myhelper: RD Sharma solution class 7 chapter 4 Rational numbers Objective Type Questions

RD Sharma solution class 7 chapter 4 Rational numbers Objective Type Questions

Objective Type Questions

Question 1:

Mark the correct alternative in each of the following:

$\frac{44}{- 77}$ in standard form is

(a)

$\frac{4}{- 7}$ (b)

$- \frac{4}{7}$ (c)

$- \frac{44}{77}$ (d) None of these

Answer 1:

The denominator of

$\frac{44}{- 77}$ is negative.

Firstly, multiply the numerator and denominator by −1 to make it positive.

$\frac{44}{- 77} = \frac{44 \times (- 1)}{- 77 \times (- 1)} = \frac{- 44}{77}$

Now,

HCF of 44 and 77 = 11

Dividing the numerator and denominator of

$\frac{- 44}{77}$ by 11, we have

$\frac{- 44}{77} = \frac{- 44 \div 11}{77 \div 11} = \frac{- 4}{7} = - \frac{4}{7}$

Thus, the standard form of

$\frac{44}{- 77}$ is

$- \frac{4}{7}$ .

Hence, the correct answer is option (b).

Question 2:

Mark the correct alternative in each of the following:

$- \frac{102}{119}$ in standard form is

(a)

$- \frac{6}{7}$ (b)

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

$- \frac{6}{17}$ (d) None of these

Answer 2:

The denominator of the rational number

$- \frac{102}{119}$ is positive.

In order to write the rational number in standard form, divide its numerator and denominator by the HCF of 102 and 119.

HCF of 102 and 119 = 17

Dividing the numerator and denominator of

$- \frac{102}{119}$ by 17, we have

$- \frac{102}{119} = - \frac{102 \div 17}{119 \div 17} = - \frac{6}{7}$

Thus, the standard form of

$- \frac{102}{119}$ is

$- \frac{6}{7}$ .

Hence, the correct answer is option (a).

Question 3:

Mark the correct alternative in each of the following:

A rational number equal to

$\frac{- 2}{3}$ is

(a)

$\frac{- 10}{25}$ (b)

$\frac{10}{- 15}$ (c)

$\frac{- 9}{6}$ (d) None of these

Answer 3:

We know that two rational numbers are equal if they have the same standard form.

The rational number

$\frac{- 2}{3}$ is in its standard form.

Consider the rational number

$\frac{10}{- 15}$ .

This rational number can be expressed in standard form as follows:

$\frac{10}{- 15} = \frac{10 \times (- 1)}{- 15 \times (- 1)} = \frac{- 10}{15}$ (Multiplying numerator and denominator by −1 to make denominator positive)

HCF of 10 and 15 = 5

Dividing the numerator and denominator of

$\frac{- 10}{15}$ by 5, we have

$\frac{- 10}{15} = \frac{- 10 \div 5}{15 \div 5} = \frac{- 2}{3}$

Thus, the standard form of

$\frac{- 10}{15}$ is

$\frac{- 2}{3}$ , which is same as the given rational number.

So, the rational number equal to

$\frac{- 2}{3}$ is

$\frac{10}{- 15}$ .

Let us check why options (a) and (c) are not correct.

The standard form of

$\frac{- 10}{25}$ is

$\frac{- 2}{5}$ .

HCF of 10 and 25 = 5

Dividing the numerator and denominator of

$\frac{- 10}{25}$ by 5, we have

$\frac{- 10}{25} = \frac{- 10 \div 5}{25 \div 5} = \frac{- 2}{5}$

The standard form of

$\frac{- 9}{6}$ is

$\frac{- 3}{2}$ .

HCF of 6 and 9 = 3

Dividing the numerator and denominator of

$\frac{- 9}{6}$ by 3, we have

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

Hence, the correct answer is option (b).

Question 4:

Mark the correct alternative in each of the following:

If

$\frac{- 3}{7} = \frac{x}{35}$ , then x =

(a) 15 (b) 21 (c) −15 (d) −21

Answer 4:

Firstly, write

$\frac{- 3}{7}$ as a rational number with denominator 35.

Multiplying the numerator and denominator of

$\frac{- 3}{7}$ by 5, we have

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

$∴ \frac{- 3}{7} = \frac{x}{35} \Rightarrow \frac{- 15}{35} = \frac{x}{35} \Rightarrow x = - 15$

Hence, the correct answer is option (c).

Question 5:

Mark the correct alternative in each of the following:

Which of the following is correct?

(a)

$\frac{5}{9} > \frac{- 3}{- 8}$ (b)

$\frac{5}{9} < \frac{- 3}{- 8}$ (c)

$\frac{2}{- 3} < \frac{- 8}{7}$ (d)

$\frac{4}{- 3} > \frac{- 8}{7}$

Answer 5:

Consider the rational numbers

$\frac{5}{9}$ and

$\frac{- 3}{- 8}$ .

We write the rational number

$\frac{- 3}{- 8}$ with positive denominator.

$\frac{- 3}{- 8} = \frac{- 3 \times (- 1)}{- 8 \times (- 1)} = \frac{3}{8}$

Now, we write the rational numbers so that they have a common denominator.

LCM of 8 and 9 = 72

So,

$\frac{5}{9} = \frac{5 \times 8}{9 \times 8} = \frac{40}{72}$ and

$\frac{3}{8} = \frac{3 \times 9}{8 \times 9} = \frac{27}{72}$

Now,

$40 > 27 \Rightarrow \frac{40}{72} > \frac{27}{72} \Rightarrow \frac{5}{9} > \frac{3}{8} \Rightarrow \frac{5}{9} > \frac{- 3}{- 8}$

It can also be checked that

$\frac{2}{- 3} > \frac{- 8}{7}$ and

$\frac{4}{- 3} < \frac{- 8}{7}$ .

Hence, the correct answer is option (a).

Question 6:

Mark the correct alternative in each of the following:

If the rational numbers

$\frac{- 2}{3}$ and

$\frac{4}{x}$ represent a pair of equivalent rational numbers, then x =

(a) 6 (b) −6 (c) 3 (d) −3

Answer 6:

It is given that the rational numbers

$\frac{- 2}{3}$ and

$\frac{4}{x}$ represent a pair of equivalent rational numbers.

We know that the values of two equivalent rational numbers is equal.

$∴ \frac{- 2}{3} = \frac{4}{x} \Rightarrow - 2 \times x = 4 \times 3 (\frac{a}{b} = \frac{c}{d} \Rightarrow a d = b c) \Rightarrow - 2 x = 12 \Rightarrow \frac{- 2 x}{- 2} = \frac{12}{- 2} (Dividing both sides by - 2) \Rightarrow x = - 6$

Hence, the correct answer is option (b).

Question 7:

Mark the correct alternative in each of the following:

What is the additive identity element in the set of whole numbers?

(a) 0 (b) 1 (c) −1 (d) None of these

Answer 7:

If a is a whole number then a + 0 = a = 0 + a.

Therefore, 0 is the additive identity element for addition of whole number because it does not change the identity or value of the whole number during the operation of addition.

Hence, the correct answer is option (a).

Question 8:

Mark the correct alternative in each of the following:

What is the multiplicative identity element in the set of whole numbers?

(a) 0 (b) 1 (c) −1 (d) None of these

Answer 8:

We know that if a is a whole number, then a × 1 = a = 1 × a.

Therefore, 1 is the multiplicative identity element for multiplication of whole numbers because it does not change the identity or value of the whole number during the operation of multiplication.

Hence, the correct answer is option (b).

Question 9:

Mark the correct alternative in each of the following:

Which of the following is not zero?

(a) 0 × 0 (b)

$\frac{0}{3}$ (c)

$\frac{7 - 7}{3}$ (d) 9 ÷ 0

Answer 9:

If any number is multiplied by 0, the product is 0.

∴ 0 × 0 = 0

If 0 is divided by any number (≠ 0), the quotient is always 0.

∴

$\frac{0}{3} = 0$ and

$\frac{7 - 7}{3} = \frac{0}{3} = 0$

Division of any number by 0 is meaningless and is not defined.

∴ 9 ÷ 0 is not defined.

Hence, the correct answer is option (d).

Question 10:

Mark the correct alternative in each of the following:

The whole number nearest to 457 and divisible by 11 is

(a) 450 (b) 451 (c) 460 (d) 462

Answer 10:

The numbers 450 and 460 are not divisible by 11.

Now, both the numbers 451 and 462 are divisible by 11.

Distance between 457 and 451 on the number line = 457 − 451 = 6

Distance between 457 and 462 on the number line = 462 − 457 = 5

Thus, the whole number nearest to 457 and divisible by 11 is 462.

Hence, the correct answer is option (d).

Question 11:

Mark the correct alternative in each of the following:

If

$- \frac{3}{8}$ and

$\frac{x}{- 24}$ are equivalent rational numbers, then x =

(a) 3 (b) 6 (c) 9 (d) 12

Answer 11:

It is given that the rational numbers

$- \frac{3}{8}$ and

$\frac{x}{- 24}$ are equivalent rational numbers.

We know that the values of two equivalent rational numbers is equal.

$∴ \frac{x}{- 24} = - \frac{3}{8} \Rightarrow x \times 8 = - 3 \times (- 24) (\frac{a}{b} = \frac{c}{d} \Rightarrow a d = b c) \Rightarrow 8 x = 72 \Rightarrow \frac{8 x}{8} = \frac{72}{8} (Dividing both sides by 8) \Rightarrow x = 9$

Hence, the correct answer is option (c).

Question 12:

Mark the correct alternative in each of the following:

If

$\frac{27}{- 45}$ is expressed as a rational number with denominator 5, then the numerator is

(a) 3 (b) −3 (c) 6 (d) −6

Answer 12:

In order to express

$\frac{27}{- 45}$ as a rational number with denominator 5, firstly find a number which gives 5 when −45 is divided by it.

This number is −45 ÷ 5 = −9.

Dividing the numerator and denominator of

$\frac{27}{- 45}$ by −9, we have

$\frac{27}{- 45} = \frac{27 \div (- 9)}{- 45 \div (- 9)} = \frac{- 3}{5}$

Thus, the numerator is −3.

Hence, the correct answer is option (b).

Question 13:

Mark the correct alternative in each of the following:

Which of the following pairs of rational numbers are on the opposite side of the zero on the number line?

(a)

$\frac{3}{7}$ and

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

$- \frac{3}{7}$ and

$\frac{- 5}{12}$ (c)

$\frac{3}{7}$ and

$\frac{- 5}{12}$ (d) None of these

Answer 13:

The rational numbers

$\frac{3}{7}$ and

$\frac{5}{12}$ are positive rational numbers. We know that every positive rational number is greater than 0, so both the rational numbers

$\frac{3}{7}$ and

$\frac{5}{12}$ are represented by points on the right of the zero on the number line.

The rational numbers

$- \frac{3}{7}$ and

$\frac{- 5}{12}$ are negative rational numbers. We know that every negative rational number is less than 0, so both the rational numbers

$\frac{3}{7}$ and

$\frac{5}{12}$ are represented by points on the left of the zero on the number line.

The rational numbers

$\frac{3}{7}$ is a positive rational number whereas the rational number

$\frac{- 5}{12}$ is a negative rational numbers. We know that every negative rational number is less than 0 and every positive rational number is greater than 0, so the rational number

$\frac{3}{7}$ is represented by point on the right of the zero and

$\frac{- 5}{12}$ is represented by point on the left of the zero on the number line.
Thus, the rational numbers

$\frac{3}{7}$ and

$\frac{- 5}{12}$ are on the opposite side of the zero on the number line.

Hence, the correct answer is option (c).

Question 14:

Mark the correct alternative in each of the following:

The rational number equal to

$\frac{2}{- 3}$ is

(a)

$\frac{14}{- 18}$ (b)

$\frac{- 6}{9}$ (c)

$\frac{- 8}{- 12}$ (d)

$\frac{3}{- 2}$

Answer 14:

We know that two rational numbers are equal if they have the same standard form.

$\frac{2}{- 3} = \frac{2 \times (- 1)}{- 3 \times (- 1)} = \frac{- 2}{3}$

The standard form of

$\frac{2}{- 3}$ is

$\frac{- 2}{3}$ .

Consider the rational number

$\frac{- 6}{9}$ .

HCF of 6 and 9 = 3

Dividing the numerator and denominator of

$\frac{- 6}{9}$ by 3, we have

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

Thus, the standard form of

$\frac{- 6}{9}$ is

$\frac{- 2}{3}$ .

So, the rational number

$\frac{- 6}{9}$ is equal to

$\frac{2}{- 3}$ .

It can be checked that

Standard form of

$\frac{14}{- 18}$ =

$\frac{- 7}{9}$

Standard form of

$\frac{- 8}{- 12}$ =

$\frac{2}{3}$

Standard form of

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

$\frac{- 3}{2}$

Hence, the correct answer is option (b).

Question 15:

Mark the correct alternative in each of the following:

If

$- \frac{3}{4} = \frac{6}{x}$ , then x =

(a) −8 (b) 4 (c) −4 (d) 8

Answer 15:

$- \frac{3}{4} = \frac{6}{x} \Rightarrow - 3 \times x = 6 \times 4 (\frac{a}{b} = \frac{c}{d} \Rightarrow a d = b c) \Rightarrow - 3 x = 24 \Rightarrow \frac{- 3 x}{- 3} = \frac{24}{- 3} (Dividing both sides by - 3) \Rightarrow x = - 8$

Hence, the correct answer is option (a).

RD Sharma solution class 7 chapter 4 Rational numbers Objective Type Questions

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