Rational Numbers — Exercise 1(E)

A number of the form $\dfrac{p}{q}$ is said to be a rational number if

1. $p \text{ and } q$ are whole numbers and $q ≠ 0$.

2. $p$ is a whole number and $q$ is a natural number.

3. $p \text{ and } q$ are integers and $q ≠ 0$.

4. $p \text{ and } q$ are integers and $p ≠ 0, q ≠ 0$.

Answer

The numbers which can be expressed in the form $\dfrac{p}{q}$, where $p \text{ and } q$ are integers and $q ≠ 0$, are called rational numbers.

The numerator $p$ may also be 0, as $\dfrac{0}{q} = 0$ is a rational number, but the denominator $q$ can never be 0 since division by 0 is not defined.

Hence, Option 3 is the correct option.

The additive inverse of $\dfrac{-5}{12}$ is

1. $\dfrac{5}{12}$

2. $\dfrac{12}{5}$

3. $\dfrac{-12}{5}$

4. $\dfrac{2}{15}$

Answer

To find the additive inverse of any rational number, change its sign.

Additive inverse of $\dfrac{-5}{12} = \dfrac{5}{12}$

Hence, Option 1 is the correct option.

The multiplicative inverse of $\dfrac{-5}{12}$ is

1. $\dfrac{5}{12}$

2. $\dfrac{12}{5}$

3. $\dfrac{-12}{5}$

4. $\dfrac{2}{15}$

Answer

To find the multiplicative inverse of a non-zero rational number, interchange its numerator and denominator.

Multiplicative inverse of $\dfrac{-5}{12} = \dfrac{12}{-5} = \dfrac{-12}{5}$

Hence, Option 3 is the correct option.

The product of additive inverse and multiplicative inverse of $\dfrac{-1}{4}$ is

1. 1

2. -1

3. 0

4. $\dfrac{-1}{16}$

Answer

To find the additive inverse, change the sign of the number, and to find the multiplicative inverse, interchange its numerator and denominator.

Additive inverse of $\dfrac{-1}{4} = \dfrac{1}{4}$

Multiplicative inverse of $\dfrac{-1}{4} = \dfrac{4}{-1} = -4$

Multiplying both the inverses,

$\Rightarrow \dfrac{1}{4} \times (-4) \\[1em] \Rightarrow \dfrac{1 \times (-4)}{4} \\[1em] \Rightarrow \dfrac{-4}{4} = -1$

Hence, Option 2 is the correct option.

The reciprocal of a negative rational number

1. is a positive rational number

2. is a negative rational number

3. can be either a positive or a negative rational number

4. does not exist

Answer

The reciprocal is obtained by interchanging the numerator and the denominator, so the sign of the rational number does not change.

For example, the reciprocal of $\dfrac{-4}{7} \text{ is } \dfrac{7}{-4} = \dfrac{-7}{4}$, which is again a negative rational number.

Hence, Option 2 is the correct option.

Which rational numbers are equal to their reciprocals?

1. 0, 1

2. -1, 0

3. -1, 1

4. -1, 0

Answer

A number is equal to its own reciprocal only when its product with itself is 1.

⇒ 1 × 1 = 1

⇒ (-1) × (-1) = 1

Also, 0 has no reciprocal at all.

Hence, Option 3 is the correct option.

Which of the following expressions shows that rational numbers are associative under multiplication?

1. $\dfrac{4}{5} \times \left(\dfrac{-6}{7} \times \dfrac{8}{9}\right) = \left(\dfrac{4}{5} \times \dfrac{-6}{7}\right) \times \dfrac{8}{9}$

2. $\dfrac{4}{5} \times \left(\dfrac{-6}{7} \times \dfrac{8}{9}\right) = \dfrac{4}{5} \times \left(\dfrac{8}{9} \times \dfrac{-6}{7}\right)$

3. $\dfrac{4}{5} \times \left(\dfrac{6}{7} \times \dfrac{8}{9}\right) = \left(\dfrac{8}{9} \times \dfrac{4}{5}\right) \times \dfrac{-6}{7}$

4. $\left(\dfrac{4}{5} \times \dfrac{-6}{7}\right) \times \dfrac{8}{9} = \left(\dfrac{-6}{7} \times \dfrac{4}{5}\right) \times \dfrac{8}{9}$

Answer

If $\dfrac{a}{b}, \dfrac{c}{d} \text{ and } \dfrac{e}{f}$ are any three rational numbers, then the associative law of multiplication states that $\dfrac{a}{b} \times \left(\dfrac{c}{d} \times \dfrac{e}{f}\right) = \left(\dfrac{a}{b} \times \dfrac{c}{d}\right) \times \dfrac{e}{f}$

In Option 1, only the grouping of the same three rational numbers, taken in the same order, has been changed.

In Options 2 and 4, the order of the rational numbers has been changed, which shows the commutative law.

In Option 3, the order of the rational numbers has been changed and $\dfrac{6}{7}$ on the left side has been replaced by $\dfrac{-6}{7}$ on the right side, so the two sides are not equal at all.

Hence, Option 1 is the correct option.

$\dfrac{-1}{4} \times \left[\dfrac{2}{5} + \left(\dfrac{-5}{6}\right)\right]$ = ?

1. $\left[\dfrac{-1}{4} \times \dfrac{2}{5}\right] + \left[\dfrac{-1}{4} \times \left(\dfrac{-5}{6}\right)\right]$

2. $\left[\dfrac{-1}{4} \times \dfrac{2}{5}\right] - \left(\dfrac{-5}{6}\right)$

3. $\dfrac{2}{5} + \left(\dfrac{-1}{4}\right) \times \left(\dfrac{-5}{6}\right)$

4. $\left[\dfrac{2}{5} + \left(\dfrac{-5}{6}\right)\right] - \dfrac{1}{4}$

Answer

By the distributive law of multiplication over addition, for any three rational numbers $\dfrac{a}{b}, \dfrac{c}{d} \text{ and } \dfrac{e}{f}$, we have

$\dfrac{a}{b} \times \left(\dfrac{c}{d} + \dfrac{e}{f}\right) = \left(\dfrac{a}{b} \times \dfrac{c}{d}\right) + \left(\dfrac{a}{b} \times \dfrac{e}{f}\right)$

So, $\dfrac{-1}{4}$ has to be multiplied by each of the two rational numbers inside the bracket separately.

Expanding the given expression by this law,

$\dfrac{-1}{4} \times \left[\dfrac{2}{5} + \left(\dfrac{-5}{6}\right)\right] = \left[\dfrac{-1}{4} \times \dfrac{2}{5}\right] + \left[\dfrac{-1}{4} \times \left(\dfrac{-5}{6}\right)\right]$

Hence, Option 1 is the correct option.

By what rational number should $\dfrac{-5}{24}$ be multiplied to get 10?

1. 12

2. -12

3. 48

4. -48

Answer

Let the required rational number be $x$.

According to the given condition,

$\Rightarrow \dfrac{-5}{24} \times x = 10 \\[1em] \Rightarrow x = 10 \div \dfrac{-5}{24} \\[1em] \Rightarrow x = \dfrac{10}{1} \times \dfrac{24}{-5} \\[1em] \Rightarrow x = \dfrac{10 \times 24}{-5} \\[1em] \Rightarrow x = \dfrac{240}{-5} = -48$

Hence, Option 4 is the correct option.

What should be added to $\dfrac{-3}{4}$ to get $\dfrac{7}{6}$?

1. $\dfrac{11}{24}$

2. $\dfrac{9}{24}$

3. $\dfrac{10}{25}$

4. $\dfrac{23}{12}$

Answer

Let the required number be $x$.

According to the given condition,

$\Rightarrow \dfrac{-3}{4} + x = \dfrac{7}{6} \\[1em] \Rightarrow x = \dfrac{7}{6} - \left(\dfrac{-3}{4}\right) \\[1em] \Rightarrow x = \dfrac{7}{6} + \dfrac{3}{4} \\[1em] [\text{ LCM of } 6 \text{ and } 4 = 12] \\[1em] \Rightarrow x = \dfrac{7 \times 2 + 3 \times 3}{12} \\[1em] \Rightarrow x = \dfrac{14 + 9}{12} = \dfrac{23}{12}$

Hence, Option 4 is the correct option.

What should be subtracted from $-\dfrac{7}{8}$ so as to get $\dfrac{5}{12}$?

1. $\dfrac{31}{24}$

2. $\dfrac{8}{5}$

3. $\dfrac{-31}{24}$

4. $\dfrac{-6}{8}$

Answer

Let the required number be $x$.

According to the given condition,

$\Rightarrow \dfrac{-7}{8} - x = \dfrac{5}{12} \\[1em] \Rightarrow x = \dfrac{-7}{8} - \dfrac{5}{12} \\[1em] [\text{ LCM of } 8 \text{ and } 12 = 24] \\[1em] \Rightarrow x = \dfrac{(-7) \times 3 - 5 \times 2}{24} \\[1em] \Rightarrow x = \dfrac{-21 - 10}{24} = \dfrac{-31}{24}$

Hence, Option 3 is the correct option.

The sum of two rational numbers is -3. If one of the numbers is $\dfrac{-7}{5}$, then the other number is

1. $\dfrac{-8}{5}$

2. $\dfrac{8}{5}$

3. $\dfrac{-6}{5}$

4. $\dfrac{6}{5}$

Answer

Let the other rational number be $x$.

As the sum of the two rational numbers is -3,

$\Rightarrow \dfrac{-7}{5} + x = -3 \\[1em] \Rightarrow x = -3 - \left(\dfrac{-7}{5}\right) \\[1em] \Rightarrow x = -3 + \dfrac{7}{5} \\[1em] \Rightarrow x = \dfrac{(-3) \times 5 + 7}{5} \\[1em] \Rightarrow x = \dfrac{-15 + 7}{5} = \dfrac{-8}{5}$

Hence, Option 1 is the correct option.

The product of two numbers is $\dfrac{-16}{35}$. If one of the numbers is $\dfrac{-15}{14}$, then the other is

1. $\dfrac{-2}{5}$

2. $\dfrac{-8}{3}$

3. $\dfrac{8}{15}$

4. $\dfrac{32}{75}$

Answer

Let the other rational number be $x$.

As the product of the two rational numbers is $\dfrac{-16}{35}$,

$\Rightarrow \dfrac{-15}{14} \times x = \dfrac{-16}{35} \\[1em] \Rightarrow x = \dfrac{-16}{35} \div \dfrac{-15}{14} \\[1em] \Rightarrow x = \dfrac{-16}{35} \times \dfrac{14}{-15} \\[1em] \Rightarrow x = \dfrac{(-16) \times 14}{35 \times (-15)} \\[1em] \Rightarrow x = \dfrac{-224}{-525} = \dfrac{32}{75}$

Hence, Option 4 is the correct option.

Which of the following numbers is in standard form?

1. $\dfrac{-12}{26}$

2. $\dfrac{-49}{70}$

3. $\dfrac{-9}{16}$

4. $\dfrac{28}{-105}$

Answer

A rational number $\dfrac{p}{q}$ is in standard form if $q \gt 0 \text{ and } p, q$ have no common divisor other than 1.

In Option 1, 12 and 26 have the common divisor 2, so $\dfrac{-12}{26} = \dfrac{-6}{13}$

In Option 2, 49 and 70 have the common divisor 7, so $\dfrac{-49}{70} = \dfrac{-7}{10}$

In Option 3, the denominator 16 is positive and the HCF of 9 and 16 is 1.

In Option 4, the denominator is negative, so $\dfrac{28}{-105} = \dfrac{-28}{105} = \dfrac{-4}{15}$

Hence, Option 3 is the correct option.

$\left(\dfrac{-9}{16} \times \dfrac{8}{15}\right)$ = ?

1. $\dfrac{-3}{10}$

2. $\dfrac{-4}{15}$

3. $\dfrac{-9}{25}$

4. $\dfrac{-2}{5}$

Answer

Multiplying the two rational numbers,

$\dfrac{-9}{16} \times \dfrac{8}{15} = \dfrac{(-9) \times 8}{16 \times 15} \\[1em] = \dfrac{-72}{240} \\[1em] = \dfrac{-3}{10}$

Hence, Option 1 is the correct option.

$\left(\dfrac{-5}{9} \div \dfrac{2}{3}\right)$ = ?

1. $\dfrac{-5}{2}$

2. $\dfrac{-5}{6}$

3. $\dfrac{-10}{27}$

4. $\dfrac{-6}{5}$

Answer

To divide one rational number by another, multiply the dividend by the reciprocal of the divisor.

$\dfrac{-5}{9} \div \dfrac{2}{3} = \dfrac{-5}{9} \times \dfrac{3}{2} \\[1em] = \dfrac{(-5) \times 3}{9 \times 2} \\[1em] = \dfrac{-15}{18} = \dfrac{-5}{6}$

Hence, Option 2 is the correct option.

$\dfrac{4}{9} \div\ ? = \dfrac{-8}{15}$

1. $\dfrac{-32}{45}$

2. $\dfrac{-8}{5}$

3. $\dfrac{-9}{10}$

4. $\dfrac{-5}{6}$

Answer

Let the required rational number be $x$.

According to the given condition,

$\Rightarrow \dfrac{4}{9} \div x = \dfrac{-8}{15} \\[1em] \Rightarrow x = \dfrac{4}{9} \div \dfrac{-8}{15} \\[1em] \Rightarrow x = \dfrac{4}{9} \times \dfrac{15}{-8} \\[1em] \Rightarrow x = \dfrac{4 \times 15}{9 \times (-8)} \\[1em] \Rightarrow x = \dfrac{60}{-72} = \dfrac{-5}{6}$

Hence, Option 4 is the correct option.

Between any two given rational numbers we can find

1. one and only one rational number

2. only two rational numbers

3. only ten rational numbers

4. infinitely many rational numbers

Answer

If $x$ and $y$ are two rational numbers, then $\dfrac{1}{2}(x + y)$ is a rational number lying between them.

This process of inserting the mean can be repeated as many times as we want.

Hence, Option 4 is the correct option.

A rational number between $\dfrac{1}{5} \text{ and } \dfrac{1}{2}$ is

1. $\dfrac{1}{10}$

2. $\dfrac{1}{20}$

3. $\dfrac{7}{10}$

4. $\dfrac{7}{20}$

Answer

Finding the average of the two given rational numbers,

$\Rightarrow \dfrac{1}{2} \times \left(\dfrac{1}{5} + \dfrac{1}{2}\right) \\[1em] [\text{ LCM of } 5 \text{ and } 2 = 10] \\[1em] \Rightarrow \dfrac{1}{2} \times \dfrac{2 + 5}{10} \\[1em] \Rightarrow \dfrac{1}{2} \times \dfrac{7}{10} = \dfrac{7}{20}$

As $\dfrac{1}{5} = \dfrac{4}{20} \text{ and } \dfrac{1}{2} = \dfrac{10}{20}$, we see that $\dfrac{4}{20} \lt \dfrac{7}{20} \lt \dfrac{10}{20}$

Hence, Option 4 is the correct option.

Identify a rational number between $\dfrac{1}{3} \text{ and } \dfrac{4}{5}$.

1. $\dfrac{1}{4}$

2. $\dfrac{9}{10}$

3. $\dfrac{17}{30}$

4. $1\dfrac{7}{10}$

Answer

Write the given rational numbers with the LCM of their denominators as the common denominator and compare them with the given options.

$[\text{ LCM of } 3 \text{ and } 5 = 30] \\[1em] \Rightarrow \dfrac{1}{3} = \dfrac{1 \times 10}{3 \times 10} = \dfrac{10}{30} \\[1em] \Rightarrow \dfrac{4}{5} = \dfrac{4 \times 6}{5 \times 6} = \dfrac{24}{30}$

As 10 < 17 < 24, so $\dfrac{10}{30} \lt \dfrac{17}{30} \lt \dfrac{24}{30}$

Hence, Option 3 is the correct option.

The arrangement of rational numbers $\dfrac{-7}{10}, \dfrac{5}{-8}, \dfrac{2}{-3}$, in ascending order is

1. $\dfrac{2}{-3}, \dfrac{5}{-8}, \dfrac{-7}{10}$

2. $\dfrac{-7}{10}, \dfrac{2}{-3}, \dfrac{5}{-8}$

3. $\dfrac{5}{-8}, \dfrac{-7}{10}, \dfrac{2}{-3}$

4. $\dfrac{-7}{10}, \dfrac{5}{-8}, \dfrac{2}{-3}$

Answer

First write each rational number with a positive denominator.

$\Rightarrow \dfrac{5}{-8} = \dfrac{5 \times (-1)}{-8 \times (-1)} = \dfrac{-5}{8} \\[1em] \Rightarrow \dfrac{2}{-3} = \dfrac{2 \times (-1)}{-3 \times (-1)} = \dfrac{-2}{3}$

Now write each rational number with the LCM of the denominators as the common denominator.

$[\text{ LCM of } 10, 8 \text{ and } 3 = 120] \\[1em] \Rightarrow \dfrac{-7}{10} = \dfrac{(-7) \times 12}{10 \times 12} = \dfrac{-84}{120} \\[1em] \Rightarrow \dfrac{-5}{8} = \dfrac{(-5) \times 15}{8 \times 15} = \dfrac{-75}{120} \\[1em] \Rightarrow \dfrac{-2}{3} = \dfrac{(-2) \times 40}{3 \times 40} = \dfrac{-80}{120}$

As -84 < -80 < -75, so $\dfrac{-84}{120} \lt \dfrac{-80}{120} \lt \dfrac{-75}{120}$.

$\Rightarrow \dfrac{-7}{10} \lt \dfrac{2}{-3} \lt \dfrac{5}{-8}$

Hence, Option 2 is the correct option.

Fill in the blanks:

(i) $-\dfrac{7}{8} + \dfrac{2}{11} = \dfrac{2}{11} + \left(-\dfrac{7}{8}\right)$ depicts ......... property of addition of rational numbers.

(ii) The multiplicative inverse of a rational number is also called its .........

(iii) The multiplicative inverse of a negative rational number is a ......... rational number.

(iv) The numbers ......... and ......... are their own reciprocals.

(v) $\left(-\dfrac{10}{11}\right) \div$ ......... = 1

(vi) $\left(-\dfrac{7}{9}\right) \times$ ......... = 1

(vii) Among $-\dfrac{5}{11}, \dfrac{-5}{13} \text{ and } \dfrac{-5}{7}$ the greatest rational number is .........

Answer

(i) Here the order of the two rational numbers has been changed but the sum remains the same.

If $\dfrac{a}{b} \text{ and } \dfrac{c}{d}$ are any two rational numbers, then $\dfrac{a}{b} + \dfrac{c}{d} = \dfrac{c}{d} + \dfrac{a}{b}$.

Hence, the given statement depicts the commutative property of addition of rational numbers.

(ii) For every non-zero rational number $\dfrac{a}{b}$, the rational number $\dfrac{b}{a}$ satisfies $\dfrac{a}{b} \times \dfrac{b}{a} = 1$

Hence, the multiplicative inverse of a rational number is also called its reciprocal.

(iii) The multiplicative inverse is obtained by interchanging the numerator and the denominator, so the sign of the rational number does not change. For example, the multiplicative inverse of $\dfrac{-2}{9} \text{ is } \dfrac{9}{-2} = \dfrac{-9}{2}$

Hence, the multiplicative inverse of a negative rational number is a negative rational number.

(iv) As 1 × 1 = 1 and (-1) × (-1) = 1, each of these numbers is its own reciprocal.

Hence, the numbers 1 and -1 are their own reciprocals.

(v) For every non-zero rational number $\dfrac{a}{b}$, we have $\dfrac{a}{b} \div \dfrac{a}{b} = 1$

$\left(\dfrac{-10}{11}\right) \div \left(\dfrac{-10}{11}\right) = \dfrac{-10}{11} \times \dfrac{11}{-10} = 1$

Hence, $\mathbf{\left(-\dfrac{10}{11}\right) \div \left(-\dfrac{10}{11}\right) = 1}$.

(vi) The product of a non-zero rational number and its reciprocal is 1, so the required number is the reciprocal of $-\dfrac{7}{9}$.

$\Rightarrow \text{Reciprocal of } \dfrac{-7}{9} = \dfrac{9}{-7} = \dfrac{-9}{7} \\[1em] \Rightarrow \dfrac{-7}{9} \times \dfrac{-9}{7} = 1$

Hence, $\mathbf{\left(-\dfrac{7}{9}\right) \times \left(\dfrac{-9}{7}\right) = 1}$.

(vii) All three rational numbers are negative and have the same numerator -5, so write them with a common denominator and compare.

$[\text{ LCM of } 11, 13 \text{ and } 7 = 1001] \\[1em] \Rightarrow \dfrac{-5}{11} = \dfrac{(-5) \times 91}{11 \times 91} = \dfrac{-455}{1001} \\[1em] \Rightarrow \dfrac{-5}{13} = \dfrac{(-5) \times 77}{13 \times 77} = \dfrac{-385}{1001} \\[1em] \Rightarrow \dfrac{-5}{7} = \dfrac{(-5) \times 143}{7 \times 143} = \dfrac{-715}{1001}$

As -715 < -455 < -385, so $\dfrac{-5}{7} \lt \dfrac{-5}{11} \lt \dfrac{-5}{13}$.

Hence, the greatest rational number is $\mathbf{\dfrac{-5}{13}}$.

Write true (T) or false (F):

(i) Every rational number has a reciprocal.

(ii) Subtraction of rational numbers is commutative.

(iii) The additive inverse of $\dfrac{1}{4}$ is -4.

(iv) $\dfrac{5}{0}$ is a rational number.

(v) 0 is the smallest rational number.

(vi) The multiplicative inverse of $\dfrac{-4}{-9}$ is $\dfrac{9}{4}$.

(vii) There are countless rational numbers between $\dfrac{2}{3} \text{ and } \dfrac{5}{6}$.

Answer

(i) There is no rational number which when multiplied by 0 gives 1. So the rational number 0 has no reciprocal.

Hence, the given statement is false.

(ii) For any two rational numbers $\dfrac{a}{b} \text{ and } \dfrac{c}{d}$, we have $\dfrac{a}{b} - \dfrac{c}{d} ≠ \dfrac{c}{d} - \dfrac{a}{b}$.

For example, $\dfrac{1}{2} - \dfrac{1}{3} = \dfrac{1}{6}$, but $\dfrac{1}{3} - \dfrac{1}{2} = \dfrac{-1}{6}$.

Hence, the given statement is false.

(iii) The additive inverse is obtained by changing the sign of the rational number. So the additive inverse of $\dfrac{1}{4} \text{ is } \dfrac{-1}{4}$ and not -4.

$[\text{Check : } \dfrac{1}{4} + \dfrac{-1}{4} = 0 \text{ but } \dfrac{1}{4} + (-4) = \dfrac{-15}{4}]$

Hence, the given statement is false.

(iv) A rational number is of the form $\dfrac{p}{q}$ where $q ≠ 0$. Here the denominator is 0 and division by 0 is not defined.

Hence, the given statement is false.

(v) Every negative rational number is smaller than 0, and there is no smallest rational number as we can always find a rational number smaller than any given one. For example, -1 < 0 and -2 < -1 and so on.

Hence, the given statement is false.

(vi) First express the rational number with a positive denominator.

$\dfrac{-4}{-9} = \dfrac{-4 \times (-1)}{-9 \times (-1)} = \dfrac{4}{9}$

To find the multiplicative inverse of a non-zero rational number, interchange its numerator and denominator.

Interchanging the numerator and the denominator of $\dfrac{4}{9}$, we get $\dfrac{9}{4}$, which is the same as the multiplicative inverse given in the statement.

$[\text{Check : } \dfrac{4}{9} \times \dfrac{9}{4} = 1]$

Hence, the given statement is true.

(vii) Between any two different rational numbers we can always insert their mean, and this process can be repeated as many times as we want.

Hence, the given statement is true.

Assertion (A): Rational numbers are closed under addition and multiplication but not under subtraction.

Reason (R): For any two rational numbers a and b,

a + b is a rational number

a - b is a rational number

a × b is a rational number.

1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).

2. Both Assertion (A) and Reason (R) are true and Reason (R) is not the correct explanation of Assertion (A).

3. Assertion (A) is true but Reason (R) is false.

4. Assertion (A) is false but Reason (R) is true.

Answer

For any two rational numbers a and b, each of a + b, a - b and a × b is again a rational number, so rational numbers are closed under addition, subtraction and multiplication.

So, the Reason (R) is true.

But the Assertion (A) claims that rational numbers are not closed under subtraction, which is wrong. For example, $\dfrac{2}{3} - \dfrac{1}{4} = \dfrac{5}{12}$, which is a rational number.

So, the Assertion (A) is false.

Hence, Option 4 is the correct option.

Assertion (A): The negative (additive inverse) of -5 is 5 and the negative of 5 is -5.

Reason (R): The negative of a negative rational number is negative and the negative of a positive rational number is negative.

1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).

2. Both Assertion (A) and Reason (R) are true and Reason (R) is not the correct explanation of Assertion (A).

3. Assertion (A) is true but Reason (R) is false.

4. Assertion (A) is false but Reason (R) is true.

Answer

The negative (additive inverse) of a rational number is obtained by changing its sign.

⇒ -5 + 5 = 0

⇒ 5 + (-5) = 0

So, the negative of -5 is 5 and the negative of 5 is -5, i.e. the Assertion (A) is true.

The Reason (R) states that the negative of a negative rational number is negative, which is wrong, as the negative of a negative rational number is a positive rational number.

So, the Reason (R) is false.

Hence, Option 3 is the correct option.

Assertion (A): For any rational numbers x, y, z, we have, x × (y + z) = x × y + x × z.

Reason (R): Rational numbers are associative under addition and multiplication.

1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).

2. Both Assertion (A) and Reason (R) are true and Reason (R) is not the correct explanation of Assertion (A).

3. Assertion (A) is true but Reason (R) is false.

4. Assertion (A) is false but Reason (R) is true.

Answer

For any three rational numbers x, y and z, we have x × (y + z) = x × y + x × z, which is the distributive law of multiplication over addition.

So, the Assertion (A) is true.

Also, for any three rational numbers, (x + y) + z = x + (y + z) and (x × y) × z = x × (y × z), so rational numbers are associative under addition as well as under multiplication.

So, the Reason (R) is also true, but it states the associative law, whereas the Assertion (A) is based on the distributive law.

So, the Reason (R) is not the correct explanation of the Assertion (A).

Hence, Option 2 is the correct option.

Ramesh earns ₹ 40,000 per month. He spends $\dfrac{3}{8}$ of income on food, $\dfrac{1}{5}$ of the remaining on LIC premium and $\dfrac{1}{2}$ of the remaining on other expenses. Find how much money is left with him?

1. ₹8,000

2. ₹10,000

3. ₹15,000

4. ₹16,000

Answer

Subtract each expense one by one from the amount remaining at that stage.

Monthly income of Ramesh = ₹ 40,000

Finding the money spent on food,

$\dfrac{3}{8} \times 40000 = 3 \times 5000 = ₹ 15000$

Money left after buying food = ₹ 40000 - ₹ 15000 = ₹ 25000

Finding the money spent on LIC premium,

$\dfrac{1}{5} \times 25000 = ₹ 5000$

Money left after paying the LIC premium = ₹ 25000 - ₹ 5000 = ₹ 20000

Finding the money spent on other expenses,

$\dfrac{1}{2} \times 20000 = ₹ 10000$

Money left with him = ₹ 20000 - ₹ 10000 = ₹ 10000

Hence, Option 2 is the correct option.

A, B, C, D and E went to a restaurant for dinner. A paid $\dfrac{1}{2}$ of the bill, B paid $\dfrac{1}{5}$ of the bill and rest of the bill was shared equally by C, D and E. What fraction of the bill was paid by each?

1. $\dfrac{3}{5}$

2. $\dfrac{2}{5}$

3. $\dfrac{3}{10}$

4. $\dfrac{1}{10}$

Answer

First find the fraction of the bill left after A and B have paid, then divide it equally among C, D and E.

Finding the fraction of the bill paid by A and B together,

LCM of 2 and 5 is 10.

$\dfrac{1}{2} + \dfrac{1}{5} = \dfrac{5 + 2}{10} \\[1em] = \dfrac{7}{10}$

Finding the fraction of the bill left,

$1 - \dfrac{7}{10} = \dfrac{10 - 7}{10} \\[1em] = \dfrac{3}{10}$

Finding the fraction of the bill paid by each of C, D and E,

$\dfrac{3}{10} \div 3 = \dfrac{3}{10} \times \dfrac{1}{3} \\[1em] = \dfrac{1}{10}$

Hence, Option 4 is the correct option.

$\dfrac{2}{5}$ of total number of students of a school come by a car, while $\dfrac{1}{4}$ of students come by bus to school. All the other students walk to school, of which $\dfrac{1}{3}$ walk on their own and rest are escorted by their parents. If 224 students come to school walking on their own, how many students study in that school?

1. 2000

2. 1970

3. 1920

4. 1904

Answer

First find the fraction of the students who walk on their own, then use it to find the total number of students.

Finding the fraction of the students who come by car or by bus,

LCM of 5 and 4 = 20

$\dfrac{2}{5} + \dfrac{1}{4} = \dfrac{2 \times 4 + 1 \times 5}{20} \\[1em] = \dfrac{8 + 5}{20} = \dfrac{13}{20}$

Finding the fraction of the students who walk to school,

$1 - \dfrac{13}{20} = \dfrac{20 - 13}{20} = \dfrac{7}{20}$

Finding the fraction of the students who walk on their own,

$\dfrac{1}{3} \times \dfrac{7}{20} = \dfrac{7}{60}$

Let the total number of students in the school be $x$.

$\Rightarrow \dfrac{7}{60} \times x = 224 \\[1em] \Rightarrow x = 224 \times \dfrac{60}{7} \\[1em] \Rightarrow x = 32 \times 60 \\[1em] \Rightarrow x = 1920$

Hence, Option 3 is the correct option.

The rational number that must be added to the sum of $\dfrac{3}{7}, -\dfrac{8}{21} \text{ and } -\dfrac{3}{14}$ to get the result $\dfrac{-5}{14}$ is:

1. $\dfrac{-4}{21}$

2. $\dfrac{4}{21}$

3. $\dfrac{-5}{21}$

4. $\dfrac{-1}{6}$

Answer

First find the sum of the three given rational numbers.

LCM of 7, 21 and 14 = 42

$\dfrac{3}{7} + \left(\dfrac{-8}{21}\right) + \left(\dfrac{-3}{14}\right) = \dfrac{3 \times 6 + (-8) \times 2 + (-3) \times 3}{42} \\[1em] = \dfrac{18 - 16 - 9}{42} \\[1em] = \dfrac{-7}{42} = \dfrac{-1}{6}$

Let the required rational number be $x$.

According to the given condition,

LCM of 14 and 6 = 42

$\Rightarrow \dfrac{-1}{6} + x = \dfrac{-5}{14} \\[1em] \Rightarrow x = \dfrac{-5}{14} - \left(\dfrac{-1}{6}\right) \\[1em] \Rightarrow x = \dfrac{-5}{14} + \dfrac{1}{6} \\[1em] \Rightarrow x = \dfrac{(-5) \times 3 + 1 \times 7}{42} \\[1em] \Rightarrow x = \dfrac{-15 + 7}{42} \\[1em] \Rightarrow x = \dfrac{-8}{42} = \dfrac{-4}{21}$

Hence, Option 1 is the correct option.

$p$ and $q$ are two non-zero unequal rational numbers. Which of the following can't be equal to 1?

(i) $\dfrac{p - q}{q - p}$

(ii) $\dfrac{p + q}{q + p}$

(iii) $\dfrac{pq}{qp}$

(iv) $\dfrac{p}{q} \div \dfrac{q}{p}$

1. only (i)

2. only (iv)

3. (i) and (ii)

4. (i) and (iv)

Answer

Examine each expression separately, keeping in mind that $p$ and $q$ are non-zero and $p ≠ q$.

(i) As $p ≠ q$, the numerator $p - q$ is non-zero and $q - p = -(p - q)$

$\Rightarrow \dfrac{p - q}{q - p} = \dfrac{p - q}{-(p - q)} = -1$

So, this expression is always equal to -1 and can never be equal to 1.

(ii) Here the numerator and the denominator are the same, as $p + q = q + p$ by the commutative law of addition.

$\Rightarrow \dfrac{p + q}{q + p} = 1$

So, this expression is equal to 1.

(iii) Here the numerator and the denominator are the same, as $pq = qp$ by the commutative law of multiplication.

$\Rightarrow \dfrac{pq}{qp} = 1$

So, this expression is equal to 1.

(iv) Dividing,

$\Rightarrow \dfrac{p}{q} \div \dfrac{q}{p} \\[1em] \Rightarrow \dfrac{p}{q} \times \dfrac{p}{q} \\[1em] \Rightarrow \dfrac{p^2}{q^2}$

This is equal to 1 whenever $p^2 = q^2$, which happens for $p = -q$. For example, taking $p = 2$ and $q = -2$, we get $\dfrac{p^2}{q^2} = \dfrac{4}{4} = 1$

So, this expression can be equal to 1.

Thus, only expression (i) can never be equal to 1.

Hence, Option 1 is the correct option.

a and b are two rational numbers such that a + b = 3. When we divide a by b, the closure property of rational numbers does not hold true. The values of ab and a - b respectively are:

1. 0, -3

2. 0, 0

3. 0, 3

4. 3, 3

Answer

Rational numbers are closed under division only when the divisor is a non-zero rational number, as division by 0 is not defined.

As a ÷ b does not satisfy the closure property, the divisor b must be 0.

⇒ b = 0

⇒ a + b = 3

⇒ a + 0 = 3

⇒ a = 3

Finding the values of ab and a - b,

⇒ ab = 3 × 0 = 0

⇒ a - b = 3 - 0 = 3

Hence, Option 3 is the correct option.