myhelper: RD Sharma solution class 8 chapter 2 Powers Exercise 2.1

Exercise 2.1

Page-2.8

Question 1:

Express each of the following as a rational number of the form $\frac{p}{q},$ where p and q are integers and q ≠ 0.
(i) 2⁻³
(ii) (−4)⁻²
(iii) $\frac{1}{3^{- 2}}$
(iv) ${(\frac{1}{2})}^{- 5}$
(v) ${(\frac{2}{3})}^{- 2}$

Answer 1:

We know that $a^{- n} = \frac{1}{a^{n}}$ . Therefore,

(i)
$2^{-3} = frac{1}{2^{3}}=frac{1}{8}$

(ii)

(iii)
$frac{1}{3^{-2}} = 3^2=9$

(iv)
$left (frac{1}{2} right )^{-5} = 2^5=32$

(v)
$left (frac{2}{3} right )^{-2} = left (frac{3}{2} right )^{2}=frac{9}{4}$

Question 2:

Fiind the value of each of the following:
(i) 3⁻¹ + 4⁻¹
(ii) (3⁰ + 4⁻¹) × 2²
(iii) (3⁻¹ + 4⁻¹ + 5⁻¹)⁰
(iv) ${\{{(\frac{1}{3})}^{- 1} - {(\frac{1}{4})}^{- 1}\}}^{- 1}$

Answer 2:

(i) We know from the property of powers that for every natural number a, a⁻¹ = 1/a. Then:
$3^{-1}+4^{-1}=frac{1}{3}+frac{1}{4}$           ---> (a⁻¹ = 1/a)
                       $=frac{4+3}{12}$
                       $=frac{7}{12}$

(ii) We know from the property of powers that for every natural number a, a⁻¹ = 1/a.
Moreover, a⁰ is 1 for every natural number a not equal to 0. Then:

$(3^{0} + 4^{- 1}) \times 2^{2} = (1 + \frac{1}{4}) \times 4 [as, a^{- 1} = \frac{1}{a}; a^{0} = 1] = \frac{5}{4} \times 4 = 5$

(iii) We know from the property of powers that for every natural number a, a⁻¹ = 1/a.
Moreover, a⁰ is 1 for every natural number a not equal to 0. Then:
$(3^{- 1} + 4^{- 1} + 5^{- 1}) = 1$           ---> (Ignore the expression inside the bracket and use a⁰ = 1 immediately.)

(iv) We know from the property of powers that for every natural number a, a⁻¹ = 1/a. Then:
$left (left (frac{1}{3} right )^{-1}-left (frac{1}{4} right )^{-1} right )^{-1}=left ( 3-4 right )^{-1}$           ---> (a⁻¹ = 1/a)
                                                   $=left ( -1 right )^{-1}$
                                                  = $- 1$                       ---> (a⁻¹ = 1/a)

Question 3:

Find the value of each of the following:
(i) ${(\frac{1}{2})}^{- 1} + {(\frac{1}{3})}^{- 1} + {(\frac{1}{4})}^{- 1}$
(ii) ${(\frac{1}{2})}^{- 2} + {(\frac{1}{3})}^{- 2} + {(\frac{1}{4})}^{- 2}$
(iii) (2⁻¹ × 4⁻¹) ÷ 2⁻²
(iv) (5⁻¹ × 2⁻¹) ÷ 6⁻¹

Answer 3:

(i)
${(\frac{1}{2})}^{- 1} + {(\frac{1}{3})}^{- 1} + {(\frac{1}{4})}^{- 1} = \frac{1}{1 / 2} + \frac{1}{1 / 3} + \frac{1}{1 / 4}$           --> (a⁻¹ = 1/a)
                                                           =2+3+4
                                                           =12

(ii)
${(\frac{1}{2})}^{- 2} + (\frac{1}{3})$ ^-2+14-2=11/22+11/32+11/42        --> (a⁻ⁿ = 1/(aⁿ))
                                                         = $\frac{1}{1 / 4} + \frac{1}{1 / 9} + \frac{1}{1 / 16}$               --> ((a/b)ⁿ = (aⁿ/bⁿ))
                                                         = 4+9+16
                                                          =29

(iii)
$(2^{- 1} \times 4^{- 1}) \div 2^{- 2} = (\frac{1}{2} \times \frac{1}{4}) \div \frac{1}{2^{2}}$    --> (a⁻ⁿ = 1/(aⁿ))

                                       = $\frac{1}{8} \times 4$
                                       = 2

(iv)
$(5^{- 1} \times 2^{- 1}) \div 6^{- 1} = (\frac{1}{5} \times \frac{1}{2}) \div \frac{1}{6}$              --> (a⁻ⁿ = 1/(aⁿ))
                                       $= \frac{1}{10} \times 6$

                                       $= \frac{3}{5}$

Question 4:

Simplify:
(i) ${(4^{- 1} \times 3^{- 1})}^{2}$
(ii) ${(5^{- 1} \div 6^{- 1})}^{3}$
(iii) ${(2^{- 1} + 3^{- 1})}^{- 1}$
(iv) ${(3^{- 1} \times 4^{- 1})}^{- 1} \times 5^{- 1}$

Answer 4:

(i)
$left (4^{-1}times 3^{-1} right )^2=left (frac{1}{4}times frac{1}{3} right )^2$           ---> (a⁻¹ = 1/a)
                             $=left (frac{1}{12} right )^2$
                             $=frac{1^2}{12^2}$                           ---> ((a/b)ⁿ = (aⁿ)/(bⁿ) )
                             $= \frac{1}{24}$

(ii)
$left (5^{-1}div 6^{-1} right )^3=left (frac{1}{5}div frac{1}{6} right )^3$           ---> (a⁻¹ = 1/a)
                            $=left (frac{1}{5}times6 right )^3$
                            = ${(\frac{6}{5})}^{3}$
                            = $\frac{{(6)}^{3}}{{(5)}^{3}}$                          ---> ((a/b)ⁿ = (aⁿ)/(bⁿ) )

                            = $\frac{216}{125}$

(iii)
$(2^{- 1} + 3^{- 1})^{- 1} = {(\frac{1}{2} + \frac{1}{3})}^{- 1}$           ---> (a⁻¹ = 1/a)
                             = ${(\frac{5}{6})}^{- 1}$
                               $= \frac{1}{5 / 6}$                           ---> (a⁻¹ = 1/a)
                               $= \frac{6}{5}$

(iv)
$left (3^{-1}times 4^{-1} right )^{-1}times 5^{-1}=left (frac{1}{3}times frac{1}{4} right )^{-1}times frac{1}{5}$           ---> (a⁻¹ = 1/a)
                                             $=left (frac{1}{12}right )^{-1}times frac{1}{5}$
                                             $= \frac{12}{5}$                                       ---> (a⁻¹ = 1/a)

Question 5:

Simplify:
(i) $(3^{2} + 2^{2}) \times {(\frac{1}{2})}^{3}$
(ii) $(3^{2} - 2^{2}) \times {(\frac{2}{3})}^{- 3}$
(iii) $[{(\frac{1}{3})}^{- 3} - {(\frac{1}{2})}^{- 3}] \div {(\frac{1}{4})}^{- 3}$
(iv) $(2^{2} + 3^{2} - 4^{2}) \div {(\frac{3}{2})}^{2}$

Answer 5:

(i)
$(3^{2} + 2^{2}) \times {(\frac{1}{2})}^{3} = (9 + 4) \times \frac{1}{8} = \frac{13}{8}$

(ii)
$left ( 3^2-2^2 right )times left(frac{2}{3} right )^{-3}=left(9-4 right )times frac{1}{left (2/3 right )^{3}}$             ---> (a⁻¹=1/(aⁿ))
                                           $=5times frac{1}{8/27}$                               ---> ((a/b)ⁿ = (aⁿ)/(bⁿ))
                                           $= 5 \times \frac{27}{8}$
                                           $= \frac{135}{8}$

(iii)
$left( left(frac{1}{3} right )^{-3}-left(frac{1}{2} right )^{-3}right )div left( frac{1}{4}right )^{-3}=left(3^3-2^3 right )div 4^3$             --->(a^-n = 1/(aⁿ))
                                                                  = $(27 - 8) \div 64$
                                                                   $= 19 \times \frac{1}{64}$
                                                                   $= \frac{19}{64}$

(iv)
$(2^{2} + 3^{2} - 4^{2}) \div {(\frac{3}{2})}^{2} = (4 + 9 - 16) \times \frac{9}{4}$                     ---> ((a/b)ⁿ = (aⁿ)/(bⁿ))

                                                   $= - 3 \times \frac{9}{4}$
                                                  $= \frac{- 27}{4}$

Question 6:

By what number should 5⁻¹ be multiplied so that the product may be equal to (−7)⁻¹?

Answer 6:

Using the property a⁻¹ = 1/a for every natural number a, we have 5⁻¹ = 1/5 and (−7)⁻¹ = −1/7. We have to find a number x such that
$\frac{1}{5} \times x = \frac{- 1}{7}$
Multiplying both sides by 5, we get:
$x = \frac{- 5}{7}$
Hence, the required number is −5/7.

Question 7:

By what number should ${(\frac{1}{2})}^{- 1}$ be multiplied so that the product may be equal to ${(- \frac{4}{7})}^{- 1} ?$

Answer 7:

Using the property a⁻¹ = 1/a for every natural number a, we have (1/2)⁻¹ = 2 and (−4/7)⁻¹ = −7/4. We have to find a number x such that
$2 x = \frac{- 7}{4}$
Dividing both sides by 2, we get:
$x = \frac{- 7}{8}$
Hence, the required number is −7/8.

Question 8:

By what number should (−15)⁻¹ be divided so that the quotient may be equal to (−5)⁻¹?

Answer 8:

Using the property a⁻¹ = 1/a for every natural number a, we have (−15)⁻¹ = −1/15 and (−5)⁻¹ = −1/5. We have to find a number x such that
$\frac{\frac{- 1}{15}}{\frac{x}{1}} = \frac{- 1}{5} or \frac{- 1}{15} \times \frac{1}{x} = \frac{- 1}{5} or x = \frac{1}{3}$
Hence, (−15)⁻¹should be divided by $\frac{1}{3}$ to obtain (−5)⁻¹.

RD Sharma solution class 8 chapter 2 Powers Exercise 2.1