Rational Numbers Class 8 Selina Solutions

✦ Class 8 Mathematics · Selina

Master Rational Numbers

Complete, step-by-step solutions for Exercises 1(A) to 1(E), with interactive answer panels, searchable questions and clean mathematical notation.

Start Exercise 1(A) ↓
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Mathematics equations and rational number learning
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Complete Solutions

Exercise 1(A)

Tap any answer panel to reveal the full working.

Exercise 1(A)

Question 1(i)

A number which is not rational is called

a natural number
an integer
an irrational number
a whole number
Show step-by-step answer+

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

Hence, Option 3 is the correct option.

Exercise 1(A)

Question 1(ii)

If x ≠ 0 then value of
\(\frac{0}{x}\)
is :
a rational number
not a rational number
not equal to zero
none of these.
Show step-by-step answer+

If x ≠ 0 then value of
\(\frac{0}{x}\)
is a rational number.

Hence, Option 1 is the correct option.

Exercise 1(A)

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 integer
Show step-by-step answer+
Given,

5x + 7 = 0

⇒ 5x = -7

⇒ x =
\(\frac{- 7}{5}\)

As x is the form of
\(\frac{p}{q}\)
where p and q both are integers and q ≠ 0,

∴ x is a rational number.

Hence, Option 3 is the correct option.

Exercise 1(A)

Question 1(iv)

Rational number
\(\frac{p}{q}\)
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.
Show step-by-step answer+

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

Hence, Option 3 is the correct option.

Exercise 1(A)

Question 1(v)

The addition of two rational numbers
\(\frac{a}{b}\)
and
\(\frac{c}{d}\)
is commutative, if
\(\frac{a}{b} + \frac{c}{d}\)
is a rational number
\(\frac{a + c}{b + c}\)
is a rational number.
\(\frac{a}{b} + \frac{c}{d}\)
=
\(\frac{c}{d} + \frac{a}{b}\)
\(\frac{a}{b} - \frac{c}{d}\)
=
\(\frac{c}{d} - \frac{a}{b}\)
Show step-by-step answer+
The addition of two rational number
\(\frac{a}{b}\)
and
\(\frac{c}{d}\)
is commutative, if
\(\frac{a}{b} + \frac{c}{d}\)
=
\(\frac{c}{d} + \frac{a}{b}\)

Hence, Option 3 is the correct option.

Exercise 1(A)

Question 1(vi)

\(- \frac{3}{7}\)
+ additive inverse of
\(- \frac{3}{7}\)
is

1

0

\(\frac{6}{7}\)
\(- \frac{6}{7}\)
Show step-by-step answer+

Hence, Option 2 is the correct option.

Exercise 1(A)

Question 2(i)

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

\(\frac{- 5}{8}\)
and
\(\frac{3}{8}\)
Show step-by-step answer+
\(\frac{- 5}{8} + \frac{3}{8} = \frac{- 5 + 3}{8} = \frac{- 2}{8}\)

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


\(\frac{- 2}{8}\)
is a rational number.
Exercise 1(A)

Question 2(ii)

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

\(\frac{- 8}{13}\)
and
\(\frac{- 4}{13}\)
Show step-by-step answer+
\(\frac{- 8}{13} + \frac{- 4}{13} = \frac{- 8 + ( - 4 )}{13} = \frac{- 12}{13}\)

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


\(\frac{- 12}{13}\)
is a rational number.
Exercise 1(A)

Question 2(iii)

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

\(\frac{6}{11}\)
and
\(\frac{- 9}{11}\)
Show step-by-step answer+
\(\frac{6}{11} + \frac{- 9}{11} = \frac{6 + ( - 9 )}{11} = \frac{- 3}{11}\)

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


\(\frac{- 3}{11}\)
is a rational number.
Exercise 1(A)

Question 2(iv)

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

\(\frac{5}{- 26}\)
and
\(\frac{8}{39}\)
Show step-by-step answer+
\(\frac{5}{- 26} + \frac{8}{39} = \frac{- 5}{26} + \frac{8}{39}\)
💡 LCM of 26 and 39 is 2 x 3 x 13 = 78
\(\frac{- 5 \times 3}{26 \times 3} + \frac{8 \times 2}{39 \times 2} = \frac{- 15}{78} + \frac{16}{78} = \frac{- 15 + 16}{78} = \frac{1}{78}\)

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


\(\frac{1}{78}\)
is a rational number.
Exercise 1(A)

Question 2(v)

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

\(\frac{5}{- 6}\)
and
\(\frac{2}{3}\)
Show step-by-step answer+
\(\frac{5}{- 6} + \frac{2}{3} = \frac{- 5}{6} + \frac{2}{3}\)
💡 LCM of 6 and 3 is 2 x 3 = 6
\(\frac{- 5 \times 1}{6 \times 1} + \frac{2 \times 2}{3 \times 2} = \frac{- 5}{6} + \frac{4}{6} = \frac{- 5 + 4}{6} = \frac{- 1}{6}\)

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


\(\frac{- 1}{6}\)
is a rational number.
Exercise 1(A)

Question 2(vi)

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

-2 and
\(\frac{2}{5}\)
Show step-by-step answer+
\(- 2 + \frac{2}{5} = \frac{- 2}{1} + \frac{2}{5}\)
💡 LCM of 1 and 5 is 5
\(\frac{- 2 \times 5}{1 \times 5} + \frac{2 \times 1}{5 \times 1} = \frac{- 10}{5} + \frac{2}{5} = \frac{- 10 + 2}{5} = \frac{- 8}{5}\)

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


\(\frac{- 8}{5}\)
is a rational number.
Exercise 1(A)

Question 2(vii)

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

\(\frac{9}{- 4}\)
and
\(\frac{- 3}{8}\)
Show step-by-step answer+
\(\frac{9}{- 4} + \frac{- 3}{8} = \frac{- 9}{4} + \frac{- 3}{8}\)
💡 LCM of 4 and 8 is 2 x 2 x 2 = 8
\(\frac{- 9 \times 2}{4 \times 2} + \frac{- 3 \times 1}{8 \times 1} = \frac{- 18}{8} + \frac{- 3}{8} = \frac{- 18 + ( - 3 )}{8} = \frac{- 21}{8}\)

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


\(\frac{- 21}{8}\)
is a rational number.
Exercise 1(A)

Question 2(viii)

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

\(\frac{7}{- 18}\)
and
\(\frac{8}{27}\)
Show step-by-step answer+
\(\frac{7}{- 18} + \frac{8}{27} = \frac{- 7}{18} + \frac{8}{27}\)
💡 LCM of 18 and 27 is 2 x 3 x 3 x 3 = 54
\(\frac{- 7 \times 3}{18 \times 3} + \frac{8 \times 2}{27 \times 2} = \frac{- 21}{54} + \frac{16}{54} = \frac{- 21 + 16}{54} = \frac{- 5}{54}\)

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


\(\frac{- 5}{54}\)
is a rational number.
Exercise 1(A)

Question 3(i)

Evaluate:

\(\frac{5}{9} + \frac{- 7}{6}\)
Show step-by-step answer+
💡 LCM of 9 and 6 is 2 x 3 x 3 = 18
\(\frac{5 \times 2}{9 \times 2} + \frac{- 7 \times 3}{6 \times 3} = \frac{10}{18} + \frac{- 21}{18} = \frac{10 + ( - 21 )}{18} = \frac{- 11}{18}\)

\(\frac{5}{9} + \frac{- 7}{6}\)
=
\(\frac{- 11}{18}\)
Exercise 1(A)

Question 3(ii)

Evaluate:

\(4 + \frac{3}{- 5}\)
Show step-by-step answer+
\(\frac{4}{1} + \frac{- 3}{5}\)
💡 LCM of 1 and 5 is 5
\(\frac{4 \times 5}{1 \times 5} + \frac{- 3 \times 1}{5 \times 1} = \frac{20}{5} + \frac{- 3}{5} = \frac{20 + ( - 3 )}{5} = \frac{17}{5} = 3 \frac{2}{5}\)

\(4 + \frac{3}{- 5} = 3 \frac{2}{5}\)
Exercise 1(A)

Question 3(iii)

Evaluate:

\(\frac{1}{- 15} + \frac{5}{- 12}\)
Show step-by-step answer+
\(\frac{- 1}{15} + \frac{- 5}{12}\)
💡 LCM of 15 and 12 is 2 x 2 x 3 x 5 = 60
\(\frac{- 1 \times 4}{15 \times 4} + \frac{- 5 \times 5}{12 \times 5} = \frac{- 4}{60} + \frac{- 25}{60} = \frac{- 4 + ( - 25 )}{60} = \frac{- 29}{60}\)

\(\frac{1}{- 15} + \frac{5}{- 12} = \frac{- 29}{60}\)
Exercise 1(A)

Question 3(iv)

Evaluate:

\(\frac{5}{9} + \frac{3}{- 4}\)
Show step-by-step answer+
\(\frac{5}{9} + \frac{- 3}{4}\)
💡 LCM of 9 and 4 is 2 x 2 x 3 x 3 = 36
\(\frac{5 \times 4}{9 \times 4} + \frac{- 3 \times 9}{4 \times 9} = \frac{20}{36} + \frac{- 27}{36} = \frac{20 + ( - 27 )}{36} = \frac{- 7}{36}\)

\(\frac{5}{9} + \frac{3}{- 4}\)
=
\(\frac{- 7}{36}\)
Exercise 1(A)

Question 3(v)

Evaluate:

\(\frac{- 8}{9} + \frac{- 5}{12}\)
Show step-by-step answer+
\(\frac{- 8}{9} + \frac{- 5}{12}\)
💡 LCM of 9 and 12 is 2 x 2 x 3 x 3 = 36
\(\frac{- 8 \times 4}{9 \times 4} + \frac{- 5 \times 3}{12 \times 3} = \frac{- 32}{36} + \frac{- 15}{36} = \frac{- 32 + ( - 15 )}{36} = \frac{- 47}{36}\)

\(\frac{- 8}{9}\)
+
\(\frac{- 5}{12}\)
=
\(\frac{- 47}{36}\)
Exercise 1(A)

Question 3(vi)

Evaluate:

\(0 + \frac{- 2}{7}\)
Show step-by-step answer+
\(\frac{0}{1} + \frac{- 2}{7}\)
💡 LCM of 1 and 7 is 7
\(\frac{0 \times 7}{1 \times 7} + \frac{- 2 \times 1}{7 \times 1} = \frac{0}{7} + \frac{- 2}{7} = \frac{0 + ( - 2 )}{7} = \frac{- 2}{7}\)
∴ 0 +
\(\frac{- 2}{7}\)
=
\(\frac{- 2}{7}\)
Exercise 1(A)

Question 3(vii)

Evaluate:

\(\frac{5}{- 11} + 0\)
+0
Show step-by-step answer+
\(\frac{- 5}{11} + \frac{0}{1}\)
💡 LCM of 11 and 1 is 11
\(\frac{- 5 \times 1}{11 \times 1} + \frac{0 \times 11}{1 \times 11} = \frac{- 5}{11} + \frac{0}{11} = \frac{- 5 + 0}{11} = \frac{- 5}{11}\)

\(\frac{5}{- 11}\)
+ 0 =
\(\frac{- 5}{11}\)
Exercise 1(A)

Question 3(viii)

Evaluate:

\(2 + \frac{- 3}{5}\)
Show step-by-step answer+
\(\frac{2}{1} + \frac{- 3}{5}\)
💡 LCM of 1 and 5 is 5
\(\frac{2 \times 5}{1 \times 5} + \frac{- 3 \times 1}{5 \times 1} = \frac{10}{5} + \frac{- 3}{5} = \frac{10 + ( - 3 )}{5} = \frac{7}{5}\)
∴ 2 +
\(\frac{- 3}{5}\)
=
\(\frac{7}{5}\)
Exercise 1(A)

Question 3(ix)

Evaluate:

\(\frac{4}{- 9} + 1\)
+1
Show step-by-step answer+
\(\frac{- 4}{9} + \frac{1}{1}\)
💡 LCM of 9 and 1 is 3 x 3 = 9
\(\frac{- 4 \times 1}{9 \times 1} + \frac{1 \times 9}{1 \times 9} = \frac{- 4}{9} + \frac{9}{9} = \frac{- 4 + 9}{9} = \frac{5}{9}\)

\(\frac{4}{- 9}\)
+ 1 =
\(\frac{5}{9}\)
Exercise 1(A)

Question 4(i)

Evaluate:

\(\frac{3}{7} + \frac{- 4}{9} + \frac{- 11}{7} + \frac{7}{9}\)
Show step-by-step answer+
💡 LCM of 7 and 9 is 3 x 3 x 7 = 63
\(\frac{3 \times 9}{7 \times 9} + \frac{- 4 \times 7}{9 \times 7} + \frac{- 11 \times 9}{7 \times 9} + \frac{7 \times 7}{9 \times 7} = \frac{27}{63} + \frac{- 28}{63} + \frac{- 99}{63} + \frac{49}{63} = \frac{27 + ( - 28 ) + ( - 99 ) + 49}{63} = \frac{- 51}{63}\)

\(\frac{3}{7} + \frac{- 4}{9} + \frac{- 11}{7} + \frac{7}{9} = \frac{- 51}{63}\)
Exercise 1(A)

Question 4(ii)

Evaluate:

\(\frac{2}{3} + \frac{- 4}{5} + \frac{1}{3} + \frac{2}{5}\)
Show step-by-step answer+
💡 LCM of 3 and 5 is 3 x 5 = 15
\(\frac{2 \times 5}{3 \times 5} + \frac{- 4 \times 3}{5 \times 3} + \frac{1 \times 5}{3 \times 5} + \frac{2 \times 3}{5 \times 3} = \frac{10}{15} + \frac{- 12}{15} + \frac{5}{15} + \frac{6}{15} = \frac{10 + ( - 12 ) + 5 + 6}{15} = \frac{9}{15}\)

\(\frac{2}{3} + \frac{- 4}{5} + \frac{1}{3} + \frac{2}{5} = \frac{9}{15}\)
Exercise 1(A)

Question 4(iii)

Evaluate:

\(\frac{4}{7} + 0 + \frac{- 8}{9} + \frac{- 13}{7} + \frac{17}{9}\)
Show step-by-step answer+
\(\frac{4}{7} + \frac{0}{1} + \frac{- 8}{9} + \frac{- 13}{7} + \frac{17}{9}\)
💡 LCM of 7 ,1 and 9 is 3 x 3 x 7 = 63
\(\frac{4 \times 9}{7 \times 9} + \frac{0 \times 63}{1 \times 63} + \frac{- 8 \times 7}{9 \times 7} + \frac{- 13 \times 9}{7 \times 9} + \frac{17 \times 7}{9 \times 7} = \frac{36}{63} + \frac{0}{63} + \frac{- 56}{63} + \frac{- 117}{63} + \frac{119}{63} = \frac{36 + 0 + ( - 56 ) + ( - 117 ) + 119}{63} = \frac{- 18}{63}\)

\(\frac{4}{7} + 0 + \frac{- 8}{9} + \frac{- 13}{7} + \frac{17}{9} = \frac{- 18}{63}\)
Exercise 1(A)

Question 4(iv)

Evaluate:

\(\frac{3}{8} + \frac{- 5}{12} + \frac{3}{7} + \frac{3}{12} + \frac{- 5}{8} + \frac{- 2}{7}\)
Show step-by-step answer+
\(( \frac{3}{8} + \frac{- 5}{8} ) + ( \frac{- 5}{12} + \frac{3}{12} ) + ( \frac{3}{7} + \frac{- 2}{7} ) = \frac{- 2}{8} + \frac{- 2}{12} + \frac{1}{7} = \frac{- 1}{4} + \frac{- 1}{6} + \frac{1}{7}\)
💡 LCM of 4 ,6 and 7 is 2 x 2 x 3 x 7 = 84
\(\frac{- 1 \times 21}{4 \times 21} + \frac{- 1 \times 14}{6 \times 14} + \frac{1 \times 12}{7 \times 12} = \frac{- 21}{84} + \frac{- 14}{84} + \frac{12}{84} = \frac{( - 21 ) + ( - 14 ) + 12}{84} = \frac{- 23}{84}\)

\(\frac{3}{8} + \frac{- 5}{12} + \frac{3}{7} + \frac{3}{12} + \frac{- 5}{8} + \frac{- 2}{7} = \frac{- 23}{84}\)
Exercise 1(A)

Question 5(i)

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

\(\frac{- 8}{7}\)
and
\(\frac{5}{14}\)
Show step-by-step answer+
To prove:
\(\frac{- 8}{7} + \frac{5}{14} = \frac{5}{14} + \frac{- 8}{7}\)
Taking LHS:
\(\frac{- 8}{7} + \frac{5}{14}\)
💡 LCM of 7 and 14 is 2 x 7 = \(14 = \frac{- 8 \times 2}{7 \times 2} + \frac{5 \times 1}{14 \times 1} = \frac{- 16}{14} + \frac{5}{14} = \frac{- 16 + 5}{14} = - 11\)
Taking RHS:
\(\frac{5}{14} + \frac{- 8}{7}\)
💡 LCM of 14 and 7 is 2 x 7 = \(14 = \frac{5 \times 1}{14 \times 1} + \frac{- 8 \times 2}{7 \times 2} = \frac{5}{14} + \frac{- 16}{14} = \frac{5 + ( - 16 )}{14} = - 11\)

∴ LHS = RHS

Hence,
\(\frac{- 8}{7} + \frac{5}{14} = \frac{5}{14} + \frac{- 8}{7}\)

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

Exercise 1(A)

Question 5(ii)

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

\(\frac{5}{9}\)
and
\(\frac{5}{- 12}\)
Show step-by-step answer+
To prove:
\(\frac{5}{9} + \frac{5}{- 12} = \frac{5}{- 12} + \frac{5}{9}\)

Taking LHS:
\(\frac{5}{9} + \frac{5}{- 12} = \frac{5}{9} + \frac{- 5}{12}\)

💡 LCM of 9 and 12 is 2 x 2 x 3 x 3 = \(36 = \frac{5 \times 4}{9 \times 4} + \frac{- 5 \times 3}{12 \times 3} = \frac{20}{36} + \frac{- 15}{36} = \frac{20 + ( - 15 )}{36} = 5\)

Taking RHS:
\(\frac{5}{- 12} + \frac{5}{9} = \frac{- 5}{12} + \frac{5}{9}\)

💡 LCM of 12 and 9 is 2 x 2 x 3 x 3 = \(36 = \frac{- 5 \times 3}{12 \times 3} + \frac{5 \times 4}{9 \times 4} = \frac{- 15}{36} + \frac{20}{36} = \frac{( - 15 ) + 20}{36} = 5\)

∴ LHS = RHS

Hence,
\(\frac{5}{9} + \frac{5}{- 12} = \frac{5}{- 12} + \frac{5}{9}\)

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

Exercise 1(A)

Question 5(iii)

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

\(\frac{- 4}{5}\)
and
\(\frac{- 13}{- 15}\)
Show step-by-step answer+
To prove:
\(\frac{- 4}{5} + \frac{- 13}{- 15} = \frac{- 13}{- 15} + \frac{- 4}{5}\)

Taking LHS:
\(\frac{- 4}{5} + \frac{- 13}{- 15} = \frac{- 4}{5} + \frac{13}{15}\)

💡 LCM of 5 and 15 is 3 x 5 = \(15 = \frac{- 4 \times 3}{5 \times 3} + \frac{13 \times 1}{15 \times 1} = \frac{- 12}{15} + \frac{13}{15} = \frac{- 12 + 13}{15} = 1\)

Taking RHS:
\(\frac{- 13}{- 15} + \frac{- 4}{5} = \frac{13}{15} + \frac{- 4}{5}\)

💡 LCM of 15 and 5 is 3 x 5 = \(15 = \frac{13 \times 1}{15 \times 1} + \frac{- 4 \times 3}{5 \times 3} = \frac{13}{15} + \frac{- 12}{15} = \frac{13 + ( - 12 )}{15} = 1\)

∴ LHS = RHS

Hence,
\(\frac{- 4}{5} + \frac{- 13}{- 15} = \frac{- 13}{- 15} + \frac{- 4}{5}\)

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

Exercise 1(A)

Question 5(iv)

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

\(\frac{2}{- 5}\)
and
\(\frac{11}{- 15}\)
Show step-by-step answer+
To prove:
\(\frac{2}{- 5} + \frac{11}{- 15} = \frac{11}{- 15} + \frac{2}{- 5}\)

Taking LHS:
\(\frac{2}{- 5} + \frac{11}{- 15} = \frac{- 2}{5} + \frac{- 11}{15}\)

💡 LCM of 5 and 15 is 3 x 5 = \(15 = \frac{- 2 \times 3}{5 \times 3} + \frac{- 11 \times 1}{15 \times 1} = \frac{- 6}{15} + \frac{- 11}{15} = \frac{- 6 + ( - 11 )}{15} = - 17\)

Taking RHS:
\(\frac{11}{- 15} + \frac{2}{- 5} = \frac{- 11}{15} + \frac{- 2}{5}\)

💡 LCM of 15 and 5 is 3 x 5 = \(15 = \frac{- 11 \times 1}{15 \times 1} + \frac{- 2 \times 3}{5 \times 3} = \frac{- 11}{15} + \frac{- 6}{15} = \frac{( - 11 ) + ( - 6 )}{15} = - 17\)

∴ LHS = RHS

Hence,
\(\frac{2}{- 5} + \frac{11}{- 15} = \frac{11}{- 15} + \frac{2}{- 5}\)

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

Exercise 1(A)

Question 5(v)

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

3 and
\(\frac{- 2}{7}\)
Show step-by-step answer+
To prove:

\(3 + \frac{- 2}{7} = \frac{- 2}{7} + 3\)
+3

Taking LHS:
\(3 + \frac{- 2}{7} = \frac{3}{1} + \frac{- 2}{7}\)

💡 LCM of 1 and 7 is \(7 = \frac{3 \times 7}{1 \times 7} + \frac{- 2 \times 1}{7 \times 1} = \frac{21}{7} + \frac{- 2}{7} = \frac{21 + ( - 2 )}{7} = 19\)

Taking RHS:
\(\frac{- 2}{7} + 3 = \frac{- 2}{7} + \frac{3}{1}\)

💡 LCM of 7 and 1 is \(7 = \frac{- 2 \times 1}{7 \times 1} + \frac{3 \times 7}{1 \times 7} = \frac{- 2}{7} + \frac{21}{7} = \frac{( - 2 ) + 21}{7} = 19\)

∴ LHS = RHS

Hence,
\(3 + \frac{- 2}{7} = \frac{- 2}{7} + 3\)
+3

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

Exercise 1(A)

Question 5(vi)

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

-2 and
\(\frac{3}{- 5}\)
Show step-by-step answer+
To prove:

\(- 2 + \frac{3}{- 5} = \frac{3}{- 5} + - 2\)
+−2

Taking LHS:
\(- 2 + \frac{3}{- 5} = \frac{- 2}{1} + \frac{- 3}{5}\)

💡 LCM of 1 and 5 is \(5 = \frac{- 2 \times 5}{1 \times 5} + \frac{- 3 \times 1}{5 \times 1} = \frac{- 10}{5} + \frac{- 3}{5} = \frac{- 10 + ( - 3 )}{5} = - 13\)

Taking RHS:
\(\frac{3}{- 5} + - 2 = \frac{- 3}{5} + \frac{- 2}{1}\)

💡 LCM of 5 and 1 is \(5 = \frac{- 3 \times 1}{5 \times 1} + \frac{- 2 \times 5}{1 \times 5} = \frac{- 3}{5} + \frac{- 10}{5} = \frac{( - 3 ) + ( - 10 )}{5} = - 13\)

∴ LHS = RHS

Hence,
\(- 2 + \frac{3}{- 5} = \frac{3}{- 5} + - 2\)
+−2

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

Exercise 1(A)

Question 6(i)

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

\(\frac{1}{2} , \frac{2}{3}\)
and
\(\frac{- 1}{6}\)
Show step-by-step answer+
To prove:

\(( \frac{1}{2} + \frac{2}{3} ) + \frac{- 1}{6} = \frac{1}{2} + ( \frac{2}{3} + \frac{- 1}{6} )\)
)

Taking LHS:
\(( \frac{1}{2} + \frac{2}{3} ) + \frac{- 1}{6}\)
💡 LCM of 2 and 3 is 2 x 3 = 6
\(= ( \frac{1 \times 3}{2 \times 3} + \frac{2 \times 2}{3 \times 2} ) + \frac{- 1}{6} = ( \frac{3}{6} + \frac{4}{6} ) + \frac{- 1}{6} = ( \frac{3 + 4}{6} ) + \frac{- 1}{6} = \frac{7}{6} + \frac{- 1}{6} = \frac{7 - 1}{6} = \frac{6}{6} = 1\)

=1

Taking RHS:
\(\frac{1}{2} + ( \frac{2}{3} + \frac{- 1}{6} )\)
)

💡 LCM of 3 and 6 is 2 x 3 = 6
\(\frac{1}{2} + ( \frac{2 \times 2}{3 \times 2} + \frac{- 1 \times 1}{6 \times 1} ) = \frac{1}{2} + ( \frac{4}{6} + \frac{- 1}{6} ) = \frac{1}{2} + ( \frac{4 + ( - 1 )}{6} ) = \frac{1}{2} + \frac{3}{6}\)
💡 LCM of 2 and 6 is 2 x 3 = 6
\(= \frac{1 \times 3}{2 \times 3} + \frac{3 \times 1}{6 \times 1} = \frac{3}{6} + \frac{3}{6} = \frac{3 + 3}{6} = \frac{6}{6} = 1\)

=1

∴ LHS = RHS

\(( \frac{1}{2} + \frac{2}{3} ) + \frac{- 1}{6} = \frac{1}{2} + ( \frac{2}{3} + \frac{- 1}{6} )\)
)

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

Exercise 1(A)

Question 6(ii)

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

\(\frac{- 2}{5} , \frac{4}{15}\)
and
\(\frac{- 7}{10}\)
Show step-by-step answer+
To prove:

\(( \frac{- 2}{5} + \frac{4}{15} ) + \frac{- 7}{10} = \frac{- 2}{5} + ( \frac{4}{15} + \frac{- 7}{10} )\)
)

Taking LHS:
\(( \frac{- 2}{5} + \frac{4}{15} ) + \frac{- 7}{10}\)
💡 LCM of 5 and 15 is 3 x 5 = 15
\(= ( \frac{- 2 \times 3}{5 \times 3} + \frac{4 \times 1}{15 \times 1} ) + \frac{- 7}{10} = ( \frac{- 6}{15} + \frac{4}{15} ) + \frac{- 7}{10} = ( \frac{- 6 + 4}{15} ) + \frac{- 7}{10} = \frac{- 2}{15} + \frac{- 7}{10}\)
💡 LCM of 15 and 10 is 2 x 3 x 5 = 30
\(= \frac{- 2 \times 2}{15 \times 2} + \frac{- 7 \times 3}{10 \times 3} = \frac{- 4}{30} + \frac{- 21}{30} = \frac{- 4 + ( - 21 )}{30} = \frac{- 25}{30} = \frac{- 5}{6}\)

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

💡 LCM of 15 and 10 is 2 x 3 x 5 = 30
\(\frac{- 2}{5} + ( \frac{4 \times 2}{15 \times 2} + \frac{- 7 \times 3}{10 \times 3} ) = \frac{- 2}{5} + ( \frac{8}{30} + \frac{- 21}{30} ) = \frac{- 2}{5} + ( \frac{8 + ( - 21 )}{30} ) = \frac{- 2}{5} + \frac{- 13}{30}\)
💡 LCM of 5 and 30 is 2 x 3 x 5 = 30
\(= \frac{- 2 \times 6}{5 \times 6} + \frac{- 13 \times 1}{30 \times 1} = \frac{- 12}{30} + \frac{- 13}{30} = \frac{- 12 + ( - 13 )}{30} = \frac{- 25}{30} = \frac{- 5}{6}\)

∴ LHS = RHS

\(( \frac{- 2}{5} + \frac{4}{15} ) + \frac{- 7}{10} = \frac{- 2}{5} + ( \frac{4}{15} + \frac{- 7}{10} )\)
)

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

Exercise 1(A)

Question 6(iii)

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

\(\frac{- 7}{9} , \frac{2}{- 3}\)
and
\(\frac{- 5}{18}\)
Show step-by-step answer+

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

Taking LHS:
\(( \frac{- 7}{9} + \frac{2}{- 3} ) + \frac{- 5}{18} = ( \frac{- 7}{9} + \frac{- 2}{3} ) + \frac{- 5}{18}\)
💡 LCM of 9 and 3 is 3 x 3 = 9
\(= ( \frac{- 7 \times 1}{9 \times 1} + \frac{- 2 \times 3}{3 \times 3} ) + \frac{- 5}{18} = ( \frac{- 7}{9} + \frac{- 6}{9} ) + \frac{- 5}{18} = ( \frac{- 7 + ( - 6 )}{9} ) + \frac{- 5}{18} = \frac{- 13}{9} + \frac{- 5}{18}\)
💡 LCM of 9 and 18 is 2 x 9 = \(18 = \frac{- 13 \times 2}{9 \times 2} + \frac{- 5 \times 1}{18 \times 1} = \frac{- 26}{18} + \frac{- 5}{18} = \frac{- 26 + ( - 5 )}{18} = - 31\)
Taking RHS:

\(\frac{- 7}{9} + ( \frac{2}{- 3} + \frac{- 5}{18} ) = \frac{- 7}{9} + ( \frac{- 2}{3} + \frac{- 5}{18} )\)
)

💡 LCM of 3 and 18 is 2 x 3 x 9 = 18
\(\frac{- 7}{9} + ( \frac{- 2 \times 6}{3 \times 6} + \frac{- 5 \times 1}{18 \times 1} ) = \frac{- 7}{9} + ( \frac{- 12}{18} + \frac{- 5}{18} ) = \frac{- 7}{9} + ( \frac{- 12 + ( - 5 )}{18} ) = \frac{- 7}{9} + \frac{- 17}{18}\)
💡 LCM of 9 and 18 is 2 x 3 x 3 = \(18 = \frac{- 7 \times 2}{9 \times 2} + \frac{- 17 \times 1}{18 \times 1} = \frac{- 14}{18} + \frac{- 17}{18} = \frac{- 14 + ( - 17 )}{18} = - 31\)

∴ LHS = RHS

\(( \frac{- 7}{9} + \frac{2}{- 3} ) + \frac{- 5}{18} = \frac{- 7}{9} + ( \frac{2}{- 3} + \frac{- 5}{18} )\)
)

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

Exercise 1(A)

Question 6(iv)

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

\(- 1 , \frac{5}{6}\)
and
\(\frac{- 2}{3}\)
Show step-by-step answer+

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

Taking LHS:
\(( - 1 + \frac{5}{6} ) + \frac{- 2}{3} = ( \frac{- 1}{1} + \frac{5}{6} ) + \frac{- 2}{3}\)
💡 LCM of 1 and 6 is 2 x 3 = 6
\(= ( \frac{- 1 \times 6}{1 \times 6} + \frac{5 \times 1}{6 \times 1} ) + \frac{- 2}{3} = ( \frac{- 6}{6} + \frac{5}{6} ) + \frac{- 2}{3} = ( \frac{- 6 + 5}{6} ) + \frac{- 2}{3} = \frac{- 1}{6} + \frac{- 2}{3}\)
💡 LCM of 6 and 3 is 2 x 3 = \(6 = \frac{- 1 \times 1}{6 \times 1} + \frac{- 2 \times 2}{3 \times 2} = \frac{- 1}{6} + \frac{- 4}{6} = \frac{- 1 + ( - 4 )}{6} = - 5\)

Taking RHS:
\(- 1 + ( \frac{5}{6} + \frac{- 2}{3} ) = \frac{- 1}{1} + ( \frac{5}{6} + \frac{- 2}{3} )\)
)

💡 LCM of 6 and 3 is 2 x 3 = 6
\(\frac{- 1}{1} + ( \frac{5 \times 1}{6 \times 1} + \frac{- 2 \times 2}{3 \times 2} ) = \frac{- 1}{1} + ( \frac{5}{6} + \frac{- 4}{6} ) = \frac{- 1}{1} + ( \frac{5 + ( - 4 )}{6} ) = \frac{- 1}{1} + \frac{1}{6}\)
💡 LCM of 1 and 6 is 2 x 3 = \(6 = \frac{- 1 \times 6}{1 \times 6} + \frac{1 \times 1}{6 \times 1} = \frac{- 6}{6} + \frac{1}{6} = \frac{- 6 + 1}{6} = - 5\)

∴ LHS = RHS

\(( - 1 + \frac{5}{6} ) + \frac{- 2}{3} = - 1 + ( \frac{5}{6} + \frac{- 2}{3} )\)
)

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

Exercise 1(A)

Question 7(i)

Write the additive inverse (negative) of:

\(\frac{- 3}{8}\)
Show step-by-step answer+
Additive inverse of
\(\frac{- 3}{8}\)
=
\(- ( \frac{- 3}{8} )\)
)
=
\(\frac{3}{8}\)
Exercise 1(A)

Question 7(ii)

Write the additive inverse (negative) of:

\(\frac{4}{- 9}\)
Show step-by-step answer+
\(\frac{4}{- 9}\)
=
\(\frac{- 4}{9}\)
Additive inverse of
\(\frac{- 4}{9}\)
=
\(- ( \frac{- 4}{9} )\)
)
=
\(\frac{4}{9}\)
Exercise 1(A)

Question 7(iii)

Write the additive inverse (negative) of:

\(\frac{- 4}{- 13}\)
Show step-by-step answer+
\(\frac{- 4}{- 13}\)
=
\(\frac{4}{13}\)
Additive inverse of
\(\frac{4}{13}\)
=
\(- ( \frac{4}{13} )\)
)
=
\(- \frac{4}{13}\)
Exercise 1(A)

Question 7(iv)

Write the additive inverse (negative) of:

0

Show step-by-step answer+
0 =
\(\frac{0}{1}\)
Additive inverse of
\(\frac{0}{1}\)
=
\(- ( \frac{0}{1} )\)
)

= 0

Exercise 1(A)

Question 7(v)

Write the additive inverse (negative) of:

-2

Show step-by-step answer+
-2 =
\(\frac{- 2}{1}\)
Additive inverse of
\(\frac{- 2}{1}\)
=
\(- ( \frac{- 2}{1} )\)
)
=
\(\frac{2}{1} = 2\)
=2
Exercise 1(A)

Question 7(vi)

Write the additive inverse (negative) of:

1

Show step-by-step answer+
1 =
\(\frac{1}{1}\)
Additive inverse of
\(\frac{1}{1}\)
=
\(- ( \frac{1}{1} )\)
)
= -
\(\frac{1}{1} = - 1\)
=−1
Exercise 1(A)

Question 8

Fill in the blanks:

(i) Additive inverse of
\(\frac{- 5}{- 12}\)
= ............... .
(ii)
\(\frac{- 5}{- 12}\)
+ its additive inverse = ............... .
(iii) If
\(\frac{a}{b}\)
is the additive inverse of
\(\frac{- c}{d}\)
, then
\(\frac{- c}{d}\)
is the additive inverse of ............... .
And, so
\(\frac{a}{b} + \frac{- c}{d} = \frac{- c}{d} + \frac{a}{b}\)
= ............... .
Show step-by-step answer+

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

(ii)
\(\frac{- 5}{- 12}\)
+ its additive inverse = 0
(iii) If
\(\frac{a}{b}\)
is the additive inverse of
\(\frac{- c}{d}\)
, then
\(\frac{- c}{d}\)
is the additive inverse of
\(\frac{a}{b}\)
.
And, so
\(\frac{a}{b} + \frac{- c}{d} = \frac{- c}{d} + \frac{a}{b} = 0\)
=0.
Explanation

(i)
\(\frac{- 5}{- 12} = \frac{5}{12}\)

Additive inverse of
\(\frac{5}{12} = - \frac{5}{12}\)

(ii\() \frac{5}{12} + ( - \frac{5}{12}\)
)

\(= \frac{5 - 5}{12} = 0\)

=0

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

Exercise 1(A)

Question 9

State, true or false:

(i)
\(\frac{7}{9} = \frac{7 + 5}{9 + 5}\)
(ii)
\(\frac{7}{9} = \frac{7 - 5}{9 - 5}\)
(iii)
\(\frac{7}{9} = \frac{7 \times 5}{9 \times 5}\)
(iv)
\(\frac{7}{9} = \frac{7 \div 5}{9 \div 5}\)
(v)
\(\frac{- 5}{- 12}\)
is a negative rational number.
(vi)
\(\frac{- 13}{25}\)
is smaller than
\(\frac{- 25}{13}\)
Show step-by-step answer+

(i) False.

Reason:

\(\frac{7 + 5}{9 + 5} = \frac{12}{14}\)

\(\frac{7}{9}\)

(ii) False

Reason:

\(\frac{7 - 5}{9 - 5} = \frac{2}{4}\)

\(\frac{7}{9}\)

(iii) True

Reason:

\(\frac{7\times5}{9\times5}=\frac79\)

\(\frac{7}{9}\)

(iv) True

Reason:
\(\frac{7\div5}{9\div5}=\frac{7\times\frac15}{9\times\frac15}=\frac79\)
\(\frac{7}{9}\)

(v) False

Reason:

\(\frac{- 5}{- 12} = \frac{5}{12}\)
is a positive rational number.

(vi) False

Reason:

We need to check if
\(\frac{- 13}{25}\)
is smaller than
\(\frac{- 25}{13}\)

💡 LCM of 25 and 13 is 325
\(\frac{- 13}{25} = \frac{- 13 \times 13}{25 \times 13} = \frac{- 169}{325}\)
\(\frac{- 25}{13} = \frac{- 25 \times 25}{13 \times 25} = \frac{- 625}{325}\)

And,
\(\frac{- 169}{325}\)
>
\(\frac{- 625}{325}\)

Hence,
\(\frac{- 13}{25}\)
>
\(\frac{- 25}{13}\)

Exercise 1(A)

Question 10

The weight of an empty fruit basket is
\(2\,\frac{1}{3}\)
kg. It contains
\(5\,\frac{5}{6}\)
kg grapes and
\(8\,\frac{3}{8}\)
kg mangoes. Find the total weight of basket with fruits.
Show step-by-step answer+
The weight of an empty fruit basket =
\(2\,\frac{1}{3}\)
kg
The weight of grapes =
\(5\,\frac{5}{6}\)
kg
The weight of mangoes =
\(8\,\frac{3}{8}\)
kg

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

\(= 2 \frac{1}{3} \mathrm{kg} + 5 \frac{5}{6} \mathrm{kg} + 8 \frac{3}{8} \mathrm{kg} = \frac{7}{3} \mathrm{kg} + \frac{35}{6} \mathrm{kg} + \frac{67}{8} \mathrm{kg}\)
kg

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

\(= \frac{7 \times 8}{3 \times 8} + \frac{35 \times 4}{6 \times 4} + \frac{67 \times 3}{8 \times 3} \mathrm{kg} = \frac{56}{24} + \frac{140}{24} + \frac{201}{24} \mathrm{kg} = \frac{56 + 140 + 201}{24} \mathrm{kg} = \frac{397}{24} \mathrm{kg} = 16 \frac{13}{24} \mathrm{kg}\)
kg

Total weight of basket with fruits =
\(16\,\frac{13}{24}\)
kg.
Complete Solutions

Exercise 1(B)

Tap any answer panel to reveal the full working.

Exercise 1(B)

Question 1(i)

The sum of two rational numbers is 8, if one of them is
\(2\,\frac{3}{4}\)
, the other number is
\(6\,\frac{3}{4}\)
\(6\,\frac{1}{4}\)
\(5\,\frac{1}{4}\)
\(5\,\frac{3}{4}\)
Show step-by-step answer+

Let
x be the other number.
\(2\,\frac{3}{4} + x = 8 \Rightarrow \frac{11}{4} + x = 8 \Rightarrow x = 8 - \frac{11}{4} \Rightarrow x = \frac{8}{1} - \frac{11}{4}\)

💡 LCM of 1 and 4 is 2 x 2 = \(4 \Rightarrow x = \frac{8 \times 4}{1 \times 4} - \frac{11 \times 1}{4 \times 1} \Rightarrow x = \frac{32}{4} - \frac{11}{4} \Rightarrow x = \frac{32 - 11}{4} \Rightarrow x = \frac{21}{4} \Rightarrow x = 5\,\frac{1}{4}\)

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

Hence, option 3 is correct option.

Exercise 1(B)

Question 1(ii)

For three rational numbers
\(\frac{a}{b}\)
,
\(\frac{c}{d}\)
and
\(\frac{e}{f}\)
, we have:
\(\frac{a}{b} - ( \frac{c}{d} - \frac{e}{f} ) = \frac{a}{b} - \frac{c}{d} + \frac{e}{f}\)
\(\frac{a}{b} - ( \frac{c}{d} - \frac{e}{f} ) = \frac{a}{b} - \frac{c}{d} - \frac{e}{f}\)
\(\frac{a}{b} + ( \frac{c}{d} + \frac{e}{f} )\)
) ≠
\(( \frac{a}{b} + \frac{c}{d} ) + \frac{e}{f}\)
\(\frac{a}{b} + ( \frac{c}{d} + \frac{e}{f} ) = ( \frac{a}{b} + \frac{c}{d} ) + \frac{a}{b} + \frac{e}{f}\)
Show step-by-step answer+
\(\frac{a}{b} - ( \frac{c}{d} - \frac{e}{f} ) = \frac{a}{b} - \frac{c}{d} + \frac{e}{f}\)

Hence, option 1 is correct option.

Exercise 1(B)

Question 1(iii)

The sum of two rational numbers is -6. If one of them is
\(4\,\frac{1}{2}\)
, the other number is:
\(2\,\frac{1}{2}\)
\(1\,\frac{1}{2}\)
-
\(1\,\frac{1}{2}\)
-
\(10\,\frac{1}{2}\)
Show step-by-step answer+

Let
x be the other number.
\(4\,\frac{1}{2} + x = - 6 \Rightarrow \frac{9}{2} + x = - 6 \Rightarrow x = - 6 - \frac{9}{2} \Rightarrow x = \frac{- 6}{1} - \frac{9}{2}\)

💡 LCM of 1 and 2 is \(2 \Rightarrow x = - \frac{6 \times 2}{1 \times 2} - \frac{9 \times 1}{2 \times 1} \Rightarrow x = - \frac{12}{2} - \frac{9}{2} \Rightarrow x = \frac{- 12 - 9}{2} \Rightarrow x = \frac{- 21}{2} \Rightarrow x = -10\,\frac{1}{2}\)

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

Hence, option 4 is correct option.

Exercise 1(B)

Question 1(iv)

The number subtracted from
\(5\,\frac{2}{3}\)
to get -
\(1\,\frac{2}{3}\)
is :

4

-
\(7\,\frac{1}{3}\)
\(6\,\frac{1}{3}\)
\(7\,\frac{1}{3}\)
Show step-by-step answer+

Let
x be subtracted from
\(5\,\frac{2}{3}\)
.

\(5\,\frac{2}{3} - x = - 1 \frac{2}{3} \Rightarrow \frac{17}{3} - x = - \frac{5}{3} \Rightarrow x = \frac{17}{3} + \frac{5}{3} \Rightarrow x = \frac{17 + 5}{3} \Rightarrow x = \frac{22}{3} \Rightarrow x = 7 \frac{1}{3}\)

The number subtracted from
\(5\,\frac{2}{3}\)
to get -
\(1\,\frac{2}{3}\)
is
\(7\,\frac{1}{3}\)
.

Hence, option 4 is correct option.

Exercise 1(B)

Question 1(v)

The number added to
\(5\,\frac{2}{3}\)
to get -
\(1\,\frac{2}{3}\)
is :

4

-
\(7\,\frac{1}{3}\)
\(6\,\frac{1}{3}\)
\(7\,\frac{1}{3}\)
Show step-by-step answer+

Let
x be added to
\(5\,\frac{2}{3}\)
.
\(5\,\frac{2}{3} + x = - 1 \frac{2}{3} \Rightarrow \frac{17}{3} + x = - \frac{5}{3} \Rightarrow x = \frac{- 5}{3} - \frac{17}{3} \Rightarrow x = \frac{- 5 - 17}{3} \Rightarrow x = \frac{- 22}{3} \Rightarrow x = - 7 \frac{1}{3}\)

The number added to
\(5\,\frac{2}{3}\)
to get -1
\(\frac{2}{3}\)
is -
\(7\,\frac{1}{3}\)
.

Hence, option 2 is correct option.

Exercise 1(B)

Question 2(i)

Evaluate:

\(\frac{2}{3} - \frac{4}{5}\)
Show step-by-step answer+
\(\frac{2}{3} - \frac{4}{5}\)
💡 LCM of 3 and 5 is 3 x 5 = 15
\(\frac{2 \times 5}{3 \times 5} - \frac{4 \times 3}{5 \times 3} = \frac{10}{15} - \frac{12}{15} = \frac{10 - 12}{15} = \frac{- 2}{15}\)
Exercise 1(B)

Question 2(ii)

Evaluate:

\(\frac{- 4}{9} - \frac{2}{- 3}\)
Show step-by-step answer+
\(\frac{- 4}{9} - \frac{2}{- 3} = \frac{- 4}{9} - \frac{- 2}{3}\)
💡 LCM of 9 and 3 is 3 x 3 = 9
\(\frac{- 4 \times 1}{9 \times 1} - \frac{- 2 \times 3}{3 \times 3} = \frac{- 4}{9} - \frac{- 6}{9} = \frac{- 4 - ( - 6 )}{9} = \frac{- 4 + 6}{9} = \frac{2}{9}\)
Exercise 1(B)

Question 2(iii)

Evaluate:

\(- 1 - \frac{4}{9}\)
Show step-by-step answer+
\(- 1 - \frac{4}{9} = \frac{- 1}{1} - \frac{4}{9}\)
💡 LCM of 1 and 9 is 3 x 3 = 9
\(\frac{- 1 \times 9}{1 \times 9} - \frac{4 \times 1}{9 \times 1} = \frac{- 9}{9} - \frac{4}{9} = \frac{- 9 - 4}{9} = \frac{- 13}{9} = - 1 \frac{4}{9}\)
Exercise 1(B)

Question 2(iv)

Evaluate:

\(\frac{- 2}{7} - \frac{3}{- 14}\)
Show step-by-step answer+
\(\frac{- 2}{7} - \frac{3}{- 14} = \frac{- 2}{7} - \frac{- 3}{14}\)
💡 LCM of 7 and 14 is 2 x 14 = 14
\(\frac{- 2 \times 2}{7 \times 2} - \frac{- 3 \times 1}{14 \times 1} = \frac{- 4}{14} - \frac{- 3}{14} = \frac{- 4 - ( - 3 )}{14} = \frac{- 4 + 3}{14} = \frac{- 1}{14}\)
Exercise 1(B)

Question 2(v)

Evaluate:

\(\frac{- 5}{18} - \frac{- 2}{9}\)
Show step-by-step answer+
\(\frac{- 5}{18} - \frac{- 2}{9}\)
💡 LCM of 18 and 9 is 2 x 3 x 3 = 18
\(\frac{- 5 \times 1}{18 \times 1} - \frac{- 2 \times 2}{9 \times 2} = \frac{- 5}{18} - \frac{- 4}{18} = \frac{- 5 - ( - 4 )}{18} = \frac{- 5 + 4}{18} = \frac{- 1}{18}\)
Exercise 1(B)

Question 2(vi)

Evaluate:

\(\frac{5}{21} - \frac{- 13}{42}\)
Show step-by-step answer+
\(\frac{5}{21} - \frac{- 13}{42}\)
💡 LCM of 21 and 42 is 2 x 3 x 7 = 42
\(\frac{5 \times 2}{21 \times 2} - \frac{- 13 \times 1}{42 \times 1} = \frac{10}{42} - \frac{- 13}{42} = \frac{10 - ( - 13 )}{42} = \frac{10 + 13}{42} = \frac{23}{42}\)
Exercise 1(B)

Question 3(i)

Subtract:

\(\frac{5}{8}\)
from
\(\frac{- 3}{8}\)
Show step-by-step answer+
\(\frac{- 3}{8} - \frac{5}{8} = \frac{- 3 - 5}{8} = \frac{- 8}{8} = - 1\)

=−1

Exercise 1(B)

Question 3(ii)

Subtract:

\(\frac{- 8}{11}\)
from
\(\frac{4}{11}\)
Show step-by-step answer+
\(\frac{4}{11} - \frac{- 8}{11} = \frac{4 - ( - 8 )}{11} = \frac{4 + 8}{11} = \frac{12}{11} = 1 \frac{1}{11}\)
Exercise 1(B)

Question 3(iii)

Subtract:

\(\frac{4}{9}\)
from
\(\frac{- 5}{9}\)
Show step-by-step answer+
\(\frac{- 5}{9} - \frac{4}{9} = \frac{- 5 - 4}{9} = \frac{- 9}{9} = - 1\)
=−1
Exercise 1(B)

Question 3(iv)

Subtract:

\(\frac{1}{4}\)
from
\(\frac{- 3}{8}\)
Show step-by-step answer+
\(\frac{- 3}{8} - \frac{1}{4}\)
💡 LCM of 8 and 4 is 2 x 2 x 2 = 8
\(\frac{- 3 \times 1}{8 \times 1} - \frac{1 \times 2}{4 \times 2} = \frac{- 3}{8} - \frac{2}{8} = \frac{- 3 - 2}{8} = - \frac{5}{8}\)
Exercise 1(B)

Question 3(v)

Subtract:

\(\frac{- 5}{8}\)
from
\(\frac{- 13}{16}\)
Show step-by-step answer+
\(\frac{- 13}{16} - \frac{- 5}{8}\)
💡 LCM of 16 and 8 is 2 x 2 x 2 x 2 = 16
\(\frac{- 13 \times 1}{16 \times 1} - \frac{- 5 \times 2}{8 \times 2} = \frac{- 13}{16} - \frac{- 10}{16} = \frac{- 13 - ( - 10 )}{16} = \frac{- 13 + 10}{16} = \frac{- 3}{16}\)
Exercise 1(B)

Question 3(vi)

Subtract:

\(\frac{- 9}{22}\)
from
\(\frac{5}{33}\)
Show step-by-step answer+
\(\frac{5}{33} - \frac{- 9}{22}\)
💡 LCM of 33 and 22 is 2 x 3 x 11 = 66
\(\frac{5 \times 2}{33 \times 2} - \frac{- 9 \times 3}{22 \times 3} = \frac{10}{66} - \frac{- 27}{66} = \frac{10 - ( - 27 )}{66} = \frac{10 + 27}{66} = \frac{37}{66}\)
Exercise 1(B)

Question 4

The sum of two rational numbers is
\(\frac{9}{20}\)
. If one of them is
\(\frac{2}{5}\)
, find the other.
Show step-by-step answer+

Let
x be the other number.
\(\frac{2}{5} + x = \frac{9}{20} \Rightarrow x = \frac{9}{20} - \frac{2}{5}\)

💡 LCM of 20 and 5 is 2 x 2 x 5 = \(20 \Rightarrow x = \frac{9 \times 1}{20 \times 1} - \frac{2 \times 4}{5 \times 4} \Rightarrow x = \frac{9}{20} - \frac{8}{20} \Rightarrow x = \frac{9 - 8}{20} \Rightarrow x = 1\)

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

Exercise 1(B)

Question 5

The sum of two rational numbers is
\(\frac{- 2}{3}\)
. If one of them is
\(\frac{- 8}{15}\)
, find the other.
Show step-by-step answer+

Let
x be the other number.
\(\frac{- 8}{15} + x = \frac{- 2}{3} \Rightarrow x = \frac{- 2}{3} - \frac{- 8}{15}\)

💡 LCM of 3 and 15 is 3 x 5 = \(15 \Rightarrow x = \frac{- 2 \times 5}{3 \times 5} - \frac{- 8 \times 1}{15 \times 1} \Rightarrow x = \frac{- 10}{15} - \frac{- 8}{15} \Rightarrow x = \frac{- 10 - ( - 8 )}{15} \Rightarrow x = \frac{- 10 + 8}{15} \Rightarrow x = - 2\)

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

Exercise 1(B)

Question 6

The sum of two rational numbers is -6. If one of them is
\(\frac{- 8}{5}\)
, find the other.
Show step-by-step answer+

Let
x be the other number.
\(\frac{- 8}{5} + x = - 6 \Rightarrow x = \frac{- 6}{1} - \frac{- 8}{5}\)

💡 LCM of 1 and 5 is \(5 \Rightarrow x = \frac{- 6 \times 5}{1 \times 5} - \frac{- 8 \times 1}{5 \times 1} \Rightarrow x = \frac{- 30}{5} - \frac{- 8}{5} \Rightarrow x = \frac{- 30 - ( - 8 )}{5} \Rightarrow x = \frac{- 30 + 8}{5} \Rightarrow x = \frac{- 22}{5} \Rightarrow x = -4\,\frac{2}{5}\)

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

Exercise 1(B)

Question 7

Which rational number should be added to
\(\frac{- 7}{8}\)
to get
\(\frac{5}{9}\)
?
Show step-by-step answer+

Let
x be added to
\(\frac{- 7}{8}\)
.

\(\frac{- 7}{8} + x = \frac{5}{9} \Rightarrow x = \frac{5}{9} - \frac{- 7}{8}\)
💡 LCM of 9 and 8 is 2 x 2 x 2 x 3 x 3 = \(72 \Rightarrow x = \frac{5 \times 8}{9 \times 8} - \frac{- 7 \times 9}{8 \times 9} \Rightarrow x = \frac{40}{72} - \frac{- 63}{72} \Rightarrow x = \frac{40 - ( - 63 )}{72} \Rightarrow x = \frac{40 + 63}{72} \Rightarrow x = \frac{103}{72} \Rightarrow x = 1\,\frac{31}{72}\)

The number added to
\(\frac{- 7}{8}\)
to get
\(\frac{5}{9}\)
is
\(1\,\frac{31}{72}\)
.

Exercise 1(B)

Question 8

Which rational number should be added to
\(\frac{- 5}{9}\)
to get
\(\frac{- 2}{3}\)
?
Show step-by-step answer+

Let
x be added to
\(\frac{- 5}{9}\)
.

\(\frac{- 5}{9} + x = \frac{- 2}{3} \Rightarrow x = \frac{- 2}{3} - \frac{- 5}{9}\)
💡 LCM of 3 and 9 is 3 x 3 = \(9 \Rightarrow x = \frac{- 2 \times 3}{3 \times 3} - \frac{- 5 \times 1}{9 \times 1} \Rightarrow x = \frac{- 6}{9} - \frac{- 5}{9} \Rightarrow x = \frac{- 6 - ( - 5 )}{9} \Rightarrow x = \frac{- 6 + 5}{9} \Rightarrow x = - 1\)

The number added to
\(\frac{- 5}{9}\)
to get
\(\frac{- 2}{3}\)
is
\(\frac{- 1}{9}\)
.

Exercise 1(B)

Question 9

Which rational number should be subtracted from
\(\frac{- 5}{6}\)
to get
\(\frac{4}{9}\)
?
Show step-by-step answer+

Let
x be subtracted from
\(\frac{- 5}{6}\)
.

\(\frac{- 5}{6} - x = \frac{4}{9} \Rightarrow x = \frac{- 5}{6} - \frac{4}{9}\)
💡 LCM of 6 and 9 is 2 x 3 x 3 = \(18 \Rightarrow x = \frac{- 5 \times 3}{6 \times 3} - \frac{4 \times 2}{9 \times 2} \Rightarrow x = \frac{- 15}{18} - \frac{8}{18} \Rightarrow x = \frac{- 15 - 8}{18} \Rightarrow x = \frac{- 23}{18} \Rightarrow x = -1\,\frac{5}{18}\)

The number subtracted from
\(\frac{- 5}{6}\)
to get
\(\frac{4}{9}\)
is -
\(1\,\frac{5}{18}\)
.

Exercise 1(B)

Question 10(i)

What should be subtracted from -2 to get
\(\frac{3}{8}\)
?
Show step-by-step answer+

Let
x be subtracted from -2.

\(- 2 - x = \frac{3}{8} \Rightarrow \frac{- 2}{1} - x = \frac{3}{8} \Rightarrow x = \frac{- 2}{1} - \frac{3}{8}\)
💡 LCM of 1 and 8 is 2 x 2 x 2 = \(8 \Rightarrow x = \frac{- 2 \times 8}{1 \times 8} - \frac{3 \times 1}{8 \times 1} \Rightarrow x = \frac{- 16}{8} - \frac{3}{8} \Rightarrow x = \frac{- 16 - 3}{8} \Rightarrow x = \frac{- 19}{8} \Rightarrow x = -2\,\frac{3}{8}\)

The number subtracted from -2 to get
\(\frac{3}{8}\)
is -
\(2\,\frac{3}{8}\)
.

Exercise 1(B)

Question 10(ii)

What should be added to -2 to get
\(\frac{3}{8}\)
?
Show step-by-step answer+

Let
x be added to -2.

\(- 2 + x = \frac{3}{8} \Rightarrow \frac{- 2}{1} + x = \frac{3}{8} \Rightarrow x = \frac{3}{8} - \frac{- 2}{1}\)
💡 LCM of 8 and 1 is 2 x 2 x 2 = \(8 \Rightarrow x = \frac{3 \times 1}{8 \times 1} - \frac{- 2 \times 8}{1 \times 8} \Rightarrow x = \frac{3}{8} - \frac{- 16}{8} \Rightarrow x = \frac{3 - ( - 16 )}{8} \Rightarrow x = \frac{3 + 16}{8} \Rightarrow x = \frac{19}{8} \Rightarrow x = 2\,\frac{3}{8}\)

The number added to -2 to get
\(\frac{3}{8}\)
is
\(2\,\frac{3}{8}\)
.

Exercise 1(B)

Question 11(i)

Evaluate:

\(\frac{3}{7} + \frac{- 4}{9} - \frac{- 11}{7} - \frac{7}{9}\)
Show step-by-step answer+

\(( \frac{3}{7} - \frac{- 11}{7} ) + ( \frac{- 4}{9} - \frac{7}{9} ) = ( \frac{3 - ( - 11 )}{7} ) + ( \frac{- 4 - 7}{9} ) = ( \frac{14}{7} ) + ( \frac{- 11}{9} ) = ( \frac{2}{1} ) + ( \frac{- 11}{9} )\)
)

💡 LCM of 1 and 9 is 3 x 3 = \(9 = \frac{2 \times 9}{1 \times 9} + \frac{- 11 \times 1}{9 \times 1} = \frac{18}{9} + \frac{- 11}{9} = \frac{18 + ( - 11 )}{9} = 7\)
Exercise 1(B)

Question 11(ii)

Evaluate:

\(\frac{2}{3} + \frac{- 4}{5} - \frac{1}{3} - \frac{2}{5}\)
Show step-by-step answer+

\(( \frac{2}{3} - \frac{1}{3} ) + ( \frac{- 4}{5} - \frac{2}{5} ) = ( \frac{2 - 1}{3} ) + ( \frac{- 4 - 2}{5} ) = ( \frac{1}{3} ) + ( \frac{- 6}{5} )\)
)

💡 LCM of 3 and 5 is 3 x 5 = \(15 = \frac{1 \times 5}{3 \times 5} + \frac{- 6 \times 3}{5 \times 3} = \frac{5}{15} + \frac{- 18}{15} = \frac{5 + ( - 18 )}{15} = - 13\)
Exercise 1(B)

Question 11(iii)

Evaluate

\(\frac{4}{7} - \frac{- 8}{9} - \frac{- 13}{7} + \frac{17}{9}\)
Show step-by-step answer+

\(( \frac{4}{7} - \frac{- 13}{7} ) + ( - \frac{- 8}{9} + \frac{17}{9} ) = ( \frac{4 - ( - 13 )}{7} ) + ( \frac{8 + 17}{9} ) = ( \frac{4 + 13}{7} ) + ( \frac{25}{9} ) = ( \frac{17}{7} ) + ( \frac{25}{9} )\)
)

💡 LCM of 7 and 9 is 3 x 3 x 7 = \(63 = \frac{17 \times 9}{7 \times 9} + \frac{25 \times 7}{9 \times 7} = \frac{153}{63} + \frac{175}{63} = \frac{153 + 175}{63} = \frac{328}{63} = 5\,\frac{13}{63}\)
Complete Solutions

Exercise 1(C)

Tap any answer panel to reveal the full working.

Exercise 1(C)

Question 1(i)

The number which on multiplying with
\(5\,\frac{2}{3}\)
gives -
\(1\,\frac{2}{3}\)
is:
\(- \frac{5}{7}\)
\(\frac{5}{7}\)
\(- \frac{5}{17}\)
\(\frac{17}{5}\)
Show step-by-step answer+

Let the number be
x.
\(5\,\frac{2}{3} \times x = - 1 \frac{2}{3} \Rightarrow \frac{17}{3} \times x = - \frac{5}{3} \Rightarrow x = - \frac{5}{3} \div \frac{17}{3} \Rightarrow x = - \frac{5}{3} \times \frac{3}{17} \Rightarrow x = - \frac{5 \times 3}{3 \times 17} \Rightarrow x = - \frac{15}{51} \Rightarrow x = - \frac{5}{17}\)

Hence, option 3 is the correct option.

Exercise 1(C)

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
Show step-by-step 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.

Exercise 1(C)

Question 1(iii)

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

0
-1
1
none
Show step-by-step answer+

Let a positive rational number be
\(\frac{a}{b}\)
.

\(\frac{a}{b} \times \frac{b}{a} = \frac{a \times b}{b \times a} = a b a b = 1\)\(\frac{ab}{ab}\)

=1

Hence, option 3 is the correct option.

Exercise 1(C)

Question 1(iv)

The area of a rectangular paper is
\(7\,\frac{1}{3}\)
cm2. If its length
\(4\,\frac{2}{5}\)
cm, its breadth is:
\(1\,\frac{3}{5}\)
cm
\(1\,\frac{2}{3}\)
cm
\(\frac{3}{5}\)
cm
\(\frac{25}{60}\)
cm
Show step-by-step answer+

Let breadth be b

Length =
\(4\,\frac{2}{5}\)
cm =
\(\frac{22}{5}\)
cm.
Area =
\(7\,\frac{1}{3}\)
cm2 =
\(\frac{22}{3}\)
cm2.
Area of rectangle = length x breadth
\(\frac{22}{3} = \frac{22}{5} \times b \Rightarrow b = \frac{22}{3} \div \frac{22}{5} \Rightarrow b = \frac{22}{3} \times \frac{5}{22} \Rightarrow b = \frac{22 \times 5}{3 \times 22} \Rightarrow b = \frac{110}{66} \Rightarrow b = \frac{5}{3} \Rightarrow b = 1 \frac{2}{3}\)

Hence, option 2 is the correct option.

Exercise 1(C)

Question 1(v)

The product of a rational number
\(\frac{2}{7}\)
and its additive inverse is:
\(\frac{4}{49}\)
\(\frac{- 4}{49}\)

0

1

Show step-by-step answer+
Rational number =
\(\frac{2}{7}\)
Additive inverse =
\(- \frac{2}{7}\)

\(\frac{2}{7} \times - \frac{2}{7} = - \frac{2 \times 2}{7 \times 7} = - \frac{4}{49} .\)
.

Hence, option 2 is the correct option.

Exercise 1(C)

Question 2(i)

Evaluate:

\(\frac{- 14}{5} \times \frac{- 6}{7}\)
Show step-by-step answer+
\(\frac{- 14}{5} \times \frac{- 6}{7} = \frac{- 14 \times ( - 6 )}{5 \times 7} = \frac{84}{35} = \frac{12}{5} = 2 \frac{2}{5}\)

Hence,
\(\frac{- 14}{5} \times \frac{- 6}{7} = 2 \frac{2}{5}\)

Exercise 1(C)

Question 2(ii)

Evaluate:

\(\frac{7}{6} \times \frac{- 18}{91}\)
Show step-by-step answer+
\(\frac{7}{6} \times \frac{- 18}{91} = \frac{7 \times ( - 18 )}{6 \times 91} = \frac{- 126}{546} = \frac{- 3}{13}\)

Hence,
\(\frac{7}{6} \times \frac{- 18}{91} = \frac{- 3}{13}\)

Exercise 1(C)

Question 2(iii)

Evaluate:

\(\frac{- 125}{72} \times \frac{9}{- 5}\)
Show step-by-step answer+
\(\frac{- 125}{72} \times \frac{9}{- 5} = \frac{- 125 \times 9}{72 \times ( - 5 )} = \frac{- 1125}{- 360} = \frac{25}{8} = 3 \frac{1}{8}\)

Hence,
\(\frac{- 125}{72} \times \frac{9}{- 5} = 3 \frac{1}{8}\)

Exercise 1(C)

Question 2(iv)

Evaluate:

\(\frac{- 11}{9} \times \frac{- 51}{- 44}\)
Show step-by-step answer+
\(\frac{- 11}{9} \times \frac{- 51}{- 44} = \frac{- 11 \times ( - 51 )}{9 \times - 44} = \frac{561}{- 396} = - \frac{17}{12} = - 1 \frac{5}{12}\)

Hence,
\(\frac{- 11}{9} \times \frac{- 51}{- 44} = - 1 \frac{5}{12}\)

Exercise 1(C)

Question 2(v)

Evaluate:

\(- \frac{16}{5} \times \frac{20}{8}\)
Show step-by-step answer+
\(- \frac{16}{5} \times \frac{20}{8} = - \frac{16 \times 20}{5 \times 8} = - \frac{320}{40} = - 8\)

=−8

Hence,
\(- \frac{16}{5} \times \frac{20}{8} = - 8\)
=−8
Exercise 1(C)

Question 3(i)

Multiply

\(\frac{5}{6}\)
and
\(\frac{8}{9}\)
Show step-by-step answer+
\(\frac{5}{6} \times \frac{8}{9} = \frac{5 \times 8}{6 \times 9} = \frac{40}{54} = \frac{20}{27}\)

Hence,
\(\frac{5}{6} \times \frac{8}{9} = \frac{20}{27}\)

Exercise 1(C)

Question 3(ii)

Multiply:

\(\frac{2}{7}\)
and
\(\frac{- 14}{9}\)
Show step-by-step answer+
\(\frac{2}{7} \times \frac{- 14}{9} = \frac{2 \times ( - 14 )}{7 \times 9} = \frac{- 28}{63} = \frac{- 4}{9}\)

Hence,
\(\frac{2}{7} \times \frac{- 14}{9} = \frac{- 4}{9}\)

Exercise 1(C)

Question 3(iii)

Multiply:

\(\frac{- 7}{8}\)
and 4
Show step-by-step answer+
\(\frac{- 7}{8} \times \frac{4}{1} = \frac{- 7 \times 4}{8 \times 1} = \frac{- 28}{8} = \frac{- 7}{2} = - 3 \frac{1}{2}\)

Hence,
\(\frac{- 7}{8} \times 4 = - 3 \frac{1}{2}\)

Exercise 1(C)

Question 3(iv)

Multiply:

\(\frac{36}{- 7}\)
and
\(\frac{- 9}{28}\)
Show step-by-step answer+
\(\frac{36}{- 7} \times \frac{- 9}{28} = \frac{36 \times ( - 9 )}{- 7 \times 28} = \frac{- 324}{- 196} = \frac{81}{49} = 1 \frac{32}{49}\)

Hence,
\(\frac{36}{- 7} \times \frac{- 9}{28} = 1 \frac{32}{49}\)

Exercise 1(C)

Question 3(v)

Multiply:

\(\frac{- 7}{10}\)
and
\(\frac{- 8}{15}\)
Show step-by-step answer+
\(\frac{- 7}{10} \times \frac{- 8}{15} = \frac{- 7 \times ( - 8 )}{10 \times 15} = \frac{56}{150} = \frac{28}{75}\)

Hence,
\(\frac{- 7}{10} \times \frac{- 8}{15} = \frac{28}{75}\)

Exercise 1(C)

Question 4(i)

Evaluate:

\(( \frac{2}{- 3} \times \frac{5}{4} ) + ( \frac{5}{9} \times \frac{3}{- 10} )\)
)
Show step-by-step answer+
\(( \frac{2 \times 5}{- 3 \times 4} ) + ( \frac{5 \times 3}{9 \times ( - 10 )} ) = ( \frac{10}{- 12} ) + ( \frac{15}{- 90} ) = ( \frac{- 5}{6} ) + ( \frac{- 1}{6} ) = ( \frac{- 5 + ( - 1 )}{6} ) = ( \frac{- 6}{6} ) = - 1\)
)
=−1

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

Exercise 1(C)

Question 4(ii)

Evaluate:

\(( 2 \times \frac{1}{4} ) - ( \frac{- 18}{7} \times \frac{- 7}{15} )\)
)
Show step-by-step answer+
\(( \frac{2 \times 1}{1 \times 4} ) - ( \frac{- 18 \times - 7}{7 \times 15} ) = ( \frac{2}{4} ) - ( \frac{126}{105} ) = ( \frac{1}{2} ) - ( \frac{6}{5} )\)
)
LCM of 2 and 5 is 2 x 5 = 10
\(= ( \frac{1 \times 5}{2 \times 5} ) - ( \frac{6 \times 2}{5 \times 2} ) = ( \frac{5}{10} ) - ( \frac{12}{10} ) = ( \frac{5 - 12}{10} ) = ( \frac{- 7}{10} )\)
)

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

Exercise 1(C)

Question 4(iii)

Evaluate:

\(( - 5 \times \frac{2}{15} ) - ( - 6 \times \frac{2}{9} )\)
)
Show step-by-step answer+

\(( \frac{- 5 \times 2}{1 \times 15} ) - ( \frac{- 6 \times 2}{1 \times 9} ) = ( \frac{- 10}{15} ) - ( \frac{- 12}{9} ) = ( \frac{- 2}{3} ) - ( \frac{- 4}{3} ) = ( \frac{- 2 - ( - 4 )}{3} ) = ( \frac{- 2 + 4}{3} ) = ( \frac{2}{3} )\)
)

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

Exercise 1(C)

Question 4(iv)

Evaluate:

\(( \frac{8}{5} \times \frac{- 3}{2} ) + ( \frac{- 3}{10} \times \frac{9}{16} )\)
)
Show step-by-step answer+

\(( \frac{8 \times - 3}{5 \times 2} ) + ( \frac{- 3 \times 9}{10 \times 16} ) = ( \frac{- 24}{10} ) + ( \frac{- 27}{160} ) = ( \frac{- 12}{5} ) + ( \frac{- 27}{160} )\)
)

💡 LCM of 5 and 160 is 2 x 2 x 2 x 2 x 2 x 5 = \(160 = ( \frac{- 12 \times 32}{5 \times 32} ) + ( \frac{- 27 \times 1}{160 \times 1} ) = ( \frac{- 384}{160} ) + ( \frac{- 27}{160} ) = ( \frac{- 384 + ( - 27 )}{160} ) = ( \frac{- 411}{160} ) = -2\,\frac{91}{160}\)

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

Exercise 1(C)

Question 5(i)

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

\(\frac{7}{- 5}\)
Show step-by-step answer+
\(\frac{7}{- 5} \times 1 = \frac{7 \times 1}{- 5 \times 1} = \frac{7}{- 5}\)

Hence,
\(\frac{7}{- 5} \times 1 = \frac{7}{- 5}\)

Exercise 1(C)

Question 5(ii)

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

\(\frac{- 3}{- 4}\)
Show step-by-step answer+
\(\frac{- 3}{- 4} \times 1 = \frac{- 3 \times 1}{- 4 \times 1} = \frac{- 3}{- 4} = \frac{3}{4}\)

Hence,
\(\frac{- 3}{- 4} \times 1 = \frac{3}{4}\)

Exercise 1(C)

Question 5(iii)

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

0

Show step-by-step answer+
0
×
1
=
0
0×1
=0

Hence, 0 x 1 = 0

Exercise 1(C)

Question 5(iv)

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

\(\frac{- 8}{13}\)
Show step-by-step answer+
\(\frac{- 8}{13} \times 1 = \frac{- 8 \times 1}{13 \times 1} = \frac{- 8}{13}\)

Hence,
\(\frac{- 8}{13} \times 1 = \frac{- 8}{13}\)

Exercise 1(C)

Question 5(v)

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

\(\frac{- 6}{- 7}\)
Show step-by-step answer+
\(\frac{- 6}{- 7} \times 1 = \frac{- 6 \times 1}{- 7 \times 1} = \frac{- 6}{- 7} = \frac{6}{7}\)

Hence,
\(\frac{- 6}{- 7} \times 1 = \frac{6}{7}\)

Exercise 1(C)

Question 6(i)

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

\(\frac{- 1}{5}\)
and
\(\frac{2}{9}\)
Show step-by-step answer+
To prove:
\(\frac{- 1}{5} \times \frac{2}{9} = \frac{2}{9} \times \frac{- 1}{5}\)
Taking LHS:
\(\frac{- 1}{5} \times \frac{2}{9} = \frac{- 1 \times 2}{5 \times 9} = \frac{- 2}{45}\)
Taking RHS:
\(\frac{2}{9} \times \frac{- 1}{5} = \frac{2 \times - 1}{9 \times 5} = \frac{- 2}{45}\)

∴ LHS = RHS

\(\frac{- 1}{5} \times \frac{2}{9} = \frac{2}{9} \times \frac{- 1}{5}\)
Exercise 1(C)

Question 6(ii)

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

\(\frac{5}{- 3}\)
and
\(\frac{13}{- 11}\)
Show step-by-step answer+
To prove:
\(\frac{5}{- 3} \times \frac{13}{- 11} = \frac{13}{- 11} \times \frac{5}{- 3}\)
Taking LHS:
\(\frac{5}{- 3} \times \frac{13}{- 11} = \frac{5 \times 13}{- 3 \times - 11} = \frac{65}{33}\)
Taking RHS:
\(\frac{13}{- 11} \times \frac{5}{- 3} = \frac{13 \times 5}{- 11 \times - 3} = \frac{65}{33}\)

∴ LHS = RHS

\(\frac{5}{- 3} \times \frac{13}{- 11} = \frac{13}{- 11} \times \frac{5}{- 3}\)
Exercise 1(C)

Question 6(iii)

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

3 and
\(\frac{- 8}{9}\)
Show step-by-step answer+
To prove:

\(3 \times \frac{- 8}{9} = \frac{- 8}{9} \times 3\)
×3

Taking LHS:
\(3 \times \frac{- 8}{9} = \frac{3 \times - 8}{1 \times 9} = \frac{- 24}{9} = \frac{- 8}{3}\)
Taking RHS:
\(\frac{- 8}{9} \times 3 = \frac{- 8 \times 3}{9 \times 1} = \frac{- 24}{9} = \frac{- 8}{3}\)

∴ LHS = RHS

\(3 \times \frac{- 8}{9} = \frac{- 8}{9} \times 3\)
×3

Exercise 1(C)

Question 6(iv)

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

0 and
\(\frac{- 12}{17}\)
Show step-by-step answer+
To prove:

\(0 \times \frac{- 12}{17} = \frac{- 12}{17} \times 0\)
×0

Taking LHS:
\(0 \times \frac{- 12}{17} = \frac{0 \times - 12}{1 \times 17} = \frac{0}{17} = 0\)

=0

Taking RHS:
\(\frac{- 12}{17} \times 0 = \frac{- 12 \times 0}{17 \times 1} = \frac{0}{17} = 0\)

=0

∴ LHS = RHS

\(0 \times \frac{- 12}{17} = \frac{- 12}{17} \times 0\)
×0

Exercise 1(C)

Question 7(i)

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

5

Show step-by-step answer+
The multiplicative inverse of 5 = reciprocal of 5 =
\(\frac{1}{5}\)
.
Exercise 1(C)

Question 7(ii)

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

-3

Show step-by-step answer+
The multiplicative inverse of -3 = reciprocal of -3 =
\(- \frac{1}{3}\)
.
Exercise 1(C)

Question 7(iii)

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

\(\frac{5}{11}\)
Show step-by-step answer+
The multiplicative inverse of
\(\frac{5}{11}\)
= reciprocal of
\(\frac{5}{11}\)
=
\(\frac{11}{5} = 2 \frac{1}{5}\)
.
Exercise 1(C)

Question 7(iv)

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

\(\frac{- 7}{- 8}\)
Show step-by-step answer+
The multiplicative inverse of
\(\frac{- 7}{- 8}\)
= reciprocal of
\(\frac{7}{8}\)
=
\(\frac{8}{7} = 1 \frac{1}{7}\)
.
Exercise 1(C)

Question 7(v)

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

\(\frac{- 8}{- 7}\)
Show step-by-step answer+
The multiplicative inverse of
\(\frac{- 8}{- 7}\)
= reciprocal of
\(\frac{8}{7}\)
=
\(\frac{7}{8}\)
.
Exercise 1(C)

Question 8(i)

Find the reciprocal (multiplicative inverse) of:

\(\frac{3}{5} \times \frac{2}{3}\)
Show step-by-step answer+
\(\frac{3}{5} \times \frac{2}{3} = \frac{3 \times 2}{5 \times 3} = \frac{6}{15} = \frac{2}{5}\)
The multiplicative inverse of
\(\frac{2}{5}\)
= reciprocal of
\(\frac{2}{5} = \frac{5}{2} = 2 \frac{1}{2}\)
.
Exercise 1(C)

Question 8(ii)

Find the reciprocal (multiplicative inverse) of:

\(\frac{- 8}{3} \times \frac{13}{- 7}\)
Show step-by-step answer+
\(\frac{- 8}{3} \times \frac{13}{- 7} = \frac{- 8 \times 13}{3 \times - 7} = \frac{- 104}{- 21} = \frac{104}{21}\)
The multiplicative inverse of
\(\frac{104}{21}\)
= reciprocal of
\(\frac{104}{21}\)
=
\(\frac{21}{104}\)
.
Exercise 1(C)

Question 8(iii)

Find the reciprocal (multiplicative inverse) of:

\(\frac{- 3}{5} \times \frac{- 1}{13}\)
Show step-by-step answer+
\(\frac{- 3}{5} \times \frac{- 1}{13} = \frac{- 3 \times ( - 1 )}{5 \times 13} = \frac{3}{65}\)
The multiplicative inverse of
\(\frac{3}{65}\)
= reciprocal of
\(\frac{3}{65}\)
=
\(\frac{65}{3} = 21 \frac{2}{3}\)
.
Exercise 1(C)

Question 9(i)

Verify that
(
x
+
y
)
×
z
=
x
×
z
+
y
×
z
(x+y)×z=x×z+y×z, if
\(x = \frac{4}{5} , y = \frac{- 2}{3}\)
and
z
=

4
z=−4
Show step-by-step answer+
To prove:
(
x
+
y
)
×
z
=
x
×
z
+
y
×
z
(x+y)×z=x×z+y×z
Taking LHS:

\(( x + y ) \times z = ( \frac{4}{5} + \frac{- 2}{3} ) \times - 4\)
)×−4

💡 LCM of 5 and 3 is 3 x 5 = 15

\(= ( \frac{4 \times 3}{5 \times 3} + \frac{- 2 \times 5}{3 \times 5} ) \times - 4 = ( \frac{12}{15} + \frac{- 10}{15} ) \times - 4 = ( \frac{12 + ( - 10 )}{15} ) \times - 4 = ( \frac{2}{15} ) \times - 4 = ( \frac{2 \times - 4}{15 \times 1} ) = ( \frac{- 8}{15} )\)
)

Taking RHS:
\(x \times z + y \times z = \frac{4}{5} \times - 4 + \frac{- 2}{3} \times - 4 = \frac{4 \times - 4}{5 \times 1} + \frac{- 2 \times - 4}{3 \times 1} = \frac{- 16}{5} + \frac{8}{3}\)
💡 LCM of 5 and 3 is 3 x 5 = \(15 = \frac{- 16 \times 3}{5 \times 3} + \frac{8 \times 5}{3 \times 5} = \frac{- 48}{15} + \frac{40}{15} = \frac{- 48 + 40}{15} = - 8\)

∴ LHS = RHS

(
x
+
y
)
×
z
=
x
×
z
+
y
×
z
(x+y)×z=x×z+y×z
Exercise 1(C)

Question 9(ii)

Verify that
(
x
+
y
)
×
z
=
x
×
z
+
y
×
z
(x+y)×z=x×z+y×z, if
\(x = 2 , y = \frac{4}{5}\)
and
\(z = \frac{3}{- 10}\)
Show step-by-step answer+
To prove:
(
x
+
y
)
×
z
=
x
×
z
+
y
×
z
(x+y)×z=x×z+y×z
Taking LHS:
\(( x + y ) \times z = ( 2 + \frac{4}{5} ) \times \frac{3}{- 10} = ( \frac{2}{1} + \frac{4}{5} ) \times \frac{3}{- 10}\)
💡 LCM of 1 and 5 is 5.

\(= ( \frac{2 \times 5}{1 \times 5} + \frac{4 \times 1}{5 \times 1} ) \times \frac{3}{- 10} = ( \frac{10}{5} + \frac{4}{5} ) \times \frac{3}{- 10} = ( \frac{10 + 4}{5} ) \times \frac{3}{- 10} = ( \frac{14}{5} ) \times \frac{3}{- 10} = ( \frac{14 \times 3}{5 \times - 10} ) = ( \frac{42}{- 50} ) = ( \frac{- 21}{25} )\)
)

Taking RHS:
\(x \times z + y \times z = 2 \times \frac{3}{- 10} + \frac{4}{5} \times \frac{3}{- 10} = \frac{2 \times 3}{1 \times - 10} + \frac{4 \times 3}{5 \times - 10} = \frac{6}{- 10} + \frac{12}{- 50} = \frac{- 3}{5} + \frac{- 6}{25}\)
💡 LCM of 5 and 25 is 5 x 5 = \(25 = \frac{- 3 \times 5}{5 \times 5} + \frac{- 6 \times 1}{25 \times 1} = \frac{- 15}{25} + \frac{- 6}{25} = \frac{- 15 + ( - 6 )}{25} = - 21\)

∴ LHS = RHS

(
x
+
y
)
×
z
=
x
×
z
+
y
×
z
(x+y)×z=x×z+y×z
Exercise 1(C)

Question 10(i)

Verify that
x
×
(
y

z
)
=
x
×
y

x
×
z
x×(y−z)=x×y−x×z, if
\(x = \frac{4}{5} , y = \frac{- 7}{4}\)
and
z
=
3
z=3
Show step-by-step answer+
To prove:
x
×
(
y

z
)
=
x
×
y

x
×
z
x×(y−z)=x×y−x×z
Taking LHS:

\(x \times ( y - z ) = \frac{4}{5} \times ( \frac{- 7}{4} - 3 ) = \frac{4}{5} \times ( \frac{- 7}{4} - \frac{3}{1} )\)
)

💡 LCM of 4 and 1 is 2 x 2 = 4.

\(= \frac{4}{5} \times ( \frac{- 7 \times 1}{4 \times 1} - \frac{3 \times 4}{1 \times 4} ) = \frac{4}{5} \times ( \frac{- 7}{4} - \frac{12}{4} ) = \frac{4}{5} \times ( \frac{- 7 - 12}{4} ) = \frac{4}{5} \times ( \frac{- 19}{4} ) = ( \frac{4 \times - 19}{5 \times 4} ) = ( \frac{- 76}{20} ) = ( \frac{- 19}{5} )\)
)

Taking RHS:
\(x \times y - x \times z = \frac{4}{5} \times \frac{- 7}{4} - \frac{4}{5} \times 3 = \frac{4 \times - 7}{5 \times 4} - \frac{4 \times 3}{5 \times 1} = \frac{- 28}{20} - \frac{12}{5} = \frac{- 7}{5} - \frac{12}{5} = \frac{- 7 - 12}{5} = \frac{- 19}{5}\)

∴ LHS = RHS

x
×
(
y

z
)
=
x
×
y

x
×
z
x×(y−z)=x×y−x×z
Exercise 1(C)

Question 10(ii)

Verify that
x
×
(
y

z
)
=
x
×
y

x
×
z
x×(y−z)=x×y−x×z, if
\(x = \frac{3}{4} , y = \frac{8}{9}\)
and
z
=

5
z=−5
Show step-by-step answer+
To prove:
x
×
(
y

z
)
=
x
×
y

x
×
z
x×(y−z)=x×y−x×z
Taking LHS:

\(x \times ( y - z ) = \frac{3}{4} \times ( \frac{8}{9} - ( - 5 ) ) = \frac{3}{4} \times ( \frac{8}{9} - \frac{- 5}{1} )\)
)

💡 LCM of 9 and 1 is 3 x 3 = 9.

\(= \frac{3}{4} \times ( \frac{8 \times 1}{9 \times 1} - \frac{- 5 \times 9}{1 \times 9} ) = \frac{3}{4} \times ( \frac{8}{9} - \frac{- 45}{9} ) = \frac{3}{4} \times ( \frac{8 - ( - 45 )}{9} ) = \frac{3}{4} \times ( \frac{8 + 45}{9} ) = \frac{3}{4} \times ( \frac{53}{9} ) = ( \frac{3 \times 53}{4 \times 9} ) = ( \frac{159}{36} ) = ( \frac{53}{12} )\)
)

Taking RHS:
\(x \times y - x \times z = \frac{3}{4} \times \frac{8}{9} - \frac{3}{4} \times - 5 = \frac{3 \times 8}{4 \times 9} - \frac{3 \times - 5}{4 \times 1} = \frac{24}{36} - \frac{- 15}{4} = \frac{2}{3} - \frac{- 15}{4}\)
💡 LCM of 3 and 4 is 2 x 2 x 3 = \(12 = \frac{2 \times 4}{3 \times 4} - \frac{- 15 \times 3}{4 \times 3} = \frac{8}{12} - \frac{- 45}{12} = \frac{8 - ( - 45 )}{12} = \frac{8 + 45}{12} = 53\)

∴ LHS = RHS

x
×
(
y

z
)
=
x
×
y

x
×
z
x×(y−z)=x×y−x×z
Exercise 1(C)

Question 11

Name the multiplication property of rational numbers shown below:

(i)
\(\frac{3}{5} \times \frac{- 8}{9} = \frac{- 8}{9} \times \frac{3}{5}\)
(ii)
\(\frac{- 3}{4} \times ( \frac{5}{7} \times \frac{- 8}{15} ) = ( \frac{- 3}{4} \times \frac{5}{7} ) \times \frac{- 8}{15}\)
(iii)
\(\frac{4}{5} \times ( \frac{3}{- 8} + \frac{- 4}{7} ) = \frac{4}{5} \times \frac{3}{- 8} + \frac{4}{5} \times \frac{- 4}{7}\)
(iv)
\(\frac{- 7}{5} \times \frac{5}{- 7} = 1\)
=1
(v)
\(\frac{8}{- 9} \times 1 = 1 \times \frac{8}{- 9} = \frac{8}{- 9}\)
Show step-by-step answer+

(i) Commutativity property

Reason

If
\(\frac{a}{b}\)
and
\(\frac{c}{d}\)
are any two rational numbers, then:

\(\frac{a}{b} \times \frac{c}{d} = \frac{c}{d} \times \frac{a}{b}\)

(ii) Associativity property

Reason

If
\(\frac{a}{b} , \frac{c}{d}\)
and
\(\frac{e}{f}\)
are any three rational numbers, then:

\(\frac{a}{b} \times ( \frac{c}{d} \times \frac{e}{f} ) = ( \frac{a}{b} \times \frac{c}{d} ) \times \frac{e}{f}\)

(iii) Distributivity property

Reason

If
\(\frac{a}{b} , \frac{c}{d}\)
and
\(\frac{e}{f}\)
are any three rational numbers, then:

\(\frac{a}{b} \times ( \frac{c}{d} + \frac{e}{f} ) = ( \frac{a}{b} \times \frac{c}{d} ) + ( \frac{a}{b} \times \frac{e}{f} )\)
)

(iv) Existence of inverse

Reason

The multiplicative inverse of
\(\frac{a}{b}\)
= reciprocal of
\(\frac{a}{b} = \frac{b}{a}\)
.

(v) Existence of identity

Reason

For a rational number
\(\frac{a}{b}\)
,

\(1 \times \frac{a}{b} = \frac{a}{b} \times 1 = \frac{a}{b}\)
.

Exercise 1(C)

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 ............... .
Show step-by-step 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
\(\frac{a}{b}\)
and
\(\frac{c}{d}\)
.

Hence,

\(\frac{a}{b} \times \frac{c}{d} = \frac{a \times c}{b \times d} = a c b d\)\(\frac{ac}{bd}\)\(\frac{ac}{bd}\)
is also positive rational number.

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

Hence,

\(- \frac{a}{b} \times - \frac{c}{d} = \frac{- a \times - c}{b \times d} = a c b d\)\(\frac{ac}{bd}\)\(\frac{ac}{bd}\)
is positive rational number.

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

Hence,

\(\frac{a}{b} \times - \frac{c}{d} = \frac{a \times - c}{b \times d} = - a c b d\)\(\frac{- ac}{bd}\)\(\frac{ac}{bd}\)
is negative rational number.

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

Reciprocal of
\(\frac{a}{b} = \frac{b}{a}\)

\(\frac{b}{a}\)
is a positive rational number.

Let the negative rational number be -
\(\frac{a}{b}\)
.

Reciprocal of -
\(\frac{a}{b} = - \frac{b}{a}\)

-
\(\frac{b}{a}\)
is a negative rational number.

(v) Reciprocal of
\(\frac{0}{1} = \frac{1}{0}\)

\(\frac{1}{0}\)
is not defined.

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

Reciprocal of
\(\frac{a}{b} = \frac{b}{a}\)

\(\frac{a}{b} \times \frac{b}{a} = \frac{a \times b}{b \times a} = a b a b = 1\)\(\frac{ab}{ab}\)
=1
(vii) Reciprocal of
\(\frac{1}{1} = \frac{1}{1} = 1\)
=1.
Reciprocal of
\(\frac{- 1}{1} = \frac{1}{- 1} = - 1\)
=−1.

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

Reciprocal of
\(\frac{n}{1} = \frac{m}{1}\)

Exercise 1(C)

Question 13

The length and breadth of a rectangular piece of paper are 9 cm and
\(10\,\frac{2}{3}\)
cm respectively. Find:

(i) its area

(ii) its perimeter

Show step-by-step answer+

Length = 9 cm

Breadth =
\(10\,\frac{2}{3}\)
cm

Area = length x breadth

\(= 9 \times 10 \frac{2}{3} = 9 \times \frac{32}{3} = \frac{9 \times 32}{1 \times 3} = \frac{288}{3} = 96 cm 2\)
=96 cm
2

Area = 96 cm2

Perimeter = 2 x (length + breadth\() = 2 \times ( 9 + 10 \frac{2}{3} ) = 2 \times ( \frac{9}{1} + \frac{32}{3}\)
)
💡 LCM of 1 and 3 is 3.

\(= 2 \times ( \frac{9 \times 3}{1 \times 3} + \frac{32 \times 1}{3 \times 1} ) = 2 \times ( \frac{27}{3} + \frac{32}{3} ) = 2 \times ( \frac{27 + 32}{3} ) = 2 \times ( \frac{59}{3} ) = ( \frac{59 \times 2}{3 \times 1} ) = ( \frac{118}{3} ) = 39 ( \frac{1}{3} )\)
)

Perimeter =
\(39\,\frac{1}{3}\)
cm

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

Exercise 1(C)

Question 14

Find the area and the perimeter of a rectangular piece of land with length
\(7\,\frac{2}{5}\)
m and breadth
\(4\,\frac{1}{6}\)
m.
Show step-by-step answer+
Length =
\(7\,\frac{2}{5}\)
m =
\(\frac{37}{5}\)
m
Breadth =
\(4\,\frac{1}{6}\)
=
\(\frac{25}{6}\)
m

Area = length x breadth

\(= \frac{37}{5} \times \frac{25}{6} = \frac{37 \times 25}{5 \times 6} = \frac{925}{30} = \frac{185}{6} = 30 \frac{5}{6} m 2\)
m
2

Perimeter = 2 x (length + breadth\() = 2 \times ( \frac{37}{5} + \frac{25}{6}\)
)
💡 LCM of 5 and 6 is 2 x 3 x 5 = 30.

\(= 2 \times ( \frac{37 \times 6}{5 \times 6} + \frac{25 \times 5}{6 \times 5} ) = 2 \times ( \frac{222}{30} + \frac{125}{30} ) = 2 \times ( \frac{222 + 125}{30} ) = 2 \times ( \frac{347}{30} ) = ( \frac{347 \times 2}{30 \times 1} ) = ( \frac{694}{30} ) = ( \frac{347}{15} ) = 23 ( \frac{2}{15} )\)
)

Area =
\(30\,\frac{5}{6}\)
m2 and perimeter =
\(23\,\frac{2}{15}\)
m
Complete Solutions

Exercise 1(D)

Tap any answer panel to reveal the full working.

Exercise 1(D)

Question 1(i)

\(- \frac{4}{9}\)
divided by
\(- \frac{2}{3}\)
gives:
\(\frac{2}{3}\)
\(- \frac{2}{3}\)
\(\frac{3}{2}\)
\(- \frac{3}{2}\)
Show step-by-step answer+
\(- \frac{4}{9} \div - \frac{2}{3} = - \frac{4}{9} \times - \frac{3}{2} = \frac{4 \times 3}{9 \times 2} = \frac{12}{18} = \frac{2}{3}\)

\(- \frac{4}{9}\)
divided by
\(- \frac{2}{3}\)
gives
\(\frac{2}{3}\)

Hence, Option 1 is the correct option.

Exercise 1(D)

Question 1(ii)

The rational number by which should
\(\frac{1}{2}\)
be divided to get
\(- \frac{2}{3}\)
is:
\(\frac{3}{4}\)
\(- \frac{3}{4}\)
\(\frac{4}{3}\)
\(- \frac{4}{3}\)
Show step-by-step answer+

Let
x be the number.

\(\frac{1}{2} \div x = - \frac{2}{3} \Rightarrow \frac{1}{2} \times \frac{1}{x} = - \frac{2}{3} \Rightarrow \frac{1}{2 x} = - \frac{2}{3} \Rightarrow 2 x = - \frac{3}{2} \Rightarrow x = \frac{1}{2} \times - \frac{3}{2} \Rightarrow x = - \frac{1 \times 3}{2 \times 2} \Rightarrow x = - \frac{3}{4}\)

The rational number by which should
\(\frac{1}{2}\)
be divided to get
\(- \frac{2}{3}\)
is
\(- \frac{3}{4}\)

Hence, Option 2 is the correct option.

Exercise 1(D)

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
Show step-by-step answer+

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

∴ a ÷ (b ÷ c) ≠ a ÷ b ÷ c

Hence, Option 4 is the correct option.

Exercise 1(D)

Question 1(iv)

The product of two rational numbers is
\(- 7 \frac{2}{3}\)
. If one of them is
\(3\,\frac{5}{6}\)
, the other number is:
3
-3
2
-2
Show step-by-step answer+

Let the number be
x.

\(3\,\frac{5}{6} \times x = - 7 \frac{2}{3} \Rightarrow \frac{23}{6} \times x = - \frac{23}{3} \Rightarrow x = - \frac{23}{3} \div \frac{23}{6} \Rightarrow x = - \frac{23}{3} \times \frac{6}{23} \Rightarrow x = - \frac{23 \times 6}{3 \times 23} \Rightarrow x = - \frac{23 \times 6}{3 \times 23} \Rightarrow x = - \frac{138}{69} \Rightarrow x = - 2\)

⇒x=−2

Hence, Option 4 is the correct option.

Exercise 1(D)

Question 1(v)

(8 ÷ 3) ÷ (3 ÷ 8) is equal to:

\(\frac{64}{9}\)
\(\frac{9}{64}\)

1

none of the above

Show step-by-step answer+
\(( 8 \div 3 ) \div ( 3 \div 8 ) = \frac{8}{3} \div \frac{3}{8} = \frac{8}{3} \times \frac{8}{3} = \frac{64}{9}\)

Hence, Option 1 is the correct option.

Exercise 1(D)

Question 2(i)

Evaluate:

1 ÷
\(\frac{1}{3}\)
Show step-by-step answer+
\(1 \div \frac{1}{3} = 1 \times \frac{3}{1} = \frac{1 \times 3}{1 \times 1} = \frac{3}{1} = 3\)

=3

Hence,
\(1 \div \frac{1}{3} = 3\)
=3
Exercise 1(D)

Question 2(ii)

Evaluate:

\(3 \div \frac{3}{5}\)
Show step-by-step answer+
\(3 \div \frac{3}{5} = 3 \times \frac{5}{3} = \frac{3 \times 5}{1 \times 3} = \frac{15}{3} = 5\)

=5

Hence,
\(3 \div \frac{3}{5} = 5\)
=5
Exercise 1(D)

Question 2(iii)

Evaluate:

\(- \frac{5}{12} \div \frac{1}{16}\)
Show step-by-step answer+
\(- \frac{5}{12} \div \frac{1}{16} = - \frac{5}{12} \times \frac{16}{1} = - \frac{5 \times 16}{12 \times 1} = - \frac{80}{12} = - \frac{20}{3} = - 6 \frac{2}{3}\)

Hence,
\(- \frac{5}{12} \div \frac{1}{16} = - 6 \frac{2}{3}\)

Exercise 1(D)

Question 2(iv)

Evaluate:

\(- \frac{21}{16} \div ( \frac{- 7}{8} )\)
)
Show step-by-step answer+
\(- \frac{21}{16} \div \frac{- 7}{8} = - \frac{21}{16} \times - \frac{8}{7} = \frac{21 \times 8}{16 \times 7} = \frac{168}{112} = \frac{3}{2} = 1 \frac{1}{2}\)

Hence,
\(- \frac{21}{16} \div ( \frac{- 7}{8} ) = 1 \frac{1}{2}\)

Exercise 1(D)

Question 2(v)

Evaluate:

\(0 \div ( \frac{- 4}{7} )\)
)
Show step-by-step answer+
\(0 \div \frac{- 4}{7} = 0 \times \frac{- 7}{4} = \frac{0 \times - 7}{1 \times 4} = \frac{0}{4} = 0\)

=0

Hence,
\(0 \div ( \frac{- 4}{7} ) = 0\)
)=0
Exercise 1(D)

Question 2(vi)

Evaluate:

\(\frac{8}{- 5} \div \frac{24}{25}\)
Show step-by-step answer+
\(\frac{8}{- 5} \div \frac{24}{25} = - \frac{8}{5} \times \frac{25}{24} = - \frac{8 \times 25}{5 \times 24} = - \frac{200}{120} = - \frac{5}{3} = - 1 \frac{2}{3}\)

Hence,
\(\frac{8}{- 5} \div \frac{24}{25} = - 1 \frac{2}{3}\)

Exercise 1(D)

Question 2(vii)

Evaluate:

\(- \frac{3}{4} \div ( - 9 )\)
÷(−9)
Show step-by-step answer+
\(- \frac{3}{4} \div ( - 9 ) = - \frac{3}{4} \times \frac{- 1}{9} = \frac{3 \times 1}{4 \times 9} = \frac{3}{36} = \frac{1}{12}\)

Hence,
\(- \frac{3}{4} \div ( - 9 ) = \frac{1}{12}\)

Exercise 1(D)

Question 2(viii)

Evaluate:

\(\frac{3}{4} \div ( - \frac{5}{12} )\)
)
Show step-by-step answer+
\(\frac{3}{4} \div - \frac{5}{12} = \frac{3}{4} \times - \frac{12}{5} = - \frac{3 \times 12}{4 \times 5} = - \frac{36}{20} = - \frac{9}{5} = - 1 \frac{4}{5}\)

Hence,
\(\frac{3}{4} \div ( - \frac{5}{12} ) = - 1 \frac{4}{5}\)

Exercise 1(D)

Question 2(ix)

Evaluate:

\(- 5 \div ( - \frac{10}{11} )\)
)
Show step-by-step answer+
\(- 5 \div - \frac{10}{11} = - 5 \times - \frac{11}{10} = \frac{5 \times 11}{1 \times 10} = \frac{55}{10} = \frac{11}{2} = 5 \frac{1}{2}\)

Hence,
\(- 5 \div ( - \frac{10}{11} ) = 5 \frac{1}{2}\)

Exercise 1(D)

Question 2(x)

Evaluate:

\(\frac{- 7}{11} \div ( \frac{- 3}{44} )\)
)
Show step-by-step answer+
\(\frac{- 7}{11} \div \frac{- 3}{44} = - \frac{7}{11} \times - \frac{44}{3} = \frac{7 \times 44}{11 \times 3} = \frac{308}{33} = \frac{28}{3} = 9 \frac{1}{3}\)

Hence,
\(\frac{- 7}{11} \div ( \frac{- 3}{44} ) = 9 \frac{1}{3}\)

Exercise 1(D)

Question 3(i)

Divide:

3 by
\(\frac{1}{3}\)
Show step-by-step answer+
\(3 \div \frac{1}{3} = 3 \times \frac{3}{1} = \frac{3 \times 3}{1 \times 1} = \frac{9}{1} = 9\)

=9

Hence,
\(3 \div \frac{1}{3} = 9\)
=9
Exercise 1(D)

Question 3(ii)

Divide:

-2 by
\(- \frac{1}{2}\)
Show step-by-step answer+
\(- 2 \div - \frac{1}{2} = - 2 \times - \frac{2}{1} = \frac{2 \times 2}{1 \times 1} = \frac{4}{1} = 4\)

=4

Hence,
\(- 2 \div - \frac{1}{2} = 4\)
=4
Exercise 1(D)

Question 3(iii)

Divide:

0 by
\(\frac{7}{- 9}\)
Show step-by-step answer+
\(0 \div \frac{7}{- 9} = 0 \times \frac{- 9}{7} = \frac{0 \times - 9}{1 \times 7} = \frac{0}{7} = 0\)

=0

Hence,
\(0 \div \frac{7}{- 9} = 0\)
=0
Exercise 1(D)

Question 3(iv)

Divide:

\(\frac{- 5}{8}\)
by
\(\frac{1}{4}\)
Show step-by-step answer+
\(\frac{- 5}{8} \div \frac{1}{4} = \frac{- 5}{8} \times \frac{4}{1} = \frac{- 5 \times 4}{8 \times 1} = \frac{- 20}{8} = \frac{- 5}{2} = - 2 \frac{1}{2}\)

Hence,
\(\frac{- 5}{8} \div \frac{1}{4} = - 2 \frac{1}{2}\)

Exercise 1(D)

Question 3(v)

Divide:

\(- \frac{3}{4}\)
by
\(- \frac{9}{16}\)
Show step-by-step answer+
\(- \frac{3}{4} \div - \frac{9}{16} = - \frac{3}{4} \times - \frac{16}{9} = \frac{3 \times 16}{4 \times 9} = \frac{48}{36} = \frac{4}{3} = 1 \frac{1}{3}\)

Hence,
\(- \frac{3}{4} \div - \frac{9}{16} = 1 \frac{1}{3}\)

Exercise 1(D)

Question 4

The product of two rational numbers is -2. If one of them is
\(\frac{4}{7}\)
, find the other.
Show step-by-step answer+

Let the number be
x.

\(\frac{4}{7} \times x = - 2 \Rightarrow \frac{4}{7} \times x = - \frac{2}{1} \Rightarrow x = - \frac{2}{1} \div \frac{4}{7} \Rightarrow x = - \frac{2}{1} \times \frac{7}{4} \Rightarrow x = - \frac{2 \times 7}{1 \times 4} \Rightarrow x = - \frac{14}{4} \Rightarrow x = - \frac{7}{2} \Rightarrow x = - 3 \frac{1}{2}\)

The other number is
\(- 3 \frac{1}{2}\)
.

Exercise 1(D)

Question 5

The product of two numbers is
\(- \frac{4}{9}\)
. If one of them is
\(\frac{- 2}{27}\)
, find the other.
Show step-by-step answer+

Let the number be
x.

\(\frac{- 2}{27} \times x = - \frac{4}{9} \Rightarrow x = - \frac{4}{9} \div \frac{- 2}{27} \Rightarrow x = - \frac{4}{9} \times \frac{- 27}{2} \Rightarrow x = \frac{4 \times 27}{9 \times 2} \Rightarrow x = \frac{108}{18} \Rightarrow x = 6\)

⇒x=6

The other number is 6.

Exercise 1(D)

Question 6(i)

m and n are two rational numbers such that
\(m \times n = - \frac{25}{9}\)
.
if
\(m = \frac{5}{3}\)
, find
n.
Show step-by-step answer+
\(m \times n = - \frac{25}{9} \Rightarrow \frac{5}{3} \times n = - \frac{25}{9} \Rightarrow n = - \frac{25}{9} \div \frac{5}{3} \Rightarrow n = - \frac{25}{9} \times \frac{3}{5} \Rightarrow n = - \frac{25 \times 3}{9 \times 5} \Rightarrow n = - \frac{75}{45} \Rightarrow n = - \frac{5}{3} \Rightarrow n = - 1 \frac{2}{3}\)

if
\(m = \frac{5}{3}\)
, then
\(n = - 1 \frac{2}{3} .\)
.

Exercise 1(D)

Question 6(ii)

m and n are two rational numbers such that
\(m \times n = - \frac{25}{9}\)
.
if
\(n = - \frac{10}{9}\)
, find
m.
Show step-by-step answer+
\(m \times n = - \frac{25}{9} \Rightarrow m \times - \frac{10}{9} = - \frac{25}{9} \Rightarrow m = - \frac{25}{9} \div - \frac{10}{9} \Rightarrow m = \frac{25}{9} \times \frac{9}{10} \Rightarrow m = \frac{25 \times 9}{9 \times 10} \Rightarrow m = \frac{225}{90} \Rightarrow m = \frac{5}{2} \Rightarrow m = 2 \frac{1}{2}\)

if
\(n = - \frac{10}{9}\)
, then
\(n = 2 \frac{1}{2} .\)
.

Exercise 1(D)

Question 7

By what number must
\(- \frac{3}{4}\)
be multiplied so that the product is
\(- \frac{9}{16}\)
?
Show step-by-step answer+

Let the number be
x

\(- \frac{3}{4} \times x = - \frac{9}{16} \Rightarrow x = - \frac{9}{16} \div - \frac{3}{4} \Rightarrow x = \frac{9}{16} \times \frac{4}{3} \Rightarrow x = \frac{9 \times 4}{16 \times 3} \Rightarrow x = \frac{36}{48} \Rightarrow x = \frac{3}{4}\)

\(- \frac{3}{4}\)
must be multiplied by
\(\frac{3}{4}\)
so that the product is
\(- \frac{9}{16}\)
.

Exercise 1(D)

Question 8

By what number should
\(- \frac{8}{13}\)
be multiplied to get 16?
Show step-by-step answer+

Let the number be
x

\(- \frac{8}{13} \times x = 16 \Rightarrow x = 16 \div - \frac{8}{13} \Rightarrow x = - \frac{16}{1} \times \frac{13}{8} \Rightarrow x = - \frac{16 \times 13}{1 \times 8} \Rightarrow x = - \frac{208}{8} \Rightarrow x = - 26\)

⇒x=−26

\(- \frac{8}{13}\)
must be multiplied by -26 so that the product is 16

Exercise 1(D)

Question 9

If
\(3\,\frac{1}{2}\)
litres of milk costs ₹49, find the cost of one litre of milk?
Show step-by-step answer+

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

\(3\,\frac{1}{2} \times x = 49 \Rightarrow \frac{7}{2} \times x = 49 \Rightarrow x = 49 \div \frac{7}{2} \Rightarrow x = 49 \times \frac{2}{7} \Rightarrow x = \frac{49 \times 2}{1 \times 7} \Rightarrow x = \frac{98}{7} \Rightarrow x = 14\)

⇒x=14

The cost of one litre of milk = ₹14.

Exercise 1(D)

Question 10

Cost of
\(3\,\frac{2}{5}\)
metre of cloth is ₹
\(88\,\frac{1}{2}\)
. What is the cost of 1 metre of cloth?
Show step-by-step answer+

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

\(3\,\frac{2}{5} \times x = 88 \frac{1}{2} \Rightarrow \frac{17}{5} \times x = \frac{177}{2} \Rightarrow x = \frac{177}{2} \div \frac{17}{5} \Rightarrow x = \frac{177}{2} \times \frac{5}{17} \Rightarrow x = \frac{177 \times 5}{2 \times 17} \Rightarrow x = \frac{885}{34} \Rightarrow x = 26 \frac{1}{34}\)

Hence, The cost of 1 meter of cloth is ₹
\(26\,\frac{1}{34}\)
.

Exercise 1(D)

Question 11

Divide the sum of
\(\frac{3}{7}\)
and
\(\frac{- 5}{14}\)
by
\(- \frac{1}{2}\)
.
Show step-by-step answer+

The sum of
\(\frac{3}{7}\)
and
\(\frac{- 5}{14}\)

\(\frac{3}{7} + \frac{- 5}{14}\)
💡 LCM of 7 and 14 is 2 x 7 = \(14 = \frac{3 \times 2}{7 \times 2} + \frac{- 5 \times 1}{14 \times 1} = \frac{6}{14} + \frac{- 5}{14} = \frac{6 + ( - 5 )}{14} = 1\)

Dividing the sum of
\(\frac{3}{7}\)
and
\(\frac{- 5}{14}\)
by
\(- \frac{1}{2}\)

\(\frac{1}{14} \div - \frac{1}{2} = \frac{1}{14} \times - \frac{2}{1} = - \frac{1 \times 2}{14 \times 1} = - \frac{2}{14} = - \frac{1}{7}\)

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

Exercise 1(D)

Question 12(i)

Find
(
m
+
n
)
÷
(
m
n
)
(m+n)÷(m−n), if;
\(m = \frac{2}{3}\)
and
\(n = \frac{3}{2}\)
Show step-by-step answer+

\(( m + n ) \div ( m - n ) = ( \frac{2}{3} + \frac{3}{2} ) \div ( \frac{2}{3} - \frac{3}{2} )\)
)

💡 LCM of 3 and 2 is 2 x 3 = 6
\(= ( \frac{2 \times 2}{3 \times 2} + \frac{3 \times 3}{2 \times 3} ) \div ( \frac{2 \times 2}{3 \times 2} - \frac{3 \times 3}{2 \times 3} ) = ( \frac{4}{6} + \frac{9}{6} ) \div ( \frac{4}{6} - \frac{9}{6} ) = ( \frac{4 + 9}{6} ) \div ( \frac{4 - 9}{6} ) = ( \frac{13}{6} ) \div ( \frac{- 5}{6} ) = \frac{13}{6} \times \frac{6}{- 5} = \frac{13 \times 6}{6 \times - 5} = - \frac{78}{30} = - \frac{13}{5}\)
If
m =
\(\frac{2}{3}\)
and
n =
\(\frac{3}{2}\)
then
\(( m + n ) \div ( m - n ) = - \frac{13}{5}\)
.
Exercise 1(D)

Question 12(ii)

Find
(
m
+
n
)
÷
(
m
n
)
(m+n)÷(m−n), if;
\(m = \frac{3}{4}\)
and
\(n = \frac{4}{3}\)
Show step-by-step answer+

\(( m + n ) \div ( m - n ) = ( \frac{3}{4} + \frac{4}{3} ) \div ( \frac{3}{4} - \frac{4}{3} )\)
)

💡 LCM of 4 and 3 is 2 x 2 x 3 = 12
\(= ( \frac{3 \times 3}{4 \times 3} + \frac{4 \times 4}{3 \times 4} ) \div ( \frac{3 \times 3}{4 \times 3} - \frac{4 \times 4}{3 \times 4} ) = ( \frac{9}{12} + \frac{16}{12} ) \div ( \frac{9}{12} - \frac{16}{12} ) = ( \frac{9 + 16}{12} ) \div ( \frac{9 - 16}{12} ) = ( \frac{25}{12} ) \div ( \frac{- 7}{12} ) = \frac{25}{12} \times \frac{12}{- 7} = \frac{25 \times 12}{12 \times - 7} = - \frac{300}{84} = - \frac{25}{7}\)
If
m =
\(\frac{3}{4}\)
and
n =
\(\frac{4}{3}\)
then
\(( m + n ) \div ( m - n ) = - \frac{25}{7}\)
.
Exercise 1(D)

Question 12(iii)

Find
(
m
+
n
)
÷
(
m
n
)
(m+n)÷(m−n), if;
\(m = \frac{4}{5}\)
and
\(n = - \frac{3}{10}\)
Show step-by-step answer+

\(( m + n ) \div ( m - n ) = [ \frac{4}{5} + ( - \frac{3}{10} ) ] \div [ \frac{4}{5} - ( - \frac{3}{10} ) ]\)
)]

💡 LCM of 5 and 10 is 2 x 5 = 10
\(= [ \frac{4 \times 2}{5 \times 2} + ( - \frac{3 \times 1}{10 \times 1} ) ] \div [ \frac{4 \times 2}{5 \times 2} - ( - \frac{3 \times 1}{10 \times 1} ) ] = [ \frac{8}{10} + ( - \frac{3}{10} ) ] \div [ \frac{8}{10} - ( - \frac{3}{10} ) ] = [ \frac{8 + ( - 3 )}{10} ] \div [ \frac{8 - ( - 3 )}{10} ] = [ \frac{5}{10} ] \div [ \frac{11}{10} ] = \frac{5}{10} \times \frac{10}{11} = \frac{5 \times 10}{10 \times 11} = \frac{50}{110} = \frac{5}{11}\)
If
m =
\(\frac{4}{5}\)
and
n = -
\(\frac{3}{10}\)
then
\(( m + n ) \div ( m - n ) = \frac{5}{11}\)
.
Exercise 1(D)

Question 13

The product of two rational numbers is -5. If one of these numbers is
\(\frac{- 7}{15}\)
, find the other.
Show step-by-step answer+

Let the number be
x.

\(\frac{- 7}{15} \times x = - 5 \Rightarrow x = - 5 \div \frac{- 7}{15} \Rightarrow x = - \frac{5}{1} \times \frac{- 15}{7} \Rightarrow x = \frac{5 \times 15}{1 \times 7} \Rightarrow x = \frac{75}{7} \Rightarrow x = 10 \frac{5}{7}\)

The other number is
\(10\,\frac{5}{7}\)
.

Exercise 1(D)

Question 14

Divide the sum of
\(\frac{5}{8}\)
and
\(\frac{- 11}{12}\)
by the difference of
\(\frac{3}{7}\)
and
\(\frac{5}{14}\)
.
Show step-by-step answer+

The sum of
\(\frac{5}{8}\)
and
\(\frac{- 11}{12}\)

\(\frac{5}{8} + \frac{- 11}{12}\)
💡 LCM of 8 and 12 is 2 x 2 x 2 x 3 = \(24 = \frac{5 \times 3}{8 \times 3} + \frac{- 11 \times 2}{12 \times 2} = \frac{15}{24} + \frac{- 22}{24} = \frac{15 + ( - 22 )}{24} = - 7\)

The difference of
\(\frac{3}{7}\)
and
\(\frac{5}{14}\)

\(\frac{3}{7} - \frac{5}{14}\)
💡 LCM of 7 and 14 is 2 x 7 = \(14 = \frac{3 \times 2}{7 \times 2} - \frac{5 \times 1}{14 \times 1} = \frac{6}{14} - \frac{5}{14} = \frac{6 - 5}{14} = 1\)

Dividing the sum of
\(\frac{5}{8}\)
and
\(\frac{- 11}{12}\)
by the difference of
\(\frac{3}{7}\)
and
\(\frac{5}{14}\)
,

\(\frac{- 7}{24} \div \frac{1}{14} = \frac{- 7}{24} \times \frac{14}{1} = - \frac{7 \times 14}{24 \times 1} = - \frac{98}{24} = - \frac{49}{12} = - 4 \frac{1}{12}\)

\(( \frac{5}{8} + \frac{- 11}{12} ) \div ( \frac{3}{7} - \frac{5}{14} ) = - 4 \frac{1}{12}\)
.

Exercise 1(D)

Question 15

The area of a rectangular plate is
\(5\,\frac{5}{7}\)
m2 and its length is
\(3\,\frac{3}{4}\)
m, find its breadth and its perimeter.
Show step-by-step answer+
Area of a rectangular plate =
\(5\,\frac{5}{7}\)
m2 =
\(\frac{40}{7}\)
m2
Length of a rectangular plate =
\(3\,\frac{3}{4}\)
m =
\(\frac{15}{4}\)
m

Let the breadth of the rectangular plate be b.

Area = length x breadth

\(\frac{40}{7} = \frac{15}{4} \times b \Rightarrow b = \frac{40}{7} \div \frac{15}{4} \Rightarrow b = \frac{40}{7} \times \frac{4}{15} \Rightarrow b = \frac{40 \times 4}{7 \times 15} \Rightarrow b = \frac{160}{105} \Rightarrow b = \frac{32}{21} \Rightarrow b = 1 \frac{11}{21}\)
Perimeter = 2(length + breadth\() = 2 \times ( \frac{15}{4} + \frac{32}{21}\)
)
💡 LCM of 4 and 21 is 2 x 2 x 3 x 7 = 84
\(= 2 \times ( \frac{15 \times 21}{4 \times 21} + \frac{32 \times 4}{21 \times 4} ) = 2 \times ( \frac{315}{84} + \frac{128}{84} ) = 2 \times ( \frac{315 + 128}{84} ) = 2 \times ( \frac{443}{84} ) = ( \frac{443 \times 2}{84} ) = ( \frac{886}{84} ) = ( \frac{443}{42} ) = 10 \frac{23}{42}\)
Hence, breadth =
\(1\,\frac{11}{21}\)
and perimeter =
\(10\,\frac{23}{42}\)
Exercise 1(D)

Question 16

The area of a piece of paper is
\(7\,\frac{3}{26}\)
cm2 and its breadth is
\(2\,\frac{9}{13}\)
cm. Find its length and perimeter.
Show step-by-step answer+
Area of a rectangular paper =
\(7\,\frac{3}{26}\)
cm2 =
\(\frac{185}{26}\)
cm2
Breadth of a rectangular paper =
\(2\,\frac{9}{13}\)
m =
\(\frac{35}{13}\)
cm

Let the length of the piece of paper be l.

Area = length x breadth

\(\frac{185}{26} = l \times \frac{35}{13} \Rightarrow l = \frac{185}{26} \div \frac{35}{13} \Rightarrow l = \frac{185}{26} \times \frac{13}{35} \Rightarrow l = \frac{185 \times 13}{26 \times 35} \Rightarrow l = \frac{2405}{910} \Rightarrow l = \frac{37}{14} \Rightarrow l = 2 \frac{9}{14}\)
Perimeter = 2(length + breadth\() = 2 \times ( \frac{37}{14} + \frac{35}{13}\)
)
💡 LCM of 14 and 13 is 2 x 7 x 13 = 182
\(= 2 \times ( \frac{37 \times 13}{14 \times 13} + \frac{35 \times 14}{13 \times 14} ) = 2 \times ( \frac{481}{182} + \frac{490}{182} ) = 2 \times ( \frac{481 + 490}{182} ) = 2 \times ( \frac{971}{182} ) = ( \frac{971 \times 2}{182} ) = ( \frac{1942}{182} ) = ( \frac{971}{91} ) = 10 \frac{61}{91}\)
Length =
\(2\,\frac{9}{14}\)
and perimeter =
\(10\,\frac{61}{91}\)
Complete Solutions

Exercise 1(E)

Tap any answer panel to reveal the full working.

Exercise 1(E)

Question 1(i)

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

In the following number line, points A and B represent: Rational Numbers, Concise Mathematics Solutions ICSE Class 8.
\(- 2 \frac{1}{5}\)
and
\(2\,\frac{3}{5}\)

\(2\,\frac{1}{5}\)
and
\(2\,\frac{3}{5}\)
\(- 1 \frac{4}{5}\)
and
\(2\,\frac{3}{5}\)
\(- \frac{1}{5}\)
and
\(\frac{2}{5}\)
Show step-by-step answer+

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

Point A is 4 step left from -1. So, A =
\(- 1 \frac{4}{5}\)
Point B is 3 step right from 2. So, B =
\(2\,\frac{3}{5}\)

Hence, Option 3 is the correct option.

Exercise 1(E)

Question 1(ii)

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

Using the number line, given below; the length of line segment AB is: Rational Numbers, Concise Mathematics Solutions ICSE Class 8.
\(\frac{13}{5}\)
=
\(2\,\frac{3}{5}\)
\(- \frac{13}{5}\)
=
\(- 2 \frac{3}{5}\)
\(4\,\frac{2}{5}\)
\(\frac{14}{15}\)
Show step-by-step answer+
As we know, A =
\(- 1 \frac{4}{5}\)
and B =
\(2\,\frac{3}{5}\)
.
Length =
\(\frac{9}{5} + \frac{13}{5}\)
\(= \frac{9 + 13}{5} = \frac{22}{5} = 4 \frac{2}{5}\)

Hence, Option 3 is the correct option.

Exercise 1(E)

Question 1(iii)

The rational number between
\(\frac{a}{b}\)
and
\(\frac{c}{d}\)
is:
\(\frac{1}{2} ( \frac{a}{b} - \frac{c}{d} )\)
\() ( \frac{a - c}{b - d}\)
\() ( \frac{a + c}{b + d}\)
\() ( \frac{a + d}{b + c}\)
)
Show step-by-step answer+

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

Hence, Option 3 is the correct option.

Exercise 1(E)

Question 1(iv)

Two rational numbers between
\(\frac{1}{3}\)
and
\(\frac{1}{2}\)
are:
\(\frac{3}{7}\)
and
\(\frac{3}{8}\)
\(\frac{2}{5}\)
and 0
\(\frac{1}{6}\)
and
\(\frac{2}{3}\)
\(\frac{2}{3}\)
and
\(\frac{3}{2}\)
Show step-by-step answer+

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

The rational number between
\(\frac{1}{3}\)
and
\(\frac{1}{2}\)
is
\(( \frac{1 + 1}{3 + 2} )\)
)

=
\(( \frac{2}{5} )\)
)

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

=
\(( \frac{3}{8} )\)
)

The rational number between
\(\frac{1}{2}\)
and
\(\frac{2}{5}\)
is
\(( \frac{1 + 2}{2 + 5} )\)
)

=
\(( \frac{3}{7} )\)
)

\(\frac{3}{7}\)
and
\(\frac{3}{8}\)
are two rational number between
\(\frac{1}{3}\)
and
\(\frac{1}{2}\)

Hence, option 1 is the correct option.

Exercise 1(E)

Question 1(v)

The rational numbers
\(- \frac{7}{4}\)
and
\(\frac{3}{4}\)
are represented by:

B and E respectively

C and D respectively

C and E respectively

B and F respectively

Show step-by-step 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
\(- \frac{1}{4}\)
.

Point C is 3 step left from -1. So, C =
\(- 1 \frac{3}{4} = - \frac{7}{4}\)
Point E is 3 step right from 0. So, E =
\(\frac{3}{4}\)

\(- \frac{7}{4}\)
and
\(\frac{3}{4}\)
are represented by C and E, respectively.

Hence, option 3 is the correct option.

Exercise 1(E)

Question 2

Draw a number line and mark

\(\frac{3}{4} , \frac{7}{4} , \frac{- 3}{4}\)
and
\(\frac{- 7}{4}\)
on it.
Show step-by-step 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
\(\frac{1}{4}\)
, move one step towards the right side of 0 to reach P as shown.

As, OA = 1 unit, therefore OP =
\(\frac{1}{4}\)
unit.

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

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

So, S represent
\(\frac{7}{4}\)
and T represent
\(\frac{- 7}{4}\)
.

Exercise 1(E)

Question 3

On a number line mark the points

\(\frac{2}{3} , \frac{- 8}{3} , \frac{7}{3} , \frac{- 2}{3}\)
and -2.
Show step-by-step 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
\(\frac{1}{3}\)
, move one step towards the right side of 0 to reach P as shown.

As, OA = 1 unit, therefore OP =
\(\frac{1}{3}\)
unit.

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

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

Similarly, S represents
\(\frac{- 8}{3}\)
, T represents
\(\frac{7}{3}\)
and B' represents -2.

Exercise 1(E)

Question 4(i)

Insert one rational number between

\(\frac{3}{5}\)
and
\(\frac{5}{8}\)
Show step-by-step answer+

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

The rational number between
\(\frac{3}{5}\)
and
\(\frac{5}{8}\)
is
\(( \frac{3 + 5}{5 + 8} )\)
)

=
\(( \frac{8}{13} )\)
)

Hence, one rational number between
\(\frac{3}{5}\)
and
\(\frac{5}{8}\)
is
\(\frac{8}{13}\)
.

Exercise 1(E)

Question 4(ii)

Insert one rational number between

\(\frac{1}{2}\)
and 2
Show step-by-step answer+

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

The rational number between
\(\frac{1}{2}\)
and
\(\frac{2}{1}\)
is
\(( \frac{1 + 2}{2 + 1} )\)
)

=
\(( \frac{3}{3} )\)
)

=1

Hence, one rational number between
\(\frac{1}{2}\)
and 2 is 1.

Exercise 1(E)

Question 5

Insert two rational numbers between:

\(\frac{5}{7}\)
and
\(\frac{3}{8}\)
Show step-by-step answer+

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

Given numbers =
\(\frac{5}{7}\)
and
\(\frac{3}{8}\)
\(= \frac{5}{7} , \frac{5 + 3}{7 + 8} , \frac{3}{8} = \frac{5}{7} , \frac{8}{15} , \frac{3}{8} = \frac{5}{7} , \frac{5 + 8}{7 + 15} , \frac{8}{15} , \frac{3}{8} = \frac{5}{7} , \frac{13}{22} , \frac{8}{15} , \frac{3}{8}\)

Hence, required rational numbers between
\(\frac{5}{7}\)
and
\(\frac{3}{8}\)
are :
\(\frac{13}{22}\)
and
\(\frac{8}{15}\)

Exercise 1(E)

Question 6

Insert three rational numbers between:

\(\frac{8}{11}\)
and
\(\frac{4}{9}\)
Show step-by-step answer+

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

Given numbers =
\(\frac{8}{11}\)
and
\(\frac{4}{9}\)
\(= \frac{8}{11} , \frac{8 + 4}{11 + 9} , \frac{4}{9} = \frac{8}{11} , \frac{12}{20} , \frac{4}{9} = \frac{8}{11} , \frac{3}{5} , \frac{4}{9} = \frac{8}{11} , \frac{8 + 3}{11 + 5} , \frac{3}{5} , \frac{4}{9} = \frac{8}{11} , \frac{11}{16} , \frac{3}{5} , \frac{4}{9} = \frac{8}{11} , \frac{11}{16} , \frac{3}{5} , \frac{3 + 4}{5 + 9} , \frac{4}{9} = \frac{8}{11} , \frac{11}{16} , \frac{3}{5} , \frac{7}{14} , \frac{4}{9} = \frac{8}{11} , \frac{11}{16} , \frac{3}{5} , \frac{1}{2} , \frac{4}{9}\)

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

Exercise 1(E)

Question 7

Insert five rational numbers between
\(\frac{3}{5}\)
and
\(\frac{2}{3}\)
Show step-by-step answer+
💡 LCM of 5 and 3 is 3 x 5 = 15

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

\(\frac{3}{5} = \frac{3 \times 3}{5 \times 3} = \frac{9}{15}\)

and

\(\frac{2}{3} = \frac{2 \times 5}{3 \times 5} = \frac{10}{15}\)
Since five rational numbers are required between
\(\frac{3}{5}\)
and
\(\frac{2}{3}\)
; multiply the numerator and the denominator of each rational number by 5 + 1 = 6.

\(\frac{9}{15} = \frac{9 \times 6}{15 \times 6} = \frac{54}{90} \therefore\)

and

\(\frac{10}{15} = \frac{10 \times 6}{15 \times 6} = \frac{60}{90}\)
⇒ Required rational numbers between
\(\frac{3}{5}\)
and
\(\frac{2}{3}\)
are :
\(\frac{54}{90} , \frac{55}{90} , \frac{56}{90} , \frac{57}{90} , \frac{58}{90} , \frac{59}{90} , \frac{60}{90}\)
=
\(\frac{3}{5} , \frac{11}{18} , \frac{28}{45} , \frac{19}{30} , \frac{29}{45} , \frac{59}{90} , \frac{2}{3}\)

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

Exercise 1(E)

Question 8

Insert six rational numbers between
\(\frac{5}{6}\)
and
\(\frac{8}{9}\)
.
Show step-by-step 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).

\(\frac{5}{6} = \frac{5 \times 3}{6 \times 3} = \frac{15}{18}\)

and

\(\frac{8}{9} = \frac{8 \times 2}{9 \times 2} = \frac{16}{18}\)
Since six rational numbers are required between
\(\frac{5}{6}\)
and
\(\frac{8}{9}\)
; multiply the numerator and the denominator of each rational number by 6 + 1 = 7.
\(\frac{15}{18} = \frac{15 \times 7}{18 \times 7} = \frac{105}{126}\)

and

\(\frac{16}{18} = \frac{16 \times 7}{18 \times 7} = \frac{112}{126}\)

Required rational numbers between
\(\frac{5}{6}\)
and
\(\frac{8}{9}\)
are :
\(\frac{105}{126} , \frac{106}{126} , \frac{107}{126} , \frac{108}{126} , \frac{109}{126} , \frac{110}{126} , \frac{111}{126} , \frac{112}{126}\)

=
\(\frac{5}{6} , \frac{53}{63} , \frac{107}{126} , \frac{6}{7} , \frac{109}{126} , \frac{55}{63} , \frac{37}{42} , \frac{8}{9}\)

Hence,
\(\frac{5}{6} , \frac{53}{63} , \frac{107}{126} , \frac{6}{7} , \frac{109}{126}\)
and
\(\frac{37}{42}\)
lie between
\(\frac{5}{6}\)
and
\(\frac{8}{9}\)
.

Exercise 1(E)

Question 9

Insert seven rational numbers between 2 and 3.

Show step-by-step answer+

Write the endpoints with denominator 1:

\(2=\frac21\qquad\text{and}\qquad3=\frac31\)

To insert seven rational numbers, multiply both fractions by \(\frac88\).

\(2=\frac{16}{8}\qquad\text{and}\qquad3=\frac{24}{8}\)

The seven rational numbers between 2 and 3 are:

\(\frac{17}{8},\;\frac{18}{8},\;\frac{19}{8},\;\frac{20}{8},\;\frac{21}{8},\;\frac{22}{8},\;\frac{23}{8}\)

In simplified or mixed-number form:

\(2\,\frac18,\;2\,\frac14,\;2\,\frac38,\;2\,\frac12,\;2\,\frac58,\;2\,\frac34,\;2\,\frac78\)
Exercise 1(E)

Question 1(i)

\(\frac{0}{5}\)
is a rational number,
\(\frac{0}{8}\)
is a rational number, then
\(\frac{0}{5} \div \frac{0}{8}\)
is:

an irrational number

a rational number

0

undefined

Show step-by-step answer+
\(\frac{0}{5} \div \frac{0}{8} = \frac{0}{5} \times \frac{8}{0} = \frac{0 \times 8}{5 \times 0} = \frac{0}{0}\)

\(\frac{0}{0}\)
is undefined.

Hence, option 4 is the correct option.

Exercise 1(E)

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

Show step-by-step 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

Exercise 1(E)

Question 1(iii)

The product of rational number
\(\frac{3}{8}\)
and its additive inverse is :

1

0

\(\frac{9}{64}\)
\(- \frac{9}{64}\)
Show step-by-step answer+
Additive inverse of
\(\frac{3}{8}\)
=
\(- \frac{3}{8}\)
\(\frac{3}{8} \times - \frac{3}{8} = - \frac{3 \times 3}{8 \times 8} = - \frac{9}{64}\)

Hence, option 4 is the correct option.

Exercise 1(E)

Question 1(iv)

The sum of rational number
\(\frac{2}{3}\)
and its reciprocal is:

1

\(2\,\frac{1}{6}\)

0

\(\frac{5}{6}\)
Show step-by-step answer+
Reciprocal of
\(\frac{2}{3}\)
=
\(\frac{3}{2}\)

We need to find the sum of
\(\frac{2}{3}\)
and
\(\frac{3}{2}\)

\(\frac{2}{3} + \frac{3}{2}\)
💡 LCM of 3 and 2 is 2 x 3 = 6
\(\frac{2 \times 2}{3 \times 2} + \frac{3 \times 3}{2 \times 3} = \frac{4}{6} + \frac{9}{6} = \frac{4 + 9}{6} = \frac{13}{6} = 2 \frac{1}{6}\)

Hence, option 2 is the correct option.

Exercise 1(E)

Question 1(v)

The product of two rational numbers is -1, if one of them is
\(\frac{2}{5}\)
, then the other is :
\(- \frac{2}{5}\)
\(\frac{5}{2}\)
\(- \frac{5}{2}\)

none of these

Show step-by-step answer+

Let the number be
x.

\(\frac{2}{5} \times x = - 1 \Rightarrow \frac{2}{5} \times x = - \frac{1}{1} \Rightarrow x = - \frac{1}{1} \div \frac{2}{5} \Rightarrow x = - \frac{1}{1} \times \frac{5}{2} \Rightarrow x = - \frac{1 \times 5}{1 \times 2} \Rightarrow x = - \frac{5}{2}\)

Hence, option 3 is the correct option.

Exercise 1(E)

Question 1(vi)

Statement 1: For a rational number
\(\frac{7}{9} , \frac{7}{9} - 0 = \frac{7}{9} but 0 - \frac{7}{9} = - \frac{7}{9}\)
. 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.

Show step-by-step answer+

For a rational number
\(\frac{7}{9} , \frac{7}{9} - 0 = \frac{7}{9} and 0 - \frac{7}{9} = - \frac{7}{9}\)

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.

Exercise 1(E)

Question 1(vii)

Assertion (A) : Additive inverse of
\(\frac{2}{5}\)
is
\(- \frac{5}{2}\)
.

Reason (R) : For every non-zero rational number 'a', its additive inverse is '-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.

Show step-by-step 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
\(\frac{2}{5}\)
is
\(- \frac{5}{2}\)
.

\(\Rightarrow \frac{2}{5} + ( - \frac{5}{2} ) \Rightarrow \frac{2}{5} - \frac{5}{2} \Rightarrow \frac{4}{10} - \frac{25}{10} \Rightarrow \frac{4 - 25}{10} \Rightarrow \frac{- 21}{10} \ne 0\)

=0

So, assertion (A) is false.

Hence, option 4 is the correct option.

Exercise 1(E)

Question 1(viii)

Assertion (A): The multiplicative inverse of \(-\frac75\) is \(-\frac57\).

Reason (R): For every non-zero rational number a, there is a rational number \(\frac1a\) such that \(a\times\frac1a=1\).

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.
Show step-by-step answer+

We know that the multiplicative inverse of every non-zero rational number a is its reciprocal \(\frac1a\). Therefore, Reason (R) is true.

\((-\frac75)\times(-\frac57)=1\)

Hence, the multiplicative inverse of \(-\frac75\) is \(-\frac57\). Assertion (A) is also true, and Reason (R) correctly explains it.

Therefore, option 1 is correct.

Exercise 1(E)

Question 1(ix)

Assertion (A) :
\(\frac{1}{2} + 2 = \frac{5}{2}\)
, which is a rational number.
Reason (R) : If
\(\frac{p}{q} and \frac{r}{s}\)
are any two rational numbers then
\(\frac{p}{q} + \frac{r}{s} = \frac{r}{s} + \frac{p}{q}\)
.

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.

Show step-by-step answer+

According to Assertion:

\(\Rightarrow \frac{1}{2} + 2 \Rightarrow \frac{1}{2} + \frac{4}{2} \Rightarrow \frac{1 + 4}{2} \Rightarrow \frac{5}{2}\)

A number is rational if it can be written in the form
\(\frac{p}{q}\)
, where p and q are integers.

Since,
\(\frac{5}{2}\)
is in the form of
\(\frac{p}{q}\)
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
\(\frac{p}{q} and \frac{r}{s}\)
are any two rational numbers then
\(\frac{p}{q} + \frac{r}{s} = \frac{r}{s} + \frac{p}{q}\)
, 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.

Exercise 1(E)

Question 1(x)

Assertion (A) : 0 and
\(\frac{11}{12}\)
are two rational numbers and
\(\frac{11}{12} \ne 0\)

=0 then 0 ÷
\(\frac{11}{12} = 0\)
=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.

Show step-by-step answer+

When 0 is divided by any non-zero number, the result is 0.

⇒ 0 ÷
\(\frac{11}{12} = 0\)
=0
Since, 0 =
\(\frac{0}{1}\)
is in the form of
\(\frac{p}{q}\)
.

So, assertion (A) is true.

The division of a rational number
\(\frac{a}{b}\)
by another non-zero rational number
\(\frac{c}{d}\)
is:

\(\Rightarrow \frac{a}{b} \div \frac{c}{d} \Rightarrow \frac{a}{b} \times \frac{d}{c} \Rightarrow a d b c\)\(\frac{ad}{bc}\)\(\frac{ad}{bc}\)
is in the form of
\(\frac{p}{q}\)
.

So, reason is true. But it does not explains about assertion.

Hence, option 2 is the correct option.

Exercise 1(E)

Question 2(i)

Write the rational number that does not have a reciprocal.

Show step-by-step answer+

The rational number is
\(\frac{0}{1}\)

The reciprocal of
\(\frac{0}{1} = \frac{1}{0}\)
(not defined).

Hence, 0 is the rational number that does not have a reciprocal.

Exercise 1(E)

Question 2(ii)

Write the rational numbers that are equal to their reciprocal.

Show step-by-step answer+
Reciprocal of
\(\frac{1}{1} = \frac{1}{1} = 1\)
=1.
Reciprocal of
\(\frac{- 1}{1} = \frac{1}{- 1} = - 1\)
=−1.

The numbers 1 and -1 are their own reciprocal.

Exercise 1(E)

Question 2(iii)

Write the reciprocal of
\(- \frac{8}{17} + ( \frac{- 8}{17} )\)
).
Show step-by-step answer+
\(- \frac{8}{17} + \frac{- 8}{17} = \frac{- 8 + ( - 8 )}{17} = \frac{- 16}{17}\)

The reciprocal of
\(\frac{- 16}{17} = - \frac{17}{16}\)
.

Exercise 1(E)

Question 3

Write five rational numbers between
\(- \frac{3}{2}\)
and
\(\frac{5}{3}\)
Show step-by-step answer+
💡 LCM of 2 and 3 is 2 x 3 = 6

Make denominator of each given rational number equal to 6 (the LCM).

\(- \frac{3}{2} = - \frac{3 \times 3}{2 \times 3} = - \frac{9}{6}\)

and

\(\frac{5}{3} = \frac{5 \times 2}{3 \times 2} = \frac{10}{6}\)

The rational number between
\(- \frac{9}{6}\)
and
\(\frac{10}{6}\)
are
\(- \frac{8}{6} , - \frac{7}{6} , - \frac{6}{6} , - \frac{5}{6} , - \frac{4}{6} , - \frac{3}{6} , - \frac{2}{6} , - \frac{1}{6} , \frac{0}{6} , \frac{1}{6} , \frac{2}{6} , \frac{3}{6} , \frac{4}{6} , \frac{5}{6} , \frac{6}{6} , \frac{7}{6} , \frac{8}{6} , \frac{9}{6}\)

From these rational numbers we can take any five rational number.

Hence, required rational numbers between
\(- \frac{3}{2}\)
and
\(\frac{5}{3}\)
are :

\(- \frac{8}{6} , - \frac{6}{6} , - \frac{4}{6} , - \frac{2}{6} , \frac{2}{6}\)
=
\(- \frac{4}{3} , - \frac{1}{1} , - \frac{2}{3} , - \frac{1}{3} , \frac{1}{3}\)

Hence,
\(- 1 \frac{1}{3} , - 1 , - \frac{2}{3} , - \frac{1}{3}\)
and
\(\frac{1}{3}\)
lies between
\(- \frac{3}{2}\)
and
\(\frac{5}{3}\)
.

Exercise 1(E)

Question 4

Write five rational numbers greater than -4.

Show step-by-step answer+

There are infinite many rational number between -4 and ∞.

Hence, -3, -2, -1, 1, 2 are five rational number greater than -4.

Exercise 1(E)

Question 5

What should be added to
\(- 2 \frac{1}{2}\)
to get
\(- 3 \frac{1}{3}\)
?
Show step-by-step answer+

Let
x be added to
\(- 2 \frac{1}{2}\)
.

\(- 2 \frac{1}{2} + x = - 3 \frac{1}{3} \Rightarrow - \frac{5}{2} + x = - \frac{10}{3} \Rightarrow x = - \frac{10}{3} - \frac{- 5}{2}\)
💡 LCM of 3 and 2 is 2 x 3 = \(6 \Rightarrow x = \frac{- 10 \times 2}{3 \times 2} - \frac{- 5 \times 3}{2 \times 3} \Rightarrow x = \frac{- 20}{6} - \frac{- 15}{6} \Rightarrow x = \frac{- 20 - ( - 15 )}{6} \Rightarrow x = \frac{- 20 + 15}{6} \Rightarrow x = - 5\)

The number added to
\(- 2 \frac{1}{2}\)
to get
\(- 3 \frac{1}{3}\)
is
\(\frac{- 5}{6}\)
.

Exercise 1(E)

Question 6

What should be subtracted from
\(2\,\frac{1}{2}\)
to get
\(- 3 \frac{1}{3}\)
?
Show step-by-step answer+

Let
x be subtracted from
\(2\,\frac{1}{2}\)
.

\(2\,\frac{1}{2} - x = - 3 \frac{1}{3} \Rightarrow \frac{5}{2} - x = - \frac{10}{3} \Rightarrow x = \frac{5}{2} - \frac{- 10}{3}\)
💡 LCM of 2 and 3 is 2 x 3 = \(6 \Rightarrow x = \frac{5 \times 3}{2 \times 3} - \frac{- 10 \times 2}{3 \times 2} \Rightarrow x = \frac{15}{6} - \frac{- 20}{6} \Rightarrow x = \frac{15 - ( - 20 )}{6} \Rightarrow x = \frac{15 + 20}{6} \Rightarrow x = \frac{35}{6} \Rightarrow x = 5\,\frac{5}{6}\)

The number subtracted from
\(2\,\frac{1}{2}\)
to get
\(- 3 \frac{1}{3}\)
is
\(5\,\frac{5}{6}\)
.

Exercise 1(E)

Question 7(i)

If
\(m = - \frac{7}{9}\)
and
\(n = \frac{5}{6}\)
, verify that:

m - n ≠ n - m

Show step-by-step answer+
To prove:

m - n ≠ n - m

LHS:

\(m - n - \frac{7}{9} - \frac{5}{6}\)
💡 LCM of 9 and 6 is 2 x 3 x 3 = \(18 - \frac{7 \times 2}{9 \times 2} - \frac{5 \times 3}{6 \times 3} = - \frac{14}{18} - \frac{15}{18} = \frac{- 14 - 15}{18} = \frac{- 29}{18} = -1\,\frac{11}{18}\)

RHS:

\(n - m \frac{5}{6} - ( - \frac{7}{9} ) = \frac{5}{6} + \frac{7}{9}\)
💡 LCM of 6 and 9 is 2 x 3 x 3 = \(18 = \frac{5 \times 3}{6 \times 3} + \frac{7 \times 2}{9 \times 2} = \frac{15}{18} + \frac{14}{18} = \frac{15 + 14}{18} = \frac{29}{18} = 1\,\frac{11}{18}\)

Hence, LHS ≠ RHS

m - n ≠ n - m

Exercise 1(E)

Question 7(ii)

If
\(m = - \frac{7}{9}\)
and
\(n = \frac{5}{6}\)
, verify that:

-(m + n) = (-m) + (-n)

Show step-by-step answer+
To prove:

-(m + n) = (-m) + (-n)

LHS:

\(- ( m + n ) - ( - \frac{7}{9} + \frac{5}{6} )\)
)

💡 LCM of 9 and 6 is 2 x 3 x 3 = 18

\(= - ( - \frac{7 \times 2}{9 \times 2} + \frac{5 \times 3}{6 \times 3} ) = - ( - \frac{14}{18} + \frac{15}{18} ) = - ( \frac{- 14 + 15}{18} ) = - ( \frac{1}{18} )\)
)

RHS:

\(( - m ) + ( - n ) = - ( - \frac{7}{9} ) + ( - \frac{5}{6} ) = ( \frac{7}{9} ) + ( - \frac{5}{6} )\)
)

💡 LCM of 9 and 6 is 2 x 3 x 3 = 18

\(= ( \frac{7 \times 2}{9 \times 2} ) + ( - \frac{5 \times 3}{6 \times 3} ) = ( \frac{14}{18} ) + ( - \frac{15}{18} ) = ( \frac{14 + ( - 15 )}{18} ) = ( \frac{- 1}{18} )\)
)

Hence, LHS = RHS



(
m
+
n
)
=
(

m
)
+
(

n
)
∴−(m+n)=(−m)+(−n)
Exercise 1(E)

Question 8

Represent rational numbers
\(- \frac{7}{3}\)
and
\(\frac{7}{4}\)
on the same number line.
Show step-by-step answer+

The rational numbers are
\(- \frac{7}{3}\)
and
\(\frac{7}{4}\)
.

💡 LCM of 3 and 4 is 2 x 2 x 3 = 12

Hence, the rational number will be-

\(- \frac{7 \times 4}{3 \times 4} , \frac{7 \times 3}{4 \times 3} = - \frac{28}{12} , \frac{21}{12} = - 2 \frac{4}{12} , 1 \frac{9}{12}\)

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
\(\frac{1}{12}\)
, moves one step towards the right side of O to reach point P as shown.

Hence, OA = 1 unit , therefore OP =
\(\frac{1}{12}\)
unit and so P represents
\(\frac{1}{12}\)
.

In the same way, to represent
\(- 2 \frac{4}{12}\)
, move 4 steps toward the left side of B' to reach point Q. Clearly, Q represents
\(- \frac{7}{3}\)
.

Similarly, to represent
\(1\,\frac{9}{12}\)
, move 9 steps toward the right side of A to reach point R. Clearly, R represents
\(\frac{7}{4}\)
.

Exercise 1(E)

Question 9(i)

Add:
\(- \frac{5}{6}\)
and
\(\frac{3}{8}\)
Show step-by-step answer+
💡 LCM of 6 and 8 is 2 x 2 x 2 x 3 = \(24 - \frac{5 \times 4}{6 \times 4} + \frac{3 \times 3}{8 \times 3} = - \frac{20}{24} + \frac{9}{24} = \frac{- 20 + 9}{24} = - 11\)

\(- \frac{5}{6} + \frac{3}{8} = - \frac{11}{24}\)
Exercise 1(E)

Question 9(ii)

Subtract:
\(\frac{3}{8}\)
from
\(- \frac{5}{6}\)
Show step-by-step answer+
💡 LCM of 8 and 6 is 2 x 2 x 2 x 3 = \(24 - \frac{5 \times 4}{6 \times 4} - \frac{3 \times 3}{8 \times 3} = - \frac{20}{24} - \frac{9}{24} = \frac{- 20 - 9}{24} = \frac{- 29}{24} = -1\,\frac{5}{24}\)

\(- \frac{5}{6} - \frac{3}{8} = - 1 \frac{5}{24}\)
Exercise 1(E)

Question 9(iii)

Multiply:
\(- \frac{5}{6}\)
and
\(\frac{3}{8}\)
Show step-by-step answer+
\(- \frac{5}{6} \times \frac{3}{8} = \frac{- 5 \times 3}{6 \times 8} = \frac{- 15}{48} = - \frac{5}{16}\)
\(- \frac{5}{6} \times \frac{3}{8} = - \frac{5}{16}\)
Exercise 1(E)

Question 9(iv)

Divide:
\(\frac{3}{8}\)
by
\(- \frac{5}{6}\)
Show step-by-step answer+
\(\frac{3}{8} \div - \frac{5}{6} = \frac{3}{8} \times - \frac{6}{5} = \frac{3 \times ( - 6 )}{8 \times 5} = \frac{- 18}{40} = - \frac{9}{20}\)
\(\frac{3}{8} \div - \frac{5}{6} = - \frac{9}{20}\)
Exercise 1(E)

Question 10(i)

By what number should
\(\frac{12}{17}\)
be multiplied to get
\(- \frac{4}{7}\)
?
Show step-by-step answer+

Let the number be
x

\(\frac{12}{17} \times x = - \frac{4}{7} \Rightarrow x = - \frac{4}{7} \div \frac{12}{17} \Rightarrow x = - \frac{4}{7} \times \frac{17}{12} \Rightarrow x = - \frac{4 \times 17}{7 \times 12} \Rightarrow x = - \frac{68}{84} \Rightarrow x = - \frac{17}{21}\)

\(- \frac{17}{21}\)
must be multiplied by
\(\frac{12}{17}\)
so that the product is
\(- \frac{4}{7}\)
.

Exercise 1(E)

Question 10(ii)

By what number should
\(\frac{12}{17}\)
be divided to get
\(- \frac{4}{7}\)
?
Show step-by-step answer+

Let the number be
x

\(\frac{12}{17} \div x = - \frac{4}{7} \Rightarrow \frac{12}{17} \times \frac{1}{x} = - \frac{4}{7} \Rightarrow x = \frac{12}{17} \times - \frac{7}{4} \Rightarrow x = - \frac{12 \times 7}{17 \times 4} \Rightarrow x = - \frac{84}{68} \Rightarrow x = - \frac{21}{17} \Rightarrow x = - 1 \frac{4}{17}\)

\(- 1 \frac{4}{17}\)
must be divided by
\(\frac{12}{17}\)
so that the product is
\(- \frac{4}{7}\)
.

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Frequently Asked Questions

Key ideas from the Rational Numbers chapter.

What is a rational number?

A rational number can be written in the form p/q, where p and q are integers and q is not zero.

How are two rational numbers added?

Use a common denominator, add the numerators, keep the common denominator and simplify the resulting fraction.

What is the additive inverse of a rational number?

The additive inverse of a number is the same number with the opposite sign. Their sum is zero.

Is addition of rational numbers commutative and associative?

Yes. For rational numbers a, b and c: a + b = b + a, and (a + b) + c = a + (b + c).

How can rational numbers be shown on a number line?

Divide the interval between consecutive integers into equal parts according to the denominator, then count the required parts from zero.

Concept by Teacher Ritu, designed by Shaleen Shekhar.
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