RD Sharma 2019 2020 Solutions for Class 8 Maths Chapter 2 Powers are provided here with simple step-by-step explanations. These solutions for Powers are extremely popular among class 8 students for Maths Powers Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the RD Sharma 2019 2020 Book of class 8 Maths Chapter 2 are provided here for you for free. You will also love the ad-free experience on Meritnation’s RD Sharma 2019 2020 Solutions. All RD Sharma 2019 2020 Solutions for class 8 Maths are prepared by experts and are 100% accurate.
Page No 2.18:
Question 1:
Write each of the following in exponential form:
(i)
(ii)
Answer:
Page No 2.18:
Question 2:
Evaluate:
(i) 5−2
(ii) (−3)−2
(iii)
(iv)
Answer:
---> (a−n = 1/(an))
---> (a−n = 1/(an))
(iii) ---> (a−n = 1/(an))
= 81
---> (a−1 = 1/(a))
Page No 2.18:
Question 3:
Express each of the following as a rational number in the form
(i) 6−1
(ii) (−7)−1
(iii)
(iv)
(v)
Answer:
---> (a−1 = 1/a)
---> (a−1 = 1/a)
=
---> (a−1 = 1/a)
(iv) ---> (a−1 = 1/a)
![(v)left(frac{3}{5} right )^{-1}times left(frac{5}{2} right )^{-1}=frac{1}{3/5}times frac{1}{5/2}](/img/site_content/ask-answer/loader.gif)
Page No 2.18:
Question 4:
Simplify:
(i)
(ii)
(iii)
(iv)
(v)
Answer:
![(i)left(4^{-1}times3^{-1} right )^2=left(frac{1}{4}times frac{1}{3}right )^2](/img/site_content/ask-answer/loader.gif)
--->((a/b)n = (an)/(bn))
![(ii)left(5^{-1}div6^{-1} right )^3=left(frac{1}{5}divfrac{1}{6} right)^3](/img/site_content/ask-answer/loader.gif)
--->((a/b)n = (an)/(bn))
![(iv)left(3^{-1}times4^{-1} right )^{-1}times5^{-1}=left(frac{1}{3}timesfrac{1}{4} right )^{-1}timesfrac{1}{5}](/img/site_content/ask-answer/loader.gif)
---> (a−1 = 1/a)
![(v)left(4^{-1}-5^{-1} right )div 3^{-1}=left(frac{1}{4}-frac{1}{5} right )div frac{1}{3}](/img/site_content/ask-answer/loader.gif)
![=left(frac{5-4}{20} right )times 3](/img/site_content/ask-answer/loader.gif)
Page No 2.18:
Question 5:
Express each of the following rational numbers with a negative exponent:
(i)
(ii)
(iii)
(iv)
(v)
Answer:
Page No 2.19:
Question 6:
Express each of the following rational numbers with a positive exponent:
(i)
(ii)
(iii)
(iv)
(v)
Answer:
---> (a−1 = 1/a)
---> (a−1 = 1/a)
---> (am x an = am+n)
---> ((am)n = amn)
---> ((am)n = amn)
Page No 2.19:
Question 7:
Simplify:
(i)
(ii)
(iii)
(iv)
(v)
Answer:
![(i)left(left(frac{1}{3} right )^{-3}- left(frac{1}{2} right )^{-3}right )div left(frac{1}{4} right )^{-3}=left(frac{1}{(1/3)^3}- frac{1}{left (1/2 right )^3}right )divfrac{1}{(1/4)^{3}}](/img/site_content/ask-answer/loader.gif)
![=left(frac{1}{(1/27)}- frac{1}{left (1/8 right )}right )divfrac{1}{(1/64)}](/img/site_content/ask-answer/loader.gif)
![=left(frac{27}{1}- frac{8}{1}right )div64](/img/site_content/ask-answer/loader.gif)
![=left(19right )timesfrac{1}{64}](/img/site_content/ask-answer/loader.gif)
![=frac{19}{64}](/img/site_content/ask-answer/loader.gif)
![(ii)left(3^2-2^2right )times left(frac{2}{3} right )^{-3}=left(9-4 right )timesfrac{1}{(2/3)^3}](/img/site_content/ask-answer/loader.gif)
![=5timesfrac{1}{8/27}](/img/site_content/ask-answer/loader.gif)
![=5timesfrac{27}{8}](/img/site_content/ask-answer/loader.gif)
![=frac{135}{8}](/img/site_content/ask-answer/loader.gif)
![(iii)left(left(frac{1}{2}right )^{-1}times left(-4 right )^{-1}right)^{-1}=left (left(frac{1}{1/2}right )times left(frac{1}{-4} right ) right )^{-1}](/img/site_content/ask-answer/loader.gif)
![=left (2times left (frac{1}{-4} right ) right )^{-1}](/img/site_content/ask-answer/loader.gif)
![=left (frac{1}{-2} right )^{-1}](/img/site_content/ask-answer/loader.gif)
![=frac{1}{1/(-2)}](/img/site_content/ask-answer/loader.gif)
=-2
![(iv)left(left(left(frac{-1}{4} right )^2 right )^{-2} right )^{-1}=left (left (frac{(-1)^2}{4^2} right )^{-2} right )^{-1}](/img/site_content/ask-answer/loader.gif)
![=left (left (frac{1}{16} right )^{-2} right )^{-1}](/img/site_content/ask-answer/loader.gif)
![=left (left (frac{1}{(1/16)^2} right ) right )^{-1}](/img/site_content/ask-answer/loader.gif)
![=left (frac{1}{(1/256)} right )^{-1}](/img/site_content/ask-answer/loader.gif)
![=256^{-1}](/img/site_content/ask-answer/loader.gif)
![=frac{1}{256}](/img/site_content/ask-answer/loader.gif)
![(v)left(left(frac{2}{3} right )^2 right )^3timesleft(frac{1}{3} right )^{-4}times3^{-1}times6^{-1} =left (frac{2^2}{3^2} right )^3timesfrac{1}{(1/3)^4}timesfrac{1}{3}timesfrac{1}{6}](/img/site_content/ask-answer/loader.gif)
![=left (frac{4}{9} right )^3timesfrac{1}{(1/81)}timesfrac{1}{3}timesfrac{1}{6}](/img/site_content/ask-answer/loader.gif)
![=frac{4^3}{9^3}times81timesfrac{1}{18}](/img/site_content/ask-answer/loader.gif)
![=frac{64}{729}times81timesfrac{1}{18}](/img/site_content/ask-answer/loader.gif)
![=frac{64}{9}timesfrac{1}{18}](/img/site_content/ask-answer/loader.gif)
![=64timesfrac{1}{162}](/img/site_content/ask-answer/loader.gif)
![=frac{64}{162}](/img/site_content/ask-answer/loader.gif)
![=frac{32}{81}](/img/site_content/ask-answer/loader.gif)
Page No 2.19:
Question 8:
By what number should 5−1 be multiplied so that the product may be equal to (−7)−1?
Answer:
Expressing in fraction form, we get:
5−1 = 1/5 (using the property a−1 = 1/a)
and
(−7)−1 = −1/7 (using the property a−1 = 1/a).
We have to find a number x such that
Multiplying both sides by 5, we get:
Hence, 5−1 should be multiplied by −5/7 to obtain (−7)−1.
Page No 2.19:
Question 9:
By what number should be multiplied so that the product may be equal to
Answer:
Expressing in fractional form, we get:
(1/2)−1 = 2, ---> (a−1 = 1/a)
and
(−4/7)−1 = −7/4 ---> (a−1 = 1/a)
We have to find a number x such that
Dividing both sides by 2, we get:
Hence, (1/2)−1should be multiplied by −7/8 to obtain (−4/7)−1.
Page No 2.19:
Question 10:
By what number should (−15)−1 be divided so that the quotient may be equal to (−5)−1?
Answer:
Expressing in fractional form, we get:
(−15)−1 = −1/15, ---> (a−1 = 1/a)
and
(−5)−1 = −1/5 ---> (a−1 = 1/a)
We have to find a number x such that
Solving this equation, we get:
Hence, (−15)−1should be divided by 1/3 to obtain (−5)−1.
Page No 2.19:
Question 11:
By what number should be multiplied so that the product may be
Answer:
Expressing as a positive exponent, we have: ---> (a−1 = 1/a)
---> ((a/b)n = (an)/(bn))
and
(7/3) −1 = 3/7. ---> (a−1 = 1/a)
We have to find a number x such that
Multiplying both sides by 25/9, we get:
Hence, (5/3)−2 should be multiplied by 25/21 to obtain (7/3)−1.
Page No 2.19:
Question 12:
Find x, if
(i)
(ii)
(iii)
(iv)
(v)
(vi)
Answer:
(i) We have:
x = 3
(ii) We have:
x = 6
(iii) We have:
x = 1/2
(iv) We have:
x = 10/3
(v) We have:
x = −1
(vi) We have:
x = −4
Page No 2.19:
Question 13:
(i) If , find the value of x−2.
(ii) If , find the value of x−1.
Answer:
(i) First, we have to find x.
--->(a−1 = 1/a)
Hence, x−2 is:
--->(a−1 = 1/a)
(ii) First, we have to find x. ---> ((a/b)n = (an)/(bn))
---> (a0 = 1)
Hence, the value of x−1 is:
--->(a−1 = 1/a)
--->(a−1 = 1/a)
Page No 2.19:
Question 14:
Find the value of x for which 52x ÷ 5−3 = 55.
Answer:
We have:
--->
Hence, x is 1.
Page No 2.22:
Question 1:
Express the following numbers in standard form:
(i) 6020000000000000
(ii) 0.00000000000943
(iii) 0.00000000085
(iv) 846 × 107
(v) 3759 × 10−4
(vi) 0.00072984
(vii) 0.000437 × 104
(viii) 4 ÷ 100000
Answer:
To express a number in the standard form, move the decimal point such that there is only one digit to the left of the decimal point.
(i) 6020000000000000 = 6.02 x 1015 (The decimal point is moved 15 places to the left.)
(ii) 0.0000000000943 = 9.43 x 10−12 (The decimal point is moved 12 places to the right.)
(iii) 0.00000000085 = 8.5 x 10−10 (The decimal point is moved 10 places to the right.)
(iv) 846 x 107 = 8.46 x 102 x 107 = 8.46 x 109 (The decimal point is moved two places to the left.)
(v) 3759 x 10−4 = 3.759 x 103 x 10−4 = 3.759 x 10−1 (The decimal point is moved three places to the left.)
(vi) 0.00072984 = 7.984 x 10−4 (The decimal point is moved four places to the right.)
(vii) 0.000437 x 104 = 4.37 x 10−4 x 104 = 4.37 x 100 = 4.37 (The decimal point is moved four places to the right.)
(viii) 4/100000 = 4 x 100000−1 = 4 x 10−5 (Just count the number of zeros in 1,00,000 to determine the exponent of 10.)
Page No 2.22:
Question 2:
Write the following numbers in the usual form:
(i) 4.83 × 107
(ii) 3.02 × 10−6
(iii) 4.5 × 104
(iv) 3 × 10−8
(v) 1.0001 × 109
(vi) 5.8 × 102
(vii) 3.61492 × 106
(viii) 3.25 × 10−7
Answer:
(i) 4.83 x 107 = 4.83 x 1,00,00,000 = 4,83,00,000
(ii) 3.02 x 10−6 = 3.02/106 = 3.02/10,00,000 = 0.00000302
(iii) 4.5 x 104 = 4.5 x 10,000 = 45,000
(iv) 3 x 10−8 = 3/108 = 3/10,00,00,000 = 0.00000003
(v) 1.0001 x 109 = 1.0001 x 1,00,00,00,000 = 1,00,01,00,000
(vi) 5.8 x 102 = 5.8 x 100 = 580
(vii) 3.61492 x 106 = 3.61492 x 10,00,000 = 3614920
(viii) 3.25 x 10−7 = 3.25/107 = 3.25/1,00,00,000 = 0.000000325
Page No 2.22:
Question 1:
Square of is
(a)
(b)
(c)
(d)
Answer:
(d) 4/9
To square a number is to raise it to the power of 2. Hence, the square of (−2/3) is
---> ( (a/b)n = (an)/(bn) )
Page No 2.22:
Question 2:
Cube of is
(a)
(b)
(c)
(d)
Answer:
(c) -1/8
The cube of a number is the number raised to the power of 3. Hence the cube of −1/2 is
---> ( (a/b)n = (an)/(bn)
Page No 2.23:
Question 3:
Which of the following is not equal to
(a)
(b)
(c)
(d)
Answer:
(c) −(34/54)
.
Page No 2.23:
Question 4:
Which of the following is not reciprocal of
(a)
(b)
(c)
(d)
Answer:
(c) (3/2)−4
The reciprocal of is
.
Therefore, option (a) is the correct answer.
Option (b) is just re-expressing the number with a negative exponent.
Option (d) is obtained by working out the exponent.
Hence,option (c) is not the reciprocal of .
Page No 2.23:
Question 5:
Which of the following numbers is not equal to
(a)
(b)
(c)
(d)
Answer:
(a) (2/3)-3
We can write as
. It can be written in the forms given below.
---> work out the minuses
Hence, option (b) is equal to .
We can also write:
Hence, option (c) is also equal to .
We can also write:
Hence, option (d) is also equal to .
This leaves out option (a) as the one not equal to .
Page No 2.23:
Question 6:
is equal to
(a)
(b)
(c)
(d)
Answer:
(b)
Rearrange (2/3)−5 to get a positive exponent.
Page No 2.23:
Question 7:
is equal to
(a)
(b)
(c)
(d)
Answer:
(a) (−1/2)8
We have:
Page No 2.23:
Question 8:
is equal to
(a)
(b)
(c)
(d)
Answer:
(c) (−5)5
We have:
Page No 2.23:
Question 9:
is equal to
(a)
(b)
(c)
(d)
Answer:
(a) 4/25
We have:
Page No 2.23:
Question 10:
is equal to
(a)
(b)
(c)
(d)
Answer:
(b) (1/3)8
We have: ---> ( (am)n = amxn)
Page No 2.24:
Question 11:
is equal to
(a) 0
(b)
(c) 1
(d) 5
Answer:
(c) 1
We have: ---> (a0 = 1, for every non-zero rational number a.)
Page No 2.24:
Question 12:
is equal to
(a)
(b)
(c)
(d) none of these
Answer:
![(b)-frac{2}{3}](/img/site_content/ask-answer/loader.gif)
We have:
![left (frac{-3}{2} right )^{-1}=frac{1}{(-3)/2}](/img/site_content/ask-answer/loader.gif)
![=frac{2}{-3}](/img/site_content/ask-answer/loader.gif)
Page No 2.24:
Question 13:
is equal to
(a)
(b)
(c)
(d)
Answer:
![(b)left(frac{2}{3}times frac{5}{7} right ) ^{-5}](/img/site_content/ask-answer/loader.gif)
We have:
![left(frac{2}{3} right ) ^{-5}times left(frac{5}{7} right ) ^{-5} =left(frac{2}{3}times frac{5}{7} right ) ^{-5}](/img/site_content/ask-answer/loader.gif)
Page No 2.24:
Question 14:
is equal to
Answer:
(a)
We have: --->
Page No 2.24:
Question 15:
For any two non-zero rational numbers a and b, a4 ÷ b4 is equal to
(a) (a ÷ b)1
(b) (a ÷ b)0
(c) (a ÷ b)4
(d) (a ÷ b)8
Answer:
![(c)left(adiv b right )^4](/img/site_content/ask-answer/loader.gif)
This is one of the basic exponential formulae, i.e.
![left(adiv b right )^n = a^n div b^n](/img/site_content/ask-answer/loader.gif)
Page No 2.24:
Question 16:
For any two rational numbers a and b, a5 × b5 is equal to
(a) (a × b)0
(b) (a × b)10
(c) (a × b)5
(d) (a × b)25
Answer:
(c) (a x b)5
an x bn = (a x b)n
Hence,
a5 x b5 = (a x b)5
Page No 2.24:
Question 17:
For a non-zero rational number a, a7 ÷ a12 is equal to
(a) a5
(b) a−19
(c) a−5
(d) a19
Answer:
(c) a−5
Hence,
Page No 2.24:
Question 18:
For a non zero rational number a, (a3)−2 is equal to
(a) a9
(b) a−6
(c) a−9
(d) a1
Answer:
(b) a−6
We have: ---> ((am)n = am x n)
Page No 2.8:
Question 1:
Express each of the following as a rational number of the form where p and q are integers and q ≠ 0.
(i) 2−3
(ii) (−4)−2
(iii)
(iv)
(v)
Answer:
We know that . Therefore,
(i)
(ii)
(iii)
(iv)
(v)
Page No 2.8:
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)
Answer:
(i) We know from the property of powers that for every natural number a, a−1 = 1/a. Then: ---> (a−1 = 1/a)
(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:
(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:
---> (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: ---> (a−1 = 1/a)
= ---> (a−1 = 1/a)
Page No 2.8:
Question 3:
Find the value of each of the following:
(i)
(ii)
(iii) (2−1 × 4−1) ÷ 2−2
(iv) (5−1 × 2−1) ÷ 6−1
Answer:
(i)
--> (a−1 = 1/a)
=2+3+4
=12
(ii)
= --> ((a/b)n = (an/bn))
= 4+9+16
=29
(iii)
--> (a−n = 1/(an))
=
= 2
(iv)
--> (a−n = 1/(an))
Page No 2.8:
Question 4:
Simplify:
(i)
(ii)
(iii)
(iv)
Answer:
(i) ---> (a−1 = 1/a)
---> ((a/b)n = (an)/(bn) )
(ii) ---> (a−1 = 1/a)
=
= ---> ((a/b)n = (an)/(bn) )
=
(iii)
---> (a−1 = 1/a)
=
---> (a−1 = 1/a)
(iv) ---> (a−1 = 1/a)
---> (a−1 = 1/a)
Page No 2.8:
Question 5:
Simplify:
(i)
(ii)
(iii)
(iv)
Answer:
(i)
(ii) ---> (a−1=1/(an))
---> ((a/b)n = (an)/(bn))
(iii) --->(a-n = 1/(an))
=
(iv)
---> ((a/b)n = (an)/(bn))
Page No 2.8:
Question 6:
By what number should 5−1 be multiplied so that the product may be equal to (−7)−1?
Answer:
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
Multiplying both sides by 5, we get:
Hence, the required number is −5/7.
Page No 2.8:
Question 7:
By what number should be multiplied so that the product may be equal to
Answer:
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
Dividing both sides by 2, we get:
Hence, the required number is −7/8.
Page No 2.8:
Question 8:
By what number should (−15)−1 be divided so that the quotient may be equal to (−5)−1?
Answer:
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
Hence, (−15)−1 should be divided by to obtain (−5)−1.
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