myhelper: RD Sharma solution class 7 chapter 5 Operations On Rational numbers Objective Type Questions

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RD Sharma solution class 7 chapter 5 Operations On Rational numbers Objective Type Questions

Objective Type Questions

Question 1:

Mark the correct alternative in each of the following:

What should be added to

$\frac{- 7}{9}$ to get 2?

(a)

$\frac{11}{9}$ (b)

$\frac{- 11}{9}$ (c)

$\frac{- 25}{9}$ (d)

$\frac{25}{9}$

Answer 1:

Sum of the given number and the required number = 2

Given number =

$\frac{- 7}{9}$

∴ Required number = Sum of the numbers − Given number

$= 2 - (\frac{- 7}{9}) = \frac{2}{1} + \frac{7}{9} = \frac{2 \times 9 + 7 \times 1}{9} = \frac{18 + 7}{9} = \frac{25}{9}$

Hence, the correct answer is option (d).

Question 2:

Mark the correct alternative in each of the following:

What should be subtracted from

$\frac{- 2}{3}$ to get

$\frac{4}{5}$ ?

(a)

$\frac{22}{15}$ (b)

$\frac{- 22}{15}$ (c)

$\frac{15}{22}$ (d)

$\frac{- 15}{22}$

Answer 2:

Difference of the given number and required number =

$\frac{4}{5}$

Given number =

$\frac{- 2}{3}$

∴ Required number = Given number − Difference of the numbers

$= \frac{- 2}{3} - \frac{4}{5} = \frac{- 2}{3} + (\frac{- 4}{5}) = \frac{- 2 \times 5 + (- 4) \times 3}{15} = \frac{- 10 + (- 12)}{15} = \frac{- 22}{15}$

Hence, the correct answer is option (b).

Question 3:

Mark the correct alternative in each of the following:

Reciprocal of

$\frac{- 3}{4}$ is

(a)

$\frac{3}{4}$ (b)

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

$\frac{- 4}{3}$ (d) None of these

Answer 3:

We know that the reciprocal of the rational number

$\frac{a}{b}$ is

${(\frac{a}{b})}^{- 1} = \frac{b}{a}$ .

∴ Reciprocal of

$\frac{- 3}{4}$

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

Hence, the correct answer is option (c).

Question 4:

Mark the correct alternative in each of the following:

The multiplicative inverse of

$\frac{4}{- 5}$ is

(a)

$- \frac{4}{5}$ (b)

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

$\frac{5}{- 4}$ (d)

$\frac{- 5}{- 4}$

Answer 4:

We know that the multiplicative inverse of the rational number

$\frac{a}{b}$ is

$\frac{b}{a}$ .

∴ Multiplicative inverse of

$\frac{4}{- 5}$

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

Hence, the correct answer is option (c).

Question 5:

Mark the correct alternative in each of the following:

$1 \div \frac{- 5}{7} =$

(a)

$\frac{2}{7}$ (b)

$\frac{5}{7}$ (c)

$- \frac{2}{7}$ (d)

$\frac{- 7}{5}$

Answer 5:

$1 \div \frac{- 5}{7} = 1 \times \frac{7}{- 5} (x \div y = x \times \frac{1}{y}) = \frac{7}{- 5} = \frac{7 \times (- 1)}{- 5 \times (- 1)} = \frac{- 7}{5}$

Hence, the correct answer is option (d).

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Question 6:

Mark the correct alternative in each of the following:

$\frac{- 5}{13} + ? = - 1$

(a) $\frac{8}{13}$ (b) $\frac{- 8}{13}$ (c) $\frac{- 18}{13}$ (d) $\frac{18}{13}$

Answer 6:

$\frac{- 5}{13} + ? = - 1 \Rightarrow ? = - 1 - (\frac{- 5}{13}) \Rightarrow ? = - 1 + \frac{5}{13} [- (\frac{- 5}{13}) = \frac{5}{13}] \Rightarrow ? = \frac{- 1 \times 13 + 5}{13}$
$\Rightarrow ? = \frac{- 13 + 5}{13} \Rightarrow ? = \frac{- 8}{13}$

Hence, the correct answer is option (b).

Question 7:

Mark the correct alternative in each of the following:

$0 \div \frac{3}{5} =$

(a) 0 (b) $\frac{5}{3}$ (c) $\frac{3}{5}$ (d) $- \frac{3}{5}$

Answer 7:

We know that 0 divided by any non-zero rational number is always 0.

$∴ 0 \div \frac{3}{5} = 0 (0 \div \frac{a}{b} = 0)$

Hence, the correct answer is option (a).

Question 8:

Mark the correct alternative in each of the following:

$- 2 \frac{3}{7} + 4 = ?$

(a) $\frac{- 11}{7}$ (b) $\frac{11}{7}$ (c) $\frac{- 45}{7}$ (d) $\frac{45}{7}$

Answer 8:

$? = - 2 \frac{3}{7} + 4 \Rightarrow ? = - \frac{17}{7} + \frac{4}{1} \Rightarrow ? = \frac{- 17 \times 1 + 4 \times 7}{7} \Rightarrow ? = \frac{- 17 + 28}{7}$
$\Rightarrow ? = \frac{11}{7}$

Hence, the correct answer is option (b).

Question 9:

Mark the correct alternative in each of the following:

If the product of two non-zero rational numbers is 1, then they are

(a) additve inverse of each other (b) multiplicative inverse of each other

(c) reciprocal of each other (d) both (b) and (c)

Answer 9:

For every non-zero rational number $\frac{a}{b}$ there exists a rational number $\frac{b}{a}$ such that

$\frac{a}{b} \times \frac{b}{a} = 1$

Here, the rational number $\frac{b}{a}$ is called the multiplicative inverse or reciprocal of $\frac{a}{b}$ .

Thus, if the product of two non-zero rational numbers is 1, then they are multiplicative inverse or reciprocal of each other.

Hence, the correct answer is option (d).

Question 10:

Mark the correct alternative in each of the following:

The product $3 \frac{1}{7} \times 1 \frac{5}{6} \times 1 \frac{2}{5} \times 1 \frac{1}{11}$ is equal to

(a) $5 \frac{8}{5}$ (b) $5 \frac{4}{5}$ (c) $8 \frac{4}{5}$ (d) $7 \frac{4}{5}$

Answer 10:

$3 \frac{1}{7} \times 1 \frac{5}{6} \times 1 \frac{2}{5} \times 1 \frac{1}{11} = \frac{22}{7} \times \frac{11}{6} \times \frac{7}{5} \times \frac{12}{11} = \frac{22 \times 11 \times 7 \times 12}{7 \times 6 \times 5 \times 11} (\frac{a}{b} \times \frac{c}{d} = \frac{a \times c}{b \times d}) = \frac{44}{5} = \frac{8 \times 5 + 4}{5} = 8 \frac{4}{5}$

Hence, the correct answer is option (c).

Question 11:

Mark the correct alternative in each of the following:

$\frac{- 7}{13} - (\frac{- 8}{15}) =$

(a) $- \frac{239}{195}$ (b) $\frac{29}{195}$ (c) $\frac{- 29}{195}$ (d) None of these

Answer 11:

$\frac{- 7}{13} - (\frac{- 8}{15}) = \frac{- 7}{13} + \frac{8}{15} [- (\frac{- 8}{15}) = \frac{8}{15}] = \frac{- 7 \times 15 + 8 \times 13}{195} (LCM of 13 and 15 = 195) = \frac{- 105 + 104}{195} = \frac{- 1}{195}$

Hence, the correct answer is option (d).

Question 12:

Mark the correct alternative in each of the following:

$1 \div \frac{1}{3} =$

(a) $\frac{1}{3}$ (b) 3 (c) $1 \frac{1}{3}$ (d) $3 \frac{1}{3}$

Answer 12:

$1 \div \frac{1}{3} = 1 \times 3 (x \div y = x \times \frac{1}{y}) = 3$

Hence, the correct answer is option (b).

Question 13:

Mark the correct alternative in each of the following:

$(- 2) \div (- \frac{5}{3}) =$

(a) $\frac{5}{6}$ (b) $- \frac{5}{6}$ (c) $\frac{6}{5}$ (d) $\frac{- 6}{5}$

Answer 13:

$(- 2) \div (- \frac{5}{3}) = - 2 \times (- \frac{3}{5}) (x \div y = x \times \frac{1}{y}) = \frac{- 2}{1} \times \frac{(- 3)}{5} = \frac{- 2 \times (- 3)}{1 \times 5} (\frac{a}{b} \times \frac{c}{d} = \frac{a \times c}{b \times d}) = \frac{6}{5}$

Hence, the correct answer is option (c).

Question 14:

Mark the correct alternative in each of the following:

If $\frac{x}{2} + \frac{1}{3} = 1$ , then x =

(a) $\frac{3}{4}$ (b) $\frac{4}{3}$ (c) $- \frac{3}{4}$ (d) $\frac{- 4}{3}$

Answer 14:

$\frac{x}{2} + \frac{1}{3} = 1 \Rightarrow \frac{x}{2} = 1 - \frac{1}{3} \Rightarrow \frac{x}{2} = \frac{3 \times 1 - 1}{3} \Rightarrow \frac{x}{2} = \frac{3 - 1}{3} \Rightarrow \frac{x}{2} = \frac{2}{3}$
$\Rightarrow \frac{2 x}{2} = \frac{2 \times 2}{3} (Multiplying both sides by 2) \Rightarrow x = \frac{4}{3}$

Hence, the correct answer is option (b).

Question 15:

Mark the correct alternative in each of the following:

$\frac{5}{4} - \frac{7}{6} - \frac{- 2}{3} =$

(a) $\frac{3}{4}$ (b) $- \frac{3}{4}$ (c) $\frac{- 7}{12}$ (d) $\frac{7}{12}$

Answer 15:

$\frac{5}{4} - \frac{7}{6} - \frac{- 2}{3} = \frac{5}{4} + (\frac{- 7}{6}) + \frac{2}{3} [- (\frac{- 2}{3}) = \frac{2}{3}] = \frac{5 \times 3 + (- 7) \times 2 + 2 \times 4}{12} (LCM of 3, 4 and 6 = 12) = \frac{15 - 14 + 8}{12} = \frac{9}{12}$

$= \frac{9 \div 3}{12 \div 3} (Dividing numerator and denominator by 3) = \frac{3}{4}$

Hence, the correct answer is option (a).