Chapter 1: Rational Numbers

Exercise 1(A)

Question 1(i)

A number which is not rational is called

a natural number
an integers
an irrational number
a whole number

Answer

A number which is not rational is called an irrational number.

Hence, Option 3 is the correct option.

Question 1(ii)

If x ≠ 0 then value of $\dfrac{0}{x}$ is :

a rational number
not a rational number
not equal to zero
none of these.

Answer
If x ≠ 0 then value of $\dfrac{0}{x}$ is a rational number.

Hence, Option 1 is the correct option.

Question 1(iii)

The equation 5x + 7 = 0, gives the value of x which is

an irrational number
a whole number
a rational number
an integers

Answer

Given,

5x + 7 = 0

⇒ 5x = -7

⇒ x = $\dfrac{-7}{5}$

As x is the form of $\dfrac{p}{q}$ where p and q both are integers and q ≠ 0,

∴ x is a rational number.

Hence, Option 3 is the correct option.

Question 1(iv)

Rational number $\dfrac{p}{q}$ is in standard form, if:

p and q have no common factor and p ≠ 0
p and q have at least one common factor other than 1 and q ≠ 0
p and q have no common factor and q ≠ 0
p is divisible by q completely.

Answer

Rational number $\dfrac{p}{q}$ is in standard form, if p and q have no common factor and q ≠ 0

Hence, Option 3 is the correct option.

Question 1(v)

The addition of two rational number $\dfrac{a}{b}$ and $\dfrac{c}{d}$ is commutative, if

$\dfrac{a}{b} + \dfrac{c}{d}$ is rational number
$\dfrac{a+c}{b +d}$ is a rational number.
$\dfrac{a}{b} + \dfrac{c}{d}$ = $\dfrac{c}{d} + \dfrac{a}{b}$
$\dfrac{a}{b} - \dfrac{c}{d}$ = $\dfrac{c}{d} - \dfrac{a}{b}$

Answer

The addition of two rational number $\dfrac{a}{b}$ and $\dfrac{c}{d}$ is commutative, if $\dfrac{a}{b} + \dfrac{c}{d}$ = $\dfrac{c}{d} + \dfrac{a}{b}$

Hence, Option 3 is the correct option.

Question 1(vi)

$-\dfrac{3}{7}$ + additive inverse of $-\dfrac{3}{7}$ is

1
0
$\dfrac{6}{7}$
$-\dfrac{6}{7}$

Answer

Hence, Option 2 is the correct option.

Question 2(i)

Add each pair of rational numbers, given below, and show that their addition (sum) is also a rational number:

$\dfrac{-5}{8}$ and $\dfrac{3}{8}$

Answer

$\dfrac{-5}{8} + \dfrac{3}{8} \\[1em] =\dfrac{-5+3}{8} \\[1em] =\dfrac{-2}{8}$

As $\dfrac{-2}{8}$ is in the form of $\dfrac{p}{q}$ where p and q both are integers and q ≠ 0,

∴ $\dfrac{-2}{8}$ is a rational number.

Question 2(ii)

Add each pair of rational numbers, given below, and show that their addition (sum) is also a rational number:

$\dfrac{-8}{13}$ and $\dfrac{-4}{13}$

Answer

$\dfrac{-8}{13} + \dfrac{-4}{13} \\[1em] =\dfrac{-8+(-4)}{13} \\[1em] =\dfrac{-12}{13}$

As $\dfrac{-12}{13}$ is in the form of $\dfrac{p}{q}$ where p and q both are integers and q ≠ 0,

∴ $\dfrac{-12}{13}$ is a rational number.

Question 2(iii)

Add each pair of rational numbers, given below, and show that their addition (sum) is also a rational number:

$\dfrac{6}{11}$ and $\dfrac{-9}{11}$

Answer

$\dfrac{6}{11} + \dfrac{-9}{11} \\[1em] =\dfrac{6+(-9)}{11} \\[1em] =\dfrac{-3}{11}$

As $\dfrac{-3}{11}$ is in the form of $\dfrac{p}{q}$ where p and q both are integers and q ≠ 0,

∴ $\dfrac{-3}{11}$ is a rational number.

Question 2(iv)

Add each pair of rational numbers, given below, and show that their addition (sum) is also a rational number:

$\dfrac{5}{-26}$ and $\dfrac{8}{39}$

Answer

$\dfrac{5}{-26} + \dfrac{8}{39} \\[1em] \dfrac{-5}{26} + \dfrac{8}{39} \\[1em]$

LCM of 26 and 39 is 2 x 3 x 13 = 78

$\dfrac{-5 \times 3}{26 \times 3} + \dfrac{8 \times 2}{39 \times 2} \\[1em] =\dfrac{-15}{78} + \dfrac{16}{78} \\[1em] =\dfrac{-15+16}{78} \\[1em] =\dfrac{1}{78}$

As $\dfrac{1}{78}$ is in the form of $\dfrac{p}{q}$ where p and q both are integers and q ≠ 0,

∴ $\dfrac{1}{78}$ is a rational number.

Question 2(v)

Add each pair of rational numbers, given below, and show that their addition (sum) is also a rational number:

$\dfrac{5}{-6}$ and $\dfrac{2}{3}$

Answer

$\dfrac{5}{-6} + \dfrac{2}{3} \\[1em] = \dfrac{-5}{6} + \dfrac{2}{3} \\[1em]$

LCM of 6 and 3 is 2 x 3 = 6

$\dfrac{-5 \times 1}{6 \times 1} + \dfrac{2 \times 2}{3 \times 2} \\[1em] =\dfrac{-5}{6} + \dfrac{4}{6} \\[1em] =\dfrac{-5+4}{6} \\[1em] =\dfrac{-1}{6}$

As $\dfrac{-1}{6}$ is in the form of $\dfrac{p}{q}$ where p and q both are integers and q ≠ 0,

∴ $\dfrac{-1}{6}$ is a rational number.

Question 2(vi)

Add each pair of rational numbers, given below, and show that their addition (sum) is also a rational number:

-2 and $\dfrac{2}{5}$

Answer

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

LCM of 1 and 5 is 5

$\dfrac{-2 \times 5}{1 \times 5} + \dfrac{2 \times 1}{5 \times 1} \\[1em] =\dfrac{-10}{5} + \dfrac{2}{5} \\[1em] =\dfrac{-10+2}{5} \\[1em] =\dfrac{-8}{5}$

As $\dfrac{-8}{5}$ is in the form of $\dfrac{p}{q}$ where p and q both are integers and q ≠ 0,

∴ $\dfrac{-8}{5}$ is a rational number.

Question 2(vii)

Add each pair of rational numbers, given below, and show that their addition (sum) is also a rational number:

$\dfrac{9}{-4}$ and $\dfrac{-3}{8}$

Answer

$\dfrac{9}{-4} + \dfrac{-3}{8} \\[1em] \dfrac{-9}{4} + \dfrac{-3}{8} \\[1em]$

LCM of 4 and 8 is 2 x 2 x 2 = 8

$\dfrac{-9 \times 2}{4 \times 2} + \dfrac{-3 \times 1}{8 \times 1} \\[1em] =\dfrac{-18}{8} + \dfrac{-3}{8} \\[1em] =\dfrac{-18+(-3)}{8} \\[1em] =\dfrac{-21}{8}$

As $\dfrac{-21}{8}$ is in the form of $\dfrac{p}{q}$ where p and q both are integers and q ≠ 0,

∴ $\dfrac{-21}{8}$ is a rational number.

Question 2(viii)

Add each pair of rational numbers, given below, and show that their addition (sum) is also a rational number:

$\dfrac{7}{-18}$ and $\dfrac{8}{27}$

Answer $\dfrac{7}{-18} + \dfrac{8}{27} \\[1em] \dfrac{-7}{18} + \dfrac{8}{27} \\[1em]$

LCM of 18 and 27 is 2 x 3 x 3 x 3 = 54

$\dfrac{-7 \times 3}{18 \times 3} + \dfrac{8 \times 2}{27 \times 2} \\[1em] =\dfrac{-21}{54} + \dfrac{16}{54} \\[1em] =\dfrac{-21+16}{54} \\[1em] =\dfrac{-5}{54}$

As $\dfrac{-5}{54}$ is in the form of $\dfrac{p}{q}$ where p and q both are integers and q ≠ 0,

∴ $\dfrac{-5}{54}$ is a rational number.

Question 3(i)

Evaluate:

$\dfrac{5}{9} + \dfrac{-7}{6}$

Answer

LCM of 9 and 6 is 2 x 3 x 3 = 18 $\dfrac{5 \times 2}{9 \times 2} + \dfrac{-7 \times 3}{6 \times 3} \\[1em] =\dfrac{10}{18} + \dfrac{-21}{18} \\[1em] =\dfrac{10 + (-21)}{18} \\[1em] =\dfrac{-11}{18}$

∴ $\dfrac{5}{9} + \dfrac{-7}{6}$ = $\dfrac{-11}{18}$

Question 3(ii)

Evaluate:

$4 + \dfrac{3}{-5}$

Answer

$\dfrac{4}{1} + \dfrac{-3}{5} \\[1em]$

LCM of 1 and 5 is 5

$\dfrac{4 \times 5}{1 \times 5} + \dfrac{-3 \times 1}{5 \times 1} \\[1em] = \dfrac{20}{5} + \dfrac{-3}{5} \\[1em] = \dfrac{20 + (-3)}{5} \\[1em] = \dfrac{17}{5} \\[1em] = 3\dfrac{2}{5}$

∴ $4 + \dfrac{3}{-5} = 3\dfrac{2}{5}$

Question 3(iii)

Evaluate:

$\dfrac{1}{-15} + \dfrac{5}{-12}$

Answer

$\dfrac{-1}{15} + \dfrac{-5}{12} \\[1em]$

LCM of 15 and 12 is 2 x 2 x 3 x 5 = 60

$\dfrac{-1 \times 4}{15 \times 4} + \dfrac{-5 \times 5}{12 \times 5} \\[1em] =\dfrac{-4}{60} + \dfrac{-25}{60} \\[1em] =\dfrac{-4 + (-25)}{60} \\[1em] =\dfrac{-29}{60}$

∴ $\dfrac{1}{-15} + \dfrac{5}{-12} = \dfrac{-29}{60}$

Question 3(iv)

Evaluate:

$\dfrac{5}{9} + \dfrac{3}{-4}$

Answer

$\dfrac{5}{9} + \dfrac{-3}{4} \\[1em]$

LCM of 9 and 4 is 2 x 2 x 3 x 3 = 36

$\dfrac{5 \times 4}{9 \times 4} + \dfrac{-3 \times 9}{4 \times 9} \\[1em] =\dfrac{20}{36} + \dfrac{-27}{36} \\[1em] =\dfrac{20 + (-27)}{36} \\[1em] =\dfrac{-7}{36}$

∴ $\dfrac{5}{9} + \dfrac{3}{-4}$ = $\dfrac{-7}{36}$

Question 3(v)

Evaluate:

$\dfrac{-8}{9} + \dfrac{-5}{12}$

Answer

$\dfrac{-8}{9} + \dfrac{-5}{12} \\[1em]$

LCM of 9 and 12 is 2 x 2 x 3 x 3 = 36

$\dfrac{-8 \times 4}{9 \times 4} + \dfrac{-5 \times 3}{12 \times 3} \\[1em] =\dfrac{-32}{36} + \dfrac{-15}{36} \\[1em] =\dfrac{-32 + (-15)}{36} \\[1em] =\dfrac{-47}{36}$

∴ $\dfrac{-8}{9}$ + $\dfrac{-5}{12}$ = $\dfrac{-47}{36}$

Question 3(vi)

Evaluate:

$0 + \dfrac{-2}{7}$

Answer

$\dfrac{0}{1} + \dfrac{-2}{7}$

LCM of 1 and 7 is 7

$\dfrac{0 \times 7}{1 \times 7} + \dfrac{-2 \times 1}{7 \times 1} \\[1em] =\dfrac{0}{7} + \dfrac{-2}{7} \\[1em] =\dfrac{0 + (-2)}{7} \\[1em] =\dfrac{-2}{7}$

∴ 0 + $\dfrac{-2}{7}$ = $\dfrac{-2}{7}$

Question 3(vii)

Evaluate:

$\dfrac{5}{-11} + 0$

Answer

$\dfrac{-5}{11} + \dfrac{0}{1}$

LCM of 11 and 1 is 11

$\dfrac{-5 \times 1}{11 \times 1} + \dfrac{0 \times 11}{1 \times 11} \\[1em] =\dfrac{-5}{11} + \dfrac{0}{11} \\[1em] =\dfrac{-5 + 0}{11} \\[1em] =\dfrac{-5}{11}$

∴ $\dfrac{5}{-11}$ + 0 = $\dfrac{-5}{11}$

Question 3(viii)

Evaluate:

$2 + \dfrac{-3}{5}$

Answer

$\dfrac{2}{1} + \dfrac{-3}{5}$

LCM of 1 and 5 is 5

$\dfrac{2 \times 5}{1 \times 5} + \dfrac{-3 \times 1}{5 \times 1} \\[1em] =\dfrac{10}{5} + \dfrac{-3}{5} \\[1em] =\dfrac{10 + (-3)}{5} \\[1em] =\dfrac{7}{5}$

∴ 2 + $\dfrac{-3}{5}$ = $\dfrac{7}{5}$

Question 3(ix)

Evaluate:

$\dfrac{4}{-9} + 1$

Answer

$\dfrac{-4}{9} + \dfrac{1}{1}$

LCM of 9 and 1 is 3 x 3 = 9

$\dfrac{-4 \times 1}{9 \times 1} + \dfrac{1 \times 9}{1 \times 9} \\[1em] =\dfrac{-4}{9} + \dfrac{9}{9} \\[1em] =\dfrac{-4 + 9}{9} \\[1em] =\dfrac{5}{9}$

∴ $\dfrac{4}{-9}$ + 1 = $\dfrac{5}{9}$

Question 4(i)

Evaluate:

$\dfrac{3}{7} + \dfrac{-4}{9} + \dfrac{-11}{7} + \dfrac{7}{9}$

Answer

LCM of 7 and 9 is 3 x 3 x 7 = 63

$\dfrac{3 \times 9}{7 \times 9} + \dfrac{-4 \times 7}{9 \times 7} + \dfrac{-11 \times 9}{7 \times 9} + \dfrac{7 \times 7}{9 \times 7} \\[1em] = \dfrac{27}{63} + \dfrac{-28}{63} + \dfrac{-99}{63} + \dfrac{49}{63} \\[1em] = \dfrac{27 + (-28) + (-99) + 49}{63} \\[1em] = \dfrac{-51}{63} \\[1em]$

∴ $\dfrac{3}{7} + \dfrac{-4}{9} + \dfrac{-11}{7} + \dfrac{7}{9} = \dfrac{-51}{63}$

Question 4(ii)

Evaluate:

$\dfrac{2}{3} + \dfrac{-4}{5} + \dfrac{1}{3} + \dfrac{2}{5}$

Answer

LCM of 3 and 5 is 3 x 5 = 15

$\dfrac{2 \times 5}{3 \times 5} + \dfrac{-4 \times 3}{5 \times 3} + \dfrac{1 \times 5}{3 \times 5} + \dfrac{2 \times 3}{5 \times 3} \\[1em] = \dfrac{10}{15} + \dfrac{-12}{15} + \dfrac{5}{15} + \dfrac{6}{15} \\[1em] = \dfrac{10 + (-12) + 5 + 6}{15} \\[1em] = \dfrac{9}{15} \\[1em]$

∴ $\dfrac{2}{3} + \dfrac{-4}{5} + \dfrac{1}{3} + \dfrac{2}{5} = \dfrac{9}{15}$

Question 4(iii)

Evaluate:

$\dfrac{4}{7} + 0 + \dfrac{-8}{9} + \dfrac{-13}{7} + \dfrac{17}{9}$

Answer $\dfrac{4}{7} + \dfrac{0}{1} + \dfrac{-8}{9} + \dfrac{-13}{7} + \dfrac{17}{9}$

LCM of 7 ,1 and 9 is 3 x 3 x 7 = 63

$\dfrac{4 \times 9}{7 \times 9} + \dfrac{0 \times 63}{1 \times 63} + \dfrac{-8 \times 7}{9 \times 7} + \dfrac{-13 \times 9}{7 \times 9} + \dfrac{17 \times 7}{9 \times 7} \\[1em] = \dfrac{36}{63} + \dfrac{0}{63} + \dfrac{-56}{63} + \dfrac{-117}{63} + \dfrac{119}{63} \\[1em] = \dfrac{36 + 0 + (-56) + (-117) + 119}{63} \\[1em] = \dfrac{-18}{63} \\[1em]$

∴ $\dfrac{4}{7} + 0 + \dfrac{-8}{9} + \dfrac{-13}{7} + \dfrac{17}{9} = \dfrac{-18}{63}$

Question 4(iv)

Evaluate:

$\dfrac{3}{8} + \dfrac{-5}{12} + \dfrac{3}{7} + \dfrac{3}{12} + \dfrac{-5}{8} + \dfrac{-2}{7}$

Answer $\Big(\dfrac{3}{8} + \dfrac{-5}{8}\Big) + \Big(\dfrac{-5}{12} + \dfrac{3}{12}\Big) + \Big(\dfrac{3}{7} + \dfrac{-2}{7}\Big) \\[1em] = \dfrac{-2}{8} + \dfrac{-2}{12} + \dfrac{1}{7} \\[1em] = \dfrac{-1}{4} + \dfrac{-1}{6} + \dfrac{1}{7} \\[1em]$

LCM of 4 ,6 and 7 is 2 x 2 x 3 x 7 = 84

$\dfrac{-1 \times 21}{4 \times 21} + \dfrac{-1 \times 14}{6 \times 14} + \dfrac{1 \times 12}{7 \times 12} \\[1em] = \dfrac{-21}{84} + \dfrac{-14}{84} + \dfrac{12}{84} \\[1em] = \dfrac{(-21) + (-14) + 12}{84} \\[1em] = \dfrac{-23}{84} \\[1em]$

∴ $\dfrac{3}{8} + \dfrac{-5}{12} + \dfrac{3}{7} + \dfrac{3}{12} + \dfrac{-5}{8} + \dfrac{-2}{7} = \dfrac{-23}{84}$

Question 5(i)

For each pair of rational number, verify commutative property of addition of rational numbers.

$\dfrac{-8}{7}$ and $\dfrac{5}{14}$

Answer

To prove:

$\dfrac{-8}{7} + \dfrac{5}{14} = \dfrac{5}{14} + \dfrac{-8}{7}$

Taking LHS:

$\dfrac{-8}{7} + \dfrac{5}{14}$

LCM of 7 and 14 is 2 x 7 = 14

$= \dfrac{-8 \times 2}{7 \times 2} + \dfrac{5 \times 1}{14 \times 1} \\[1em] = \dfrac{-16}{14} + \dfrac{5}{14} \\[1em] = \dfrac{-16 + 5}{14} \\[1em] = \dfrac{-11}{14} \\[1em]$

Taking RHS:

$\dfrac{5}{14} + \dfrac{-8}{7}$

LCM of 14 and 7 is 2 x 7 = 14

$= \dfrac{5 \times 1}{14 \times 1} + \dfrac{-8 \times 2}{7 \times 2} \\[1em] = \dfrac{5}{14} + \dfrac{-16}{14} \\[1em] = \dfrac{5 + (-16)}{14} \\[1em] = \dfrac{-11}{14} \\[1em]$

∴ LHS = RHS

Hence, $\dfrac{-8}{7} + \dfrac{5}{14} = \dfrac{5}{14} + \dfrac{-8}{7}$

So, the commutative property for the addition of the rational number is verified.

Question 5(ii)

For each pair of rational number, verify commutative property of addition of rational numbers.

$\dfrac{5}{9}$ and $\dfrac{5}{-12}$

Answer

To prove:

$\dfrac{5}{9} + \dfrac{5}{-12} = \dfrac{5}{-12} + \dfrac{5}{9}$

Taking LHS: $\dfrac{5}{9} + \dfrac{5}{-12} \\[1em] = \dfrac{5}{9} + \dfrac{-5}{12} \\[1em]$

LCM of 9 and 12 is 2 x 2 x 3 x 3 = 36

$= \dfrac{5 \times 4}{9 \times 4} + \dfrac{-5 \times 3}{12 \times 3} \\[1em] = \dfrac{20}{36} + \dfrac{-15}{36} \\[1em] = \dfrac{20 + (-15)}{36} \\[1em] = \dfrac{5}{36} \\[1em]$

Taking RHS: $\dfrac{5}{-12} + \dfrac{5}{9} \\[1em] = \dfrac{-5}{12} + \dfrac{5}{9} \\[1em]$

LCM of 12 and 9 is 2 x 2 x 3 x 3 = 36

$= \dfrac{-5 \times 3}{12 \times 3} + \dfrac{5 \times 4}{9 \times 4} \\[1em] = \dfrac{-15}{36} + \dfrac{20}{36} \\[1em] = \dfrac{(-15) + 20}{36} \\[1em] = \dfrac{5}{36} \\[1em]$

∴ LHS = RHS

Hence, $\dfrac{5}{9} + \dfrac{5}{-12} = \dfrac{5}{-12} + \dfrac{5}{9}$

So, the commutative property for the addition of the rational number is verified.

Question 5(iii)

For each pair of rational number, verify commutative property of addition of rational numbers.

$\dfrac{-4}{5}$ and $\dfrac{-13}{-15}$

Answer

To prove:

$\dfrac{-4}{5} + \dfrac{-13}{-15} = \dfrac{-13}{-15} + \dfrac{-4}{5}$

Taking LHS: $\dfrac{-4}{5} + \dfrac{-13}{-15} \\[1em] = \dfrac{-4}{5} + \dfrac{13}{15} \\[1em]$

LCM of 5 and 15 is 3 x 5 = 15

$= \dfrac{-4 \times 3}{5 \times 3} + \dfrac{13 \times 1}{15 \times 1} \\[1em] = \dfrac{-12}{15} + \dfrac{13}{15} \\[1em] = \dfrac{-12 + 13}{15} \\[1em] = \dfrac{1}{15} \\[1em]$

Taking RHS: $\dfrac{-13}{-15} + \dfrac{-4}{5} \\[1em] = \dfrac{13}{15} + \dfrac{-4}{5} \\[1em]$

LCM of 15 and 5 is 3 x 5 = 15

$= \dfrac{13 \times 1}{15 \times 1} + \dfrac{-4 \times 3}{5 \times 3} \\[1em] = \dfrac{13}{15} + \dfrac{-12}{15} \\[1em] = \dfrac{13 + (-12)}{15} \\[1em] = \dfrac{1}{15} \\[1em]$

∴ LHS = RHS

Hence, $\dfrac{-4}{5} + \dfrac{-13}{-15} = \dfrac{-13}{-15} + \dfrac{-4}{5}$

So, the commutative property for the addition of the rational number is verified.

Question 5(iv)

For each pair of rational number, verify commutative property of addition of rational numbers.

$\dfrac{2}{-5}$ and $\dfrac{11}{-15}$

Answer

To prove:

$\dfrac{2}{-5} + \dfrac{11}{-15} = \dfrac{11}{-15} + \dfrac{2}{-5}$

Taking LHS: $\dfrac{2}{-5} + \dfrac{11}{-15} \\[1em] = \dfrac{-2}{5} + \dfrac{-11}{15} \\[1em]$

LCM of 5 and 15 is 3 x 5 = 15

$= \dfrac{-2 \times 3}{5 \times 3} + \dfrac{-11 \times 1}{15 \times 1} \\[1em] = \dfrac{-6}{15} + \dfrac{-11}{15} \\[1em] = \dfrac{-6 + (-11)}{15} \\[1em] = \dfrac{-17}{15} \\[1em]$

Taking RHS: $\dfrac{11}{-15} + \dfrac{2}{-5} \\[1em] = \dfrac{-11}{15} + \dfrac{-2}{5} \\[1em]$

LCM of 15 and 5 is 3 x 5 = 15

$= \dfrac{-11 \times 1}{15 \times 1} + \dfrac{-2 \times 3}{5 \times 3} \\[1em] = \dfrac{-11}{15} + \dfrac{-6}{15} \\[1em] = \dfrac{(-11) + (-6)}{15} \\[1em] = \dfrac{-17}{15} \\[1em]$

∴ LHS = RHS

Hence, $\dfrac{2}{-5} + \dfrac{11}{-15} = \dfrac{11}{-15} + \dfrac{2}{-5}$

So, the commutative property for the addition of the rational number is verified.

Question 5(v)

For each pair of rational number, verify commutative property of addition of rational numbers.

3 and $\dfrac{-2}{7}$

Answer

To prove:

$3 + \dfrac{-2}{7} = \dfrac{-2}{7} + 3$

Taking LHS: $3 + \dfrac{-2}{7} \\[1em] = \dfrac{3}{1} + \dfrac{-2}{7} \\[1em]$

LCM of 1 and 7 is 7

$= \dfrac{3 \times 7}{1 \times 7} + \dfrac{-2 \times 1}{7 \times 1} \\[1em] = \dfrac{21}{7} + \dfrac{-2}{7} \\[1em] = \dfrac{21 + (-2)}{7} \\[1em] = \dfrac{19}{7} \\[1em]$

Taking RHS: $\dfrac{-2}{7} + 3 \\[1em] = \dfrac{-2}{7} + \dfrac{3}{1} \\[1em]$

LCM of 7 and 1 is 7

$= \dfrac{-2 \times 1}{7 \times 1} + \dfrac{3 \times 7}{1 \times 7} \\[1em] = \dfrac{-2}{7} + \dfrac{21}{7} \\[1em] = \dfrac{(-2) + 21}{7} \\[1em] = \dfrac{19}{7} \\[1em]$

∴ LHS = RHS

Hence, $3 + \dfrac{-2}{7} = \dfrac{-2}{7} + 3$

So, the commutative property for the addition of the rational number is verified.

Question 5(vi)

For each pair of rational number, verify commutative property of addition of rational numbers.

-2 and $\dfrac{3}{-5}$

Answer

To prove:

$-2 + \dfrac{3}{-5} = \dfrac{3}{-5} + -2$

Taking LHS: $-2 + \dfrac{3}{-5} \\[1em] = \dfrac{-2}{1} + \dfrac{-3}{5}$

LCM of 1 and 5 is 5

$= \dfrac{-2 \times 5}{1 \times 5} + \dfrac{-3 \times 1}{5 \times 1} \\[1em] = \dfrac{-10}{5} + \dfrac{-3}{5} \\[1em] = \dfrac{-10 + (-3)}{5} \\[1em] = \dfrac{-13}{5}$

Taking RHS: $\dfrac{3}{-5} + -2 \\[1em] = \dfrac{-3}{5} + \dfrac{-2}{1}$

LCM of 5 and 1 is 5

$= \dfrac{-3 \times 1}{5 \times 1} + \dfrac{-2 \times 5}{1 \times 5} \\[1em] = \dfrac{-3}{5} + \dfrac{-10}{5} \\[1em] = \dfrac{(-3) + (-10)}{5} \\[1em] = \dfrac{-13}{5}$

∴ LHS = RHS

Hence, $-2 + \dfrac{3}{-5} = \dfrac{3}{-5} + -2$

So, the commutative property for the addition of the rational number is verified.

Question 6(i)

For each set of rational numbers, given below, verify the associative property of addition of rational numbers:

$\dfrac{1}{2} , \dfrac{2}{3}$ and $\dfrac{-1}{6}$

Answer

To prove:

$\Big(\dfrac{1}{2} + \dfrac{2}{3}\Big) + \dfrac{-1}{6} = \dfrac{1}{2} + \Big(\dfrac{2}{3} + \dfrac{-1}{6}\Big)$

Taking LHS:

$\Big(\dfrac{1}{2} + \dfrac{2}{3}\Big) + \dfrac{-1}{6} \\[1em]$ LCM of 2 and 3 is 2 x 3 = 6

$= \Big(\dfrac{1 \times 3}{2 \times 3} + \dfrac{2 \times 2}{3 \times 2}\Big) + \dfrac{-1}{6} \\[1em] = \Big(\dfrac{3}{6} + \dfrac{4}{6}\Big) + \dfrac{-1}{6} \\[1em] = \Big(\dfrac{3 + 4}{6}\Big) + \dfrac{-1}{6} \\[1em] = \dfrac{7}{6}+ \dfrac{-1}{6} \\[1em] = \dfrac{7-1}{6} \\[1em] = \dfrac{6}{6} \\[1em] = 1$

Taking RHS: $\dfrac{1}{2} + \Big(\dfrac{2}{3} + \dfrac{-1}{6}\Big)$

LCM of 3 and 6 is 2 x 3 = 6 $\dfrac{1}{2} + \Big(\dfrac{2 \times 2}{3 \times 2} + \dfrac{-1 \times 1}{6 \times 1}\Big) \\[1em] = \dfrac{1}{2} + \Big(\dfrac{4}{6} + \dfrac{-1}{6}\Big) \\[1em] = \dfrac{1}{2} + \Big( \dfrac{4 +(-1)}{6}\Big) \\[1em] = \dfrac{1}{2} + \dfrac{3}{6} \\[1em]$

LCM of 2 and 6 is 2 x 3 = 6

$= \dfrac{1 \times 3}{2 \times 3} + \dfrac{3 \times 1}{6 \times 1} \\[1em] = \dfrac{3}{6} + \dfrac{3}{6} \\[1em] = \dfrac{3 + 3}{6} \\[1em] = \dfrac{6}{6} \\[1em] = 1$

∴ LHS = RHS

$\Big(\dfrac{1}{2} + \dfrac{2}{3}\Big) + \dfrac{-1}{6} = \dfrac{1}{2} + \Big(\dfrac{2}{3} + \dfrac{-1}{6}\Big)$

So, the associative property for the addition of the rational number is verified.

Question 6(ii)

For each set of rational numbers, given below, verify the associative property of addition of rational numbers:

$\dfrac{-2}{5} , \dfrac{4}{15}$ and $\dfrac{-7}{10}$

Answer

To prove:

$\Big(\dfrac{-2}{5} + \dfrac{4}{15}\Big) + \dfrac{-7}{10} = \dfrac{-2}{5} + \Big(\dfrac{4}{15} + \dfrac{-7}{10}\Big)$

Taking LHS:

$\Big(\dfrac{-2}{5} + \dfrac{4}{15}\Big) + \dfrac{-7}{10} \\[1em]$ LCM of 5 and 15 is 3 x 5 = 15

$= \Big(\dfrac{-2 \times 3}{5 \times 3} + \dfrac{4 \times 1}{15 \times 1}\Big) + \dfrac{-7}{10} \\[1em] = \Big(\dfrac{-6}{15} + \dfrac{4}{15}\Big) + \dfrac{-7}{10} \\[1em] = \Big(\dfrac{-6 + 4}{15}\Big) + \dfrac{-7}{10} \\[1em] = \dfrac{-2}{15}+ \dfrac{-7}{10} \\[1em]$ LCM of 15 and 10 is 2 x 3 x 5 = 30

$= \dfrac{-2 \times 2}{15 \times 2} + \dfrac{-7 \times 3}{10 \times 3} \\[1em] = \dfrac{-4}{30} + \dfrac{-21}{30} \\[1em] = \dfrac{-4 + (-21)}{30} \\[1em] = \dfrac{-25}{30} \\[1em] = \dfrac{-5}{6} \\[1em]$

Taking RHS: $\dfrac{-2}{5} + \Big(\dfrac{4}{15} + \dfrac{-7}{10}\Big)$

LCM of 15 and 10 is 2 x 3 x 5 = 30 $\dfrac{-2}{5} + \Big(\dfrac{4 \times 2}{15 \times 2} + \dfrac{-7 \times 3}{10 \times 3}\Big) \\[1em] = \dfrac{-2}{5} + \Big(\dfrac{8}{30} + \dfrac{-21}{30}\Big) \\[1em] = \dfrac{-2}{5} + \Big( \dfrac{8 +(-21)}{30}\Big) \\[1em] = \dfrac{-2}{5} + \dfrac{-13}{30} \\[1em]$

LCM of 5 and 30 is 2 x 3 x 5 = 30

$= \dfrac{-2 \times 6}{5 \times 6} + \dfrac{-13 \times 1}{30 \times 1} \\[1em] = \dfrac{-12}{30} + \dfrac{-13}{30} \\[1em] = \dfrac{-12 + (-13)}{30} \\[1em] = \dfrac{-25}{30} \\[1em] = \dfrac{-5}{6} \\[1em]$

∴ LHS = RHS

$\Big(\dfrac{-2}{5} + \dfrac{4}{15}\Big) + \dfrac{-7}{10} = \dfrac{-2}{5} + \Big(\dfrac{4}{15} + \dfrac{-7}{10}\Big)$

So, the associative property for the addition of the rational number is verified.

Question 6(iii)

For each set of rational numbers, given below, verify the associative property of addition of rational numbers:

$\dfrac{-7}{9} , \dfrac{2}{-3}$ and $\dfrac{-5}{18}$

Answer

To prove: $\Big(\dfrac{-7}{9} + \dfrac{2}{-3}\Big) + \dfrac{-5}{18} = \dfrac{-7}{9} + \Big(\dfrac{2}{-3} + \dfrac{-5}{18}\Big)$

Taking LHS:

$\Big(\dfrac{-7}{9} + \dfrac{2}{-3}\Big) + \dfrac{-5}{18} \\[1em] = \Big(\dfrac{-7}{9} + \dfrac{-2}{3}\Big) + \dfrac{-5}{18} \\[1em]$ LCM of 9 and 3 is 3 x 3 = 9

$= \Big(\dfrac{-7 \times 1}{9 \times 1} + \dfrac{-2 \times 3}{3 \times 3}\Big) + \dfrac{-5}{18} \\[1em] = \Big(\dfrac{-7}{9} + \dfrac{-6}{9}\Big) + \dfrac{-5}{18} \\[1em] = \Big(\dfrac{-7 + (-6)}{9}\Big) + \dfrac{-5}{18} \\[1em] = \dfrac{-13}{9}+ \dfrac{-5}{18} \\[1em]$

LCM of 9 and 18 is 2 x 9 = 18

$= \dfrac{-13 \times 2}{9 \times 2} + \dfrac{-5 \times 1}{18 \times 1} \\[1em] = \dfrac{-26}{18} + \dfrac{-5}{18} \\[1em] = \dfrac{-26 + (-5)}{18} \\[1em] = \dfrac{-31}{18} \\[1em]$

Taking RHS:

$\dfrac{-7}{9} + \Big(\dfrac{2}{-3} + \dfrac{-5}{18}\Big) \\[1em] = \dfrac{-7}{9} + \Big(\dfrac{-2}{3} + \dfrac{-5}{18}\Big) \\[1em]$

LCM of 3 and 18 is 2 x 3 x 9 = 18

$\dfrac{-7}{9} + \Big(\dfrac{-2 \times 6}{3 \times 6} + \dfrac{-5 \times 1}{18 \times1}\Big) \\[1em] = \dfrac{-7}{9} + \Big(\dfrac{-12}{18} + \dfrac{-5}{18}\Big) \\[1em] = \dfrac{-7}{9} + \Big( \dfrac{-12 +(-5)}{18}\Big) \\[1em] = \dfrac{-7}{9} + \dfrac{-17}{18} \\[1em]$

LCM of 9 and 18 is 2 x 3 x 3 = 18

$= \dfrac{-7 \times 2}{9 \times 2} + \dfrac{-17 \times 1}{18 \times 1} \\[1em] = \dfrac{-14}{18} + \dfrac{-17}{18} \\[1em] = \dfrac{-14 + (-17)}{18} \\[1em] = \dfrac{-31}{18} \\[1em]$

∴ LHS = RHS

$\Big(\dfrac{-7}{9} + \dfrac{2}{-3}\Big) + \dfrac{-5}{18} = \dfrac{-7}{9} + \Big(\dfrac{2}{-3} + \dfrac{-5}{18}\Big)$

So, the associative property for the addition of the rational number is verified.

Question 6(iv)

For each set of rational numbers, given below, verify the associative property of addition of rational numbers:

$-1 , \dfrac{5}{6}$ and $\dfrac{-2}{3}$

Answer

To prove: $\Big(-1 + \dfrac{5}{6}\Big) + \dfrac{-2}{3} = -1 + \Big(\dfrac{5}{6} + \dfrac{-2}{3}\Big)$

Taking LHS:

$\Big(-1 + \dfrac{5}{6}\Big) + \dfrac{-2}{3} \\[1em] = \Big(\dfrac{-1}{1} + \dfrac{5}{6}\Big) + \dfrac{-2}{3} \\[1em]$ LCM of 1 and 6 is 2 x 3 = 6

$= \Big(\dfrac{-1 \times 6}{1 \times 6} + \dfrac{5 \times 1}{6 \times 1}\Big) + \dfrac{-2}{3} \\[1em] = \Big(\dfrac{-6}{6} + \dfrac{5}{6}\Big) + \dfrac{-2}{3} \\[1em] = \Big(\dfrac{-6 + 5}{6}\Big) + \dfrac{-2}{3} \\[1em] = \dfrac{-1}{6}+ \dfrac{-2}{3} \\[1em]$

LCM of 6 and 3 is 2 x 3 = 6

$= \dfrac{-1 \times 1}{6 \times 1} + \dfrac{-2 \times 2}{3 \times 2} \\[1em] = \dfrac{-1}{6} + \dfrac{-4}{6} \\[1em] = \dfrac{-1 + (-4)}{6} \\[1em] = \dfrac{-5}{6} \\[1em]$

Taking RHS: $-1 + \Big(\dfrac{5}{6} + \dfrac{-2}{3}\Big) \\[1em] = \dfrac{-1}{1} + \Big(\dfrac{5}{6} + \dfrac{-2}{3}\Big) \\[1em]$

LCM of 6 and 3 is 2 x 3 = 6 $\dfrac{-1}{1} + \Big(\dfrac{5 \times 1}{6 \times 1} + \dfrac{-2 \times 2}{3 \times 2}\Big) \\[1em] = \dfrac{-1}{1} + \Big(\dfrac{5}{6} + \dfrac{-4}{6}\Big) \\[1em] = \dfrac{-1}{1} + \Big( \dfrac{5 +(-4)}{6}\Big) \\[1em] = \dfrac{-1}{1} + \dfrac{1}{6} \\[1em]$

LCM of 1 and 6 is 2 x 3 = 6

$= \dfrac{-1 \times 6}{1 \times 6} + \dfrac{1 \times 1}{6 \times 1} \\[1em] = \dfrac{-6}{6} + \dfrac{1}{6} \\[1em] = \dfrac{-6 + 1}{6} \\[1em] = \dfrac{-5}{6} \\[1em]$

∴ LHS = RHS

$\Big(-1 + \dfrac{5}{6}\Big) + \dfrac{-2}{3} = -1 + \Big(\dfrac{5}{6} + \dfrac{-2}{3}\Big)$

So, the associative property for the addition of the rational number is verified.

Question 7(i)

Write the additive inverse (negative) of:

$\dfrac{-3}{8}$

Answer

Additive inverse of $\dfrac{-3}{8}$ = $-\Big(\dfrac{-3}{8}\Big)$

= $\dfrac{3}{8}$

Question 7(ii)

Write the additive inverse (negative) of:

$\dfrac{4}{-9}$

Answer

$\dfrac{4}{-9}$ = $\dfrac{-4}{9}$

Additive inverse of $\dfrac{-4}{9}$ = $-\Big(\dfrac{-4}{9}\Big)$

= $\dfrac{4}{9}$

Question 7(iii)

Write the additive inverse (negative) of:

$\dfrac{-4}{-13}$

Answer

$\dfrac{-4}{-13}$ = $\dfrac{4}{13}$

Additive inverse of $\dfrac{4}{13}$ = $-\Big(\dfrac{4}{13}\Big)$

= $-\dfrac{4}{13}$

Question 7(iv)

Write the additive inverse (negative) of:

Answer

0 = $\dfrac{0}{1}$

Additive inverse of $\dfrac{0}{1}$ = $-\Big(\dfrac{0}{1}\Big)$

= 0

Question 7(v)

Write the additive inverse (negative) of:

-2

Answer

-2 = $\dfrac{-2}{1}$

Additive inverse of $\dfrac{-2}{1}$ = $-\Big(\dfrac{-2}{1}\Big)$

= $\dfrac{2}{1} = 2$

Question 7(vi)

Write the additive inverse (negative) of:

Answer

1 = $\dfrac{1}{1}$

Additive inverse of $\dfrac{1}{1}$ = $-\Big(\dfrac{1}{1}\Big)$

= - $\dfrac{1}{1} = -1$

Question 8

Fill in the blanks:

(i) Additive inverse of $\dfrac{-5}{-12}$ = ............... .

(ii) $\dfrac{-5}{-12}$ + its additive inverse = ............... .

(iii) If $\dfrac{a}{b}$ is the additive inverse of $\dfrac{-c}{d}$ , then $\dfrac{-c}{d}$ is the additive inverse of ............... .
And, so $\dfrac{a}{b} + \dfrac{-c}{d} = \dfrac{-c}{d} + \dfrac{a}{b}$ = ............... .

Answer

(i) Additive inverse of $\dfrac{-5}{-12} = -\dfrac{5}{12}$ .

(ii) $\dfrac{-5}{-12}$ + its additive inverse = 0

(iii) If $\dfrac{a}{b}$ is the additive inverse of $\dfrac{-c}{d}$ , then $\dfrac{-c}{d}$ is the additive inverse of $\dfrac{a}{b}$ .
And, so $\dfrac{a}{b} + \dfrac{-c}{d} = \dfrac{-c}{d} + \dfrac{a}{b} = 0$ .

Explanation

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

(ii) $\dfrac{5}{12} + \Big(-\dfrac{5}{12}\Big)$

$= \dfrac{5 - 5}{12} \\[1em] = 0$

(iii) The sum of number and its additive inverse = Additive identity.

Question 9

State, true or false:

(i) $\dfrac{7}{9} = \dfrac{7 + 5}{9 + 5}$

(ii) $\dfrac{7}{9} = \dfrac{7 - 5}{9 - 5}$

(iii) $\dfrac{7}{9} = \dfrac{7 \times 5}{9 \times 5}$

(iv) $\dfrac{7}{9} = \dfrac{7 ÷ 5}{9 ÷ 5}$

(v) $\dfrac{-5}{-12}$ is a negative rational number.

(vi) $\dfrac{-13}{25}$ is smaller than $\dfrac{-25}{13}$

Answer

(i) False.

Reason:

$\dfrac{7 + 5}{9 + 5} = \dfrac{12}{14}$ ≠ $\dfrac{7}{9}$

(ii) False

Reason:

$\dfrac{7 - 5}{9 - 5} = \dfrac{2}{4}$ ≠ $\dfrac{7}{9}$

(iii) True

Reason:

$\dfrac{7 \times 5}{9 \times 5} = \dfrac{7 \times \cancel{5}}{9 \times \cancel{5}} = \dfrac{7}{9}$

(iv) True

Reason:

$\dfrac{7 ÷ 5}{9 ÷ 5} \\[1em] = \dfrac{7 \times \dfrac{1}{5}}{9 \times \dfrac{1}{5}} \\[1em] = \dfrac{7 \times \cancel{\dfrac{1}{5}}}{9 \times \cancel{\dfrac{1}{5}}} \\[1em] = \dfrac{7}{9}$

(v) False

Reason:

$\dfrac{-5}{-12} = \dfrac{5}{12}$ is a positive rational number.

(vi) False

Reason:

We need to check if $\dfrac{-13}{25}$ is smaller than $\dfrac{-25}{13}$

LCM of 25 and 13 is 325

$\dfrac{-13}{25} = \dfrac{-13 \times 13}{25 \times 13} = \dfrac{-169}{325}$

$\dfrac{-25}{13} = \dfrac{-25 \times 25}{13 \times 25} = \dfrac{-625}{325}$

And, $\dfrac{-169}{325}$ > $\dfrac{-625}{325}$

Hence, $\dfrac{-13}{25}$ > $\dfrac{-25}{13}$

Question 10

The weight of an empty fruit basket is $2\dfrac{1}{3}$ kg. It contains $5\dfrac{5}{6}$ kg grapes and $8\dfrac{3}{8}$ mangoes. Find the total weight of basket with fruits.

Answer

The weight of an empty fruit basket = $2\dfrac{1}{3}$ kg

The weight of grapes = $5\dfrac{5}{6}$ kg

The weight of mangoes = $8\dfrac{3}{8}$ kg

Total weight of basket with fruits = Weight of empty fruit basket + weight of grapes + weight of mangoes

$= 2\dfrac{1}{3} \text{ kg} + 5 \dfrac{5}{6} \text{ kg} + 8 \dfrac{3}{8} \text{ kg} \\[1em] =\dfrac{7}{3} \text{ kg} + \dfrac{35}{6} \text{ kg} + \dfrac{67}{8} \text{ kg} \\[1em]$

LCM of 3, 6 and 8 is 2 x 2 x 2 x 3 = 24

$= \dfrac{7 \times 8}{3 \times 8} + \dfrac{35 \times 4}{6 \times 4} + \dfrac{67 \times 3}{8 \times 3} \text{ kg}\\[1em] = \dfrac{56}{24} + \dfrac{140}{24} + \dfrac{201}{24} \text{ kg}\\[1em] = \dfrac{56 + 140 + 201}{24} \text{ kg}\\[1em] = \dfrac{397}{24} \text{ kg}\\[1em] = 16\dfrac{13}{24} \text{ kg}$

Total weight of basket with fruits = $16\dfrac{13}{24}$

Exercise 1(B)

Question 1(i)

The sum of two rational numbers is 8, if one of them is $2\dfrac{3}{4}$ , the other number is

$6\dfrac{3}{4}$
$6\dfrac{1}{4}$
$5\dfrac{1}{4}$
$5\dfrac{3}{4}$

Answer

Let $x$ be the other number. $2\dfrac{3}{4} + x = 8 \\[1em] ⇒\dfrac{11}{4} + x = 8 \\[1em] ⇒ x = 8 - \dfrac{11}{4} \\[1em] ⇒ x = \dfrac{8}{1} - \dfrac{11}{4}$

LCM of 1 and 4 is 2 x 2 = 4

$⇒ x = \dfrac{8 \times 4}{1 \times 4} - \dfrac{11 \times 1}{4 \times 1} \\[1em] ⇒ x = \dfrac{32}{4} - \dfrac{11}{4} \\[1em] ⇒ x = \dfrac{32 - 11}{4} \\[1em] ⇒ x = \dfrac{21}{4}\\[1em] ⇒ x = 5\dfrac{1}{4}$

The sum of two rational numbers is 8, if one of them is $2\dfrac{3}{5}$ , the other number is $5\dfrac{1}{4}$ .

Hence, option 3 is correct option.

Question 1(ii)

For three rational numbers $\dfrac{a}{b}$ , $\dfrac{c}{d}$ and $\dfrac{e}{f}$ , we have:

$\dfrac{a}{b} - \Big(\dfrac{c}{d} - \dfrac{e}{f}\Big) = \dfrac{a}{b} - \dfrac{c}{d} + \dfrac{e}{f}$
$\dfrac{a}{b} - \Big(\dfrac{c}{d} - \dfrac{e}{f}\Big) = \dfrac{a}{b} - \dfrac{c}{d} - \dfrac{e}{f}$
$\dfrac{a}{b} + \Big(\dfrac{c}{d} + \dfrac{e}{f}\Big)$ ≠ $\Big(\dfrac{a}{b} + \dfrac{c}{d}\Big) + \dfrac{e}{f}$
$\dfrac{a}{b} + \Big(\dfrac{c}{d} + \dfrac{e}{f}\Big) = \Big(\dfrac{a}{b} + \dfrac{c}{d}\Big) + \dfrac{a}{b} + \dfrac{e}{f}$

Answer

$\dfrac{a}{b} - \Big(\dfrac{c}{d} - \dfrac{e}{f}\Big) = \dfrac{a}{b} - \dfrac{c}{d} + \dfrac{e}{f}$

Hence, option 1 is correct option.

Question 1(iii)

The sum of two rational numbers is -6. If one of them is $4\dfrac{1}{2}$ , the other number is:

$2\dfrac{1}{2}$
$1\dfrac{1}{2}$
- $1\dfrac{1}{2}$
- $10\dfrac{1}{2}$

Answer

Let $x$ be the other number. $4\dfrac{1}{2} + x = -6 \\[1em] ⇒\dfrac{9}{2} + x = -6 \\[1em] ⇒ x = -6 - \dfrac{9}{2} \\[1em] ⇒ x = \dfrac{-6}{1} - \dfrac{9}{2}$

LCM of 1 and 2 is 2

$⇒ x = -\dfrac{6 \times 2}{1 \times 2} - \dfrac{9\times 1}{2 \times 1} \\[1em] ⇒ x = -\dfrac{12}{2} - \dfrac{9}{2} \\[1em] ⇒ x = \dfrac{-12 - 9}{2} \\[1em] ⇒ x = \dfrac{-21}{2} \\[1em] ⇒ x = -10\dfrac{1}{2}$

The sum of two rational numbers is -6, if one of them is $4\dfrac{1}{2}$ , the other number is - $10\dfrac{1}{2}$ .

Hence, option 4 is correct option.

Question 1(iv)

The number subtracted from $5\dfrac{2}{3}$ to get - $1\dfrac{2}{3}$ is :

4
- $7\dfrac{1}{3}$
$6\dfrac{1}{3}$
$7\dfrac{1}{3}$

Answer

Let $x$ be subtracted from $5\dfrac{2}{3}$ .

$5\dfrac{2}{3} - x = -1\dfrac{2}{3}\\[1em] ⇒\dfrac{17}{3} - x = -\dfrac{5}{3}\\[1em] ⇒ x = \dfrac{17}{3} + \dfrac{5}{3}\\[1em] ⇒ x = \dfrac{17 + 5}{3}\\[1em] ⇒ x = \dfrac{22}{3}\\[1em] ⇒ x = 7\dfrac{1}{3}$

The number subtracted from $5\dfrac{2}{3}$ to get - $1\dfrac{2}{3}$ is $7\dfrac{1}{3}$ .

Hence, option 4 is correct option.

Question 1(v)

The number added to $5\dfrac{2}{3}$ to get - $1\dfrac{2}{3}$ is :

4
- $7\dfrac{1}{3}$
$6\dfrac{1}{3}$
$7\dfrac{1}{3}$

Answer

Let $x$ be added to $5\dfrac{2}{3}$ . $5\dfrac{2}{3} + x = -1\dfrac{2}{3}\\[1em] ⇒\dfrac{17}{3} + x = -\dfrac{5}{3}\\[1em] ⇒ x = \dfrac{-5}{3} - \dfrac{17}{3}\\[1em] ⇒ x = \dfrac{-5 - 17}{3}\\[1em] ⇒ x = \dfrac{-22}{3}\\[1em] ⇒ x = -7\dfrac{1}{3}$

The number added to $5\dfrac{2}{3}$ to get -1 $\dfrac{2}{3}$ is - $7\dfrac{1}{3}$ .

Hence, option 2 is correct option.

Question 2(i)

Evaluate:

$\dfrac{2}{3} - \dfrac{4}{5}$

Answer

$\dfrac{2}{3} - \dfrac{4}{5}$

LCM of 3 and 5 is 3 x 5 = 15

$\dfrac{2 \times 5}{3 \times 5} - \dfrac{4 \times 3}{5 \times 3}\\[1em] = \dfrac{10}{15} - \dfrac{12}{15}\\[1em] = \dfrac{10 - 12}{15}\\[1em] = \dfrac{-2}{15}$

Question 2(ii)

Evaluate:

$\dfrac{-4}{9} - \dfrac{2}{-3}$

Answer

$\dfrac{-4}{9} - \dfrac{2}{-3}\\[1em] =\dfrac{-4}{9} - \dfrac{-2}{3}$

LCM of 9 and 3 is 3 x 3 = 9

$\dfrac{-4 \times 1}{9 \times 1} - \dfrac{-2 \times 3}{3 \times 3}\\[1em] = \dfrac{-4}{9} - \dfrac{-6}{9}\\[1em] = \dfrac{-4 - (-6)}{9}\\[1em] = \dfrac{-4 + 6}{9}\\[1em] = \dfrac{2}{9}$

Question 2(iii)

Evaluate:

$-1 - \dfrac{4}{9}$

Answer

$-1 - \dfrac{4}{9} \\[1em] =\dfrac{-1}{1} - \dfrac{4}{9} \\[1em]$

LCM of 1 and 9 is 3 x 3 = 9

$\dfrac{-1 \times 9}{1 \times 9} - \dfrac{4 \times 1}{9 \times 1}\\[1em] = \dfrac{-9}{9} - \dfrac{4}{9}\\[1em] = \dfrac{-9 - 4}{9}\\[1em] = \dfrac{-13}{9}\\[1em] = -1\dfrac{4}{9}$

Question 2(iv)

Evaluate:

$\dfrac{-2}{7} - \dfrac{3}{-14}$

Answer

$\dfrac{-2}{7} - \dfrac{3}{-14}\\[1em] =\dfrac{-2}{7} - \dfrac{-3}{14}$

LCM of 7 and 14 is 2 x 14 = 14

$\dfrac{-2 \times 2}{7 \times 2} - \dfrac{-3 \times 1}{14 \times 1}\\[1em] = \dfrac{-4}{14} - \dfrac{-3}{14}\\[1em] = \dfrac{-4 - (-3)}{14}\\[1em] = \dfrac{-4 + 3}{14}\\[1em] = \dfrac{-1}{14}$

Question 2(v)

Evaluate:

$\dfrac{-5}{18} - \dfrac{-2}{9}$

Answer

$\dfrac{-5}{18} - \dfrac{-2}{9}\\[1em]$

LCM of 18 and 9 is 2 x 3 x 3 = 18

$\dfrac{-5 \times 1}{18 \times 1} - \dfrac{-2 \times 2}{9 \times 2}\\[1em] = \dfrac{-5}{18} - \dfrac{-4}{18}\\[1em] = \dfrac{-5 - (-4)}{18}\\[1em] = \dfrac{-5 + 4}{18}\\[1em] = \dfrac{-1}{18}$

Question 2(vi)

Evaluate:

$\dfrac{5}{21} - \dfrac{-13}{42}$

Answer

$\dfrac{5}{21} - \dfrac{-13}{42}$

LCM of 21 and 42 is 2 x 3 x 7 = 42

$\dfrac{5 \times 2}{21 \times 2} - \dfrac{-13 \times 1}{42 \times 1}\\[1em] = \dfrac{10}{42} - \dfrac{-13}{42}\\[1em] = \dfrac{10 - (-13)}{42}\\[1em] = \dfrac{10 + 13}{42}\\[1em] = \dfrac{23}{42}$

Question 3(i)

Subtract:

$\dfrac{5}{8}$ from $\dfrac{-3}{8}$

Answer

$\dfrac{-3}{8} - \dfrac{5}{8}\\[1em] = \dfrac{-3 - 5}{8}\\[1em] = \dfrac{-8}{8}\\[1em] = -1$

Question 3(ii)

Subtract:

$\dfrac{-8}{11}$ from $\dfrac{4}{11}$

Answer

$\dfrac{4}{11} - \dfrac{-8}{11}\\[1em] = \dfrac{4 - (-8)}{11}\\[1em] = \dfrac{4 + 8}{11}\\[1em] = \dfrac{12}{11}\\[1em] = 1 \dfrac{1}{11}$

Question 3(iii)

Subtract:

$\dfrac{4}{9}$ from $\dfrac{-5}{9}$

Answer

$\dfrac{-5}{9} - \dfrac{4}{9}\\[1em] = \dfrac{-5 - 4}{9}\\[1em] = \dfrac{-9}{9} = - 1$

Question 3(iv)

Subtract:

$\dfrac{1}{4}$ from $\dfrac{-3}{8}$

Answer

$\dfrac{-3}{8} - \dfrac{1}{4}$

LCM of 8 and 4 is 2 x 2 x 2 = 8

$\dfrac{-3 \times 1}{8 \times 1} - \dfrac{1 \times 2}{4 \times 2}\\[1em] = \dfrac{-3}{8} - \dfrac{2}{8}\\[1em] = \dfrac{-3 - 2}{8}\\[1em] = -\dfrac{5}{8}$

Question 3(v)

Subtract:

$\dfrac{-5}{8}$ from $\dfrac{-13}{16}$

Answer

$\dfrac{-13}{16} - \dfrac{-5}{8}$

LCM of 16 and 8 is 2 x 2 x 2 x 2 = 16

$\dfrac{-13 \times 1}{16 \times 1} - \dfrac{-5 \times 2}{8 \times 2}\\[1em] = \dfrac{-13}{16} - \dfrac{-10}{16}\\[1em] = \dfrac{-13 - (-10)}{16}\\[1em] = \dfrac{-13 + 10}{16}\\[1em] = \dfrac{-3}{16}$

Question 3(vi)

Subtract:

$\dfrac{-9}{22}$ from $\dfrac{5}{33}$

Answer

$\dfrac{5}{33} - \dfrac{-9}{22}$

LCM of 33 and 22 is 2 x 3 x 11 = 66

$\dfrac{5 \times 2}{33 \times 2} - \dfrac{-9 \times 3}{22 \times 3}\\[1em] = \dfrac{10}{66} - \dfrac{-27}{66}\\[1em] = \dfrac{10 - (-27)}{66}\\[1em] = \dfrac{10 + 27}{66}\\[1em] = \dfrac{37}{66}$

Question 4

The sum of two rational numbers is $\dfrac{9}{20}$ . If one of them is $\dfrac{2}{5}$ , find the other.

Answer

Let $x$ be the other number. $\dfrac{2}{5} + x = \dfrac{9}{20} \\[1em] ⇒ x = \dfrac{9}{20} - \dfrac{2}{5}$

LCM of 20 and 5 is 2 x 2 x 5 = 20

$⇒ x = \dfrac{9 \times 1}{20 \times 1} - \dfrac{2 \times 4}{5 \times 4}\\[1em] ⇒ x = \dfrac{9}{20} - \dfrac{8}{20}\\[1em] ⇒ x = \dfrac{9 - 8}{20}\\[1em] ⇒ x = \dfrac{1}{20}$

The sum of two rational numbers is $\dfrac{9}{20}$ , if one of them is $\dfrac{2}{5}$ , the other number is $\dfrac{1}{20}$

Question 5

The sum of two rational numbers is $\dfrac{-2}{3}$ . If one of them is $\dfrac{-8}{15}$ , find the other.

Answer

Let $x$ be the other number. $\dfrac{-8}{15} + x = \dfrac{-2}{3}\\[1em] ⇒ x = \dfrac{-2}{3} - \dfrac{-8}{15}$

LCM of 3 and 15 is 3 x 5 = 15

$⇒ x = \dfrac{-2 \times 5}{3 \times 5} - \dfrac{-8 \times 1}{15 \times 1}\\[1em] ⇒ x = \dfrac{-10}{15} - \dfrac{-8}{15}\\[1em] ⇒ x = \dfrac{-10 - (-8)}{15}\\[1em] ⇒ x = \dfrac{-10 + 8}{15}\\[1em] ⇒ x = \dfrac{-2}{15}$

The sum of two rational numbers is $\dfrac{-2}{3}$ , if one of them is $\dfrac{-8}{15}$ , the other number is $\dfrac{-2}{15}$ .

Question 6

The sum of two rational numbers is -6. If one of them is $\dfrac{-8}{5}$ , find the other.

Answer

Let $x$ be the other number. $\dfrac{-8}{5} + x = -6 \\[1em] \Rightarrow x = \dfrac{-6}{1} - \dfrac{-8}{5}$

LCM of 1 and 5 is 5

$\Rightarrow x = \dfrac{-6 \times 5}{1 \times 5} - \dfrac{-8 \times 1}{5 \times 1}\\[1em] \Rightarrow x = \dfrac{-30}{5} - \dfrac{-8}{5}\\[1em] \Rightarrow x = \dfrac{-30 - (-8)}{5}\\[1em] \Rightarrow x = \dfrac{-30 + 8}{5}\\[1em] \Rightarrow x = \dfrac{-22}{5}\\[1em] \Rightarrow x = -4\dfrac{2}{5}$

The sum of two rational numbers is -6, if one of them is $\dfrac{-8}{5}$ , the other number is $-4\dfrac{2}{5}$ .

Question 7

Which rational number should be added to $\dfrac{-7}{8}$ to get $\dfrac{5}{9}$ ?

Answer

Let $x$ be added to $\dfrac{-7}{8}$ .

$\dfrac{-7}{8} + x = \dfrac{5}{9}\\[1em] ⇒ x = \dfrac{5}{9} - \dfrac{-7}{8}$

LCM of 9 and 8 is 2 x 2 x 2 x 3 x 3 = 72

$\Rightarrow x = \dfrac{5 \times 8}{9 \times 8} - \dfrac{-7 \times 9}{8 \times 9}\\[1em] \Rightarrow x = \dfrac{40}{72} - \dfrac{-63}{72}\\[1em] \Rightarrow x = \dfrac{40 - (-63)}{72}\\[1em] \Rightarrow x = \dfrac{40 + 63}{72}\\[1em] \Rightarrow x = \dfrac{103}{72}\\[1em] \Rightarrow x = 1 \dfrac{31}{72}$

The number added to $\dfrac{-7}{8}$ to get $\dfrac{5}{9}$ is $1\dfrac{31}{72}$ .

Question 8

Which rational number should be added to $\dfrac{-5}{9}$ to get $\dfrac{-2}{3}$ ?

Answer

Let $x$ be added to $\dfrac{-5}{9}$ .

$\dfrac{-5}{9} + x = \dfrac{-2}{3}\\[1em] ⇒ x = \dfrac{-2}{3} - \dfrac{-5}{9}$

LCM of 3 and 9 is 3 x 3 = 9

$\Rightarrow x = \dfrac{-2 \times 3}{3 \times 3} - \dfrac{-5 \times 1}{9 \times 1}\\[1em] \Rightarrow x = \dfrac{-6}{9} - \dfrac{-5}{9}\\[1em] \Rightarrow x = \dfrac{-6 - (-5)}{9}\\[1em] \Rightarrow x = \dfrac{-6 + 5}{9}\\[1em] \Rightarrow x = \dfrac{-1}{9}$

The number added to $\dfrac{-5}{9}$ to get $\dfrac{-2}{3}$ is $\dfrac{-1}{9}$ .

Question 9

Which rational number should be subtracted from $\dfrac{-5}{6}$ to get $\dfrac{4}{9}$ ?

Answer

Let $x$ be subtracted from $\dfrac{-5}{6}$ .

$\dfrac{-5}{6} - x = \dfrac{4}{9}\\[1em] \Rightarrow x = \dfrac{-5}{6} - \dfrac{4}{9}$

LCM of 6 and 9 is 2 x 3 x 3 = 18

$\Rightarrow x = \dfrac{-5 \times 3}{6 \times 3} - \dfrac{4 \times 2}{9 \times 2}\\[1em] \Rightarrow x = \dfrac{-15}{18} - \dfrac{8}{18}\\[1em] \Rightarrow x = \dfrac{-15 - 8}{18}\\[1em] \Rightarrow x = \dfrac{-23}{18}\\[1em] \Rightarrow x = -1\dfrac{5}{18}$

The number subtracted from $\dfrac{-5}{6}$ to get $\dfrac{4}{9}$ is - $1\dfrac{5}{18}$ .

Question 10(i)

What should be subtracted from -2 to get $\dfrac{3}{8}$ ?

Answer

Let $x$ be subtracted from -2.

$-2 - x = \dfrac{3}{8}\\[1em] \Rightarrow \dfrac{-2}{1} - x = \dfrac{3}{8}\\[1em] \Rightarrow x = \dfrac{-2}{1} - \dfrac{3}{8}$

LCM of 1 and 8 is 2 x 2 x 2 = 8

$\Rightarrow x = \dfrac{-2 \times 8}{1 \times 8} - \dfrac{3 \times 1}{8 \times 1}\\[1em] \Rightarrow x = \dfrac{-16}{8} - \dfrac{3}{8}\\[1em] \Rightarrow x = \dfrac{-16 - 3}{8}\\[1em] \Rightarrow x = \dfrac{-19}{8}\\[1em] \Rightarrow x = -2 \dfrac{3}{8}$

The number subtracted from -2 to get $\dfrac{3}{8}$ is - $2\dfrac{3}{8}$ .

Question 10(ii)

What should be added to -2 to get $\dfrac{3}{8}$ ?

Answer

Let $x$ be added to -2.

$-2 + x = \dfrac{3}{8}\\[1em] \Rightarrow \dfrac{-2}{1} + x = \dfrac{3}{8}\\[1em] \Rightarrow x = \dfrac{3}{8} - \dfrac{-2}{1}$

LCM of 8 and 1 is 2 x 2 x 2 = 8

$\Rightarrow x = \dfrac{3 \times 1}{8 \times 1} - \dfrac{-2 \times 8}{1 \times 8}\\[1em] \Rightarrow x = \dfrac{3}{8} - \dfrac{-16}{8}\\[1em] \Rightarrow x = \dfrac{3 - (-16)}{8}\\[1em] \Rightarrow x = \dfrac{3 + 16}{8}\\[1em] \Rightarrow x = \dfrac{19}{8}\\[1em] \Rightarrow x = 2\dfrac{3}{8}$

The number added to -2 to get $\dfrac{3}{8}$ is $2\dfrac{3}{8}$ .

Question 11(i)

Evaluate:

$\dfrac{3}{7} + \dfrac{-4}{9} - \dfrac{-11}{7} - \dfrac{7}{9}$

Answer

$\Big(\dfrac{3}{7} - \dfrac{-11}{7}\Big) + \Big(\dfrac{-4}{9} - \dfrac{7}{9}\Big)\\[1em] =\Big(\dfrac{3 - (-11)}{7}\Big) + \Big(\dfrac{-4 - 7}{9}\Big)\\[1em] =\Big(\dfrac{14}{7}\Big) + \Big(\dfrac{-11}{9}\Big)\\[1em] =\Big(\dfrac{2}{1}\Big) + \Big(\dfrac{-11}{9}\Big)$

LCM of 1 and 9 is 3 x 3 = 9

$=\dfrac{2 \times 9}{1 \times 9} + \dfrac{-11 \times 1}{9 \times 1}\\[1em] =\dfrac{18}{9} + \dfrac{-11}{9}\\[1em] =\dfrac{18 + (-11)}{9}\\[1em] =\dfrac{7}{9}$

Question 11(ii)

Evaluate:

$\dfrac{2}{3} + \dfrac{-4}{5} - \dfrac{1}{3} - \dfrac{2}{5}$

Answer

$\Big(\dfrac{2}{3} - \dfrac{1}{3}\Big) + \Big(\dfrac{-4}{5} - \dfrac{2}{5}\Big)\\[1em] =\Big(\dfrac{2 - 1}{3}\Big) + \Big(\dfrac{-4 - 2}{5}\Big)\\[1em] =\Big(\dfrac{1}{3}\Big) + \Big(\dfrac{-6}{5}\Big)$

LCM of 3 and 5 is 3 x 5 = 15

$=\dfrac{1 \times 5}{3 \times 5} + \dfrac{-6 \times 3}{5 \times 3}\\[1em] =\dfrac{5}{15} + \dfrac{-18}{15}\\[1em] =\dfrac{5 + (-18)}{15}\\[1em] =\dfrac{-13}{15}$

Question 11(iii)

Evaluate

$\dfrac{4}{7} - \dfrac{-8}{9} - \dfrac{-13}{7} + \dfrac{17}{9}$

Answer

$\Big(\dfrac{4}{7} - \dfrac{-13}{7}\Big) + \Big(-\dfrac{-8}{9} + \dfrac{17}{9}\Big)\\[1em] =\Big(\dfrac{4 - (-13)}{7}\Big) + \Big(\dfrac{8 + 17}{9}\Big)\\[1em] =\Big(\dfrac{4 + 13}{7}\Big) + \Big(\dfrac{25}{9}\Big)\\[1em] =\Big(\dfrac{17}{7}\Big) + \Big(\dfrac{25}{9}\Big)$

LCM of 7 and 9 is 3 x 3 x 7 = 63

$=\dfrac{17 \times 9}{7 \times 9} + \dfrac{25 \times 7}{9 \times 7}\\[1em] =\dfrac{153}{63} + \dfrac{175}{63}\\[1em] =\dfrac{153 + 175}{63}\\[1em] =\dfrac{328}{63}\\[1em] =5\dfrac{13}{63}$

Exercise 1(C)

Question 1(i)

The number which on multiplying with $5\dfrac{2}{3}$ gives - $1\dfrac{2}{3}$ is:

$-\dfrac{5}{7}$
$\dfrac{5}{7}$
$-\dfrac{5}{17}$
$\dfrac{17}{5}$

Answer

Let the number be $x$ . $5\dfrac{2}{3} × x = -1\dfrac{2}{3}\\[1em] ⇒ \dfrac{17}{3} × x = -\dfrac{5}{3}\\[1em] ⇒ x = -\dfrac{5}{3} ÷ \dfrac{17}{3}\\[1em] ⇒ x = -\dfrac{5}{3} × \dfrac{3}{17}\\[1em] ⇒ x = -\dfrac{5 \times 3}{3 \times 17}\\[1em] ⇒ x = -\dfrac{15}{51}\\[1em] ⇒ x = -\dfrac{5}{17}$

Hence, option 3 is the correct option.

Question 1(ii)

If a, b and c are three rational numbers, we have:

(a x b) x c = (a x c) x (b x c)
(a + b) x c = (a + c) x (b + c)
a x (b - c) = a x b - a x c
a x (b - c) = a x b - c

Answer

We know that, multiplication of rational numbers is distributive over their addition/subtraction.

∴ a x (b - c) = a x b - a x c

Hence, option 3 is the correct option.

Question 1(iii)

The product of a positive rational number and its reciprocal is:

0
-1
1
none

Answer

Let a positive rational number be $\dfrac{a}{b}$ .

$\dfrac{a}{b} \times \dfrac{b}{a}\\[1em] = \dfrac{a \times b}{b \times a}\\[1em] = \dfrac{ab}{ab}\\[1em] = 1$

Hence, option 3 is the correct option.

Question 1(iv)

The area of a rectangular paper is $7\dfrac{1}{3}$ cm². If its length $4\dfrac{2}{5}$ cm, its breadth is:

$1\dfrac{3}{5}$ cm
$1\dfrac{2}{3}$ cm
$\dfrac{3}{5}$ cm
$\dfrac{25}{60}$ cm

Answer

Let breadth be b

Length = $4\dfrac{2}{5}$ cm = $\dfrac{22}{5}$ cm.

Area = $7\dfrac{1}{3}$ cm² = $\dfrac{22}{3}$ cm².

Area of rectangle = length x breadth $\dfrac{22}{3} = \dfrac{22}{5} \times b\\[1em] ⇒ b = \dfrac{22}{3} ÷ \dfrac{22}{5}\\[1em] ⇒ b = \dfrac{22}{3} \times \dfrac{5}{22}\\[1em] ⇒ b = \dfrac{22 \times 5}{3 \times 22} \\[1em] ⇒ b = \dfrac{110}{66}\\[1em] ⇒ b = \dfrac{5}{3}\\[1em] ⇒ b = 1\dfrac{2}{3}$

Hence, option 2 is the correct option.

Question 1(v)

The product of a rational number $\dfrac{2}{7}$ and its additive inverse is:

$\dfrac{4}{49}$
$\dfrac{-4}{7}$
0
1

Answer

Rational number = $\dfrac{2}{7}$

Additive inverse = $-\dfrac{2}{7}$

$\dfrac{2}{7} \times -\dfrac{2}{7}\\[1em] =-\dfrac{2 \times 2}{7 \times 7}\\[1em] =-\dfrac{4}{49}$

Hence, none of the options are correct.

Question 2(i)

Evaluate:

$\dfrac{-14}{5} \times \dfrac{-6}{7}$

Answer

$\dfrac{-14}{5} \times \dfrac{-6}{7}\\[1em] =\dfrac{-14 \times (-6)}{5 \times 7}\\[1em] =\dfrac{84}{35}\\[1em] =\dfrac{12}{5}\\[1em] =2\dfrac{2}{5}$

Hence, $\dfrac{-14}{5} \times \dfrac{-6}{7} = 2\dfrac{2}{5}$

Question 2(ii)

Evaluate:

$\dfrac{7}{6} \times \dfrac{-18}{91}$

Answer

$\dfrac{7}{6} \times \dfrac{-18}{91}\\[1em] =\dfrac{7 \times (-18)}{6 \times 91}\\[1em] =\dfrac{-126}{546}\\[1em] =\dfrac{-3}{13}$

Hence, $\dfrac{7}{6} \times \dfrac{-18}{91} = \dfrac{-3}{13}$

Question 2(iii)

Evaluate:

$\dfrac{-125}{72} \times \dfrac{9}{-5}$

Answer

$\dfrac{-125}{72} \times \dfrac{9}{-5}\\[1em] =\dfrac{-125 \times 9}{72 \times (-5)}\\[1em] =\dfrac{-1125}{-360}\\[1em] =\dfrac{25}{8}\\[1em] =3\dfrac{1}{8}$

Hence, $\dfrac{-125}{72} \times \dfrac{9}{-5} = 3\dfrac{1}{8}$

Question 2(iv)

Evaluate:

$\dfrac{-11}{9} \times \dfrac{-51}{-44}$

Answer

$\dfrac{-11}{9} \times \dfrac{-51}{-44}\\[1em] =\dfrac{-11 \times (-51)}{9 \times -44}\\[1em] =\dfrac{561}{-396}\\[1em] =-\dfrac{17}{12}\\[1em] =-1\dfrac{5}{12}$

Hence, $\dfrac{-11}{9} \times \dfrac{-51}{-44} = -1\dfrac{5}{12}$

Question 2(v)

Evaluate:

$-\dfrac{16}{5} \times \dfrac{20}{8}$

Answer

$-\dfrac{16}{5} \times \dfrac{20}{8}\\[1em] =-\dfrac{16 \times 20}{5 \times 8}\\[1em] =-\dfrac{320}{40}\\[1em] =-8$

Hence, $-\dfrac{16}{5} \times \dfrac{20}{8} = -8$

Question 3(i)

Multiply

$\dfrac{5}{6}$ and $\dfrac{8}{9}$

Answer

$\dfrac{5}{6} \times \dfrac{8}{9}\\[1em] =\dfrac{5 \times 8}{6 \times 9}\\[1em] =\dfrac{40}{54}\\[1em] =\dfrac{20}{27}$

Hence, $\dfrac{5}{6} \times \dfrac{8}{9} = \dfrac{20}{27}$

Question 3(ii)

Multiply:

$\dfrac{2}{7}$ and $\dfrac{-14}{9}$

Answer

$\dfrac{2}{7} \times \dfrac{-14}{9}\\[1em] =\dfrac{2 \times (-14)}{7 \times 9}\\[1em] =\dfrac{-28}{63}\\[1em] =\dfrac{-4}{9}$

Hence, $\dfrac{2}{7} \times \dfrac{-14}{9} = \dfrac{-4}{9}$

Question 3(iii)

Multiply:

$\dfrac{-7}{8}$ and 4

Answer

$\dfrac{-7}{8} \times \dfrac{4}{1}\\[1em] =\dfrac{-7 \times 4}{8 \times 1}\\[1em] =\dfrac{-28}{8}\\[1em] =\dfrac{-7}{2}\\[1em] =-3\dfrac{1}{2}$

Hence, $\dfrac{-7}{8} \times 4 = -3\dfrac{1}{2}$

Question 3(iv)

Multiply:

$\dfrac{36}{-7}$ and $\dfrac{-9}{28}$

Answer

$\dfrac{36}{-7} \times \dfrac{-9}{28}\\[1em] =\dfrac{36 \times (-9)}{-7 \times 28}\\[1em] =\dfrac{-324}{-196}\\[1em] =\dfrac{81}{49}\\[1em] =1\dfrac{32}{49}$

Hence, $\dfrac{36}{-7} \times \dfrac{-9}{28} = 1\dfrac{32}{49}$

Question 3(v)

Multiply:

$\dfrac{-7}{10}$ and $\dfrac{-8}{15}$

Answer

$\dfrac{-7}{10} \times \dfrac{-8}{15}\\[1em] =\dfrac{-7 \times (-8)}{10 \times 15}\\[1em] =\dfrac{56}{150}\\[1em] =\dfrac{28}{75}$

Hence, $\dfrac{-7}{10} \times \dfrac{-8}{15} = \dfrac{28}{75}$

Question 4(i)

Evaluate:

$\Big(\dfrac{2}{-3} \times \dfrac{5}{4}\Big) + \Big(\dfrac{5}{9} \times \dfrac{3}{-10}\Big)$

Answer

$\Big(\dfrac{2 \times 5}{-3 \times 4}\Big) + \Big(\dfrac{5 \times 3}{9 \times (-10)}\Big)\\[1em] =\Big(\dfrac{10}{-12}\Big) + \Big(\dfrac{15}{-90}\Big)\\[1em] =\Big(\dfrac{-5}{6}\Big) + \Big(\dfrac{-1}{6}\Big)\\[1em] =\Big(\dfrac{-5 + (-1)}{6}\Big)\\[1em] =\Big(\dfrac{-6}{6}\Big)\\[1em] =-1$

Hence, $\Big(\dfrac{2}{-3} \times \dfrac{5}{4}\Big) + \Big(\dfrac{5}{9} \times \dfrac{3}{-10}\Big) = \Big(\dfrac{-6}{6}\Big)$

Question 4(ii)

Evaluate:

$\Big(2 \times \dfrac{1}{4}\Big) - \Big(\dfrac{-18}{7} \times \dfrac{-7}{15}\Big)$

Answer

$\Big(\dfrac{2 \times 1}{1 \times 4}\Big) - \Big(\dfrac{-18 \times -7}{7 \times 15}\Big) \\[1em] =\Big(\dfrac{2}{4}\Big) - \Big(\dfrac{126}{105}\Big) \\[1em] =\Big(\dfrac{1}{2}\Big) - \Big(\dfrac{6}{5}\Big) \\[1em]$ LCM of 2 and 5 is 2 x 5 = 10 $=\Big(\dfrac{1 \times 5}{2 \times 5}\Big) - \Big(\dfrac{6 \times 2}{5 \times 2}\Big)\\[1em] =\Big(\dfrac{5}{10}\Big) - \Big(\dfrac{12}{10}\Big)\\[1em] =\Big(\dfrac{5 - 12}{10}\Big)\\[1em] =\Big(\dfrac{-7}{10}\Big)$

Hence, $\Big(2 \times \dfrac{1}{4}\Big) - \Big(\dfrac{-18}{7} \times \dfrac{-7}{15}\Big) = \Big(\dfrac{-7}{10}\Big)$

Question 4(iii)

Evaluate:

$\Big(-5 \times \dfrac{2}{15}\Big) - \Big(-6 \times \dfrac{2}{9}\Big)$

Answer

$\Big(\dfrac{-5 \times 2}{1 \times 15}\Big) - \Big(\dfrac{-6 \times 2}{1 \times 9}\Big)\\[1em] =\Big(\dfrac{-10}{15}\Big) - \Big(\dfrac{-12}{9}\Big)\\[1em] =\Big(\dfrac{-2}{3}\Big) - \Big(\dfrac{-4}{3}\Big)\\[1em] =\Big(\dfrac{-2 - (-4)}{3}\Big)\\[1em] =\Big(\dfrac{-2 + 4}{3})\\[1em] =\Big(\dfrac{2}{3}\Big)$

Hence, $\Big(-5 \times \dfrac{5}{12}\Big) - \Big(-6 \times \dfrac{2}{9}\Big) = \Big(\dfrac{2}{3}\Big)$

Question 4(iv)

Evaluate:

$\Big(\dfrac{8}{5} \times \dfrac{-3}{2}\Big) + \Big(\dfrac{-3}{10} \times \dfrac{9}{16}\Big)$

Answer

$\Big(\dfrac{8 \times -3}{5 \times 2}\Big) + \Big(\dfrac{-3 \times 9}{10 \times 16}\Big)\\[1em] =\Big(\dfrac{-24}{10}\Big) + \Big(\dfrac{-27}{160}\Big)\\[1em] =\Big(\dfrac{-12}{5}\Big) + \Big(\dfrac{-27}{160}\Big)\\[1em]$

LCM of 5 and 160 is 2 x 2 x 2 x 2 x 2 x 5 = 160

$=\Big(\dfrac{-12 \times 32}{5 \times 32}\Big) + \Big(\dfrac{-27 \times 1}{160 \times 1}\Big)\\[1em] =\Big(\dfrac{-384}{160}\Big) + \Big(\dfrac{-27}{160}\Big)\\[1em] =\Big(\dfrac{-384 + (-27)}{160}\Big)\\[1em] =\Big(\dfrac{-411}{160}\Big)\\[1em] =-2\dfrac{91}{160}$

Hence, $\Big(\dfrac{8}{5} \times \dfrac{-3}{2}\Big) + \Big(\dfrac{-3}{10} \times \dfrac{9}{16}\Big) = -2\dfrac{91}{160}$

Question 5(i)

Multiply each rational number, given below, by one (1):

$\dfrac{7}{-5}$

Answer

$\dfrac{7}{-5} \times 1\\[1em] =\dfrac{7 \times 1}{-5 \times 1}\\[1em] =\dfrac{7}{-5}$

Hence, $\dfrac{7}{-5} \times 1 = \dfrac{7}{-5}$

Question 5(ii)

Multiply each rational number, given below, by one (1):

$\dfrac{-3}{-4}$

Answer

$\dfrac{-3}{-4} \times 1\\[1em] =\dfrac{-3 \times 1}{-4 \times 1}\\[1em] =\dfrac{-3}{-4}\\[1em] =\dfrac{3}{4}$

Hence, $\dfrac{-3}{-4} \times 1 = \dfrac{3}{4}$

Question 5(iii)

Multiply each rational number, given below, by one (1):

Answer

$0 \times 1 \\[1em] = 0$

Hence, 0 x 1 = 0

Question 5(iv)

Multiply each rational number, given below, by one (1):

$\dfrac{-8}{13}$

Answer

$\dfrac{-8}{13} \times 1\\[1em] =\dfrac{-8 \times 1}{13 \times 1}\\[1em] =\dfrac{-8}{13}$

Hence, $\dfrac{-8}{13} \times 1 = \dfrac{-8}{13}$

Question 5(v)

Multiply each rational number, given below, by one (1):

$\dfrac{-6}{-7}$

Answer

$\dfrac{-6}{-7} \times 1\\[1em] =\dfrac{-6 \times 1}{-7 \times 1}\\[1em] =\dfrac{-6}{-7}\\[1em] =\dfrac{6}{7}$

Hence, $\dfrac{-6}{-7} \times 1 = \dfrac{6}{7}$

Question 6(i)

For each pair of rational numbers, given below, verify that the multiplication is commutative:

$\dfrac{-1}{5}$ and $\dfrac{2}{9}$

Answer

To prove:

$\dfrac{-1}{5} \times \dfrac{2}{9} = \dfrac{2}{9} \times \dfrac{-1}{5}$

Taking LHS:

$\dfrac{-1}{5} \times \dfrac{2}{9}\\[1em] =\dfrac{-1 \times 2}{5 \times 9}\\[1em] =\dfrac{-2}{45}\\[1em]$

Taking RHS:

$\dfrac{2}{9} \times \dfrac{-1}{5}\\[1em] =\dfrac{2 \times -1}{9 \times 5}\\[1em] =\dfrac{-2}{45}\\[1em]$

∴ LHS = RHS

$\dfrac{-1}{5} \times \dfrac{2}{9} = \dfrac{2}{9} \times \dfrac{-1}{5}$

Question 6(ii)

For each pair of rational numbers, given below, verify that the multiplication is commutative:

$\dfrac{5}{-3}$ and $\dfrac{13}{-11}$

Answer

To prove:

$\dfrac{5}{-3} \times \dfrac{13}{-11} = \dfrac{13}{-11} \times \dfrac{5}{-3}$

Taking LHS:

$\dfrac{5}{-3} \times \dfrac{13}{-11}\\[1em] =\dfrac{5 \times 13}{-3 \times -11}\\[1em] =\dfrac{65}{33}\\[1em]$

Taking RHS:

$\dfrac{13}{-11} \times \dfrac{5}{-3}\\[1em] =\dfrac{13 \times 5}{-11 \times -3}\\[1em] =\dfrac{65}{33}\\[1em]$

∴ LHS = RHS

$\dfrac{5}{-3} \times \dfrac{13}{-11} = \dfrac{13}{-11} \times \dfrac{5}{-3}$

Question 6(iii)

For each pair of rational numbers, given below, verify that the multiplication is commutative:

3 and $\dfrac{-8}{9}$

Answer

To prove:

$3 \times \dfrac{-8}{9} = \dfrac{-8}{9} \times 3$

Taking LHS:

$3 \times \dfrac{-8}{9}\\[1em] =\dfrac{3 \times -8}{1 \times 9}\\[1em] =\dfrac{-24}{9}\\[1em] =\dfrac{-8}{3}$

Taking RHS:

$\dfrac{-8}{9} \times 3\\[1em] =\dfrac{-8 \times 3}{9 \times 1}\\[1em] =\dfrac{-24}{9}\\[1em] =\dfrac{-8}{3}$

∴ LHS = RHS

$3 \times \dfrac{-8}{9} = \dfrac{-8}{9} \times 3$

Question 6(iv)

For each pair of rational numbers, given below, verify that the multiplication is commutative:

0 and $\dfrac{-12}{17}$

Answer

To prove:

$0 \times \dfrac{-12}{17} = \dfrac{-12}{17} \times 0$

Taking LHS:

$0 \times \dfrac{-12}{17}\\[1em] =\dfrac{0 \times -12}{1 \times 17}\\[1em] =\dfrac{0}{17}\\[1em] =0$

Taking RHS:

$\dfrac{-12}{17} \times 0\\[1em] =\dfrac{-12 \times 0}{17 \times 1}\\[1em] =\dfrac{0}{17}\\[1em] =0$

∴ LHS = RHS

$0 \times \dfrac{-12}{17} = \dfrac{-12}{17} \times 0$

Question 7(i)

Write the reciprocal (multiplicative inverse) of each rational number given below:

Answer

The multiplicative inverse of 5 = reciprocal of 5 = $\dfrac{1}{5}$ .

Question 7(ii)

Write the reciprocal (multiplicative inverse) of each rational number given below:

-3

Answer

The multiplicative inverse of -3 = reciprocal of -3 = $-\dfrac{1}{3}$ .

Question 7(iii)

Write the reciprocal (multiplicative inverse) of each rational number given below:

$\dfrac{5}{11}$

Answer

The multiplicative inverse of $\dfrac{5}{11}$ = reciprocal of $\dfrac{5}{11}$ = $\dfrac{11}{5} = 2\dfrac{1}{5}$ .

Question 7(iv)

Write the reciprocal (multiplicative inverse) of each rational number given below:

$\dfrac{-7}{-8}$

Answer

The multiplicative inverse of $\dfrac{-7}{-8}$ = reciprocal of $\dfrac{7}{8}$ = $\dfrac{8}{7} = 1\dfrac{1}{7}$ .

Question 7(v)

Write the reciprocal (multiplicative inverse) of each rational number given below:

$\dfrac{-8}{-7}$

Answer

The multiplicative inverse of $\dfrac{-8}{-7}$ = reciprocal of $\dfrac{8}{7}$ = $\dfrac{7}{8}$ .

Question 8(i)

Find the reciprocal (multiplicative inverse) of:

$\dfrac{3}{5} \times \dfrac{2}{3}$

Answer

$\dfrac{3}{5} \times \dfrac{2}{3}\\[1em] =\dfrac{3 \times 2}{5 \times 3}\\[1em] =\dfrac{6}{15}\\[1em] =\dfrac{2}{5}$

The multiplicative inverse of $\dfrac{2}{5}$ = reciprocal of $\dfrac{2}{5} = \dfrac{5}{2} = 2\dfrac{1}{2}$ .

Question 8(ii)

Find the reciprocal (multiplicative inverse) of:

$\dfrac{-8}{3} \times \dfrac{13}{-7}$

Answer

$\dfrac{-8}{3} \times \dfrac{13}{-7}\\[1em] =\dfrac{-8 \times 13}{3 \times -7}\\[1em] =\dfrac{-104}{-21}\\[1em] =\dfrac{104}{21}$

The multiplicative inverse of $\dfrac{104}{21}$ = reciprocal of $\dfrac{104}{21}$ = $\dfrac{21}{104}$ .

Question 8(iii)

Find the reciprocal (multiplicative inverse) of:

$\dfrac{-3}{5} \times \dfrac{-1}{13}$

Answer

$\dfrac{-3}{5} \times \dfrac{-1}{13}\\[1em] =\dfrac{-3 \times (-1)}{5 \times 13}\\[1em] =\dfrac{3}{65}$

The multiplicative inverse of $\dfrac{3}{65}$ = reciprocal of $\dfrac{3}{65}$ = $\dfrac{65}{3} = 21\dfrac{2}{3}$ .

Question 9(i)

Verify that $(x + y) \times z = x \times z + y \times z$ , if

$x = \dfrac{4}{5}, y = \dfrac{-2}{3}$ and $z = -4$

Answer

To prove:

$(x + y) \times z = x \times z + y \times z$

Taking LHS:

$(x + y) \times z\\[1em] =\Big(\dfrac{4}{5} + \dfrac{-2}{3}\Big) \times -4$

LCM of 5 and 3 is 3 x 5 = 15

$=\Big(\dfrac{4 \times 3}{5 \times 3} + \dfrac{-2 \times 5}{3 \times 5}\Big) \times -4\\[1em] =\Big(\dfrac{12}{15} + \dfrac{-10}{15}\Big) \times -4\\[1em] =\Big(\dfrac{12 + (-10)}{15}\Big) \times -4\\[1em] =\Big(\dfrac{2}{15}\Big) \times -4\\[1em] =\Big(\dfrac{2 \times -4}{15 \times 1}\Big)\\[1em] =\Big(\dfrac{-8}{15}\Big)$

Taking RHS:

$x \times z + y \times z\\[1em] =\dfrac{4}{5} \times -4 + \dfrac{-2}{3} \times -4\\[1em] = \dfrac{4 \times -4}{5 \times 1} + \dfrac{-2 \times -4}{3 \times 1}\\[1em] =\dfrac{-16}{5} + \dfrac{8}{3}\\[1em]$

LCM of 5 and 3 is 3 x 5 = 15

$=\dfrac{-16 \times 3}{5 \times 3} + \dfrac{8 \times 5}{3 \times 5}\\[1em] =\dfrac{-48}{15} + \dfrac{40}{15}\\[1em] =\dfrac{-48 + 40}{15}\\[1em] =\dfrac{-8}{15}$

∴ LHS = RHS

$(x + y) \times z = x \times z + y \times z$

Question 9(ii)

Verify that $(x + y) \times z = x \times z + y \times z$ , if

$x = 2, y = \dfrac{4}{5}$ and $z = \dfrac{3}{-10}$

Answer

To prove:

$(x + y) \times z = x \times z + y \times z$

Taking LHS:

$(x + y) \times z\\[1em] =\Big(2 + \dfrac{4}{5}\Big) \times \dfrac{3}{-10}\\[1em] =\Big(\dfrac{2}{1} + \dfrac{4}{5}\Big) \times \dfrac{3}{-10}$

LCM of 1 and 5 is 5.

$=\Big(\dfrac{2 \times 5}{1 \times 5} + \dfrac{4 \times 1}{5 \times 1}\Big) \times \dfrac{3}{-10}\\[1em] =\Big(\dfrac{10}{5} + \dfrac{4}{5}\Big) \times \dfrac{3}{-10}\\[1em] =\Big(\dfrac{10 + 4}{5}\Big) \times \dfrac{3}{-10}\\[1em] =\Big(\dfrac{14}{5}\Big) \times \dfrac{3}{-10}\\[1em] =\Big(\dfrac{14 \times 3}{5 \times -10}\Big)\\[1em] =\Big(\dfrac{42}{-50}\Big)\\[1em] =\Big(\dfrac{-21}{25}\Big)$

Taking RHS:

$x \times z + y \times z\\[1em] =2 \times \dfrac{3}{-10} + \dfrac{4}{5} \times \dfrac{3}{-10}\\[1em] = \dfrac{2 \times 3}{1 \times -10} + \dfrac{4 \times 3}{5 \times -10}\\[1em] =\dfrac{6}{-10} + \dfrac{12}{-50}\\[1em] =\dfrac{-3}{5} + \dfrac{-6}{25}$

LCM of 5 and 25 is 5 x 5 = 25

$=\dfrac{-3 \times 5}{5 \times 5} + \dfrac{-6 \times 1}{25 \times 1}\\[1em] =\dfrac{-15}{25} + \dfrac{-6}{25}\\[1em] =\dfrac{-15 + (-6)}{25}\\[1em] =\dfrac{-21}{25}$

∴ LHS = RHS

$(x + y) \times z = x \times z + y \times z$

Question 10(i)

Verify that $x \times (y - z) = x \times y - x \times z$ , if

$x = \dfrac{4}{5}, y = \dfrac{-7}{4}$ and $z = 3$

Answer

To prove:

$x \times (y - z) = x \times y - x \times z$

Taking LHS:

$x \times (y - z)\\[1em] =\dfrac{4}{5} \times \Big(\dfrac{-7}{4} - 3\Big)\\[1em] =\dfrac{4}{5} \times \Big(\dfrac{-7}{4} - \dfrac{3}{1}\Big)$

LCM of 4 and 1 is 2 x 2 = 4.

$=\dfrac{4}{5} \times \Big(\dfrac{-7 \times 1}{4 \times 1} - \dfrac{3 \times 4}{1 \times 4}\Big)\\[1em] =\dfrac{4}{5} \times \Big(\dfrac{-7}{4} - \dfrac{12}{4}\Big)\\[1em] =\dfrac{4}{5} \times \Big(\dfrac{-7 - 12}{4}\Big)\\[1em] =\dfrac{4}{5} \times \Big(\dfrac{-19}{4}\Big)\\[1em] =\Big(\dfrac{4 \times -19}{5 \times 4}\Big)\\[1em] =\Big(\dfrac{-76}{20}\Big)\\[1em] =\Big(\dfrac{-19}{5}\Big)$

Taking RHS:

$x \times y - x \times z\\[1em] =\dfrac{4}{5} \times \dfrac{-7}{4} - \dfrac{4}{5} \times 3\\[1em] = \dfrac{4 \times -7}{5 \times 4} - \dfrac{4 \times 3}{5 \times 1}\\[1em] =\dfrac{-28}{20} - \dfrac{12}{5}\\[1em] =\dfrac{-7}{5} - \dfrac{12}{5}\\[1em] =\dfrac{-7 - 12}{5}\\[1em] =\dfrac{-19}{5}$

∴ LHS = RHS

$x \times (y - z) = x \times y - x \times z$

Question 10(ii)

Verify that $x \times (y - z) = x \times y - x \times z$ , if

$x = \dfrac{3}{4}, y = \dfrac{8}{9}$ and $z = -5$

Answer

To prove:

$x \times (y - z) = x \times y - x \times z$

Taking LHS:

$x \times (y - z)\\[1em] =\dfrac{3}{4} \times \Big(\dfrac{8}{9} - (-5)\Big)\\[1em] =\dfrac{3}{4} \times \Big(\dfrac{8}{9} - \dfrac{-5}{1}\Big)$

LCM of 9 and 1 is 3 x 3 = 9.

$=\dfrac{3}{4} \times \Big(\dfrac{8 \times 1}{9 \times 1} - \dfrac{-5 \times 9}{1 \times 9}\Big)\\[1em] =\dfrac{3}{4} \times \Big(\dfrac{8}{9} - \dfrac{-45}{9}\Big)\\[1em] =\dfrac{3}{4} \times \Big(\dfrac{8 - (-45)}{9}\Big)\\[1em] =\dfrac{3}{4} \times \Big(\dfrac{8 + 45}{9}\Big)\\[1em] =\dfrac{3}{4} \times \Big(\dfrac{53}{9}\Big)\\[1em] =\Big(\dfrac{3 \times 53}{4 \times 9}\Big)\\[1em] =\Big(\dfrac{159}{36}\Big)\\[1em] =\Big(\dfrac{53}{12}\Big)$

Taking RHS:

$x \times y - x \times z\\[1em] =\dfrac{3}{4} \times \dfrac{8}{9} - \dfrac{3}{4} \times -5\\[1em] = \dfrac{3 \times 8}{4 \times 9} - \dfrac{3 \times -5}{4 \times 1}\\[1em] =\dfrac{24}{36} - \dfrac{-15}{4}\\[1em] =\dfrac{2}{3} - \dfrac{-15}{4}$

LCM of 3 and 4 is 2 x 2 x 3 = 12

$=\dfrac{2 \times 4}{3 \times 4} - \dfrac{-15 \times 3}{4 \times 3}\\[1em] =\dfrac{8}{12} - \dfrac{-45}{12}\\[1em] =\dfrac{8 - (-45)}{12}\\[1em] =\dfrac{8 + 45}{12}\\[1em] =\dfrac{53}{12}$

∴ LHS = RHS

$x \times (y - z) = x \times y - x \times z$

Question 11

Name the multiplication property of rational numbers shown below:

(i) $\dfrac{3}{5} \times \dfrac{-8}{9} = \dfrac{-8}{9} \times \dfrac{3}{5}$

(ii) $\dfrac{-3}{4} \times \Big(\dfrac{5}{7} \times \dfrac{-8}{15}\Big) = \Big(\dfrac{-3}{4} \times \dfrac{5}{7}\Big) \times \dfrac{-8}{15}$

(iii) $\dfrac{4}{5} \times \Big(\dfrac{3}{-8} + \dfrac{-4}{7}\Big) = \dfrac{4}{5} \times \dfrac{3}{-8} + \dfrac{4}{5} \times \dfrac{-4}{7}$

(iv) $\dfrac{-7}{5} \times \dfrac{5}{-7} = 1$

(v) $\dfrac{8}{-9} \times 1 = 1 \times \dfrac{8}{-9} = \dfrac{8}{-9}$

Answer

(i) Commutativity property

Reason

If $\dfrac{a}{b}$ and $\dfrac{c}{d}$ are any two rational numbers, then:

$\dfrac{a}{b} \times \dfrac{c}{d} = \dfrac{c}{d} \times \dfrac{a}{b}$

(ii) Associativity property

Reason

If $\dfrac{a}{b}, \dfrac{c}{d}$ and $\dfrac{e}{f}$ are any three rational numbers, then:

$\dfrac{a}{b} \times \Big(\dfrac{c}{d} \times \dfrac{e}{f}\Big) = \Big(\dfrac{a}{b} \times \dfrac{c}{d}\Big) \times \dfrac{e}{f}$

(iii) Distributivity property

Reason

If $\dfrac{a}{b}, \dfrac{c}{d}$ and $\dfrac{e}{f}$ are any three rational numbers, then:

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

(iv) Existence of inverse

Reason

The multiplicative inverse of $\dfrac{a}{b}$ = reciprocal of $\dfrac{a}{b} = \dfrac{b}{a}$ .

(v) Existence of identity

Reason

For a rational number $\dfrac{a}{b}$ ,

$1 \times \dfrac{a}{b} = \dfrac{a}{b} \times 1 = \dfrac{a}{b}$ .

Question 12

Fill in the blanks:

(i) The product of two positive rational numbers is always ............... .

(ii) The product of two negative rational numbers is always ............... .

(iii) If two rational numbers have opposite signs then their product is always ............... .

(iv) The reciprocal of a positive rational number is ............... and the reciprocal of a negative rational number is ............... .

(v) Rational number 0 has ............... reciprocal.

(vi) The product of a non-zero rational number and its reciprocal is ............... .

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

(viii) If $m$ is reciprocal of $n$ , then the reciprocal of $n$ is ............... .

Answer

(i) The product of two positive rational numbers is always positive.

(ii) The product of two negative rational numbers is always positive.

(iii) If two rational numbers have opposite signs then their product is always negative.

(iv) The reciprocal of a positive rational number is positive and the reciprocal of a negative rational number is negative.

(v) Rational number 0 has no reciprocal.

(vi) The product of a non-zero rational number and its reciprocal is 1.

(vii) The numbers 1 and -1 are their own reciprocal.

(viii) If m is reciprocal of n, then the reciprocal of n is m.

Explanation

(i) Let 2 positive rational numbers be $\dfrac{a}{b}$ and $\dfrac{c}{d}$ .

Hence,

$\dfrac{a}{b} \times \dfrac{c}{d}\\[1em] = \dfrac{a \times c}{b \times d}\\[1em] = \dfrac{ac}{bd}$

$\dfrac{ac}{bd}$ is also positive rational number.

(ii) Let 2 negative rational numbers be - $\dfrac{a}{b}$ and - $\dfrac{c}{d}$ .

Hence,

$-\dfrac{a}{b} \times -\dfrac{c}{d}\\[1em] = \dfrac{-a \times -c}{b \times d}\\[1em] = \dfrac{ac}{bd}$

$\dfrac{ac}{bd}$ is positive rational number.

(iii) Let 2 rational numbers be $\dfrac{a}{b}$ and - $\dfrac{c}{d}$ .

Hence,

$\dfrac{a}{b} \times -\dfrac{c}{d}\\[1em] = \dfrac{a \times -c}{b \times d}\\[1em] = \dfrac{-ac}{bd}$

- $\dfrac{ac}{bd}$ is negative rational number.

(iv) Let the positive rational number be $\dfrac{a}{b}$ .

Reciprocal of $\dfrac{a}{b} = \dfrac{b}{a}$

$\dfrac{b}{a}$ is a positive rational number.

Let the negative rational number be - $\dfrac{a}{b}$ .

Reciprocal of - $\dfrac{a}{b} = -\dfrac{b}{a}$

- $\dfrac{b}{a}$ is a negative rational number.

(v) Reciprocal of $\dfrac{0}{1} = \dfrac{1}{0}$

$\dfrac{1}{0}$ is not defined.

(vi) Let the positive rational number be $\dfrac{a}{b}$ .

Reciprocal of $\dfrac{a}{b} = \dfrac{b}{a}$

$\dfrac{a}{b} \times \dfrac{b}{a} =\dfrac{a \times b}{b \times a} =\dfrac{ab}{ab} =1$

(vii) Reciprocal of $\dfrac{1}{1} = \dfrac{1}{1} = 1$ .

Reciprocal of $\dfrac{-1}{1} = \dfrac{1}{-1} = -1$ .

(viii) If reciprocal of $\dfrac{m}{1} = \dfrac{n}{1}$

Reciprocal of $\dfrac{n}{1} = \dfrac{m}{1}$

Question 13

The length and breadth of a rectangular piece of paper are 9 cm and $10\dfrac{2}{3}$ cm respectively. Find:

(i) its area

(ii) its perimeter

Answer

Length = 9 cm

Breadth = $10\dfrac{2}{3}$ cm

Area = length x breadth

$= 9 \times 10\dfrac{2}{3}\\[1em] = 9 \times \dfrac{32}{3}\\[1em] = \dfrac{9 \times 32}{1 \times 3}\\[1em] = \dfrac{288}{3}\\[1em] = 96 \text{ cm}^2$

Area = 96 cm²

Perimeter = 2 x (length + breadth)

$= 2 \times \Big(9 + 10\dfrac{2}{3}\Big)\\[1em] = 2 \times \Big(\dfrac{9}{1} + \dfrac{32}{3}\Big)$

LCM of 1 and 3 is 3.

$= 2 \times \Big(\dfrac{9 \times 3}{1 \times 3} + \dfrac{32 \times 1}{3 \times 1}\Big)\\[1em] = 2 \times \Big(\dfrac{27}{3} + \dfrac{32}{3}\Big)\\[1em] = 2 \times \Big(\dfrac{27 + 32}{3}\Big)\\[1em] = 2 \times \Big(\dfrac{59}{3}\Big)\\[1em] = \Big(\dfrac{59 \times 2}{3 \times 1}\Big)\\[1em] = \Big(\dfrac{118}{3}\Big)\\[1em] = 39\Big(\dfrac{1}{3}\Big)$

Perimeter = $39\dfrac{1}{3}$ cm

Hence, area of the rectangular piece of paper is 96 cm² and its perimeter is $39\dfrac{1}{3}$ cm.

Question 14

Find the area and the perimeter of a rectangular piece of land with length $7\dfrac{2}{5}$ m and breadth $4\dfrac{1}{6}$ m.

Answer

Length = $7\dfrac{2}{5}$ m = $\dfrac{37}{5}$ m

Breadth = $4\dfrac{1}{6}$ = $\dfrac{25}{6}$ m

Area = length x breadth

$= \dfrac{37}{5} \times \dfrac{25}{6}\\[1em] = \dfrac{37 \times 25}{5 \times 6}\\[1em] = \dfrac{925}{30}\\[1em] = \dfrac{185}{6}\\[1em] = 30\dfrac{5}{6} \text{ m}^2$

Perimeter = 2 x (length + breadth)

$= 2 \times \Big(\dfrac{37}{5} + \dfrac{25}{6}\Big)\\[1em]$

LCM of 5 and 6 is 2 x 3 x 5 = 30.

$= 2 \times \Big(\dfrac{37 \times 6}{5 \times 6} + \dfrac{25 \times 5}{6 \times 5}\Big)\\[1em] = 2 \times \Big(\dfrac{222}{30} + \dfrac{125}{30}\Big)\\[1em] = 2 \times \Big(\dfrac{222 + 125}{30}\Big)\\[1em] = 2 \times \Big(\dfrac{347}{30}\Big)\\[1em] = \Big(\dfrac{347 \times 2}{30 \times 1}\Big)\\[1em] = \Big(\dfrac{694}{30}\Big)\\[1em] = \Big(\dfrac{347}{15}\Big)\\[1em] = 23\Big(\dfrac{2}{15}\Big)$

Area = $30\dfrac{5}{6}$ m² and perimeter = $23\dfrac{2}{15}$ m

Exercise 1(D)

Question 1(i)

$-\dfrac{4}{9}$ divided by $-\dfrac{2}{3}$ gives:

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

Answer

$-\dfrac{4}{9} ÷ -\dfrac{2}{3}\\[1em] = -\dfrac{4}{9} \times -\dfrac{3}{2}\\[1em] = \dfrac{4 \times 3}{9 \times 2}\\[1em] = \dfrac{12}{18}\\[1em] = \dfrac{2}{3}$

$-\dfrac{4}{9}$ divided by $-\dfrac{2}{3}$ gives $\dfrac{2}{3}$

Hence, Option 1 is the correct option.

Question 1(ii)

The rational number by which should $\dfrac{1}{2}$ be divided to get $-\dfrac{2}{3}$ is:

$\dfrac{3}{4}$
$-\dfrac{3}{4}$
$\dfrac{4}{3}$
$-\dfrac{4}{3}$

Answer

Let $x$ be the number.

$\dfrac{1}{2} ÷ x = -\dfrac{2}{3}\\[1em] ⇒ \dfrac{1}{2} \times \dfrac{1}{x} = -\dfrac{2}{3} \\[1em] ⇒ \dfrac{1}{2x} = -\dfrac{2}{3} \\[1em] ⇒ 2x = -\dfrac{3}{2} \\[1em] ⇒ x = \dfrac{1}{2} \times -\dfrac{3}{2}\\[1em] ⇒ x = -\dfrac{1 \times 3}{2 \times 2}\\[1em] ⇒ x = -\dfrac{3}{4}$

The rational number by which should $\dfrac{1}{2}$ be divided to get $-\dfrac{2}{3}$ is $-\dfrac{3}{4}$

Hence, Option 2 is the correct option.

Question 1(iii)

For the three rational number a, b and c; which of the following is correct:

a ÷ b = b ÷ a
a x (b ÷ c) = (a ÷ b) x (a ÷ c)
a ÷ (b ÷ c) = (a ÷ b) ÷ (a ÷ c)
a ÷ (b ÷ c) ≠ a ÷ b ÷ c

Answer

We know that, division of rational numbers is not associative.

∴ a ÷ (b ÷ c) ≠ a ÷ b ÷ c

Hence, Option 4 is the correct option.

Question 1(iv)

The product of two rational numbers is $-7\dfrac{2}{3}$ . If one of them is $3\dfrac{5}{6}$ , the other number is:

3
-3
2
-2

Answer

Let the number be $x$ .

$3\dfrac{5}{6} \times x = -7\dfrac{2}{3}\\[1em] ⇒ \dfrac{23}{6} \times x = -\dfrac{23}{3}\\[1em] ⇒ x = -\dfrac{23}{3} ÷ \dfrac{23}{6}\\[1em] ⇒ x = -\dfrac{23}{3} \times \dfrac{6}{23}\\[1em] ⇒ x = -\dfrac{23 \times 6}{3 \times 23}\\[1em] ⇒ x = -\dfrac{23 \times 6}{3 \times 23}\\[1em] ⇒ x = -\dfrac{138}{69}\\[1em] ⇒ x = -2$

Hence, Option 4 is the correct option.

Question 1(v)

(8 ÷ 3) ÷ (3 ÷ 8) is equal to:

$\dfrac{64}{9}$
$\dfrac{9}{64}$
1
none of the above

Answer

$(8 ÷ 3) ÷ (3 ÷ 8)\\[1em] = \dfrac{8}{3} ÷ \dfrac{3}{8}\\[1em] = \dfrac{8}{3} \times \dfrac{8}{3}\\[1em] = \dfrac{64}{9}$

Hence, Option 1 is the correct option.

Question 2(i)

Evaluate:

1 ÷ $\dfrac{1}{3}$

Answer

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

Hence, $1 ÷ \dfrac{1}{3} = 3$

Question 2(ii)

Evaluate:

$3 ÷ \dfrac{3}{5}$

Answer

$3 ÷ \dfrac{3}{5}\\[1em] = 3 \times \dfrac{5}{3}\\[1em] = \dfrac{3 \times 5}{1 \times 3}\\[1em] = \dfrac{15}{3}\\[1em] = 5$

Hence, $3 ÷ \dfrac{3}{5} = 5$

Question 2(iii)

Evaluate:

$-\dfrac{5}{12} ÷ \dfrac{1}{16}$

Answer

$-\dfrac{5}{12} ÷ \dfrac{1}{16}\\[1em] = -\dfrac{5}{12} \times \dfrac{16}{1}\\[1em] = -\dfrac{5 \times 16}{12 \times 1}\\[1em] = -\dfrac{80}{12}\\[1em] = -\dfrac{20}{3}\\[1em] = -6\dfrac{2}{3}$

Hence, $-\dfrac{5}{12} ÷ \dfrac{1}{16} = -6\dfrac{2}{3}$

Question 2(iv)

Evaluate:

$-\dfrac{21}{16} ÷ \Big(\dfrac{-7}{8}\Big)$

Answer

$-\dfrac{21}{16} ÷ \dfrac{-7}{8}\\[1em] = -\dfrac{21}{16} \times -\dfrac{8}{7}\\[1em] = \dfrac{21 \times 8}{16 \times 7}\\[1em] = \dfrac{168}{112}\\[1em] = \dfrac{3}{2}\\[1em] = 1\dfrac{1}{2}$

Hence, $-\dfrac{21}{16} ÷ \Big(\dfrac{-7}{8}\Big) = 1\dfrac{1}{2}$

Question 2(v)

Evaluate:

$0 ÷ \Big(\dfrac{-4}{7}\Big)$

Answer

$0 ÷ \dfrac{-4}{7}\\[1em] = 0 \times \dfrac{-7}{4}\\[1em] = \dfrac{0 \times -7}{1 \times 4}\\[1em] = \dfrac{0}{4}\\[1em] = 0$

Hence, $0 ÷ \Big(\dfrac{-4}{7}\Big) = 0$

Question 2(vi)

Evaluate:

$\dfrac{8}{-5} ÷ \dfrac{24}{25}$

Answer

$\dfrac{8}{-5} ÷ \dfrac{24}{25}\\[1em] = -\dfrac{8}{5} \times \dfrac{25}{24}\\[1em] = -\dfrac{8 \times 25}{5 \times 24}\\[1em] = -\dfrac{200}{120}\\[1em] = -\dfrac{5}{3}\\[1em] = -1\dfrac{2}{3}$

Hence, $\dfrac{8}{-5} ÷ \dfrac{24}{25} = -1\dfrac{2}{3}$

Question 2(vii)

Evaluate:

$-\dfrac{3}{4} ÷ (-9)$

Answer

$-\dfrac{3}{4} ÷ (-9)\\[1em] = -\dfrac{3}{4} \times \dfrac{-1}{9}\\[1em] = \dfrac{3 \times 1}{4 \times 9}\\[1em] = \dfrac{3}{36}\\[1em] = \dfrac{1}{12}$

Hence, $-\dfrac{3}{4} ÷ (-9) = \dfrac{1}{12}$

Question 2(viii)

Evaluate:

$\dfrac{3}{4} ÷ \Big(-\dfrac{5}{12}\Big)$

Answer

$\dfrac{3}{4} ÷ -\dfrac{5}{12}\\[1em] = \dfrac{3}{4} \times -\dfrac{12}{5}\\[1em] = -\dfrac{3 \times 12}{4 \times 5}\\[1em] = -\dfrac{36}{20}\\[1em] = -\dfrac{9}{5}\\[1em] = -1\dfrac{4}{5}$

Hence, $\dfrac{3}{4} ÷ \Big(-\dfrac{5}{12}\Big) = -1\dfrac{4}{5}$

Question 2(ix)

Evaluate:

$-5 ÷ \Big(-\dfrac{10}{11}\Big)$

Answer

$-5 ÷ -\dfrac{10}{11}\\[1em] = -5 \times -\dfrac{11}{10}\\[1em] = \dfrac{5 \times 11}{1 \times 10}\\[1em] = \dfrac{55}{10}\\[1em] = \dfrac{11}{2}\\[1em] = 5\dfrac{1}{2}$

Hence, $-5 ÷ \Big(-\dfrac{10}{11}\Big) = 5\dfrac{1}{2}$

Question 2(x)

Evaluate:

$\dfrac{-7}{11} ÷ \Big(\dfrac{-3}{44}\Big)$

Answer

$\dfrac{-7}{11} ÷ \dfrac{-3}{44}\\[1em] = -\dfrac{7}{11} \times -\dfrac{44}{3}\\[1em] = \dfrac{7 \times 44}{11 \times 3}\\[1em] = \dfrac{308}{33}\\[1em] = \dfrac{28}{3}\\[1em] = 9\dfrac{1}{3}$

Hence, $\dfrac{-7}{11} ÷ \Big(\dfrac{-3}{44}\Big) = 9\dfrac{1}{3}$

Question 3(i)

Divide:

3 by $\dfrac{1}{3}$

Answer

$3 ÷ \dfrac{1}{3}\\[1em] = 3 \times \dfrac{3}{1}\\[1em] = \dfrac{3 \times 3}{1 \times 1}\\[1em] = \dfrac{9}{1}\\[1em] = 9$

Hence, $3 ÷ \dfrac{1}{3} = 9$

Question 3(ii)

Divide:

-2 by $-\dfrac{1}{2}$

Answer

$-2 ÷ -\dfrac{1}{2}\\[1em] = -2 \times -\dfrac{2}{1}\\[1em] = \dfrac{2 \times 2}{1 \times 1}\\[1em] = \dfrac{4}{1}\\[1em] = 4$

Hence, $-2 ÷ -\dfrac{1}{2} = 4$

Question 3(iii)

Divide:

0 by $\dfrac{7}{-9}$

Answer

$0 ÷ \dfrac{7}{-9}\\[1em] = 0 \times \dfrac{-9}{7}\\[1em] = \dfrac{0 \times -9}{1 \times 7}\\[1em] = \dfrac{0}{7}\\[1em] = 0$

Hence, $0 ÷ \dfrac{7}{-9} = 0$

Question 3(iv)

Divide:

$\dfrac{-5}{8}$ by $\dfrac{1}{4}$

Answer

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

Hence, $\dfrac{-5}{8} ÷ \dfrac{1}{4} = -2\dfrac{1}{2}$

Question 3(v)

Divide:

$-\dfrac{3}{4}$ by $-\dfrac{9}{16}$

Answer

$-\dfrac{3}{4} ÷ -\dfrac{9}{16}\\[1em] = -\dfrac{3}{4} \times -\dfrac{16}{9}\\[1em] = \dfrac{3 \times 16}{4 \times 9}\\[1em] = \dfrac{48}{36}\\[1em] = \dfrac{4}{3}\\[1em] = 1\dfrac{1}{3}$

Hence, $-\dfrac{3}{4} ÷ -\dfrac{9}{16} = 1\dfrac{1}{3}$

Question 4

The product of two rational numbers is -2. If one of them is $\dfrac{4}{7}$ , find the other.

Answer

Let the number be $x$ .

$\dfrac{4}{7} \times x = -2\\[1em] \Rightarrow \dfrac{4}{7} \times x = -\dfrac{2}{1}\\[1em] \Rightarrow x = -\dfrac{2}{1} ÷ \dfrac{4}{7}\\[1em] \Rightarrow x = -\dfrac{2}{1} \times \dfrac{7}{4}\\[1em] \Rightarrow x = -\dfrac{2 \times 7}{1 \times 4}\\[1em] \Rightarrow x = -\dfrac{14}{4}\\[1em] \Rightarrow x = -\dfrac{7}{2}\\[1em] \Rightarrow x = -3\dfrac{1}{2}$

The other number is $-3\dfrac{1}{2}$ .

Question 5

The product of two rational numbers is $-\dfrac{4}{9}$ . If one of them is $\dfrac{-2}{27}$ , find the other.

Answer

Let the number be $x$ .

$\dfrac{-2}{27} \times x = -\dfrac{4}{9}\\[1em] \Rightarrow x = -\dfrac{4}{9} ÷ \dfrac{-2}{27}\\[1em] \Rightarrow x = -\dfrac{4}{9} \times \dfrac{-27}{2}\\[1em] \Rightarrow x = \dfrac{4 \times 27}{9 \times 2}\\[1em] \Rightarrow x = \dfrac{108}{18}\\[1em] \Rightarrow x = 6$

The other number is 6.

Question 6(i)

m and n are two rational numbers such that $m \times n = -\dfrac{25}{9}$ .

if $m = \dfrac{5}{3}$ , find $n$ .

Answer

$m \times n = -\dfrac{25}{9}\\[1em] \Rightarrow \dfrac{5}{3} \times n = - \dfrac{25}{9}\\[1em] \Rightarrow n = -\dfrac{25}{9} ÷ \dfrac{5}{3}\\[1em] \Rightarrow n = -\dfrac{25}{9} \times \dfrac{3}{5}\\[1em] \Rightarrow n = -\dfrac{25 \times 3}{9 \times 5}\\[1em] \Rightarrow n = -\dfrac{75}{45}\\[1em] \Rightarrow n = -\dfrac{5}{3}\\[1em] \Rightarrow n = -1\dfrac{2}{3}$

if $m = \dfrac{5}{3}$ , then $n = -1\dfrac{2}{3}.$

Question 6(ii)

m and n are two rational numbers such that $m \times n = -\dfrac{25}{9}$ .

if $n = -\dfrac{10}{9}$ , find $m$ .

Answer

$m \times n = -\dfrac{25}{9}\\[1em] \Rightarrow m \times -\dfrac{10}{9} = - \dfrac{25}{9}\\[1em] \Rightarrow m = -\dfrac{25}{9} ÷ -\dfrac{10}{9}\\[1em] \Rightarrow m = \dfrac{25}{9} \times \dfrac{9}{10}\\[1em] \Rightarrow m = \dfrac{25 \times 9}{9 \times 10}\\[1em] \Rightarrow m = \dfrac{225}{90}\\[1em] \Rightarrow m = \dfrac{5}{2}\\[1em] \Rightarrow m = 2\dfrac{1}{2}$

if $n = -\dfrac{10}{9}$ , then $n = 2\dfrac{1}{2}.$

Question 7

By what number must $-\dfrac{3}{4}$ be multiplied so that the product is $-\dfrac{9}{16}$ ?

Answer

Let the number be $x$

$-\dfrac{3}{4} \times x = -\dfrac{9}{16}\\[1em] \Rightarrow x = -\dfrac{9}{16} ÷ -\dfrac{3}{4}\\[1em] \Rightarrow x = \dfrac{9}{16} \times \dfrac{4}{3}\\[1em] \Rightarrow x = \dfrac{9 \times 4}{16 \times 3}\\[1em] \Rightarrow x = \dfrac{36}{48}\\[1em] \Rightarrow x = \dfrac{3}{4}$

$-\dfrac{3}{4}$ must be multiplied by $\dfrac{3}{4}$ so that the product is $-\dfrac{9}{16}$ .

Question 8

By what number must be $-\dfrac{8}{13}$ multiplied to get 16?

Answer

Let the number be $x$

$-\dfrac{8}{13} \times x = 16\\[1em] \Rightarrow x = 16 ÷ -\dfrac{8}{13}\\[1em] \Rightarrow x = -\dfrac{16}{1} \times \dfrac{13}{8}\\[1em] \Rightarrow x = -\dfrac{16 \times 13}{1 \times 8}\\[1em] \Rightarrow x = -\dfrac{208}{8}\\[1em] \Rightarrow x = -26$

$-\dfrac{8}{13}$ must be multiplied by -26 so that the product is 16

Question 9

If $3\dfrac{1}{2}$ litres of milk costs ₹49, find the cost of one litre of milk?

Answer

Let the cost of one litre of milk be ₹ $x$ .

$3\dfrac{1}{2} \times x = 49\\[1em] \Rightarrow \dfrac{7}{2} \times x = 49\\[1em] \Rightarrow x = 49 ÷ \dfrac{7}{2}\\[1em] \Rightarrow x = 49 \times \dfrac{2}{7}\\[1em] \Rightarrow x = \dfrac{49 \times 2}{1 \times 7}\\[1em] \Rightarrow x = \dfrac{98}{7}\\[1em] \Rightarrow x = 14$

The cost of one litre of milk = ₹14.

Question 10

Cost of $3\dfrac{2}{5}$ metre of cloth is ₹ $88\dfrac{1}{2}$ . What is the cost of 1 metre of cloth?

Answer

Let the cost of 1 metre of cloth be ₹ $x$ .

$3\dfrac{2}{5} \times x = 88\dfrac{1}{2}\\[1em] \Rightarrow \dfrac{17}{5} \times x = \dfrac{177}{2}\\[1em] \Rightarrow x = \dfrac{177}{2} ÷ \dfrac{17}{5}\\[1em] \Rightarrow x = \dfrac{177}{2} \times \dfrac{5}{17}\\[1em] \Rightarrow x = \dfrac{177 \times 5}{2 \times 17}\\[1em] \Rightarrow x = \dfrac{885}{34}\\[1em] \Rightarrow x = 26\dfrac{1}{34}$

Hence, The cost of 1 meter of cloth is ₹ $26\dfrac{1}{34}$ .

Question 11

Divide the sum of $\dfrac{3}{7}$ and $\dfrac{-5}{14}$ by $-\dfrac{1}{2}$ .

Answer

The sum of $\dfrac{3}{7}$ and $\dfrac{-5}{14}$

$\dfrac{3}{7} + \dfrac{-5}{14}$

LCM of 7 and 14 is 2 x 7 = 14

$= \dfrac{3 \times 2}{7 \times 2} + \dfrac{-5 \times 1}{14 \times 1}\\[1em] = \dfrac{6}{14} + \dfrac{-5}{14}\\[1em] = \dfrac{6 + (-5)}{14}\\[1em] = \dfrac{1}{14}$

Dividing the sum of $\dfrac{3}{7}$ and $\dfrac{-5}{14}$ by $-\dfrac{1}{2}$

$\dfrac{1}{14} ÷ -\dfrac{1}{2}\\[1em] = \dfrac{1}{14} \times -\dfrac{2}{1}\\[1em] = -\dfrac{1 \times 2}{14 \times 1}\\[1em] = -\dfrac{2}{14}\\[1em] = -\dfrac{1}{7}$

On dividing the sum of $\dfrac{3}{7}$ and $\dfrac{-5}{14}$ by $-\dfrac{1}{2}$ we get $-\dfrac{1}{7}$ .

Question 12(i)

Find $(m + n) ÷ (m - n)$ , if;

$m = \dfrac{2}{3}$ and $n = \dfrac{3}{2}$

Answer

$(m + n) ÷ (m - n)\\[1em] = \Big(\dfrac{2}{3} + \dfrac{3}{2}\Big) ÷ \Big(\dfrac{2}{3} - \dfrac{3}{2}\Big)$

LCM of 3 and 2 is 2 x 3 = 6

$= \Big(\dfrac{2 \times 2}{3 \times 2} + \dfrac{3 \times 3}{2 \times 3}\Big) ÷ \Big(\dfrac{2 \times 2}{3 \times 2} - \dfrac{3 \times 3}{2 \times 3}\Big)\\[1em] = \Big(\dfrac{4}{6} + \dfrac{9}{6}\Big) ÷ \Big(\dfrac{4}{6} - \dfrac{9}{6}\Big)\\[1em] = \Big(\dfrac{4 + 9}{6}\Big) ÷ \Big(\dfrac{4 - 9}{6}\Big)\\[1em] = \Big(\dfrac{13}{6}\Big) ÷ \Big(\dfrac{-5}{6}\Big)\\[1em] = \dfrac{13}{6} \times \dfrac{6}{-5}\\[1em] = \dfrac{13 \times 6}{6 \times -5}\\[1em] = -\dfrac{78}{30}\\[1em] = -\dfrac{13}{5}$

If $m$ = $\dfrac{2}{3}$ and $n$ = $\dfrac{3}{2}$ then $(m + n) ÷ (m - n) = -\dfrac{13}{5}$ .

Question 12(ii)

Find $(m + n) ÷ (m - n)$ , if;

$m = \dfrac{3}{4}$ and $n = \dfrac{4}{3}$

Answer

$(m + n) ÷ (m - n)\\[1em] = \Big(\dfrac{3}{4} + \dfrac{4}{3}\Big) ÷ \Big(\dfrac{3}{4} - \dfrac{4}{3}\Big)$

LCM of 4 and 3 is 2 x 2 x 3 = 12

$= \Big(\dfrac{3 \times 3}{4 \times 3} + \dfrac{4 \times 4}{3 \times 4}\Big) ÷ \Big(\dfrac{3 \times 3}{4 \times 3} - \dfrac{4 \times 4}{3 \times 4}\Big)\\[1em] = \Big(\dfrac{9}{12} + \dfrac{16}{12}\Big) ÷ \Big(\dfrac{9}{12} - \dfrac{16}{12}\Big)\\[1em] = \Big(\dfrac{9 + 16}{12}\Big) ÷ \Big(\dfrac{9 - 16}{12}\Big)\\[1em] = \Big(\dfrac{25}{12}\Big) ÷ \Big(\dfrac{-7}{12}\Big)\\[1em] = \dfrac{25}{12} \times \dfrac{12}{-7}\\[1em] = \dfrac{25 \times 12}{12 \times -7}\\[1em] = -\dfrac{300}{84}\\[1em] = -\dfrac{25}{7}$

If $m$ = $\dfrac{3}{4}$ and $n$ = $\dfrac{4}{3}$ then $(m + n) ÷ (m - n) = -\dfrac{25}{7}$ .

Question 12(iii)

Find $(m + n) ÷ (m - n)$ , if;

$m = \dfrac{4}{5}$ and $n = -\dfrac{3}{10}$

Answer

$(m + n) ÷ (m - n)\\[1em] =\Big[\dfrac{4}{5} + \Big(-\dfrac{3}{10}\Big)\Big] ÷ \Big[\dfrac{4}{5} - \Big(-\dfrac{3}{10}\Big)\Big]$

LCM of 5 and 10 is 2 x 5 = 10

$= \Big[\dfrac{4 \times 2}{5 \times 2} + \Big(-\dfrac{3 \times 1}{10 \times 1}\Big)\Big] ÷ \Big[\dfrac{4 \times 2}{5 \times 2} - \Big(-\dfrac{3 \times 1}{10 \times 1}\Big)\Big]\\[1em] = \Big[\dfrac{8}{10} + \Big(-\dfrac{3}{10}\Big)\Big] ÷ \Big[\dfrac{8}{10} - \Big(-\dfrac{3}{10}\Big)]\\[1em] = \Big[\dfrac{8 + (-3)}{10}\Big] ÷ \Big[\dfrac{8 - (-3)}{10}\Big]\\[1em] = \Big[\dfrac{5}{10}\Big] ÷ \Big[\dfrac{11}{10}\Big]\\[1em] = \dfrac{5}{10} \times \dfrac{10}{11}\\[1em] = \dfrac{5 \times 10}{10 \times 11}\\[1em] = \dfrac{50}{110}\\[1em] = \dfrac{5}{11}$

If $m$ = $\dfrac{4}{5}$ and $n$ = - $\dfrac{3}{10}$ then $(m + n) ÷ (m - n) = \dfrac{5}{11}$ .

Question 13

The product of two rational numbers is -5. If one of these numbers is $\dfrac{-7}{15}$ , find the other.

Answer

Let the number be $x$ .

$\dfrac{-7}{15} \times x = -5\\[1em] \Rightarrow x = -5 ÷ \dfrac{-7}{15}\\[1em] \Rightarrow x = -\dfrac{5}{1} \times \dfrac{-15}{7}\\[1em] \Rightarrow x = \dfrac{5 \times 15}{1 \times 7}\\[1em] \Rightarrow x = \dfrac{75}{7}\\[1em] \Rightarrow x = 10\dfrac{5}{7}$

The other number is $10\dfrac{5}{7}$ .

Question 14

Divide the sum of $\dfrac{5}{8}$ and $\dfrac{-11}{12}$ by the difference of $\dfrac{3}{7}$ and $\dfrac{5}{14}$ .

Answer

The sum of $\dfrac{5}{8}$ and $\dfrac{-11}{12}$

$\dfrac{5}{8} + \dfrac{-11}{12}$

LCM of 8 and 12 is 2 x 2 x 2 x 3 = 24

$= \dfrac{5 \times 3}{8 \times 3} + \dfrac{-11 \times 2}{12 \times 2}\\[1em] = \dfrac{15}{24} + \dfrac{-22}{24}\\[1em] = \dfrac{15 + (-22)}{24}\\[1em] = \dfrac{-7}{24}$

The difference of $\dfrac{3}{7}$ and $\dfrac{5}{14}$

$\dfrac{3}{7} - \dfrac{5}{14}$

LCM of 7 and 14 is 2 x 7 = 14

$= \dfrac{3 \times 2}{7 \times 2} - \dfrac{5 \times 1}{14 \times 1}\\[1em] = \dfrac{6}{14} - \dfrac{5}{14}\\[1em] = \dfrac{6 - 5}{14}\\[1em] = \dfrac{1}{14}$

Dividing the sum of $\dfrac{5}{8}$ and $\dfrac{-11}{12}$ by the difference of $\dfrac{3}{7}$ and $\dfrac{5}{14}$ ,

$\dfrac{-7}{24} ÷ \dfrac{1}{14}\\[1em] = \dfrac{-7}{24} \times \dfrac{14}{1}\\[1em] = -\dfrac{7 \times 14}{24 \times 1}\\[1em] = -\dfrac{98}{24}\\[1em] = -\dfrac{49}{12}\\[1em] = -4\dfrac{1}{12}$

$\Big(\dfrac{5}{8} + \dfrac{-11}{12}\Big) ÷ \Big(\dfrac{3}{7} - \dfrac{5}{14}\Big) = -4\dfrac{1}{12}$ .

Question 15

The area of a rectangular plate is $5\dfrac{5}{7}$ m² and its length is $3\dfrac{3}{4}$ m, find its breadth and its perimeter.

Answer

Area of a rectangular plate = $5\dfrac{5}{7}$ m² = $\dfrac{40}{7}$ m²

Length of a rectangular plate = $3\dfrac{3}{4}$ m = $\dfrac{15}{4}$ m

Let the breadth of the rectangular plate be b.

Area = length x breadth

$\dfrac{40}{7} = \dfrac{15}{4} \times b\\[1em] \Rightarrow b = \dfrac{40}{7} ÷ \dfrac{15}{4}\\[1em] \Rightarrow b = \dfrac{40}{7} \times \dfrac{4}{15}\\[1em] \Rightarrow b = \dfrac{40 \times 4}{7 \times 15}\\[1em] \Rightarrow b = \dfrac{160}{105}\\[1em] \Rightarrow b = \dfrac{32}{21}\\[1em] \Rightarrow b = 1\dfrac{11}{21}$

Perimeter = 2(length + breadth)

$= 2 \times \Big(\dfrac{15}{4} + \dfrac{32}{21}\Big)$

LCM of 4 and 21 is 2 x 2 x 3 x 7 = 84

$= 2 \times \Big(\dfrac{15 \times 21}{4 \times 21} + \dfrac{32 \times 4}{21 \times 4}\Big)\\[1em] = 2 \times \Big(\dfrac{315}{84} + \dfrac{128}{84}\Big)\\[1em] = 2 \times \Big(\dfrac{315 + 128}{84}\Big)\\[1em] = 2 \times \Big(\dfrac{443}{84}\Big)\\[1em] = \Big(\dfrac{443 \times 2}{84}\Big)\\[1em] = \Big(\dfrac{886}{84}\Big)\\[1em] = \Big(\dfrac{443}{42}\Big)\\[1em] = 10\dfrac{23}{42}$

Hence, breadth = $1\dfrac{11}{21}$ and perimeter = $10\dfrac{23}{42}$

Question 16

The area of a piece of paper is $7\dfrac{3}{26}$ cm² and its breadth is $2\dfrac{9}{13}$ cm. Find its length and perimeter.

Answer

Area of a rectangular paper = $7\dfrac{3}{26}$ cm² = $\dfrac{185}{26}$ cm²

Breadth of a rectangular paper = $2\dfrac{9}{13}$ m = $\dfrac{35}{13}$ cm

Let the length of the piece of paper be l.

Area = length x breadth

$\dfrac{185}{26} = l \times \dfrac{35}{13}\\[1em] \Rightarrow l = \dfrac{185}{26} ÷ \dfrac{35}{13}\\[1em] \Rightarrow l = \dfrac{185}{26} \times \dfrac{13}{35}\\[1em] \Rightarrow l = \dfrac{185 \times 13}{26 \times 35}\\[1em] \Rightarrow l = \dfrac{2405}{910}\\[1em] \Rightarrow l = \dfrac{37}{14}\\[1em] \Rightarrow l = 2\dfrac{9}{14}$

Perimeter = 2(length + breadth)

$= 2 \times \Big(\dfrac{37}{14} + \dfrac{35}{13}\Big)$

LCM of 14 and 13 is 2 x 7 x 13 = 182

$= 2 \times \Big(\dfrac{37 \times 13}{14 \times 13} + \dfrac{35 \times 14}{13 \times 14}\Big)\\[1em] = 2 \times \Big(\dfrac{481}{182} + \dfrac{490}{182}\Big)\\[1em] = 2 \times \Big(\dfrac{481 + 490}{182}\Big)\\[1em] = 2 \times \Big(\dfrac{971}{182}\Big)\\[1em] = \Big(\dfrac{971 \times 2}{182}\Big)\\[1em] = \Big(\dfrac{1942}{182}\Big)\\[1em] = \Big(\dfrac{971}{91}\Big)\\[1em] = 10\dfrac{61}{91}$

Length = $2\dfrac{9}{14}$ and perimeter = $10\dfrac{61}{91}$

Exercise 1(E)

Question 1(i)

In the following number line, points A and B represent:

$-2\dfrac{1}{5}$ and $2\dfrac{3}{5}$
$2\dfrac{1}{5}$ and $2\dfrac{3}{5}$
$-1\dfrac{4}{5}$ and $2\dfrac{3}{5}$
$-\dfrac{1}{5}$ and $\dfrac{2}{5}$

Answer

In this number line, there are 5 small lines between every 2 consecutive integers, which means moving one step left from 0 gives $-\dfrac{1}{5}$ .

Point A is 4 step left from -1. So, A = $-1\dfrac{4}{5}$

Point B is 3 step right from 2. So, B = $2\dfrac{3}{5}$

Hence, Option 3 is the correct option.

Question 1(ii)

Using the number line, given below; the length of line segment AB is:

$\dfrac{13}{5}$ = $2\dfrac{3}{5}$
$-\dfrac{13}{5}$ = $-2\dfrac{3}{5}$
$4\dfrac{2}{5}$
$\dfrac{14}{15}$

Answer

As we know, A = $-1\dfrac{4}{5}$ and B = $2\dfrac{3}{5}$ .

Length = $\dfrac{9}{5} + \dfrac{13}{5}$ $= \dfrac{9 + 13}{5}\\[1em] = \dfrac{22}{5}\\[1em] = 4\dfrac{2}{5}$

Hence, Option 3 is the correct option.

Question 1(iii)

The rational number between $\dfrac{a}{b}$ and $\dfrac{c}{d}$ is:

$\dfrac{1}{2}\Big(\dfrac{a}{b} - \dfrac{c}{d}\Big)$
$\Big(\dfrac{a - c}{b - d}\Big)$
$\Big(\dfrac{a + c}{b + d}\Big)$
$\Big(\dfrac{a + d}{b + c}\Big)$

Answer

For any two rational numbers $\dfrac{a}{b}$ and $\dfrac{c}{d}$ , $\Big(\dfrac{a + c}{b + d}\Big)$ is also a rational number with its value lying between $\dfrac{a}{b}$ and $\dfrac{c}{d}$ .

Hence, Option 3 is the correct option.

Question 1(iv)

The rational number between $\dfrac{1}{3}$ and $\dfrac{1}{2}$ is:

$\dfrac{3}{7}$ and $\dfrac{3}{8}$
$\dfrac{2}{5}$ and 0
$\dfrac{1}{6}$ and $\dfrac{2}{3}$
$\dfrac{2}{3}$ and $\dfrac{3}{2}$

Answer

As we know that, for any two rational numbers $\dfrac{a}{b}$ and $\dfrac{c}{d}$ , $\Big(\dfrac{a + c}{b + d}\Big)$ is also a rational number with its value lying between $\dfrac{a}{b}$ and $\dfrac{c}{d}$ .

The rational number between $\dfrac{1}{3}$ and $\dfrac{1}{2}$ is $\Big(\dfrac{1 + 1}{3 + 2}\Big)$

= $\Big(\dfrac{2}{5}\Big)$

The rational number between $\dfrac{1}{3}$ and $\dfrac{2}{5}$ is $\Big(\dfrac{1 + 2}{3 + 5}\Big)$

= $\Big(\dfrac{3}{8}\Big)$

The rational number between $\dfrac{1}{2}$ and $\dfrac{2}{5}$ is $\Big(\dfrac{1 + 2}{2 + 5}\Big)$

= $\Big(\dfrac{3}{7}\Big)$

$\dfrac{3}{7}$ and $\dfrac{3}{8}$ are two rational number between $\dfrac{1}{3}$ and $\dfrac{1}{2}$

Hence, option 1 is the correct option.

Question 1(v)

The rational numbers $-\dfrac{7}{4}$ and $\dfrac{3}{4}$ are represented by:

B and E respectively
C and D respectively
C and E respectively
B and F respectively

Answer

In this number line, there are 4 small lines between every 2 consecutive integers, which means moving one step towards left from 0 gives $-\dfrac{1}{4}$ .

Point C is 3 step left from -1. So, C = $-1\dfrac{3}{4} = -\dfrac{7}{4}$

Point E is 3 step right from 0. So, E = $\dfrac{3}{4}$

∴ $-\dfrac{7}{4}$ and $\dfrac{3}{4}$ are represented by C and E, respectively.

Hence, option 3 is the correct option.

Question 2

Draw a number line and mark

$\dfrac{3}{4}, \dfrac{7}{4}, \dfrac{-3}{4}$ and $\dfrac{-7}{4}$ on it.

Answer

Draw a number line as shown below:

Draw a number line and mark. Rational Numbers, Concise Mathematics Solutions ICSE Class 8.

In this number line

OA = AB = ...............= OA' = A'B' = 1 unit

Since the denominator of each given rational number is 4, divide each OA, AB, BC, OA', A'B',etc into four equal parts.

To represent $\dfrac{1}{4}$ , move one step towards the right side of 0 to reach P as shown.

As, OA = 1 unit, therefore OP = $\dfrac{1}{4}$ unit.

Hence, to represent $\dfrac{3}{4}$ , move 3 steps towards the right side of 0 to reach point Q. So, Q represent $\dfrac{3}{4}$ .

In the same way to represent $\dfrac{-3}{4}$ , move 3 steps towards the left side of 0 to reach point R. So, R represent $\dfrac{-3}{4}$ .

So, S represent $\dfrac{7}{4}$ and T represent $\dfrac{-7}{4}$ .

Question 3

On a number line mark the points

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

Answer

Draw a number line as shown below:

On a number line mark the points. Rational Numbers, Concise Mathematics Solutions ICSE Class 8.

In this number line

OA = AB = ...............= OA' = A'B' = 1 unit

Since the denominator of each given rational number is 3, divide each OA, AB, BC, OA', A'B', etc into three equal parts.

To represent $\dfrac{1}{3}$ , move one step towards the right side of 0 to reach P as shown.

As, OA = 1 unit, therefore OP = $\dfrac{1}{3}$ unit.

Hence, to represent $\dfrac{2}{3}$ , move 2 steps towards the right side of 0 to reach point Q. So, Q represents $\dfrac{2}{3}$ .

In the same way to represent $\dfrac{-2}{3}$ , move 2 steps towards the left side of 0 to reach point R. So, R represents $\dfrac{-2}{3}$ .

Similarly, S represents $\dfrac{-8}{3}$ , T represents $\dfrac{7}{3}$ and B' represents -2.

Question 4(i)

Insert one rational number between

$\dfrac{3}{5}$ and $\dfrac{5}{8}$

Answer

The rational number between $\dfrac{3}{5}$ and $\dfrac{5}{8}$ is $\Big(\dfrac{3 + 5}{5 + 8}\Big)$

= $\Big(\dfrac{8}{13}\Big)$

Hence, one rational number between $\dfrac{3}{5}$ and $\dfrac{5}{8}$ is $\dfrac{8}{13}$ .

Question 4(ii)

Insert one rational number between

$\dfrac{1}{2}$ and 2

Answer

The rational number between $\dfrac{1}{2}$ and $\dfrac{2}{1}$ is $\Big(\dfrac{1 + 2}{2 + 1}\Big)$

= $\Big(\dfrac{3}{3}\Big)$

Hence, one rational number between $\dfrac{1}{2}$ and 2 is 1.

Question 5

Insert two rational numbers between:

$\dfrac{5}{7}$ and $\dfrac{3}{8}$

Answer

Given numbers = $\dfrac{5}{7}$ and $\dfrac{3}{8}$

$= \dfrac{5}{7}, \dfrac{5 + 3}{7 + 8} ,\dfrac{3}{8} \\[1em] = \dfrac{5}{7}, \dfrac{8}{15}, \dfrac{3}{8} \\[1em] = \dfrac{5}{7}, \dfrac{5 + 8}{7 + 15}, \dfrac{8}{15}, \dfrac{3}{8} \\[1em] = \dfrac{5}{7}, \dfrac{13}{22}, \dfrac{8}{15}, \dfrac{3}{8}$

Hence, required rational numbers between $\dfrac{5}{7}$ and $\dfrac{3}{8}$ are : $\dfrac{13}{22}$ and $\dfrac{8}{15}$

Question 6

Insert three rational numbers between:

$\dfrac{8}{11}$ and $\dfrac{4}{9}$

Answer

Given numbers = $\dfrac{8}{11}$ and $\dfrac{4}{9}$

$= \dfrac{8}{11}, \dfrac{8 + 4}{11 + 9} ,\dfrac{4}{9} \\[1em] = \dfrac{8}{11}, \dfrac{12}{20}, \dfrac{4}{9} \\[1em] = \dfrac{8}{11}, \dfrac{3}{5}, \dfrac{4}{9} \\[1em] = \dfrac{8}{11}, \dfrac{8 + 3}{11 + 5}, \dfrac{3}{5}, \dfrac{4}{9} \\[1em] = \dfrac{8}{11}, \dfrac{11}{16}, \dfrac{3}{5}, \dfrac{4}{9}\\[1em] = \dfrac{8}{11}, \dfrac{11}{16}, \dfrac{3}{5},\dfrac{3 + 4}{5 + 9}, \dfrac{4}{9}\\[1em] = \dfrac{8}{11}, \dfrac{11}{16}, \dfrac{3}{5},\dfrac{7}{14}, \dfrac{4}{9}\\[1em] = \dfrac{8}{11}, \dfrac{11}{16}, \dfrac{3}{5},\dfrac{1}{2}, \dfrac{4}{9}$

Hence, required rational numbers between $\dfrac{8}{11}$ and $\dfrac{4}{9}$ are : $\dfrac{11}{16},\dfrac{3}{5}$ and $\dfrac{1}{2}$

Question 7

Insert five rational numbers between $\dfrac{3}{5}$ and $\dfrac{2}{3}$

Answer

LCM of 5 and 3 is 3 x 5 = 15

Make denominator of each given rational number equal to 15 (the LCM).

$\dfrac{3}{5} = \dfrac{3 \times 3}{5 \times 3} = \dfrac{9}{15}$

and

$\dfrac{2}{3} = \dfrac{2 \times 5}{3 \times 5} = \dfrac{10}{15}$

Since five rational numbers are required between $\dfrac{3}{5}$ and $\dfrac{2}{3}$ ; multiply the numerator and the denominator of each rational number by 5 + 1 = 6.

$\therefore \dfrac{9}{15} = \dfrac{9 \times 6}{15 \times 6} = \dfrac{54}{90}$

and

$\dfrac{10}{15} = \dfrac{10 \times 6}{15 \times 6} = \dfrac{60}{90}$

⇒ Required rational numbers between $\dfrac{3}{5}$ and $\dfrac{2}{3}$ are : $\dfrac{54}{90} , \dfrac{55}{90} , \dfrac{56}{90} , \dfrac{57}{90} , \dfrac{58}{90} , \dfrac{59}{90} , \dfrac{60}{90}$

= $\dfrac{3}{5} , \dfrac{11}{18} , \dfrac{28}{45} , \dfrac{19}{30} , \dfrac{29}{45} , \dfrac{59}{90} , \dfrac{2}{3}$

Hence, $\dfrac{11}{18} , \dfrac{28}{45} , \dfrac{19}{30} , \dfrac{29}{45}$ and $\dfrac{59}{90}$ lie between $\dfrac{3}{5}$ and $\dfrac{2}{3}$ .

Question 8

Insert six rational numbers between $\dfrac{5}{6}$ and $\dfrac{8}{9}$ .

Answer

LCM of 6 and 9 is 2 x 3 x 3 = 18

Make denominator of each given rational number equal to 18 (the LCM).

$\dfrac{5}{6} = \dfrac{5 \times 3}{6 \times 3} = \dfrac{15}{18}$

and

$\dfrac{8}{9} = \dfrac{8 \times 2}{9 \times 2} = \dfrac{16}{18}$

Since six rational numbers are required between $\dfrac{5}{6}$ and $\dfrac{8}{9}$ ; multiply the numerator and the denominator of each rational number by 6 + 1 = 7.

$\dfrac{15}{18} = \dfrac{15 \times 7}{18 \times 7} = \dfrac{105}{126}$

and

$\dfrac{16}{18} = \dfrac{16 \times 7}{18 \times 7} = \dfrac{112}{126}$

Required rational numbers between $\dfrac{5}{6}$ and $\dfrac{8}{9}$ are : $\dfrac{105}{126} , \dfrac{106}{126} , \dfrac{107}{126} , \dfrac{108}{126} , \dfrac{109}{126} , \dfrac{110}{126} , \dfrac{111}{126} , \dfrac{112}{126}$

= $\dfrac{5}{6} , \dfrac{53}{63} , \dfrac{107}{126} , \dfrac{6}{7} , \dfrac{109}{126} , \dfrac{55}{63} , \dfrac{37}{42} , \dfrac{8}{9}$

Hence, $\dfrac{5}{6} , \dfrac{53}{63} , \dfrac{107}{126} , \dfrac{6}{7} , \dfrac{109}{126}$ and $\dfrac{37}{42}$ lie between $\dfrac{5}{6}$ and $\dfrac{8}{9}$ .

Question 9

Insert seven rational numbers between 2 and 3.

Answer

2 = $\dfrac{2}{1}$

3 = $\dfrac{3}{1}$

Since seven rational numbers are required between $\dfrac{2}{1}$ and $\dfrac{3}{1}$ ; multiply the numerator and the denominator of each rational number by 7 + 1 = 8.

$\dfrac{2}{1} = \dfrac{2 \times 8}{1 \times 8} = \dfrac{16}{8}$

and

$\dfrac{3}{1} = \dfrac{3 \times 8}{1 \times 8} = \dfrac{24}{8}$

Required rational numbers between $\dfrac{2}{1}$ and $\dfrac{3}{1}$ are : $\dfrac{16}{8} , \dfrac{17}{8} , \dfrac{18}{8} , \dfrac{19}{8} , \dfrac{20}{8} , \dfrac{21}{8} , \dfrac{22}{8} , \dfrac{23}{8} , \dfrac{24}{8}$

= $\dfrac{2}{1} , \dfrac{17}{8} , \dfrac{9}{4} , \dfrac{19}{8} , \dfrac{5}{2} , \dfrac{21}{8} , \dfrac{11}{4} , \dfrac{23}{8} , \dfrac{3}{1}$

= $2 , 2\dfrac{1}{8} , 2\dfrac{1}{4} , 2\dfrac{3}{8} , 2\dfrac{1}{2} , 2\dfrac{5}{8} , 2\dfrac{3}{4} , 2\dfrac{7}{8} , 3$

Hence, $2\dfrac{1}{8} , 2\dfrac{1}{4} , 2\dfrac{3}{8} , 2\dfrac{1}{2} , 2\dfrac{5}{8} , 2\dfrac{3}{4}$ and $2\dfrac{7}{8}$ lie between 2 and 3.

Test Yourself

Question 1(i)

$\dfrac{0}{5}$ is a rational number, $\dfrac{0}{8}$ is a rational number, then $\dfrac{0}{5} ÷ \dfrac{0}{8}$ is:

an irrational number
a rational number
0
undefined

Answer

$\dfrac{0}{5} ÷ \dfrac{0}{8}\\[1em] = \dfrac{0}{5} \times \dfrac{8}{0}\\[1em] = \dfrac{0 \times 8}{5 \times 0}\\[1em] = \dfrac{0}{0}\\[1em]$ $\dfrac{0}{0}$ is undefined.

Hence, option 4 is the correct option.

Question 1(ii)

a and b are two rational numbers such that a + b = 0; then :

a = b
a and b are numerically equal
a and b are numerically equal but opposite in sign
none of the above

Answer

a + b = 0

a = 0 - b

a = -b

a and b are numerically equal but opposite in sign.

Hence, option 3 is the correct option

Question 1(iii)

The product of rational number $\dfrac{3}{8}$ and its additive inverse is :

1
0
$\dfrac{9}{64}$
$-\dfrac{9}{64}$

Answer

Additive inverse of $\dfrac{3}{8}$ = $-\dfrac{3}{8}$

$\dfrac{3}{8} \times -\dfrac{3}{8}\\[1em] = -\dfrac{3 \times 3}{8 \times 8}\\[1em] = -\dfrac{9}{64}$

Hence, option 4 is the correct option.

Question 1(iv)

The sum of rational number $\dfrac{2}{3}$ and its reciprocal is:

1
$2\dfrac{1}{6}$
0
$\dfrac{5}{6}$

Answer

Reciprocal of $\dfrac{2}{3}$ = $\dfrac{3}{2}$

We need to find the sum of $\dfrac{2}{3}$ and $\dfrac{3}{2}$

$\dfrac{2}{3} + \dfrac{3}{2}$

LCM of 3 and 2 is 2 x 3 = 6

$\dfrac{2 \times 2}{3 \times 2} + \dfrac{3 \times 3}{2 \times 3}\\[1em] = \dfrac{4}{6} + \dfrac{9}{6}\\[1em] = \dfrac{4 + 9}{6}\\[1em] = \dfrac{13}{6}\\[1em] = 2\dfrac{1}{6}$

Hence, option 2 is the correct option.

Question 1(v)

The product of two rational numbers is -1, if one of them is $\dfrac{2}{5}$ , then the other is :

$-\dfrac{2}{5}$
$\dfrac{5}{2}$
$-\dfrac{5}{2}$
none of these

Answer

Let the number be $x$ .

$\dfrac{2}{5} \times x = -1\\[1em] ⇒ \dfrac{2}{5} \times x = -\dfrac{1}{1}\\[1em] ⇒ x = -\dfrac{1}{1} ÷ \dfrac{2}{5}\\[1em] ⇒ x = -\dfrac{1}{1} \times \dfrac{5}{2}\\[1em] ⇒ x = -\dfrac{1 \times 5}{1 \times 2}\\[1em] ⇒ x = -\dfrac{5}{2}$

Hence, option 3 is the correct option.

Question 1(vi)

Statement 1: For a rational number $\dfrac{7}{9}, \dfrac{7}{9} - 0 = \dfrac{7}{9} \text{ and } 0 - \dfrac{7}{9} = - \dfrac{7}{9}$ . Hence, Subtraction has only right identity.

Statement 2: Subtraction has no identity.

Which of the following options is correct?

Both the statement are true.
Both the statement are false.
Statement 1 is true, and statement 2 is false.
Statement 1 is false, and statement 2 is true.

Answer

For a rational number $\dfrac{7}{9}, \dfrac{7}{9} - 0 = \dfrac{7}{9} \text{ and } 0 - \dfrac{7}{9} = - \dfrac{7}{9}$

We know that,

For any number a,

a - 0 = a and 0 - a ≠ 0

Thus, subtraction only has right identity.

So, statement 1 is true.

Subtraction has 0 as right identity element.

So, statement 2 is false.

Hence, option 3 is the correct option.

Question 1(vii)

Assertion (A) : Additive inverse of $\dfrac{2}{5}$ is $-\dfrac{5}{2}$ .

Reason (R) : For every non-zero rational number 'a', '-a' such that a + (-a) = 0.

Both A and R are correct, and R is the correct explanation for A.
Both A and R are correct, and R is not the correct explanation for A.
A is true, but R is false.
A is false, but R is true.

Answer

The additive inverse of a number a is a number -a such that :

⇒ a + (-a) = 0.

So, reason (R) is true.

According to Assertion: Additive inverse of $\dfrac{2}{5}$ is $-\dfrac{5}{2}$ .

$\Rightarrow \dfrac{2}{5} + (-\dfrac{5}{2})\\[1em] \Rightarrow \dfrac{2}{5} - \dfrac{5}{2}\\[1em] \Rightarrow \dfrac{4}{10} - \dfrac{25}{10}\\[1em] \Rightarrow \dfrac{4 - 25}{10}\\[1em] \Rightarrow \dfrac{-21}{10}\\[1em] \ne 0$

So, assertion (A) is false.

Hence, option 4 is the correct option.

Question 1(viii)

Assertion (A) : Multiplicative inverse of $-\dfrac{7}{5} \text{ is } -\dfrac{5}{7}$ .

Reason (R) : For every non-zero rational number 'a', there is a rational number $\dfrac{1}{a}$ such that $a \times \dfrac{1}{a} = 1$ .

Both A and R are correct, and R is the correct explanation for A.
Both A and R are correct, and R is not the correct explanation for A.
A is true, but R is false.
A is false, but R is true.

Answer

We know that,

The multiplicative inverse of any number a is it's reciprocal i.e. $\dfrac{1}{a}$ .

So, reason (R) is true.

The reciprocal of $-\dfrac{7}{5} = -\dfrac{1}{\dfrac{7}{5}} = -\dfrac{5}{7}$ .

So, assertion (A) is true.

∴ Both A and R are true and R is correct reason for A.

Hence, option 1 is the correct option.

Question 1(ix)

Assertion (A) : $\dfrac{1}{2} + 2 = \dfrac{5}{2}$ , which is a rational number.

Reason (R) : If $\dfrac{p}{q} \text{ and } \dfrac{r}{s}$ are any two rational numbers then $\dfrac{p}{q} + \dfrac{r}{s} = \dfrac{r}{s} + \dfrac{p}{q}$ .

Both A and R are correct, and R is the correct explanation for A.
Both A and R are correct, and R is not the correct explanation for A.
A is true, but R is false.
A is false, but R is true.

Answer

According to Assertion:

$\Rightarrow \dfrac{1}{2} + 2 \\[1em] \Rightarrow \dfrac{1}{2} + \dfrac{4}{2} \\[1em] \Rightarrow \dfrac{1 + 4}{2} \\[1em] \Rightarrow \dfrac{5}{2}$

A number is rational if it can be written in the form $\dfrac{p}{q}$ , where p and q are integers.

Since, $\dfrac{5}{2}$ is in the form of $\dfrac{p}{q}$ as well as 5 and 2 are integers.

So, assertion (A) is true.

According to commutative property of addition: When two numbers are added together, then a change in their positions does not change the result.

When $\dfrac{p}{q} \text{ and } \dfrac{r}{s}$ are any two rational numbers then $\dfrac{p}{q} + \dfrac{r}{s} = \dfrac{r}{s} + \dfrac{p}{q}$ , as addition of rational numbers is a commutative property.

So, reason (R) is true but it does not explain assertion.

Hence, option 2 is the correct option.

Question 1(x)

Assertion (A) : 0 and $\dfrac{11}{12}$ are two rational numbers and $\dfrac{11}{12} \ne 0$ then 0 ÷ $\dfrac{11}{12} = 0$ , a rational number.

Reason (R) : If a rational number is divided by some non - zero rational number, the result is always a rational number.

Both A and R are correct, and R is the correct explanation for A.
Both A and R are correct, and R is not the correct explanation for A.
A is true, but R is false.
A is false, but R is true.

Answer

When 0 is divided by any non-zero number, the result is 0.

⇒ 0 ÷ $\dfrac{11}{12} = 0$

Since, 0 = $\dfrac{0}{1}$ is in the form of $\dfrac{p}{q}$ .

So, assertion (A) is true.

The division of a rational number $\dfrac{a}{b}$ by another non-zero rational number $\dfrac{c}{d}$ is:

$\Rightarrow \dfrac{a}{b} ÷ \dfrac{c}{d}\\[1em] \Rightarrow \dfrac{a}{b} \times \dfrac{d}{c}\\[1em] \Rightarrow \dfrac{ad}{bc}$

Since, $\dfrac{ad}{bc}$ is in the form of $\dfrac{p}{q}$ .

So, reason is true. But it does not explains about assertion.

Hence, option 2 is the correct option.

Question 2(i)

Write the rational number that does not have a reciprocal.

Answer

The rational number is $\dfrac{0}{1}$

The reciprocal of $\dfrac{0}{1} = \dfrac{1}{0}$ (not defined).

Hence, 0 is the rational number that does not have a reciprocal.

Question 2(ii)

Write the rational numbers that are equal to their reciprocal.

Answer

Reciprocal of $\dfrac{1}{1} = \dfrac{1}{1} = 1$ .

Reciprocal of $\dfrac{-1}{1} = \dfrac{1}{-1} = -1$ .

The numbers 1 and -1 are their own reciprocal.

Question 2(iii)

Write the reciprocal of $-\dfrac{8}{17} + \dfrac{-8}{17}$ .

Answer

$-\dfrac{8}{17} + \dfrac{-8}{17}\\[1em] = \dfrac{-8 + (-8)}{17}\\[1em] = \dfrac{-16}{17}$

The reciprocal of $\dfrac{-16}{17} = -\dfrac{17}{16}$ .

Question 3

Write five rational numbers between $-\dfrac{3}{2}$ and $\dfrac{5}{3}$

Answer

LCM of 2 and 3 is 2 x 3 = 6

Make denominator of each given rational number equal to 6 (the LCM).

$-\dfrac{3}{2} = -\dfrac{3 \times 3}{2 \times 3} = -\dfrac{9}{6}$

and

$\dfrac{5}{3} = \dfrac{5 \times 2}{3 \times 2} = \dfrac{10}{6}$

The rational number between $-\dfrac{9}{6}$ and $\dfrac{10}{6}$ are $-\dfrac{8}{6}, -\dfrac{7}{6}, -\dfrac{6}{6}, -\dfrac{5}{6}, -\dfrac{4}{6}, -\dfrac{3}{6}, -\dfrac{2}{6}, -\dfrac{1}{6}, \dfrac{0}{6}, \dfrac{1}{6}, \dfrac{2}{6}, \dfrac{3}{6}, \dfrac{4}{6}, \dfrac{5}{6}, \dfrac{6}{6}, \dfrac{7}{6}, \dfrac{8}{6}, \dfrac{9}{6}$

From these rational numbers we can take any five rational number.

Hence, required rational numbers between $-\dfrac{3}{2}$ and $\dfrac{5}{3}$ are :

$-\dfrac{8}{6} , -\dfrac{6}{6} , -\dfrac{4}{6} , -\dfrac{2}{6} , \dfrac{2}{6}$

= $-\dfrac{4}{3} , -\dfrac{1}{1} , -\dfrac{2}{3} , -\dfrac{1}{3} , \dfrac{1}{3}$

Hence, $-1\dfrac{1}{3} , -1 , -\dfrac{2}{3} , -\dfrac{1}{3}$ and $\dfrac{1}{3}$ lies between $-\dfrac{3}{2}$ and $\dfrac{5}{3}$ .

Question 4

Write five rational number greater than -4.

Answer

There are infinite many rational number between -4 and ∞.

Hence, -3, -2, -1, 1, 2 are five rational number greater than -4.

Question 5

What should be added to $-2\dfrac{1}{2}$ to get $-3\dfrac{1}{3}$ ?

Answer

Let $x$ be added to $-2\dfrac{1}{2}$ .

$-2\dfrac{1}{2} + x = -3\dfrac{1}{3}\\[1em] \Rightarrow -\dfrac{5}{2} + x = -\dfrac{10}{3}\\[1em] \Rightarrow x = -\dfrac{10}{3} - \dfrac{-5}{2}$

LCM of 3 and 2 is 2 x 3 = 6

$\Rightarrow x = \dfrac{-10 \times 2}{3 \times 2} - \dfrac{-5 \times 3}{2 \times 3}\\[1em] \Rightarrow x = \dfrac{-20}{6} - \dfrac{-15}{6}\\[1em] \Rightarrow x = \dfrac{-20 - (-15)}{6}\\[1em] \Rightarrow x = \dfrac{-20 + 15}{6}\\[1em] \Rightarrow x = \dfrac{-5}{6}$

The number added to $-2\dfrac{1}{2}$ to get $-3\dfrac{1}{3}$ is $\dfrac{-5}{6}$ .

Question 6

Which should be subtracted from $2\dfrac{1}{2}$ to get $-3\dfrac{1}{3}$ ?

Answer

Let $x$ be subtracted from $2\dfrac{1}{2}$ .

$2\dfrac{1}{2} - x = -3\dfrac{1}{3}\\[1em] \Rightarrow \dfrac{5}{2} - x = -\dfrac{10}{3}\\[1em] \Rightarrow x = \dfrac{5}{2} - \dfrac{-10}{3}$

LCM of 2 and 3 is 2 x 3 = 6

$\Rightarrow x = \dfrac{5 \times 3}{2 \times 3} - \dfrac{-10 \times 2}{3 \times 2}\\[1em] \Rightarrow x = \dfrac{15}{6} - \dfrac{-20}{6}\\[1em] \Rightarrow x = \dfrac{15 - (-20)}{6}\\[1em] \Rightarrow x = \dfrac{15 + 20}{6}\\[1em] \Rightarrow x = \dfrac{35}{6}\\[1em] \Rightarrow x = 5\dfrac{5}{6}$

The number subtracted from $2\dfrac{1}{2}$ to get $-3\dfrac{1}{3}$ is $5\dfrac{5}{6}$ .

Question 7(i)

If $m = -\dfrac{7}{9}$ and $n = \dfrac{5}{6}$ , verify that:

m - n ≠ n - m

Answer

To prove:

m - n ≠ n - m

LHS:

$m - n\\[1em] -\dfrac{7}{9} - \dfrac{5}{6}$

LCM of 9 and 6 is 2 x 3 x 3 = 18

$-\dfrac{7 \times 2}{9 \times 2} - \dfrac{5 \times 3}{6 \times 3}\\[1em] = -\dfrac{14}{18} - \dfrac{15}{18}\\[1em] = \dfrac{-14 - 15}{18}\\[1em] = \dfrac{-29}{18}\\[1em] = -1\dfrac{11}{18}$

RHS:

$n - m\\[1em] \dfrac{5}{6} - \Big(-\dfrac{7}{9}\Big)\\[1em] = \dfrac{5}{6} + \dfrac{7}{9}$

LCM of 6 and 9 is 2 x 3 x 3 = 18

$= \dfrac{5 \times 3}{6 \times 3} + \dfrac{7 \times 2}{9 \times2}\\[1em] = \dfrac{15}{18} + \dfrac{14}{18}\\[1em] = \dfrac{15 + 14}{18}\\[1em] = \dfrac{29}{18}\\[1em] = 1\dfrac{11}{18}$

Hence, LHS ≠ RHS

m - n ≠ n - m

Question 7(ii)

If $m = -\dfrac{7}{9}$ and $n = \dfrac{5}{6}$ , verify that:

-(m + n) = (-m) + (-n)

Answer

To prove:

-(m + n) = (-m) + (-n)

LHS:

$-(m + n)\\[1em] -\Big(-\dfrac{7}{9} + \dfrac{5}{6}\Big)$

LCM of 9 and 6 is 2 x 3 x 3 = 18

$= -\Big(-\dfrac{7 \times 2}{9 \times 2} + \dfrac{5 \times 3}{6 \times 3}\Big)\\[1em] = -\Big(-\dfrac{14}{18} + \dfrac{15}{18}\Big)\\[1em] = -\Big(\dfrac{-14 + 15}{18}\Big)\\[1em] = -\Big(\dfrac{1}{18}\Big)$

RHS:

$(-m) + (-n)\\[1em] = -\Big(-\dfrac{7}{9}\Big) + \Big(-\dfrac{5}{6}\Big)\\[1em] = \Big(\dfrac{7}{9}\Big) + \Big(-\dfrac{5}{6}\Big)$

LCM of 9 and 6 is 2 x 3 x 3 = 18

$= \Big(\dfrac{7 \times 2}{9 \times 2}\Big) + \Big(-\dfrac{5 \times 3}{6 \times 3}\Big)\\[1em] = \Big(\dfrac{14}{18}\Big) + \Big(-\dfrac{15}{18}\Big)\\[1em] = \Big(\dfrac{14 + (-15)}{18}\Big)\\[1em] = \Big(\dfrac{-1}{18}\Big)$

Hence, LHS = RHS

$\therefore -(m + n) = (-m) + (-n)$

Question 8

Represent rational numbers $-\dfrac{7}{3}$ and $\dfrac{7}{4}$ on the same number line.

Answer

The rational numbers are $-\dfrac{7}{3}$ and $\dfrac{7}{4}$ .

LCM of 3 and 4 is 2 x 2 x 3 = 12

Hence, the rational number will be-

$-\dfrac{7 \times 4}{3 \times 4} , \dfrac{7 \times 3}{4 \times 3}\\[1em] = -\dfrac{28}{12} , \dfrac{21}{12}\\[1em] = -2\dfrac{4}{12} , 1\dfrac{9}{12}$

Draw a number line as shown below:

Represent rational numbers -7/3 and 7/4 on the same number line. Rational Numbers, Concise Mathematics Solutions ICSE Class 8.

In this number line OA = AB = ........... = OA' = A'B' = 1 unit

Since, the denominator of each given rational numbers is 12, divide each of OA, AB,...,OA', A'B', etc. into twelve equal parts.

To represent $\dfrac{1}{12}$ , moves one step towards the right side of O to reach point P as shown.

Hence, OA = 1 unit , therefore OP = $\dfrac{1}{12}$ unit and so P represents $\dfrac{1}{12}$ .

In the same way, to represent $-2\dfrac{4}{12}$ , move 4 steps toward the left side of B' to reach point Q. Clearly, Q represents $-\dfrac{7}{3}$ .

Similarly, to represent $1\dfrac{9}{12}$ , move 9 steps toward the right side of A to reach point R. Clearly, R represents $\dfrac{7}{4}$ .

Question 9(i)

Add: $-\dfrac{5}{6}$ and $\dfrac{3}{8}$

Answer

LCM of 6 and 8 is 2 x 2 x 2 x 3 = 24

$-\dfrac{5 \times 4}{6 \times 4} + \dfrac{3 \times 3}{8 \times 3} \\[1em] =-\dfrac{20}{24} + \dfrac{9}{24} \\[1em] =\dfrac{-20 + 9}{24} \\[1em] =\dfrac{-11}{24}$

∴ $-\dfrac{5}{6} + \dfrac{3}{8} = -\dfrac{11}{24}$

Question 9(ii)

Subtract: $\dfrac{3}{8}$ from $-\dfrac{5}{6}$

Answer

LCM of 8 and 6 is 2 x 2 x 2 x 3 = 24

$-\dfrac{5 \times 4}{6 \times 4} - \dfrac{3 \times 3}{8 \times 3} \\[1em] =-\dfrac{20}{24} - \dfrac{9}{24} \\[1em] =\dfrac{-20 - 9}{24} \\[1em] =\dfrac{-29}{24}\\[1em] =-1\dfrac{5}{24}$

∴ $-\dfrac{5}{6} - \dfrac{3}{8} = -1\dfrac{5}{24}$

Question 9(iii)

Multiply: $-\dfrac{5}{6}$ and $\dfrac{3}{8}$

Answer

$-\dfrac{5}{6} \times \dfrac{3}{8}\\[1em] = \dfrac{-5 \times 3}{6 \times 8}\\[1em] = \dfrac{-15}{48}\\[1em] = -\dfrac{5}{16}$

$-\dfrac{5}{6} \times \dfrac{3}{8} = -\dfrac{5}{16}$

Question 9(iv)

Divide: $\dfrac{3}{8}$ by $-\dfrac{5}{6}$

Answer

$\dfrac{3}{8} ÷ -\dfrac{5}{6}\\[1em] = \dfrac{3}{8} \times -\dfrac{6}{5}\\[1em] = \dfrac{3 \times (-6)}{8 \times 5}\\[1em] = \dfrac{-18}{40}\\[1em] = -\dfrac{9}{20}$

$\dfrac{3}{8} ÷ -\dfrac{5}{6} = -\dfrac{9}{20}$

Question 10(i)

By what number should $\dfrac{12}{17}$ be multiplied to get $-\dfrac{4}{7}$ ?

Answer

Let the number be $x$

$\dfrac{12}{17} \times x = -\dfrac{4}{7}\\[1em] ⇒ x = -\dfrac{4}{7} ÷ \dfrac{12}{17}\\[1em] ⇒ x = -\dfrac{4}{7} \times \dfrac{17}{12}\\[1em] ⇒ x = -\dfrac{4 \times 17}{7 \times 12}\\[1em] ⇒ x = -\dfrac{68}{84}\\[1em] ⇒ x = -\dfrac{17}{21}$

$-\dfrac{17}{21}$ must be multiplied by $\dfrac{12}{17}$ so that the product is $-\dfrac{4}{7}$ .

Question 10(ii)

By what number should $\dfrac{12}{17}$ be divided to get $-\dfrac{4}{7}$ ?

Answer

Let the number be $x$

$\dfrac{12}{17} ÷ x = -\dfrac{4}{7}\\[1em] \Rightarrow \dfrac{12}{17} \times \dfrac{1}{x} = -\dfrac{4}{7} \\[1em] \Rightarrow x = \dfrac{12}{17} \times -\dfrac{7}{4}\\[1em] \Rightarrow x = -\dfrac{12 \times 7}{17 \times 4}\\[1em] \Rightarrow x = -\dfrac{84}{68}\\[1em] \Rightarrow x = -\dfrac{21}{17}\\[1em] \Rightarrow x = -1\dfrac{4}{17}$

$-1\dfrac{4}{17}$ must be divided by $\dfrac{12}{17}$ so that the product is $-\dfrac{4}{7}$ .

Chapter 1

Rational Numbers

Class - 8 Concise Mathematics Selina

Exercise 1(A)

Question 1(i)

Question 1(ii)

Question 1(iii)

Question 1(iv)

Question 1(v)

Question 1(vi)

Question 2(i)

Question 2(ii)

Question 2(iii)

Question 2(iv)

Question 2(v)

Question 2(vi)

Question 2(vii)

Question 2(viii)

Question 3(i)

Question 3(ii)

Question 3(iii)

Question 3(iv)

Question 3(v)

Question 3(vi)

Question 3(vii)

Question 3(viii)

Question 3(ix)

Question 4(i)

Question 4(ii)

Question 4(iii)

Question 4(iv)

Question 5(i)

Question 5(ii)

Question 5(iii)

Question 5(iv)

Question 5(v)

Question 5(vi)

Question 6(i)

Question 6(ii)

Question 6(iii)

Question 6(iv)

Question 7(i)

Question 7(ii)

Question 7(iii)

Question 7(iv)

Question 7(v)

Question 7(vi)

Question 8

Question 9

Question 10

Exercise 1(B)

Question 1(i)

Question 1(ii)

Question 1(iii)

Question 1(iv)

Question 1(v)

Question 2(i)

Question 2(ii)

Question 2(iii)

Question 2(iv)

Question 2(v)

Question 2(vi)

Question 3(i)

Question 3(ii)

Question 3(iii)

Question 3(iv)

Question 3(v)

Question 3(vi)

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10(i)

Question 10(ii)

Question 11(i)

Question 11(ii)

Question 11(iii)

Exercise 1(C)