, we rewrite the division as multiplication:

$\frac{-7}{12} \times \frac{24}{21}$

---

Now, we multiply the fractions. We can simplify by cancelling common factors between the numerators and denominators before multiplying.

We can cancel $7$ from $-7$ and $21$ ($21 = 7 \times 3$).

We can cancel $12$ from $12$ and $24$ ($24 = 12 \times 2$).

Performing the cancellations:

$\frac{\cancel{-7}^{-1}}{\cancel{12}^{1}} \times \frac{\cancel{24}^{2}}{\cancel{21}^{3}}$

---

Multiply the simplified numerators and denominators:

Numerator $= (-1) \times 2 = -2$

Denominator $= 1 \times 3 = 3$

---

The result is $\frac{-2}{3}$.

This fraction is in standard form because the denominator ($3$) is positive and the greatest common divisor of $|-2|$ and $|3|$ is $1$.

---

Therefore, $\frac{-7}{12} \div \frac{21}{24} = \frac{-2}{3}$.

Let the given rational number be $x = \frac{-11}{13}$.

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Additive Inverse:

The additive inverse of a number $x$ is a number $-x$ such that $x + (-x) = 0$.

For a rational number $\frac{p}{q}$, its additive inverse is $\frac{-p}{q}$ (or equivalently, $\frac{p}{-q}$ or $-\frac{p}{q}$).

---

The additive inverse of $\frac{-11}{13}$ is $-(\frac{-11}{13})$.

$-(\frac{-11}{13}) = \frac{-(-11)}{13} = \frac{11}{13}$.

Check: $\frac{-11}{13} + \frac{11}{13} = \frac{-11 + 11}{13} = \frac{0}{13} = 0$.

The additive inverse of $\frac{-11}{13}$ is $\frac{11}{13}$.

---

Multiplicative Inverse (Reciprocal):

The multiplicative inverse (or reciprocal) of a non-zero number $x$ is a number $1/x$ such that $x \times (1/x) = 1$.

For a non-zero rational number $\frac{p}{q}$, its multiplicative inverse is $\frac{q}{p}$.

---

The multiplicative inverse of $\frac{-11}{13}$ is the reciprocal, which is $\frac{13}{-11}$.

We can write $\frac{13}{-11}$ in standard form by making the denominator positive:

$\frac{13}{-11} = \frac{13 \times (-1)}{-11 \times (-1)} = \frac{-13}{11}$.

Check: $\frac{-11}{13} \times \frac{13}{-11} = \frac{(-11) \times 13}{13 \times (-11)} = \frac{-143}{-143} = 1$.

The multiplicative inverse (reciprocal) of $\frac{-11}{13}$ is $\frac{-13}{11}$.

To compare the rational numbers $\frac{-3}{5}$ and $\frac{-2}{7}$ using a common denominator, we first find the least common multiple (LCM) of the denominators, $5$ and $7$.

---

The denominators are $5$ and $7$. Since $5$ and $7$ are prime numbers, their LCM is their product.

LCM$(5, 7) = 5 \times 7 = 35$.

The common denominator is $35$.

---

Now, we rewrite each rational number with the denominator $35$.

For $\frac{-3}{5}$: We need to multiply the denominator $5$ by $7$ to get $35$. So, we multiply the numerator and the denominator by $7$.

$\frac{-3}{5} = \frac{-3 \times 7}{5 \times 7} = \frac{-21}{35}$

---

For $\frac{-2}{7}$: We need to multiply the denominator $7$ by $5$ to get $35$. So, we multiply the numerator and the denominator by $5$.

$\frac{-2}{7} = \frac{-2 \times 5}{7 \times 5} = \frac{-10}{35}$

---

Now we compare the numerators of the rewritten fractions $\frac{-21}{35}$ and $\frac{-10}{35}$. Since the denominators are positive and the same ($35$), we compare their numerators, which are $-21$ and $-10$.

On the number line, $-21$ is to the left of $-10$. Therefore, $-21 < -10$.

---

Since the numerator $-21$ is less than the numerator $-10$ when the denominator is positive, the fraction $\frac{-21}{35}$ is less than the fraction $\frac{-10}{35}$.

Thus, $\frac{-3}{5} < \frac{-2}{7}$.

To represent the rational number $\frac{5}{7}$ on a number line, we follow these steps:

---

Step 1: Draw a number line.

Draw a straight line and mark a point on it as $0$.

---

Step 2: Locate the integers.

Mark points to the right of $0$ at equal distances to represent positive integers $1, 2, 3, \dots$ and points to the left of $0$ at equal distances to represent negative integers $-1, -2, -3, \dots$.

---

Step 3: Determine the location interval.

The given rational number is $\frac{5}{7}$. Since the numerator $5$ is less than the denominator $7$, the value of $\frac{5}{7}$ is between $0$ and $1$. Also, it is positive, so it lies to the right of $0$. Thus, $\frac{5}{7}$ lies between $0$ and $1$ on the number line.

---

Step 4: Divide the relevant segment.

We need to divide the segment between $0$ and $1$ into $7$ equal parts (as the denominator is $7$). To do this, we mark $6$ points at equal intervals between $0$ and $1$.

---

Step 5: Label the parts.

Each of these $7$ equal parts represents $\frac{1}{7}$ of the unit length. The points starting from $0$ will represent $\frac{0}{7}, \frac{1}{7}, \frac{2}{7}, \frac{3}{7}, \frac{4}{7}, \frac{5}{7}, \frac{6}{7}, \frac{7}{7}$. Note that $\frac{0}{7} = 0$ and $\frac{7}{7} = 1$.

---

Step 6: Mark the desired number.

The point representing $\frac{5}{7}$ is the fifth mark from $0$ among the $7$ divisions.

---

Visually, the number line would look like this (conceptual representation):

$\longleftrightarrow \stackrel{-1}{\bullet} \cdots \stackrel{0}{\bullet} - - - - - - - \stackrel{1}{\bullet} \cdots \longrightarrow$

Divide the segment from 0 to 1 into 7 equal parts:

$\stackrel{0}{\bullet} \quad \frac{1}{7} \quad \frac{2}{7} \quad \frac{3}{7} \quad \frac{4}{7} \quad \stackrel{\frac{5}{7}}{\bullet} \quad \frac{6}{7} \quad \stackrel{1}{\bullet}$

The point marked as $\frac{5}{7}$ is the required representation.

To represent the rational number $\frac{-2}{3}$ on a number line, we follow these steps:

---

Step 1: Draw a number line.

Draw a straight line and mark a point on it as $0$.

---

Step 2: Locate the integers.

Mark points to the right of $0$ at equal distances to represent positive integers $1, 2, \dots$ and points to the left of $0$ at equal distances to represent negative integers $-1, -2, \dots$.

---

Step 3: Determine the location interval.

The given rational number is $\frac{-2}{3}$. Since it is negative, it lies to the left of $0$. The absolute value is $|\frac{-2}{3}| = \frac{2}{3}$. Since $0 < \frac{2}{3} < 1$, the number $\frac{-2}{3}$ lies between $-1$ and $0$ on the number line.

---

Step 4: Divide the relevant segment.

We need to divide the segment between $0$ and $-1$ into $3$ equal parts (as the denominator is $3$). To do this, we mark $2$ points at equal intervals between $0$ and $-1$.

---

Step 5: Label the parts.

Each of these $3$ equal parts represents $\frac{1}{3}$ of the unit length. The points starting from $0$ towards $-1$ will represent $\frac{0}{3}, \frac{-1}{3}, \frac{-2}{3}, \frac{-3}{3}$. Note that $\frac{0}{3} = 0$ and $\frac{-3}{3} = -1$.

---

Step 6: Mark the desired number.

The point representing $\frac{-2}{3}$ is the second mark from $0$ towards the left, among the $3$ divisions.

---

Visually, the number line would look like this (conceptual representation):

$\longleftrightarrow \stackrel{-1}{\bullet} - - \stackrel{0}{\bullet} \cdots \stackrel{1}{\bullet} \longrightarrow$

Divide the segment from $-1$ to $0$ into 3 equal parts:

$\stackrel{-1}{\bullet} \quad \frac{-2}{3} \quad \frac{-1}{3} \quad \stackrel{0}{\bullet}$

The point marked as $\frac{-2}{3}$ is the required representation.

To find a rational number between two given rational numbers, one method is to find their average (mean).

---

Let the two rational numbers be $a = \frac{1}{3}$ and $b = \frac{1}{2}$.

A rational number between $a$ and $b$ can be found by calculating $\frac{a+b}{2}$.

---

First, find the sum of the two rational numbers:

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

Find the LCM of the denominators $3$ and $2$. LCM$(3, 2) = 6$.

Rewrite the fractions with the denominator $6$:

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

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

Sum $= \frac{2}{6} + \frac{3}{6} = \frac{2+3}{6} = \frac{5}{6}$.

---

Now, divide the sum by $2$ (which is the same as multiplying by $\frac{1}{2}$):

Rational number $= \frac{5}{6} \div 2 = \frac{5}{6} \times \frac{1}{2}$

$= \frac{5 \times 1}{6 \times 2} = \frac{5}{12}$

---

So, $\frac{5}{12}$ is a rational number between $\frac{1}{3}$ and $\frac{1}{2}$.

---

Verification:

We need to check if $\frac{1}{3} < \frac{5}{12} < \frac{1}{2}$.

Compare $\frac{1}{3}$ and $\frac{5}{12}$: Use a common denominator, LCM$(3, 12) = 12$.

$\frac{1}{3} = \frac{1 \times 4}{3 \times 4} = \frac{4}{12}$.

Since $\frac{4}{12} < \frac{5}{12}$, we have $\frac{1}{3} < \frac{5}{12}$.

---

Compare $\frac{5}{12}$ and $\frac{1}{2}$: Use a common denominator, LCM$(12, 2) = 12$.

$\frac{1}{2} = \frac{1 \times 6}{2 \times 6} = \frac{6}{12}$.

Since $\frac{5}{12} < \frac{6}{12}$, we have $\frac{5}{12} < \frac{1}{2}$.

Thus, $\frac{1}{3} < \frac{5}{12} < \frac{1}{2}$. The rational number $\frac{5}{12}$ is indeed between $\frac{1}{3}$ and $\frac{1}{2}$.

We are asked to evaluate the expression: $\frac{2}{5} \times (\frac{-3}{7}) - \frac{1}{14} - \frac{3}{7} \times \frac{3}{5}$ using appropriate properties.

---

The expression can be written as:

$(\frac{2}{5} \times \frac{-3}{7}) - \frac{1}{14} - (\frac{3}{7} \times \frac{3}{5})$

---

We can observe that the first term $(\frac{2}{5} \times \frac{-3}{7})$ and the third term $(\frac{3}{7} \times \frac{3}{5})$ share common factors. Let's rearrange the terms using the commutative property of addition, grouping the terms with common factors together:

$\frac{2}{5} \times (\frac{-3}{7}) - \frac{3}{7} \times \frac{3}{5} - \frac{1}{14}$

---

We can rewrite the expression to highlight the common factor $\frac{-3}{7}$. Note that $-\frac{3}{7} \times \frac{3}{5} = \frac{-3}{7} \times \frac{3}{5}$ (by associating the negative sign with the numerator or the fraction itself).

So, the expression becomes:

$\frac{2}{5} \times (\frac{-3}{7}) + (\frac{-3}{7}) \times \frac{3}{5} - \frac{1}{14}$

---

Now, we can apply the distributive property of multiplication over addition, which states that $a \times b + a \times c = a \times (b+c)$. Here, we can take $a = \frac{-3}{7}$, $b = \frac{2}{5}$, and $c = \frac{3}{5}$.

Applying the distributive property to the first two terms:

$\frac{-3}{7} \times (\frac{2}{5} + \frac{3}{5}) - \frac{1}{14}$

---

Next, we evaluate the sum inside the parenthesis:

$\frac{2}{5} + \frac{3}{5} = \frac{2+3}{5} = \frac{5}{5} = 1$

---

Substitute the result back into the expression:

$\frac{-3}{7} \times 1 - \frac{1}{14}$

---

Using the multiplicative identity property, which states that $a \times 1 = a$:

$\frac{-3}{7} - \frac{1}{14}$

---

Now, we subtract the two rational numbers. We need a common denominator, which is the LCM of $7$ and $14$.

LCM$(7, 14) = 14$.

Rewrite $\frac{-3}{7}$ with a denominator of $14$ by multiplying the numerator and denominator by $2$:

$\frac{-3}{7} = \frac{-3 \times 2}{7 \times 2} = \frac{-6}{14}$

---

Now, perform the subtraction:

$\frac{-6}{14} - \frac{1}{14} = \frac{-6 - 1}{14} = \frac{-7}{14}$

---

Finally, simplify the resulting fraction by dividing the numerator and denominator by their greatest common divisor, which is $7$.

$\frac{-7 \div 7}{14 \div 7} = \frac{-1}{2}$

---

Thus, the value of the expression is $\frac{-1}{2}$.

We need to simplify the expression: $(\frac{-3}{5} \times \frac{10}{9}) + (\frac{5}{12} \div \frac{-15}{4})$.

---

First, we evaluate the multiplication part: $\frac{-3}{5} \times \frac{10}{9}$.

We can cancel common factors diagonally.

The numerator $10$ and the denominator $5$ have a common factor of $5$. $10 = 5 \times 2$, $5 = 5 \times 1$.

The numerator $-3$ and the denominator $9$ have a common factor of $3$. $-3 = -1 \times 3$, $9 = 3 \times 3$.

$\frac{\cancel{-3}^{-1}}{\cancel{5}^{1}} \times \frac{\cancel{10}^{2}}{\cancel{9}^{3}}$

Now, multiply the simplified numerators and denominators:

Result of multiplication $= \frac{-1 \times 2}{1 \times 3} = \frac{-2}{3}$.

---

Next, we evaluate the division part: $\frac{5}{12} \div \frac{-15}{4}$.

Dividing by a fraction is the same as multiplying by its reciprocal. The reciprocal of $\frac{-15}{4}$ is $\frac{4}{-15}$.

So, the division becomes multiplication: $\frac{5}{12} \times \frac{4}{-15}$.

We can cancel common factors diagonally.

The numerator $5$ and the denominator $-15$ have a common factor of $5$. $5 = 5 \times 1$, $-15 = -3 \times 5$.

The numerator $4$ and the denominator $12$ have a common factor of $4$. $4 = 4 \times 1$, $12 = 3 \times 4$.

$\frac{\cancel{5}^{1}}{\cancel{12}^{3}} \times \frac{\cancel{4}^{1}}{\cancel{-15}^{-3}}$

Now, multiply the simplified numerators and denominators:

Result of division $= \frac{1 \times 1}{3 \times (-3)} = \frac{1}{-9}$.

We can write $\frac{1}{-9}$ in standard form as $\frac{-1}{9}$.

---

Finally, we add the results of the multiplication and division:

$\frac{-2}{3} + \frac{-1}{9}$

To add these fractions, we need a common denominator. The LCM of $3$ and $9$ is $9$.

Rewrite $\frac{-2}{3}$ with a denominator of $9$:

$\frac{-2}{3} = \frac{-2 \times 3}{3 \times 3} = \frac{-6}{9}$.

Now add:

$\frac{-6}{9} + \frac{-1}{9} = \frac{-6 + (-1)}{9} = \frac{-6 - 1}{9} = \frac{-7}{9}$.

---

The simplified value of the expression is $\frac{-7}{9}$.

Given:

The sum of two rational numbers is $\frac{3}{4}$.

One of the rational numbers is $\frac{-1}{3}$.

---

To Find:

The other rational number.

---

Solution:

Let the other rational number be $x$.

According to the problem, the sum of the two rational numbers is $\frac{3}{4}$.

So, we can write the equation:

---

To find the value of $x$, we need to isolate $x$ on one side of the equation. We can do this by subtracting $\frac{-1}{3}$ from both sides of the equation (or adding $\frac{1}{3}$ to both sides, as subtracting a negative is equivalent to adding the positive).

From equation (i):

$x = \frac{3}{4} - (\frac{-1}{3})$

$x = \frac{3}{4} + \frac{1}{3}$

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Now, we need to find the sum of $\frac{3}{4}$ and $\frac{1}{3}$. To add these fractions, we need to find a common denominator, which is the least common multiple (LCM) of $4$ and $3$.

LCM$(4, 3) = 12$.

---

Rewrite each fraction with the common denominator $12$:

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

$\frac{1}{3} = \frac{1 \times 4}{3 \times 4} = \frac{4}{12}$

---

Now, substitute these equivalent fractions back into the equation for $x$ and add:

$x = \frac{9}{12} + \frac{4}{12}$

$x = \frac{9 + 4}{12}$

$x = \frac{13}{12}$

---

The other rational number is $\frac{13}{12}$.

Given:

We are given a starting number $\frac{7}{8}$ and a target result $\frac{-1}{16}$.

---

To Find:

The number that should be subtracted from $\frac{7}{8}$ to get $\frac{-1}{16}$.

---

Solution:

Let the unknown number that should be subtracted be $x$.

According to the problem statement, when we subtract $x$ from $\frac{7}{8}$, we get $\frac{-1}{16}$.

This can be written as an equation:

---

To find the value of $x$, we can rearrange the equation. We can add $x$ to both sides and add $\frac{1}{16}$ (or subtract $\frac{-1}{16}$) to both sides of the equation.

From equation (i):

$\frac{7}{8} = \frac{-1}{16} + x$

Now, move $\frac{-1}{16}$ to the left side by subtracting it from both sides:

$x = \frac{7}{8} - (\frac{-1}{16})$

Subtracting a negative number is equivalent to adding its positive counterpart:

$x = \frac{7}{8} + \frac{1}{16}$

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Now, we need to find the sum of $\frac{7}{8}$ and $\frac{1}{16}$. To add these fractions, we need a common denominator. We find the least common multiple (LCM) of the denominators $8$ and $16$.

LCM$(8, 16) = 16$.

---

Rewrite the first fraction $\frac{7}{8}$ with the common denominator $16$. We multiply the numerator and denominator by $16 \div 8 = 2$:

$\frac{7}{8} = \frac{7 \times 2}{8 \times 2} = \frac{14}{16}$

---

The second fraction $\frac{1}{16}$ already has the required denominator.

Now, substitute these equivalent fractions back into the equation for $x$ and perform the addition:

$x = \frac{14}{16} + \frac{1}{16}$

$x = \frac{14 + 1}{16}$

$x = \frac{15}{16}$

---

The number that should be subtracted from $\frac{7}{8}$ to get $\frac{-1}{16}$ is $\frac{15}{16}$.

Given:

One rational number is $\frac{-5}{4}$.

The desired product is $\frac{7}{2}$.

---

To Find:

The number that should be multiplied by $\frac{-5}{4}$ to get $\frac{7}{2}$.

---

Solution:

Let the unknown number be $x$.

According to the problem, the product of $\frac{-5}{4}$ and $x$ is $\frac{7}{2}$.

We can write this as an equation:

---

To find the value of $x$, we need to isolate $x$. We can do this by multiplying both sides of the equation by the multiplicative inverse (reciprocal) of $\frac{-5}{4}$.

The reciprocal of $\frac{-5}{4}$ is obtained by swapping the numerator and the denominator, keeping the sign with the numerator or the fraction itself. So, the reciprocal is $\frac{4}{-5}$. We can write this in standard form as $\frac{-4}{5}$.

---

Multiply both sides of equation (i) by the reciprocal $\frac{-4}{5}$:

$(\frac{-4}{5}) \times (\frac{-5}{4} \times x) = (\frac{-4}{5}) \times \frac{7}{2}$

Using the associative property of multiplication, the left side becomes:

$((\frac{-4}{5}) \times (\frac{-5}{4})) \times x$

The product of a number and its reciprocal is $1$:

$1 \times x = x$

So, the equation simplifies to:

$x = \frac{-4}{5} \times \frac{7}{2}$

---

Now, we perform the multiplication on the right side. Multiply the numerators together and the denominators together.

$x = \frac{-4 \times 7}{5 \times 2}$

$x = \frac{-28}{10}$

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The resulting fraction is $\frac{-28}{10}$. We need to simplify this fraction to its standard form by dividing the numerator and the denominator by their greatest common divisor (GCD). The GCD of $|-28|$ and $|10|$ is $2$.

Divide both the numerator and the denominator by $2$:

$\frac{-28 \div 2}{10 \div 2} = \frac{-14}{5}$

We can also show the cancellation directly:

$x = \frac{\cancel{-4}^{-2}}{5} \times \frac{7}{\cancel{2}^{1}} = \frac{-2 \times 7}{5 \times 1} = \frac{-14}{5}$

---

The number that should be multiplied by $\frac{-5}{4}$ to get $\frac{7}{2}$ is $\frac{-14}{5}$.

Given:

A number is divided by $\frac{-3}{4}$.

The result of the division is $\frac{9}{16}$.

---

To Find:

The number that is divided by $\frac{-3}{4}$.

---

Solution:

Let the unknown number be $x$.

According to the problem statement, when $x$ is divided by $\frac{-3}{4}$, the result is $\frac{9}{16}$.

We can write this as an equation:

The left side of the equation can be rewritten as $x \div \frac{-3}{4}$. Division by a fraction is the same as multiplication by its reciprocal. The reciprocal of $\frac{-3}{4}$ is $\frac{4}{-3}$.

So, equation (i) becomes:

To find the value of $x$, we need to isolate $x$. We can do this by multiplying both sides of equation (ii) by the multiplicative inverse (reciprocal) of $\frac{4}{-3}$.

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

---

Multiply both sides of equation (ii) by $\frac{-3}{4}$:

$(x \times \frac{4}{-3}) \times \frac{-3}{4} = \frac{9}{16} \times \frac{-3}{4}$

Using the associative property of multiplication, the left side becomes:

$x \times (\frac{4}{-3} \times \frac{-3}{4})$

The product of a number and its reciprocal is $1$:

$x \times 1 = x$

So, the equation simplifies to:

$x = \frac{9}{16} \times \frac{-3}{4}$

---

Now, we perform the multiplication on the right side. Multiply the numerators together and the denominators together.

$x = \frac{9 \times (-3)}{16 \times 4}$

$x = \frac{-27}{64}$

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The resulting fraction is $\frac{-27}{64}$. This fraction is in its standard form because the denominator $64$ is positive and the greatest common divisor (GCD) of $|-27|$ and $|64|$ is $1$. ($27 = 3^3$, $64 = 2^6$, they share no common prime factors).

---

Therefore, the number that should be divided by $\frac{-3}{4}$ to get $\frac{9}{16}$ is $\frac{-27}{64}$.

Given:

The product of two rational numbers is $\frac{-8}{15}$.

One of the rational numbers is $\frac{4}{5}$.

---

To Find:

The other rational number.

---

Solution:

Let the other rational number be $x$.

According to the problem statement, the product of the two numbers is $\frac{-8}{15}$.

We can write this relationship as an equation:

---

To find the value of $x$, we need to isolate $x$ on one side of the equation. We can achieve this by dividing both sides of equation (i) by the known number, $\frac{4}{5}$.

So, we have:

$x = \frac{-8}{15} \div \frac{4}{5}$

---

Dividing by a fraction is equivalent to multiplying by its reciprocal. The reciprocal of $\frac{4}{5}$ is $\frac{5}{4}$.

Therefore, the division can be rewritten as a multiplication:

$x = \frac{-8}{15} \times \frac{5}{4}$

---

Now, we perform the multiplication of the two rational numbers. We multiply the numerators and the denominators.

Before multiplying, we can simplify by cancelling common factors between the numerators and denominators.

We can cancel $4$ from the numerator $-8$ and the denominator $4$ (since $-8 = -2 \times 4$ and $4 = 1 \times 4$).

We can cancel $5$ from the numerator $5$ and the denominator $15$ (since $5 = 1 \times 5$ and $15 = 3 \times 5$).

Performing the cancellations:

$x = \frac{\cancel{-8}^{-2}}{\cancel{15}^{3}} \times \frac{\cancel{5}^{1}}{\cancel{4}^{1}}$

---

Now, multiply the simplified numerators and denominators:

$x = \frac{-2 \times 1}{3 \times 1}$

$x = \frac{-2}{3}$

---

The other rational number is $\frac{-2}{3}$. This fraction is in its standard form as the denominator is positive and the greatest common divisor of $|-2|$ and $|3|$ is $1$.

Given:

Total length of the wire $= 5\frac{1}{2}$ meters.

The wire is cut into $11$ equal pieces.

---

To Find:

The length of each piece of wire.

---

Solution:

First, convert the mixed number representing the total length into an improper fraction.

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

---

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

Length of each piece $= \text{Total Length} \div \text{Number of Pieces}$

---

Division by a number is the same as multiplication by its reciprocal. The number $11$ can be written as $\frac{11}{1}$. The reciprocal of $\frac{11}{1}$ is $\frac{1}{11}$.

So, from equation (i):

$\text{Length of each piece} = \frac{11}{2} \times \frac{1}{11}$

---

Now, perform the multiplication. We can cancel the common factor $11$ from the numerator and the denominator before multiplying.

$\text{Length of each piece} = \frac{\cancel{11}^{1}}{2} \times \frac{1}{\cancel{11}^{1}}$

$\text{Length of each piece} = \frac{1 \times 1}{2 \times 1}$

$\text{Length of each piece} = \frac{1}{2}$

---

The length of each piece is $\frac{1}{2}$ meters.

Given:

We are given the values of $x, y,$ and $z$ as $x = \frac{1}{2}$, $y = \frac{-1}{4}$, and $z = \frac{1}{8}$.

---

To Verify:

We need to verify if the distributive property of division over addition holds, i.e., if $x \div (y+z) = x \div y + x \div z$ for the given values.

---

Verification:

Let's evaluate the Left Hand Side (LHS) and the Right Hand Side (RHS) of the equation separately.

---

Left Hand Side (LHS): $x \div (y+z)$

First, calculate $y+z$:

$y+z = \frac{-1}{4} + \frac{1}{8}$

Find the LCM of the denominators $4$ and $8$. LCM$(4, 8) = 8$.

Rewrite the first fraction with denominator $8$:

$\frac{-1}{4} = \frac{-1 \times 2}{4 \times 2} = \frac{-2}{8}$

Now, add the fractions:

$y+z = \frac{-2}{8} + \frac{1}{8} = \frac{-2 + 1}{8} = \frac{-1}{8}$

---

Now, calculate $x \div (y+z)$:

$x \div (y+z) = \frac{1}{2} \div (\frac{-1}{8})$

Divide by a fraction by multiplying by its reciprocal. The reciprocal of $\frac{-1}{8}$ is $\frac{8}{-1} = -8$.

LHS $= \frac{1}{2} \times (-8)$

LHS $= \frac{1 \times (-8)}{2} = \frac{-8}{2} = -4$.

So, LHS $= -4$.

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Right Hand Side (RHS): $x \div y + x \div z$

First, calculate $x \div y$:

$x \div y = \frac{1}{2} \div \frac{-1}{4}$

Multiply by the reciprocal of $\frac{-1}{4}$, which is $\frac{4}{-1} = -4$.

$x \div y = \frac{1}{2} \times (-4) = \frac{1 \times (-4)}{2} = \frac{-4}{2} = -2$.

---

Next, calculate $x \div z$:

$x \div z = \frac{1}{2} \div \frac{1}{8}$

Multiply by the reciprocal of $\frac{1}{8}$, which is $\frac{8}{1} = 8$.

$x \div z = \frac{1}{2} \times 8 = \frac{1 \times 8}{2} = \frac{8}{2} = 4$.

---

Now, add the results of $x \div y$ and $x \div z$:

RHS $= (x \div y) + (x \div z)$

RHS $= -2 + 4 = 2$.

So, RHS $= 2$.

---

Comparing the LHS and RHS:

LHS $= -4$

RHS $= 2$

Since $-4 \neq 2$, the LHS is not equal to the RHS.

---

Conclusion:

For the given values of $x = \frac{1}{2}, y = \frac{-1}{4}, z = \frac{1}{8}$, the property $x \div (y+z) = x \div y + x \div z$ is not verified.

This shows that division is generally not distributive over addition for rational numbers.

We are asked to verify if multiplication of rational numbers is distributive over subtraction using the expression $\frac{1}{2} \times (\frac{3}{4} - \frac{1}{5})$.

---

The distributive property of multiplication over subtraction states that for any three rational numbers $a, b,$ and $c$:

$a \times (b - c) = a \times b - a \times c$

---

Given:

The expression $\frac{1}{2} \times (\frac{3}{4} - \frac{1}{5})$. This fits the form $a \times (b-c)$ with $a = \frac{1}{2}$, $b = \frac{3}{4}$, and $c = \frac{1}{5}$.

---

To Verify:

We need to check if $\frac{1}{2} \times (\frac{3}{4} - \frac{1}{5}) = \frac{1}{2} \times \frac{3}{4} - \frac{1}{2} \times \frac{1}{5}$.

---

Verification:

We will evaluate the Left Hand Side (LHS) and the Right Hand Side (RHS) separately.

---

Left Hand Side (LHS): $\frac{1}{2} \times (\frac{3}{4} - \frac{1}{5})$

First, evaluate the subtraction inside the parenthesis: $\frac{3}{4} - \frac{1}{5}$.

Find the LCM of the denominators $4$ and $5$. LCM$(4, 5) = 20$.

Rewrite the fractions with the common denominator $20$:

$\frac{3}{4} = \frac{3 \times 5}{4 \times 5} = \frac{15}{20}$

$\frac{1}{5} = \frac{1 \times 4}{5 \times 4} = \frac{4}{20}$

Now, perform the subtraction:

$\frac{3}{4} - \frac{1}{5} = \frac{15}{20} - \frac{4}{20} = \frac{15 - 4}{20} = \frac{11}{20}$.

Now, multiply this result by $\frac{1}{2}$:

LHS $= \frac{1}{2} \times \frac{11}{20} = \frac{1 \times 11}{2 \times 20} = \frac{11}{40}$.

So, LHS $= \frac{11}{40}$.

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Right Hand Side (RHS): $\frac{1}{2} \times \frac{3}{4} - \frac{1}{2} \times \frac{1}{5}$

First, evaluate the first multiplication: $\frac{1}{2} \times \frac{3}{4}$.

$\frac{1}{2} \times \frac{3}{4} = \frac{1 \times 3}{2 \times 4} = \frac{3}{8}$.

Next, evaluate the second multiplication: $\frac{1}{2} \times \frac{1}{5}$.

$\frac{1}{2} \times \frac{1}{5} = \frac{1 \times 1}{2 \times 5} = \frac{1}{10}$.

Now, perform the subtraction of the two results: $\frac{3}{8} - \frac{1}{10}$.

Find the LCM of the denominators $8$ and $10$. LCM$(8, 10) = 40$.

Rewrite the fractions with the common denominator $40$:

$\frac{3}{8} = \frac{3 \times 5}{8 \times 5} = \frac{15}{40}$

$\frac{1}{10} = \frac{1 \times 4}{10 \times 4} = \frac{4}{40}$

Now, perform the subtraction:

RHS $= \frac{15}{40} - \frac{4}{40} = \frac{15 - 4}{40} = \frac{11}{40}$.

So, RHS $= \frac{11}{40}$.

---

Comparing the LHS and RHS:

LHS $= \frac{11}{40}$

RHS $= \frac{11}{40}$

Since LHS $=$ RHS, the property is verified for the given values.

---

Conclusion:

Yes, multiplication of rational numbers is distributive over subtraction.

Here are the properties of rational numbers under addition:

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1. Closure Property

Explanation: The closure property for addition states that when you add any two rational numbers, the result is always another rational number.

Property: For any two rational numbers $a$ and $b$, the sum $a+b$ is also a rational number.

Let $a \in \mathbb{Q}$ and $b \in \mathbb{Q}$, then $a+b \in \mathbb{Q}$.

Example:

Let $a = \frac{1}{3}$ and $b = \frac{2}{5}$. Both are rational numbers.

Their sum is $a+b = \frac{1}{3} + \frac{2}{5}$.

To add, we find a common denominator: LCM$(3, 5) = 15$.

$a+b = \frac{1 \times 5}{3 \times 5} + \frac{2 \times 3}{5 \times 3} = \frac{5}{15} + \frac{6}{15} = \frac{5+6}{15} = \frac{11}{15}$.

Since $11$ and $15$ are integers and $15 \neq 0$, $\frac{11}{15}$ is a rational number.

Thus, the sum of two rational numbers $\frac{1}{3}$ and $\frac{2}{5}$ is a rational number $\frac{11}{15}$, which verifies the closure property for this example.

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2. Commutative Property

Explanation: The commutative property for addition states that the order in which two rational numbers are added does not affect the sum.

Property: For any two rational numbers $a$ and $b$, $a+b = b+a$.

Let $a \in \mathbb{Q}$ and $b \in \mathbb{Q}$, then $a+b = b+a$.

Example:

Let $a = \frac{-2}{7}$ and $b = \frac{3}{4}$. Both are rational numbers.

Calculate $a+b$: $\frac{-2}{7} + \frac{3}{4}$. LCM$(7, 4) = 28$.

$a+b = \frac{-2 \times 4}{7 \times 4} + \frac{3 \times 7}{4 \times 7} = \frac{-8}{28} + \frac{21}{28} = \frac{-8+21}{28} = \frac{13}{28}$.

Calculate $b+a$: $\frac{3}{4} + \frac{-2}{7}$. LCM$(4, 7) = 28$.

$b+a = \frac{3 \times 7}{4 \times 7} + \frac{-2 \times 4}{7 \times 4} = \frac{21}{28} + \frac{-8}{28} = \frac{21-8}{28} = \frac{13}{28}$.

Since $a+b = \frac{13}{28}$ and $b+a = \frac{13}{28}$, we have $a+b = b+a$, which verifies the commutative property for this example.

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3. Associative Property

Explanation: The associative property for addition states that when adding three or more rational numbers, the way in which they are grouped does not affect the sum.

Property: For any three rational numbers $a, b,$ and $c$, $(a+b)+c = a+(b+c)$.

Let $a, b, c \in \mathbb{Q}$, then $(a+b)+c = a+(b+c)$.

Example:

Let $a = \frac{1}{2}$, $b = \frac{-1}{3}$, and $c = \frac{1}{6}$. All are rational numbers.

Calculate LHS: $(a+b)+c = (\frac{1}{2} + \frac{-1}{3}) + \frac{1}{6}$.

First, $\frac{1}{2} + \frac{-1}{3}$. LCM$(2, 3) = 6$.

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

Now, add $c$: $(\frac{1}{6}) + \frac{1}{6} = \frac{1+1}{6} = \frac{2}{6}$.

Simplify $\frac{2}{6} = \frac{1}{3}$. So, LHS $= \frac{1}{3}$.

Calculate RHS: $a+(b+c) = \frac{1}{2} + (\frac{-1}{3} + \frac{1}{6})$.

First, $\frac{-1}{3} + \frac{1}{6}$. LCM$(3, 6) = 6$.

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

Now, add $a$: $\frac{1}{2} + (\frac{-1}{6})$. LCM$(2, 6) = 6$.

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

Simplify $\frac{2}{6} = \frac{1}{3}$. So, RHS $= \frac{1}{3}$.

Since LHS $= \frac{1}{3}$ and RHS $= \frac{1}{3}$, we have $(a+b)+c = a+(b+c)$, which verifies the associative property for this example.

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4. Existence of Identity (Additive Identity)

Explanation: The additive identity is a number that, when added to any rational number, leaves the rational number unchanged. This number is $0$.

Property: For any rational number $a$, there exists a rational number $0$ such that $a+0 = 0+a = a$. The rational number $0$ is called the additive identity for rational numbers.

Let $a \in \mathbb{Q}$, then $a+0 = 0+a = a$.

Example:

Let $a = \frac{5}{8}$. This is a rational number.

Add $0$ to $\frac{5}{8}$: $\frac{5}{8} + 0$. We can write $0$ as $\frac{0}{8}$.

$\frac{5}{8} + \frac{0}{8} = \frac{5+0}{8} = \frac{5}{8}$.

Add $\frac{5}{8}$ to $0$: $0 + \frac{5}{8}$. We can write $0$ as $\frac{0}{8}$.

$\frac{0}{8} + \frac{5}{8} = \frac{0+5}{8} = \frac{5}{8}$.

In both cases, the sum is $\frac{5}{8}$, which is the original number $a$. This verifies the existence of the additive identity $0$ for this example.

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5. Existence of Inverse (Additive Inverse)

Explanation: For every rational number, there exists another rational number (called its additive inverse or opposite) such that their sum is the additive identity, $0$.

Property: For every rational number $a$, there exists a rational number $-a$ such that $a + (-a) = (-a) + a = 0$. The rational number $-a$ is called the additive inverse of $a$.

Let $a = \frac{p}{q} \in \mathbb{Q}$. Then its additive inverse is $-a = \frac{-p}{q}$.

Example:

Let $a = \frac{-3}{11}$. This is a rational number.

The additive inverse of $\frac{-3}{11}$ is $-(\frac{-3}{11}) = \frac{-(-3)}{11} = \frac{3}{11}$.

Let's check their sum:

$a + (-a) = \frac{-3}{11} + \frac{3}{11} = \frac{-3+3}{11} = \frac{0}{11} = 0$.

Also, $(-a) + a = \frac{3}{11} + \frac{-3}{11} = \frac{3+(-3)}{11} = \frac{3-3}{11} = \frac{0}{11} = 0$.

The sum is $0$, which is the additive identity. This verifies the existence of the additive inverse for $\frac{-3}{11}$, which is $\frac{3}{11}$.

Here are the properties of rational numbers under multiplication:

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1. Closure Property

Explanation: The closure property for multiplication states that when you multiply any two rational numbers, the result is always another rational number.

Property: For any two rational numbers $a$ and $b$, the product $a \times b$ is also a rational number.

Let $a \in \mathbb{Q}$ and $b \in \mathbb{Q}$, then $a \times b \in \mathbb{Q}$.

Example:

Let $a = \frac{-2}{3}$ and $b = \frac{5}{7}$. Both are rational numbers.

Their product is $a \times b = \frac{-2}{3} \times \frac{5}{7}$.

$a \times b = \frac{-2 \times 5}{3 \times 7} = \frac{-10}{21}$.

Since $-10$ and $21$ are integers and $21 \neq 0$, $\frac{-10}{21}$ is a rational number.

Thus, the product of two rational numbers $\frac{-2}{3}$ and $\frac{5}{7}$ is a rational number $\frac{-10}{21}$, which verifies the closure property for this example.

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2. Commutative Property

Explanation: The commutative property for multiplication states that the order in which two rational numbers are multiplied does not affect the product.

Property: For any two rational numbers $a$ and $b$, $a \times b = b \times a$.

Let $a \in \mathbb{Q}$ and $b \in \mathbb{Q}$, then $a \times b = b \times a$.

Example:

Let $a = \frac{4}{5}$ and $b = \frac{-7}{9}$. Both are rational numbers.

Calculate $a \times b$: $\frac{4}{5} \times \frac{-7}{9} = \frac{4 \times (-7)}{5 \times 9} = \frac{-28}{45}$.

Calculate $b \times a$: $\frac{-7}{9} \times \frac{4}{5} = \frac{-7 \times 4}{9 \times 5} = \frac{-28}{45}$.

Since $a \times b = \frac{-28}{45}$ and $b \times a = \frac{-28}{45}$, we have $a \times b = b \times a$, which verifies the commutative property for this example.

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3. Associative Property

Explanation: The associative property for multiplication states that when multiplying three or more rational numbers, the way in which they are grouped does not affect the product.

Property: For any three rational numbers $a, b,$ and $c$, $(a \times b) \times c = a \times (b \times c)$.

Let $a, b, c \in \mathbb{Q}$, then $(a \times b) \times c = a \times (b \times c)$.

Example:

Let $a = \frac{1}{2}$, $b = \frac{3}{4}$, and $c = \frac{-2}{5}$. All are rational numbers.

Calculate LHS: $(a \times b) \times c = (\frac{1}{2} \times \frac{3}{4}) \times \frac{-2}{5}$.

First, $\frac{1}{2} \times \frac{3}{4} = \frac{1 \times 3}{2 \times 4} = \frac{3}{8}$.

Now, multiply by $c$: $\frac{3}{8} \times \frac{-2}{5} = \frac{3 \times (-2)}{8 \times 5} = \frac{-6}{40}$.

Simplify $\frac{-6}{40}$ by dividing numerator and denominator by $2$: $\frac{-3}{20}$. So, LHS $= \frac{-3}{20}$.

Calculate RHS: $a \times (b \times c) = \frac{1}{2} \times (\frac{3}{4} \times \frac{-2}{5})$.

First, $\frac{3}{4} \times \frac{-2}{5} = \frac{3 \times (-2)}{4 \times 5} = \frac{-6}{20}$.

Simplify $\frac{-6}{20}$ by dividing numerator and denominator by $2$: $\frac{-3}{10}$.

Now, multiply by $a$: $\frac{1}{2} \times (\frac{-3}{10}) = \frac{1 \times (-3)}{2 \times 10} = \frac{-3}{20}$.

So, RHS $= \frac{-3}{20}$.

Since LHS $= \frac{-3}{20}$ and RHS $= \frac{-3}{20}$, we have $(a \times b) \times c = a \times (b \times c)$, which verifies the associative property for this example.

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4. Existence of Identity (Multiplicative Identity)

Explanation: The multiplicative identity is a number that, when multiplied by any rational number, leaves the rational number unchanged. This number is $1$.

Property: For any rational number $a$, there exists a rational number $1$ such that $a \times 1 = 1 \times a = a$. The rational number $1$ is called the multiplicative identity for rational numbers.

Let $a \in \mathbb{Q}$, then $a \times 1 = 1 \times a = a$.

Example:

Let $a = \frac{-9}{10}$. This is a rational number.

Multiply $\frac{-9}{10}$ by $1$: $\frac{-9}{10} \times 1$. We can write $1$ as $\frac{1}{1}$.

$\frac{-9}{10} \times \frac{1}{1} = \frac{-9 \times 1}{10 \times 1} = \frac{-9}{10}$.

Multiply $1$ by $\frac{-9}{10}$: $1 \times \frac{-9}{10}$. We can write $1$ as $\frac{1}{1}$.

$\frac{1}{1} \times \frac{-9}{10} = \frac{1 \times (-9)}{1 \times 10} = \frac{-9}{10}$.

In both cases, the product is $\frac{-9}{10}$, which is the original number $a$. This verifies the existence of the multiplicative identity $1$ for this example.

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5. Existence of Inverse (Multiplicative Inverse/Reciprocal)

Explanation: For every non-zero rational number, there exists another rational number (called its multiplicative inverse or reciprocal) such that their product is the multiplicative identity, $1$.

Property: For every non-zero rational number $a$, there exists a rational number $1/a$ (its reciprocal) such that $a \times (1/a) = (1/a) \times a = 1$. The rational number $1/a$ is called the multiplicative inverse of $a$.

Let $a = \frac{p}{q} \in \mathbb{Q}$ where $p \neq 0$ and $q \neq 0$. Then its multiplicative inverse is $1/a = \frac{q}{p}$.

Example:

Let $a = \frac{-5}{12}$. This is a non-zero rational number.

The multiplicative inverse (reciprocal) of $\frac{-5}{12}$ is $\frac{12}{-5}$. We can write this in standard form as $\frac{-12}{5}$.

Let's check their product:

$a \times (1/a) = \frac{-5}{12} \times \frac{-12}{5} = \frac{(-5) \times (-12)}{12 \times 5} = \frac{60}{60} = 1$.

Also, $(1/a) \times a = \frac{-12}{5} \times \frac{-5}{12} = \frac{(-12) \times (-5)}{5 \times 12} = \frac{60}{60} = 1$.

The product is $1$, which is the multiplicative identity. This verifies the existence of the multiplicative inverse for $\frac{-5}{12}$, which is $\frac{-12}{5}$.

Statement of the Distributive Property of Multiplication over Addition:

For any three rational numbers $a, b,$ and $c$, the product of $a$ and the sum of $b$ and $c$ is equal to the sum of the product of $a$ and $b$, and the product of $a$ and $c$.

Mathematically, this is expressed as:

$a \times (b + c) = a \times b + a \times c$

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Verification with the given example:

Given:

The rational numbers are $a = \frac{-3}{4}$, $b = \frac{2}{5}$, and $c = \frac{-7}{10}$.

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To Verify:

We need to verify if $\frac{-3}{4} \times (\frac{2}{5} + \frac{-7}{10}) = (\frac{-3}{4} \times \frac{2}{5}) + (\frac{-3}{4} \times \frac{-7}{10})$.

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Calculation of Left Hand Side (LHS):

LHS $= a \times (b + c) = \frac{-3}{4} \times (\frac{2}{5} + \frac{-7}{10})$

First, calculate the sum inside the parenthesis: $(\frac{2}{5} + \frac{-7}{10})$.

Find the LCM of the denominators $5$ and $10$. LCM$(5, 10) = 10$.

Rewrite $\frac{2}{5}$ with denominator $10$:

$\frac{2}{5} = \frac{2 \times 2}{5 \times 2} = \frac{4}{10}$

Now add:

$\frac{2}{5} + \frac{-7}{10} = \frac{4}{10} + \frac{-7}{10} = \frac{4 + (-7)}{10} = \frac{4 - 7}{10} = \frac{-3}{10}$.

Now, multiply the result by $a = \frac{-3}{4}$:

LHS $= \frac{-3}{4} \times (\frac{-3}{10})$

LHS $= \frac{(-3) \times (-3)}{4 \times 10} = \frac{9}{40}$.

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Calculation of Right Hand Side (RHS):

RHS $= a \times b + a \times c = (\frac{-3}{4} \times \frac{2}{5}) + (\frac{-3}{4} \times \frac{-7}{10})$

First, calculate the product $a \times b$:

$\frac{-3}{4} \times \frac{2}{5} = \frac{-3 \times 2}{4 \times 5} = \frac{-6}{20}$

This can be simplified to $\frac{-3}{10}$ by dividing numerator and denominator by $2$.

Next, calculate the product $a \times c$:

$\frac{-3}{4} \times \frac{-7}{10} = \frac{(-3) \times (-7)}{4 \times 10} = \frac{21}{40}$.

Now, add the two products:

RHS $= \frac{-6}{20} + \frac{21}{40}$.

Find the LCM of the denominators $20$ and $40$. LCM$(20, 40) = 40$.

Rewrite $\frac{-6}{20}$ with denominator $40$:

$\frac{-6}{20} = \frac{-6 \times 2}{20 \times 2} = \frac{-12}{40}$.

Now add:

RHS $= \frac{-12}{40} + \frac{21}{40} = \frac{-12 + 21}{40} = \frac{9}{40}$.

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Comparison of LHS and RHS:

LHS $= \frac{9}{40}$

RHS $= \frac{9}{40}$

Since LHS $=$ RHS, the property $a \times (b + c) = a \times b + a \times c$ is verified for the given rational numbers.

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Conclusion:

The distributive property of multiplication over addition holds for rational numbers.

Given:

Two rational numbers: $\frac{-2}{5}$ and $\frac{1}{2}$.

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To Find:

Ten rational numbers between $\frac{-2}{5}$ and $\frac{1}{2}$.

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Solution:

To find rational numbers between two given rational numbers, we can rewrite them with a common denominator. This allows us to easily identify rational numbers with the same denominator but different numerators that lie between the two given numbers.

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Step 1: Find a common denominator.

Find the least common multiple (LCM) of the denominators $5$ and $2$.

LCM$(5, 2) = 10$.

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Step 2: Rewrite the given fractions with the common denominator.

Rewrite $\frac{-2}{5}$ with a denominator of $10$:

$\frac{-2}{5} = \frac{-2 \times 2}{5 \times 2} = \frac{-4}{10}$

Rewrite $\frac{1}{2}$ with a denominator of $10$:

$\frac{1}{2} = \frac{1 \times 5}{2 \times 5} = \frac{5}{10}$

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Step 3: Check the number of integers between the numerators.

We need to find rational numbers between $\frac{-4}{10}$ and $\frac{5}{10}$. The integers between the numerators $-4$ and $5$ are $-3, -2, -1, 0, 1, 2, 3, 4$.

The rational numbers with denominator $10$ between $\frac{-4}{10}$ and $\frac{5}{10}$ are $\frac{-3}{10}, \frac{-2}{10}, \frac{-1}{10}, \frac{0}{10}, \frac{1}{10}, \frac{2}{10}, \frac{3}{10}, \frac{4}{10}$. There are $8$ such numbers.

We need to find ten rational numbers, so we need to create a larger gap between the numerators. We can do this by multiplying the current common denominator by a suitable integer. Since we need more than $8$ numbers and the current difference in numerators is $5 - (-4) = 9$, multiplying by $2$ should be sufficient ($2 \times 9 = 18$, which is more than 10). Let's use a common denominator of $10 \times 2 = 20$.

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Step 4: Rewrite the fractions with a larger common denominator.

Use the common denominator $20$.

Rewrite $\frac{-2}{5}$ with a denominator of $20$:

$\frac{-2}{5} = \frac{-2 \times 4}{5 \times 4} = \frac{-8}{20}$

Rewrite $\frac{1}{2}$ with a denominator of $20$:

$\frac{1}{2} = \frac{1 \times 10}{2 \times 10} = \frac{10}{20}$

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Step 5: List ten rational numbers between the rewritten fractions.

We need to find ten rational numbers between $\frac{-8}{20}$ and $\frac{10}{20}$. The integers between $-8$ and $10$ are $-7, -6, -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9$. There are $10 - (-8) - 1 = 17$ integers between them.

We can pick any ten of the rational numbers with denominator $20$ and numerators between $-8$ and $10$.

For example, we can list the first ten starting from $\frac{-7}{20}$:

$\frac{-7}{20}, \frac{-6}{20}, \frac{-5}{20}, \frac{-4}{20}, \frac{-3}{20}, \frac{-2}{20}, \frac{-1}{20}, \frac{0}{20}, \frac{1}{20}, \frac{2}{20}$.

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These ten rational numbers are between $\frac{-8}{20}$ (which is equal to $\frac{-2}{5}$) and $\frac{10}{20}$ (which is equal to $\frac{1}{2}$).

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Note: There are infinitely many rational numbers between any two distinct rational numbers. The numbers listed above are just one possible set of ten rational numbers.

Given:

Two rational numbers: $\frac{-5}{6}$ and $\frac{5}{8}$.

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To Find:

Six rational numbers between $\frac{-5}{6}$ and $\frac{5}{8}$.

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Solution:

To find rational numbers between two given rational numbers, we can rewrite them with a common denominator. This allows us to easily identify rational numbers with the same denominator but different numerators that lie between the two given numbers.

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Step 1: Find a common denominator.

Find the least common multiple (LCM) of the denominators $6$ and $8$.

Prime factorization of $6$ is $2 \times 3$.

Prime factorization of $8$ is $2 \times 2 \times 2 = 2^3$.

LCM$(6, 8) = 2^3 \times 3 = 8 \times 3 = 24$.

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Step 2: Rewrite the given fractions with the common denominator.

Rewrite $\frac{-5}{6}$ with a denominator of $24$:

$\frac{-5}{6} = \frac{-5 \times 4}{6 \times 4} = \frac{-20}{24}$

Rewrite $\frac{5}{8}$ with a denominator of $24$:

$\frac{5}{8} = \frac{5 \times 3}{8 \times 3} = \frac{15}{24}$

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Step 3: Check the number of integers between the numerators.

We need to find rational numbers between $\frac{-20}{24}$ and $\frac{15}{24}$. The integers between the numerators $-20$ and $15$ are $-19, -18, \dots, 14$. The number of integers between them is $15 - (-20) - 1 = 15 + 20 - 1 = 35 - 1 = 34$.

Since we need to find only six rational numbers, and there are $34$ integers between the numerators, we can simply pick any six rational numbers with denominator $24$ and numerators between $-20$ and $15$.

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Step 4: List six rational numbers between the rewritten fractions.

We can pick any six rational numbers with denominator $24$ and numerators between $-20$ and $15$. For example, we can choose the numerators $-19, -10, 0, 5, 10, 14$.

The six rational numbers are:

$\frac{-19}{24}, \frac{-10}{24}, \frac{0}{24}, \frac{5}{24}, \frac{10}{24}, \frac{14}{24}$.

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These six rational numbers are between $\frac{-20}{24}$ (which is equal to $\frac{-5}{6}$) and $\frac{15}{24}$ (which is equal to $\frac{5}{8}$).

We can also simplify some of these fractions:

$\frac{-10}{24} = \frac{-5}{12}$

$\frac{0}{24} = 0$

$\frac{10}{24} = \frac{5}{12}$

$\frac{14}{24} = \frac{7}{12}$

So, another set of six rational numbers is: $\frac{-19}{24}, \frac{-5}{12}, 0, \frac{5}{24}, \frac{5}{12}, \frac{7}{12}$.

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Any six rational numbers with a denominator of $24$ (or any multiple of $24$) and a numerator between $-20$ and $15$ would be correct.

We need to simplify the expression: $(\frac{1}{2} - \frac{1}{4}) \times (\frac{5}{6} + \frac{2}{3}) \div (\frac{3}{8})$.

We will follow the order of operations (PEMDAS/BODMAS), which means evaluating expressions inside parentheses first, then multiplication and division from left to right.

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Step 1: Evaluate the first parenthesis: $(\frac{1}{2} - \frac{1}{4})$.

Find the LCM of denominators $2$ and $4$. LCM$(2, 4) = 4$.

Rewrite the fractions with denominator $4$:

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

Now subtract:

$\frac{1}{2} - \frac{1}{4} = \frac{2}{4} - \frac{1}{4} = \frac{2 - 1}{4} = \frac{1}{4}$.

The expression now becomes: $(\frac{1}{4}) \times (\frac{5}{6} + \frac{2}{3}) \div (\frac{3}{8})$.

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Step 2: Evaluate the second parenthesis: $(\frac{5}{6} + \frac{2}{3})$.

Find the LCM of denominators $6$ and $3$. LCM$(6, 3) = 6$.

Rewrite the second fraction with denominator $6$:

$\frac{2}{3} = \frac{2 \times 2}{3 \times 2} = \frac{4}{6}$

Now add:

$\frac{5}{6} + \frac{2}{3} = \frac{5}{6} + \frac{4}{6} = \frac{5 + 4}{6} = \frac{9}{6}$.

Simplify the fraction $\frac{9}{6}$ by dividing numerator and denominator by their GCD, $3$: $\frac{9 \div 3}{6 \div 3} = \frac{3}{2}$.

The expression now becomes: $(\frac{1}{4}) \times (\frac{3}{2}) \div (\frac{3}{8})$.

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Step 3: Perform multiplication and division from left to right.

First, perform the multiplication: $\frac{1}{4} \times \frac{3}{2}$.

Multiply numerators and denominators:

$\frac{1}{4} \times \frac{3}{2} = \frac{1 \times 3}{4 \times 2} = \frac{3}{8}$.

The expression now becomes: $(\frac{3}{8}) \div (\frac{3}{8})$.

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Step 4: Perform the division: $\frac{3}{8} \div \frac{3}{8}$.

Dividing a number by itself (when the number is non-zero) results in $1$. Alternatively, divide by multiplying by the reciprocal.

The reciprocal of $\frac{3}{8}$ is $\frac{8}{3}$.

$\frac{3}{8} \div \frac{3}{8} = \frac{3}{8} \times \frac{8}{3}$.

Cancel common factors:

$\frac{\cancel{3}^{1}}{\cancel{8}^{1}} \times \frac{\cancel{8}^{1}}{\cancel{3}^{1}} = \frac{1 \times 1}{1 \times 1} = \frac{1}{1} = 1$.

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The simplified value of the expression is $1$.

Given:

Length of the rectangular park ($l$) $= 15\frac{3}{5}$ meters.

Breadth of the rectangular park ($b$) $= 8\frac{1}{2}$ meters.

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To Find:

The perimeter and the area of the rectangular park. The answers should be expressed as mixed fractions.

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Solution:

First, convert the given mixed fractions for length and breadth into improper fractions.

Length ($l$) $= 15\frac{3}{5} = \frac{(15 \times 5) + 3}{5} = \frac{75 + 3}{5} = \frac{78}{5}$ meters.

Breadth ($b$) $= 8\frac{1}{2} = \frac{(8 \times 2) + 1}{2} = \frac{16 + 1}{2} = \frac{17}{2}$ meters.

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Calculate the Perimeter:

The formula for the perimeter of a rectangle is $P = 2 \times (\text{length} + \text{breadth})$.

$P = 2 \times (l + b)$

$P = 2 \times (\frac{78}{5} + \frac{17}{2})$

First, add the fractions inside the parenthesis: $\frac{78}{5} + \frac{17}{2}$. Find the LCM of denominators $5$ and $2$. LCM$(5, 2) = 10$.

Rewrite the fractions with denominator $10$:

$\frac{78}{5} = \frac{78 \times 2}{5 \times 2} = \frac{156}{10}$

$\frac{17}{2} = \frac{17 \times 5}{2 \times 5} = \frac{85}{10}$

Now add:

$\frac{156}{10} + \frac{85}{10} = \frac{156 + 85}{10} = \frac{241}{10}$.

Now multiply the sum by $2$:

$P = 2 \times \frac{241}{10} = \frac{2}{\cancel{1}} \times \frac{241}{\cancel{10}_{5}} = \frac{1 \times 241}{1 \times 5} = \frac{241}{5}$.

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The perimeter is $\frac{241}{5}$ meters. Convert this improper fraction to a mixed fraction.

Divide $241$ by $5$.

$241 \div 5 = 48$ with a remainder of $1$.

So, $\frac{241}{5} = 48 \frac{1}{5}$.

The perimeter of the park is $48\frac{1}{5}$ meters.

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Calculate the Area:

The formula for the area of a rectangle is $A = \text{length} \times \text{breadth}$.

$A = l \times b$

$A = \frac{78}{5} \times \frac{17}{2}$

Multiply the numerators and the denominators. We can simplify by cancelling the common factor $2$ between $78$ and $2$. $78 = 39 \times 2$, $2 = 1 \times 2$.

$A = \frac{\cancel{78}^{39}}{5} \times \frac{17}{\cancel{2}^{1}}$

$A = \frac{39 \times 17}{5 \times 1} = \frac{39 \times 17}{5}$.

Calculate the product $39 \times 17$:

$\begin{array}{cc}& & 3 & 9 \\ \times & & 1 & 7 \\ \hline & 2 & 7 & 3 \\ & 3 & 9 & \times \\ \hline 6 & 6 & 3 \\ \hline \end{array}$

So, $39 \times 17 = 663$.

$A = \frac{663}{5}$.

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The area is $\frac{663}{5}$ square meters. Convert this improper fraction to a mixed fraction.

Divide $663$ by $5$.

$663 \div 5 = 132$ with a remainder of $3$.

So, $\frac{663}{5} = 132 \frac{3}{5}$.

The area of the park is $132\frac{3}{5}$ square meters.

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Final Answers:

Perimeter $= 48\frac{1}{5}$ meters.

Area $= 132\frac{3}{5}$ square meters.

Given:

One rational number is $\frac{-7}{12}$.

The desired product is $\frac{5}{2}$.

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To Find:

1. The rational number by which $\frac{-7}{12}$ should be multiplied to get $\frac{5}{2}$.

2. The reciprocal of the result obtained in step 1.

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Solution:

Part 1: Finding the unknown rational number.

Let the unknown rational number be $x$.

According to the problem, the product of $\frac{-7}{12}$ and $x$ is $\frac{5}{2}$.

We can write this as an equation:

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To find the value of $x$, we need to isolate $x$. We can do this by multiplying both sides of equation (i) by the multiplicative inverse (reciprocal) of $\frac{-7}{12}$.

The reciprocal of $\frac{-7}{12}$ is obtained by swapping the numerator and the denominator. The reciprocal is $\frac{12}{-7}$, which can be written in standard form as $\frac{-12}{7}$.

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Multiply both sides of equation (i) by $\frac{-12}{7}$:

$(\frac{-12}{7}) \times (\frac{-7}{12} \times x) = (\frac{-12}{7}) \times \frac{5}{2}$

Using the associative property on the left side:

$((\frac{-12}{7}) \times (\frac{-7}{12})) \times x$

The product of a number and its reciprocal is $1$:

$1 \times x = x$

So, the equation simplifies to:

$x = \frac{-12}{7} \times \frac{5}{2}$

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Now, perform the multiplication on the right side. Multiply the numerators together and the denominators together. We can simplify by cancelling common factors, such as $2$ from $-12$ and $2$ ($ -12 = -6 \times 2$, $2 = 1 \times 2$).

$x = \frac{\cancel{-12}^{-6}}{7} \times \frac{5}{\cancel{2}^{1}}$

$x = \frac{-6 \times 5}{7 \times 1} = \frac{-30}{7}$.

The rational number is $\frac{-30}{7}$. This is in standard form.

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Part 2: Finding the reciprocal of the result.

The result obtained in Part 1 is $x = \frac{-30}{7}$.

The reciprocal of a rational number $\frac{p}{q}$ is $\frac{q}{p}$ (for $p \neq 0$).

The reciprocal of $\frac{-30}{7}$ is $\frac{7}{-30}$.

To express this in standard form, make the denominator positive by multiplying the numerator and denominator by $-1$:

$\frac{7}{-30} = \frac{7 \times (-1)}{-30 \times (-1)} = \frac{-7}{30}$.

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Final Answers:

The rational number that should be multiplied by $\frac{-7}{12}$ to get $\frac{5}{2}$ is $\frac{-30}{7}$.

The reciprocal of this result ($\frac{-30}{7}$) is $\frac{-7}{30}$.

Given:

Total money Rahul had $= \textsf{₹}1500$.

Fraction of money spent on books $= \frac{1}{5}$ of the total money.

Fraction of remaining money spent on clothes $= \frac{3}{10}$ of the remaining money after buying books.

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To Find:

The amount of money left with Rahul.

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Solution:

Step 1: Calculate the amount spent on books.

Amount spent on books $= \frac{1}{5}$ of $\textsf{₹}1500$

Amount spent on books $= \frac{1}{5} \times 1500$

Amount spent on books $= \frac{1500}{5} = 300$.

Rahul spent $\textsf{₹}300$ on books.

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Step 2: Calculate the money remaining after buying books.

Money remaining after books $= \text{Total Money} - \text{Amount spent on books}$

Money remaining after books $= \textsf{₹}1500 - \textsf{₹}300 = \textsf{₹}1200$.

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Step 3: Calculate the amount spent on clothes.

Amount spent on clothes $= \frac{3}{10}$ of the remaining money ($\textsf{₹}1200$).

Amount spent on clothes $= \frac{3}{10} \times 1200$

Amount spent on clothes $= \frac{3 \times 1200}{10} = \frac{3600}{10} = 360$.

Rahul spent $\textsf{₹}360$ on clothes.

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Step 4: Calculate the money left with Rahul.

Money left $= \text{Money remaining after books} - \text{Amount spent on clothes}$

Money left $= \textsf{₹}1200 - \textsf{₹}360 = \textsf{₹}840$.

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The amount of money left with Rahul is $\textsf{₹}840$.

Given:

The cost of $2\frac{1}{2}$ kg of tomatoes is $\textsf{₹}62\frac{1}{2}$.

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To Find:

1. The cost of $1$ kg of tomatoes.

2. The quantity of tomatoes that can be purchased for $\textsf{₹}100$.

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Solution:

First, convert the given mixed numbers into improper fractions.

Quantity of tomatoes $= 2\frac{1}{2}$ kg $= \frac{(2 \times 2) + 1}{2} = \frac{4 + 1}{2} = \frac{5}{2}$ kg.

Cost of tomatoes $= \textsf{₹}62\frac{1}{2} = \textsf{₹}\frac{(62 \times 2) + 1}{2} = \textsf{₹}\frac{124 + 1}{2} = \textsf{₹}\frac{125}{2}$.

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Part 1: Find the cost of 1 kg of tomatoes.

The cost of $1$ kg of tomatoes is found by dividing the total cost by the total quantity.

Cost of 1 kg $= \text{Total Cost} \div \text{Total Quantity}$

Cost of 1 kg $= \textsf{₹}\frac{125}{2} \div \frac{5}{2}$ kg.

To divide by a fraction, multiply by its reciprocal. The reciprocal of $\frac{5}{2}$ is $\frac{2}{5}$.

Cost of 1 kg $= \frac{125}{2} \times \frac{2}{5}$.

Cancel common factors. $2$ from numerator and denominator, $5$ from $125$ and $5$ ($125 = 25 \times 5$, $5 = 1 \times 5$).

Cost of 1 kg $= \frac{\cancel{125}^{25}}{\cancel{2}^{1}} \times \frac{\cancel{2}^{1}}{\cancel{5}^{1}}$

Cost of 1 kg $= \frac{25 \times 1}{1 \times 1} = 25$.

The cost of $1$ kg of tomatoes is $\textsf{₹}25$.

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Part 2: Find the quantity of tomatoes for $\textsf{₹}100$.

We know the cost of $1$ kg is $\textsf{₹}25$. To find the quantity for $\textsf{₹}100$, we divide the amount of money by the cost per kg.

Quantity for $\textsf{₹}100 = \text{Amount of Money} \div \text{Cost per kg}$

Quantity for $\textsf{₹}100 = \textsf{₹}100 \div \textsf{₹}25/\text{kg}$

Quantity for $\textsf{₹}100 = \frac{100}{25}$ kg $= 4$ kg.

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Final Answers:

The cost of $1$ kg of tomatoes is $\textsf{₹}25$.

The quantity of tomatoes that can be purchased for $\textsf{₹}100$ is $4$ kg.

We need to simplify the expression: $(\frac{-4}{7} \times \frac{14}{15}) - (\frac{5}{6} \times \frac{-9}{10}) + (\frac{3}{5} \div \frac{9}{20})$.

We will evaluate each part of the expression separately before combining them according to the operations.

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Part 1: Evaluate the first multiplication: $(\frac{-4}{7} \times \frac{14}{15})$.

Multiply numerators and denominators, cancelling common factors. $7$ from $7$ and $14$ ($14 = 2 \times 7$), no common factors for $4$ and $15$.

$\frac{-4}{\cancel{7}^{1}} \times \frac{\cancel{14}^{2}}{15} = \frac{-4 \times 2}{1 \times 15} = \frac{-8}{15}$.

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Part 2: Evaluate the second multiplication: $(\frac{5}{6} \times \frac{-9}{10})$.

Multiply numerators and denominators, cancelling common factors. $5$ from $5$ and $10$ ($10 = 2 \times 5$), $3$ from $6$ and $-9$ ($6 = 2 \times 3$, $-9 = -3 \times 3$).

$\frac{\cancel{5}^{1}}{\cancel{6}^{2}} \times \frac{\cancel{-9}^{-3}}{\cancel{10}^{2}} = \frac{1 \times (-3)}{2 \times 2} = \frac{-3}{4}$.

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Part 3: Evaluate the division: $(\frac{3}{5} \div \frac{9}{20})$.

Divide by multiplying by the reciprocal of the second fraction. The reciprocal of $\frac{9}{20}$ is $\frac{20}{9}$.

$\frac{3}{5} \div \frac{9}{20} = \frac{3}{5} \times \frac{20}{9}$.

Multiply numerators and denominators, cancelling common factors. $5$ from $5$ and $20$ ($20 = 4 \times 5$), $3$ from $3$ and $9$ ($9 = 3 \times 3$).

$\frac{\cancel{3}^{1}}{\cancel{5}^{1}} \times \frac{\cancel{20}^{4}}{\cancel{9}^{3}} = \frac{1 \times 4}{1 \times 3} = \frac{4}{3}$.

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Step 4: Combine the results using the given operations.

The original expression is $(\text{Part 1}) - (\text{Part 2}) + (\text{Part 3})$.

Substitute the results from the parts:

$\frac{-8}{15} - (\frac{-3}{4}) + \frac{4}{3}$

Subtracting a negative number is the same as adding the positive number:

$\frac{-8}{15} + \frac{3}{4} + \frac{4}{3}$

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Step 5: Add the three rational numbers.

Find a common denominator for $15, 4,$ and $3$. Find the LCM$(15, 4, 3)$.

Prime factorization of $15$ is $3 \times 5$.

Prime factorization of $4$ is $2^2$.

Prime factorization of $3$ is $3$.

LCM$(15, 4, 3) = 2^2 \times 3 \times 5 = 4 \times 3 \times 5 = 60$.

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Rewrite each fraction with the denominator $60$:

$\frac{-8}{15} = \frac{-8 \times 4}{15 \times 4} = \frac{-32}{60}$ (since $60 \div 15 = 4$)

$\frac{3}{4} = \frac{3 \times 15}{4 \times 15} = \frac{45}{60}$ (since $60 \div 4 = 15$)

$\frac{4}{3} = \frac{4 \times 20}{3 \times 20} = \frac{80}{60}$ (since $60 \div 3 = 20$)

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Now, add the fractions with the common denominator:

$\frac{-32}{60} + \frac{45}{60} + \frac{80}{60} = \frac{-32 + 45 + 80}{60}$

Combine the numerators: $-32 + 45 = 13$.

$13 + 80 = 93$.

So, the sum is $\frac{93}{60}$.

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Step 6: Simplify the final result.

The fraction is $\frac{93}{60}$. Both the numerator and the denominator are divisible by $3$.

$\frac{93 \div 3}{60 \div 3} = \frac{31}{20}$.

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The simplified value of the expression is $\frac{31}{20}$. This is in standard form.

Given:

Fraction of students who are boys $= \frac{2}{3}$ of the total number of students.

Number of girls $= 200$.

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To Find:

1. The total number of students in the school.

2. The number of boys in the school.

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Solution:

Step 1: Find the fraction of students who are girls.

The total fraction of students in the school is represented by $1$.

The total number of students consists of boys and girls.

Fraction of girls $= 1 - \text{Fraction of boys}$

Fraction of girls $= 1 - \frac{2}{3}$.

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

So, $\frac{1}{3}$ of the total number of students are girls.

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Step 2: Find the total number of students.

We know that $\frac{1}{3}$ of the total number of students is equal to the number of girls, which is $200$.

Let the total number of students be $T$.

To find $T$, multiply both sides of the equation by $3$ (the reciprocal of $\frac{1}{3}$).

$3 \times (\frac{1}{3} \times T) = 3 \times 200$

$(3 \times \frac{1}{3}) \times T = 600$

$1 \times T = 600$

$T = 600$.

The total number of students in the school is $600$.

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Step 3: Find the number of boys.

The number of boys can be found in two ways:

Method 1: Using the fraction of boys.

Number of boys $= \frac{2}{3}$ of the total number of students.

Number of boys $= \frac{2}{3} \times 600$

Number of boys $= \frac{2 \times 600}{3} = \frac{1200}{3} = 400$.

Method 2: Subtracting the number of girls from the total number of students.

Number of boys $= \text{Total number of students} - \text{Number of girls}$.

Number of boys $= 600 - 200 = 400$.

Both methods give the same result.

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Final Answers:

The total number of students in the school is $600$.

The number of boys in the school is $400$.