# RD Sharma Solutions Class 8 Rational Numbers Exercise 1.6

## RD Sharma Solutions Class 8 Chapter 1 Exercise 1.6

#### Exercise 1.6

Q-1. Verify the property $x \times y = y \times x$ by taking:

(i) x = $\frac{-1}{3}$, $y = \frac{2}{7}$

(ii) $x = \frac{-3}{5} , y = \frac{-11}{13}$

(iii) $x = 2 , y = \frac{7}{-8}$

(iv) $x = 0 , y = \frac{-15}{8}$

Solution. We have to verify that, $x \times y = y \times x$

(i) $x = \frac{-1}{3}, y = \frac{2}{7}$

$x\times y = \frac{ -1 }{ 3 }\times \frac{ 2 }{7 } = \frac{ -2 }{ 21 }$

$y\times x = \frac{ 2 }{ 7 }\times \frac{ -1 }{3 } = \frac{ -2 }{ 21 }$

$∴ \frac{-1 }{ 3 } \times \frac{ 2 }{ 7 } = \frac{ 2 }{ 7 }\times \frac{ -1 }{ 3 }$

Hence, verified.

(ii) $x = \frac{-3}{5}, y = \frac{-11}{13}$

$x\times y = \frac{ -3 }{ 5 }\times \frac{ -11 }{ 13 } = \frac{ 33 }{ 65 }$

$y\times x = \frac{ -11 }{ 13 }\times \frac{ -3 }{5 } = \frac{ 33 }{ 65 }$

$∴ \frac{-3 }{ 5 } \times \frac{ -11 }{ 13 } = \frac{ -11 }{ 13 }\times \frac{ -3 }{ 5 }$

Hence, verified.

(iii) $x = 2, y = \frac{7}{-8}$

$x\times y = 2\times \frac{ 7 }{-8 } = \frac{ 7 }{ -4 }$

$y\times x = \frac{ 7 }{ -8 } \times 2 = \frac{ 7 }{ -4 }$

$∴ 2 \times \frac{ 7 }{ -8 } = \frac{ 7 }{ -8 }\times 2$

Hence, verified.

(iv) $x = 0 , y = \frac{-15}{8}$

$x\times y = 0 \times \frac{ -15 }{ 8 } = 0$

$y\times x = \frac{ -15 }{ 8 }\times 0 = 0$

$∴ 0 \times \frac{ -15 }{ 8 } = \frac{ -15 }{ 8 }\times 0 = 0$

Hence, verified.

Q-2. Verify the property: $x \times \left ( y \times z \right ) = \left ( x \times y \right ) \times z$

(i) $x = \frac{-7}{3} , y = \frac{12}{5}, z = \frac{4}{9}$

(ii) $x = 0 , y = \frac{-3}{5}, z = \frac{-9}{4}$

(iii) $x = \frac{1}{2} , y = \frac{5}{-4}, z = \frac{-7}{5}$

(iv) $x = \frac{5}{7} , y = \frac{-12}{13}, z = \frac{-7}{18}$

Solution: We have to verify that, $x \times \left ( y \times z \right ) = \left ( x \times y \right ) \times z$

(i) $x = \frac{-7}{3} , y = \frac{12}{5}, z = \frac{4}{9}$

$\Rightarrow x \times \left ( y \times z \right ) = \frac{-7 }{ 3 } \times \left ( \frac{12 }{ 5 }\times \frac{ 4 }{ 9 } \right )= \frac{ -7 }{ 3 }\times \frac{ 16 }{ 15 } = \frac{ -112 }{ 45 }$

$\Rightarrow \left ( x \times y \right )\times z = \left (\frac{-7 }{ 3 } \times \frac{ 12 }{ 5 } \right ) \times \frac{ 4 }{ 9 } = \frac{ -28 }{ 5 }\times \frac{ 4 }{ 9 } = \frac{ -112 }{ 45 }$

$∴ \frac{ -7 }{ 8 }\times \left ( \frac{ 15 }{ 5 } \times \frac{ 4 }{ 9 } \right ) = \left ( \frac{ -7 }{ 8 }\times \frac{ 15 }{ 5 } \right )\times \frac{ 4 }{ 9 }$

(ii)  $x = 0 , y = \frac{-3}{5}, z = \frac{-9}{4}$

$\Rightarrow x \times \left ( y \times z \right ) = 0 \times \left ( \frac{ -3 }{ 5 }\times \frac{ -9 }{ 4 } \right )= 0 \times \frac{ 27 }{ 20 } = 0$

$\Rightarrow \left ( x \times y \right )\times z = \left ( 0 \times \frac{ -3 }{ 5 } \right ) \times \frac{ -9 }{ 4 } = 0 \times \frac{ -9 }{ 4 } = 0$

$∴ 0 \times \left ( \frac{ -3 }{ 5 } \times \frac{ -9 }{ 4 } \right ) = \left ( 0 \times \frac{ -3 }{ 5 } \right )\times \frac{ -9 }{ 4 }$

(iii) $x = \frac{1}{2} , y = \frac{5}{-4}, z = \frac{-7}{5}$

$\Rightarrow x \times \left ( y \times z \right ) = \frac{ 1 }{ 2 } \times \left ( \frac{5 }{ -4 }\times \frac{ -7 }{ 4 } \right )= \frac{ -1 }{ 2 }\times \frac{ 35 }{ 16 } = \frac{ 35 }{ 32 }$

$\Rightarrow \left ( x \times y \right )\times z = \left (\frac{1 }{ 2 } \times \frac{ 5 }{ -4 } \right ) \times \frac{ -7 }{ 4 } = \frac{ -5 }{ -8 }\times \frac{ -7 }{ 4 } = \frac{ 35 }{ 32 }$

$∴ \frac{1 }{ 2 }\times \left ( \frac{ 5 }{ -4 } \times \frac{ -7 }{ 4 } \right ) = \left ( \frac{ 1 }{ 2 }\times \frac{ 5 }{ -4 } \right )\times \frac{ -7 }{ 4 }$

(iv) $x = \frac{5}{7} , y = \frac{-12}{13}, z = \frac{-7}{18}$

$\Rightarrow x \times \left ( y \times z \right ) = \frac{5 }{ 7 } \times \left ( \frac{ -12 }{ 13 }\times \frac{ -7 }{ 18 } \right )= \frac{ 5 }{ 7 }\times \frac{ 14 }{ 39 } = \frac{ 10 }{ 39 }$

$\Rightarrow \left ( x \times y \right )\times z = \left (\frac{ 5 }{ 7 } \times \frac{ -12 }{ 13 } \right ) \times \frac{ -7 }{ 18 } = \frac{ -60 }{ 91 }\times \frac{ -7 }{ 18 } = \frac{ 10 }{ 39 }$

$∴ \frac{ 5 }{ 7 }\times \left ( \frac{ -12 }{ 13 } \times \frac{ -7 }{ 18 } \right ) = \left ( \frac{ 5 }{ 7 }\times \frac{ -12 }{ 13 } \right )\times \frac{ -7 }{ 18 }$

Q-3. Verify the property: $x \times \left ( y \times z \right ) = x \times y + x \times z$:

(i) $x = \frac{-3}{ 7 } , y = \frac{ 12 }{ 13 }, z = \frac{ -5 }{ 6 }$

(ii) $x = \frac{ -12 }{ 5 } , y = \frac{ -15 }{ 4 }, z = \frac{ 8 }{ 3 }$

(iii) $x = \frac{ -8 }{ 3 } , y = \frac{ 5 }{ 6 }, z = \frac{ – 13 }{ 12 }$

(iv) $x = \frac{ -3 }{ 4 } , y = \frac{ -5 }{ 2 }, z = \frac{ 7 }{ 6 }$

Solution: We have to verify that, $x \times \left ( y \times z \right ) = x \times y + x \times z$

(i) $x = \frac{-3}{ 7 } , y = \frac{ 12 }{ 13 }, z = \frac{ -5 }{ 6 }$

$x\times \left ( y + z \right ) = \frac{ -3 }{ 7 } \times \left ( \frac{ 12 }{ 13 } + \frac{ -5 }{ 6 } \right ) = \frac{ -3 }{ 7 } \times \frac{ 72 – 65 }{ 78 } = \frac{ -3 }{ 7 } \times \frac{ 7 }{ 78 } = \frac{ -1 }{ 26}$

$x \times y + x \times z = \frac{ -3 }{ 7 } \times \frac{ 12 }{ 13 } + \frac{ -3 }{ 7 } \times \frac{ -5 }{ 6 } = \frac{ -36 }{ 91 } + \frac{ 5 }{ 14} = \frac{ -36 \times 2 + 5 \times 13 }{ 182} = \frac{-1 }{ 26 }$

$∴ \frac{-3 }{ 7 }\times \left ( \frac{ 12 }{ 13 } + \frac{ -5 }{ 6 } \right ) = \frac{-3 }{ 7 } \times \frac{ 12 }{ 13 } + \frac{ -3 }{ 7 } \times \frac{ -5 }{ 6 }$

Hence, verified.

(ii) $x = \frac{-12}{ 5 } , y = \frac{ -15 }{ 4 }, z = \frac{ 8 }{ 3 }$

$x\times \left ( y + z \right ) = \frac{ -12 }{ 5 } \times \left ( \frac{ -15 }{ 4 } + \frac{ 8 }{ 3 } \right ) = \frac{ -12 }{ 5 } \times \frac{ -45 + 32 }{ 12 } = \frac{ -12 }{ 5 } \times \frac{ -13 }{ 12 } = \frac{ 13 }{ 5}$

$x \times y + x \times z = \frac{ -12 }{ 5 } \times \frac{ -15 }{ 4 } + \frac{ -12 }{ 5 } \times \frac{ 8 }{ 3 } = \frac{ 9 }{ 1 } + \frac{ -32 }{ 5 } = \frac{ 45 – 32 }{ 5 } = \frac{ 13 }{ 5 }$

$∴ \frac{ -12 }{ 5 } \times \left ( \frac{ -15 }{ 4 } + \frac{ 8 }{ 3 } \right ) = \frac{ -12 }{ 5 } \times \frac{ -15 }{ 4 } + \frac{ -12 }{ 5 } \times \frac{ 8 }{ 3 }$

Hence, verified.

(iii) $x = \frac{-8}{ 3 } , y = \frac{ 5 }{ 6 }, z = \frac{ -13 }{ 12 }$

$x\times \left ( y + z \right ) = \frac{ -8 }{ 3 } \times \left ( \frac{ 5 }{ 6 } + \frac{ -13 }{ 12 } \right ) = \frac{ -8 }{ 3 } \times \frac{ 10 – 13 }{ 12 } = \frac{ -8 }{ 3 } \times \frac{ -3 }{ 12 } = \frac{ 2 }{ 3}$

$x \times y + x \times z = \frac{ -8 }{ 3 } \times \frac{ 5 }{ 6 } + \frac{ -8 }{ 3 } \times \frac{ -13 }{ 12 } = \frac{ -20 }{ 9 } + \frac{ 26 }{ 9 } = \frac{ -20 + 26 }{ 9 } = \frac{ 6 }{ 9 } = \frac{2}{3}$

$∴ \frac{-8 }{ 3 }\times \left ( \frac{ 5 }{ 6 } + \frac{ -13 }{ 12 } \right ) = \frac{-8 }{ 3 } \times \frac{ 5 }{ 6 } + \frac{ -8 }{ 3 } \times \frac{ -13 }{ 12 }$

Hence, verified.

(iv) $x = \frac{-3}{ 4 } , y = \frac{ -5 }{ 2 }, z = \frac{ 7 }{ 6 }$

$x\times \left ( y + z \right ) = \frac{ -3 }{ 4 } \times \left ( \frac{ -5 }{ 2 } + \frac{ 7 }{ 6 } \right ) = \frac{ -3 }{ 4 } \times \frac{ -15 + 7 }{ 6 } = \frac{ -3 }{ 4 } \times \frac{ -8 }{ 6 } = 1$

$x \times y + x \times z = \frac{ -3 }{ 4 } \times \frac{ -5 }{ 2 } + \frac{ -3 }{ 4 } \times \frac{ 7 }{ 6 } = \frac{ 15 }{ 8 } + \frac{ -7 }{ 8 } = \frac{ 15 – 7 }{ 8 } = 1$

$∴ \frac{-3 }{ 4 }\times \left ( \frac{ -5 }{ 2 } + \frac{ 7 }{ 6 } \right ) = \frac{-3 }{ 4 } \times \frac{ -5 }{ 2 } + \frac{ -3 }{ 4 } \times \frac{ 7 }{ 6 }$

Hence, verified.

Q-4. Use the distributivity of multiplication of rational numbers over their addition to simplify:

(i) $\frac{ 3 }{ 5 } \times \left ( \frac{ 35 }{ 24 } + \frac{ 10 }{ 1 } \right )$

(ii) $\frac{ -5 }{ 4 } \times \left ( \frac{ 8 }{ 5 } + \frac{ 16 }{ 5 } \right )$

(iii) $\frac{ 2}{ 7 } \times \left ( \frac{ 7 }{ 16 } – \frac{ 21 }{ 4 } \right )$

(iv) $\frac{ 3 }{ 4 } \times \left ( \frac{ 8 }{ 9 } – 40 \right )$

Solution:

(i) $\frac{ 3 }{ 5 } \times \left ( \frac{ 35 }{ 24 } + \frac{ 10 }{ 1 } \right ) = \frac{ 3 }{ 5 } \times \frac{ 35 }{ 24 } + \frac{ 3 }{ 5 }\times \frac{ 10 }{ 1 } = \frac{ 7 + 48 }{ 8 } = \frac{ 55 }{ 8 }$

(ii) $\frac{ -5 }{ 4 } \times \left ( \frac{ 8 }{ 5 } + \frac{ 16 }{ 5 } \right ) = \frac{ -5 }{ 4 } \times \frac{ 8 }{ 5 } + \frac{ -5 }{ 4 }\times \frac{ 16 }{ 5 } = \frac{ -2 }{ 2 } \times \frac{-4 }{ 1} = -6$

(iii) $\frac{ 2 }{ 7 } \times \left ( \frac{ 7 }{ 16 } – \frac{ 21 }{ 4 } \right ) = \frac{ 2 }{ 7 } \times \frac{ 7 }{ 16 } – \frac{ 2 }{ 7 }\times \frac{ 21 }{ 4 } = \frac{ 1 }{ 8 } – \frac{ 3 }{ 2 } = \frac{ 1 – 12 }{ 8 } = \frac{ -11 }{ 8 }$

(iv) $\frac{ 3 }{ 4 } \times \left ( \frac{ 8 }{ 9 } – 40 \right ) = \frac{ 3 }{ 4 } \times \frac{ 8 }{ 9 } – \frac{ 3 }{ 4 }\times 40 =\frac{ 2 }{ 3 } – 30 = \frac{ 2 – 90 }{ 3 } = \frac{ -88 }{ 3 }$

Q-5. Find the multiplicative inverse (reciprocal) of each of the following rational numbers:

(i) 9

(ii) -7

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

(iv) $\frac{-7}{9}$

(v) $\frac{-3}{-5}$

(vi) $\frac{ 2 }{ 3} \times \frac{ 9 }{ 4 }$

(vii) $\frac{ -5}{ 16 } { 15 } \times \frac{ -3 }{ 5 }$

(viii) $-2 \times \frac{ -3 }{ 5 }$

(ix) -1

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

(xi) 1

Solution:

(i) Multiplicative inverse ( reciprocal ) of 9 = $\frac{ 1 }{ 9 }$.

(ii) Multiplicative inverse ( reciprocal ) of -7 = $\frac{ 1 }{ -7 }$

(iii) Multiplicative inverse (reciprocal) of $\frac{ 12 }{ 5 }$ = $\frac{ 5 }{ 12 }$

(iv) Multiplicative inverse (reciprocal) of $\frac{ -7 }{ 9 }$ = $\frac{ -9 }{ 7 }$

(v) Multiplicative inverse (reciprocal) of $\frac{ -3 }{ -5 }$ = $\frac{ -5 }{ -3 } \; or \; \frac{ 5 }{ 3 }$

(vi) Multiplicative inverse (reciprocal) of $\frac{ 2 }{ 3 }\times \frac{ 9 }{ 4 }$ = $\frac{ 3 }{ 2 }\times \frac { 4 }{ 9 } = \frac{ 2 }{ 3 }$

(vii) Multiplicative inverse (reciprocal) of $\frac{ -5 }{ 8 }\times \frac{ 16 }{ 15 }$ = $\frac{ 8 }{ -5 }\times \frac { 15 }{ 16 } = \frac{ -3 }{ 2 }$

(viii) Multiplicative inverse (reciprocal) of $-2 \times \frac{ -3 }{ 5 }$ = $\frac{ 1 }{ -2 }\times \frac { 5 }{ -3 } = \frac{ 5 }{ 6 }$

(ix) Multiplicative inverse (reciprocal) of -1 = $\frac{ 1 }{ -1 } = -1$

(x) Multiplicative inverse (reciprocal) of $\frac{0}{3}$ = $\frac{3}{0}$  Undefined

(xi) Multiplicative inverse (reciprocal) of 1 = $\frac{ 1 }{ 1 }$ = 1

Q-6. Name the property of multiplication of rational numbers illustrated by the following statements:

(i) $\frac{ – 5}{ 16 }\times \frac{ 8 }{ 15 } = \frac{ 8 }{ 15 } \times \frac{ – 5}{ 16 }$

(ii) $\frac{ – 17}{ 5 }\times 9 = 9 \times \frac{ – 17}{ 5 }$

(iii) $\frac{ 7 }{ 4 } \times \left ( \frac{ -8 }{ 3 } + \frac{ -13 }{ 12 } \right ) = \frac{ 7 }{ 4 }\times \frac{ -8 }{ 3 } + \frac{ 7 }{ 4 } \times \frac{ -13 }{ 12 }$

(iv) $\frac{ -5 }{ 9 } \times \left ( \frac{ 4 }{ 15 } + \frac{ -9 }{ 8 } \right ) = \left (\frac{ -5 }{ 9 }\times \frac{ 4 }{ 15 } \right ) \times \frac{ -9 }{ 8 }$

(v) $\frac{ 13 }{ -17 } \times 1 = \frac{ 13 }{ -17 } = 1 \times \frac{ 13 }{ -17 }$

(vi) $\frac{ -11 }{ 16 } \times \frac{16 }{ -11 } = 1$

(vii) $\frac{ 2 }{ 13 } \times 0 = 0 = 0 \times \frac{ 2 }{ 13 }$

(viii) $\frac{ 3 }{-2}\times \frac{ 5 }{ 4 } + \frac{ -3 }{ 2 }\times \frac{ -7 }{ 6 } = \frac{ -3 }{ 2 } \times \left ( \frac{ 5 }{ 4 } + \frac{ -7 }{ 6 } \right )$

Solution:

(i) Commutative property

(ii) Commutative Property

(iii) Distributivity of multiplication over addition

(iv) Associativity of multiplication.

(v) The existence of identity for multiplication.

(vi) Existence of multiplicative inverse

(vii) Multiplication by 0

(viii) Distributive property

Q-7. Fill in the blanks:

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

(ii) The product of a positive rational number and a negative rational number is always ……………

(iii) The product of two negative rational numbers is always …………………

(iv) The reciprocal of a positive rational number is ……………….

(v) The reciprocal of a negative rational number is ……………….

(vi) Zero has ………… reciprocal.

(vii) The product of a rational number and its reciprocal is ……………….

(viii)  The numbers ………. and ……….. are their own reciprocals.

(ix) If a is reciprocal of b, then the reciprocal of b is …………………

(x) The number 0 is …………. The reciprocal of any number.

(xi) Reciprocal of $\frac{1}{a}, a \neq 0$ is …………………

(xii) $\left ( 17 \times 12 \right )^{-1} = \left ( 17 \right )^{-1} \times ………$

Solution:

(i) Positive

(ii) Negative

(iii) Positive

(iv) Positive

(v) Negative

(vi) No

(vii) 1

(viii) -1 and 1

(ix) a

(x) not

(xi) a

(xii) $12^{-1}$

Q-8. Fill in the blanks:

(i) $-4 \times \frac{7}{9} = \frac{7}{9} \times ………$

(ii) $\frac{5}{11} \times \frac{-3}{8} = \frac{-3}{8} \times ………$

(iii) $\frac{ 1 }{ 2 }\times \left ( \frac{ 3 }{ 4 } + \frac{ -5 }{ 12 } \right ) = \frac{ 1 }{ 2 } \times ………. + ………. \times \frac{ -5 }{ 12 }$

(iv) $\frac{-4}{5} \times \left ( \frac{5}{7} + \frac{-8}{9} \right ) = \left ( \frac{-4}{5} \times …… \right ) + \frac{-4}{5} \times \frac{-8}{9}$

Solution:

(i) As, $x \times y = y \times x$ that is commutativity

So, -4

(ii) As, $x \times y = y \times x$ that is commutativity

So, $\frac{ 5 }{ 11 }$

(iii) As, $x \times \left ( y + z \right ) = x \times y + x \times z$ that is distributivity of multiplication over addition

So, $\frac{ 3 }{ 4 }, \frac{ 1 }{ 2 }$

(iv) As, $x \times \left ( y \times z \right ) = \left( x \times y \right ) \times z$ that is associativity  of multiplication

So, $\frac{ 5 }{ 7 }$