Important Questions: Power Play | Mathematics Class 8- New NCERT (Ganita Prakash) PDF Download

Q1: Evaluate
(i) 2^-2
(ii) (-2)^-2
(iii) (3/2)^-5

Sol:
As we know that
b^-n= 1/bⁿ
(i) 2^-2 = 1/ 2² =1/4
(ii) (-2)^-2 = 1/(-2)² = 1/4
(iii) (3/2)^-5 = 3^-5/ 2^-5 =2⁵/3⁵ = 32/243

Q2: Find the value of.
(i) (4⁰ + 4^-1) × 2²
(ii) (3^-1 × 9^-1) ÷ 3^-2
(iii)(11^-1 + 12^-1 + 13^-1)⁰

Sol:
(i) (4⁰ + 4^-1) × 2² = (1+1/4) × 4 = 4 + 1 =5
(ii) (3^-1 × 9^-1) ÷ 3^-2 = [(1/3) × (1/9)] ÷ (1/9) =1/3
(iii) (11^-1 + 12^-1 + 13^-1)⁰ =1 as a⁰= 1 

Q3: Find the value of m for which 2^m ÷ 2^-4 = 4⁵

Sol:
2^m ÷ 2^-4 = 4⁵
2^m × (1/2^-4) = 2¹⁰
2^m+4 =2¹⁰
So m + 4 = 10
m = 6 

Q4: Express the following numbers in usual form.
(i) 34.02 x 10^-5
(ii) 9.5 x 10⁵
(iii) 9 x 10^-4
(iv) 2.0001 x 10⁸

Sol:
(i) 34.02 x 10^-5 = 0.0003402
(ii) 9.5 x 10⁵ = 950000
(iii) 9 x 10^-4 = 0.0009
(iv) 2.0001 x 10⁸ =200010000

Q5: Find the Multiplicative inverse of
(i) 3²⁵
(ii) 4⁻³
(iii) (2/3)⁻²

Sol:
(i) The multiplicative inverse of a number is the number that, when multiplied by the original number, equals 1.

(ii) 4^-3

(iii) (2/3)²

Q6: True & False
(i) Very small numbers can be expressed in standard form using negative exponents.
(ii) a^p × b^q = (ab)^pq
(iii) .00567 = 5.67×10⁻³
(iv) 1/(8)⁻³ = 2⁹
(v) 4⁰ = 4
(vi) (−1)³ = 1
(vii) The multiplicative inverse of (–2)^–2 is (2)².

Ans: 
(i) True

(ii) False

(iii) True

(iv) True 

(v) False

(vi) False

(vii) True as (–2)^–2 = 1/4

Q7: Match the column

Ans:

therefore correct match is : 
(p) → (ii)
(q) → (iii)
(r) → (iv)
(s) → (I)

Q8: The multiplicative inverse of (–7)^–2 ÷ (90)^–1 is
(a) −(7)² × (90)^–1 
(b) −(7)⁻² × (90)^–1
(c) −(7)² ÷ (90)^–1
(d) −(7)² × (90)¹ 

Ans: (a)
Sol: (–7)^–2 ÷ (90)^–1= −(1/7)² × 90¹. 
So Multiplication inverse is −(7)² × (90)^–1. 
Hence correct Option is (a) 

Q9: If a be any non-zero integer, then a^–1 is equal to

(a) a
(b) -a
(c) 1/a
(d) -1/a 
Ans: (c)

Q10: If 5^3x–1 ÷ 25 = 125, Then the value x is
(a) 1
(b) 2
(c) 3
(d) 4 
Ans: (b)

Sol:

Express 25 and 125 as powers of 5

• $25=5\mathrm{<sup\; style="box-sizing:\; border-box;\; font-family:\; Lato,\; sans-serif\; !important;\; letter-spacing:\; inherit\; !important;\; position:\; relative;\; font-size:\; 13px\; !important;\; line-height:\; 1.8;\; vertical-align:\; baseline;\; top:\; -0.5em;\; word-break:\; break-word;\; color:\; black">2</sup>}$

• $125=$ 5\mathrm{<sup\; style="box-sizing:\; border-box;\; font-family:\; Lato,\; sans-serif\; !important;\; letter-spacing:\; inherit\; !important;\; position:\; relative;\; font-size:\; 13px\; !important;\; line-height:\; 1.8;\; vertical-align:\; baseline;\; top:\; -0.5em;\; word-break:\; break-word;\; color:\; black">3</sup>}$$

So the equation becomes:

Apply the laws of exponents

Now, equating powers (since bases are same):3x − 3 = 3

⇒ 3x = 6
⇒x=2

Hence x = 2
Option(b) is correct

Q11: Simplify and express the result in power notation with positive exponent.
(i) (-2)⁵ ÷ (-2)⁴
(ii) (1/2)² × (2/5)²
(iii) (-5)² × (3/5) 

Sol:
(i) (-2)⁵ ÷ (-2)⁴
= (-2)⁵ / (-2)⁴
= (-2)^5-4
=(-2)

(ii) (1/2)² × (2/5)²
= (1/4) X (4/25)
= 1/25

(iii) (-5)² × (3/5)
= 25 × (3/5) 
= 15 

Q12: Find the value of x here

Sol:

Q13: Express the following numbers in standard form.
(i) 0.0000000015
(ii) 0.00000001425
(iii) 102000000000000000

Sol:
(i) 0.0000000015 = 1.5 ×10^-9
(ii) 0.00000001425 = 1.425×10^-8
(iii) 102000000000000000 = 1.02×10¹⁷

Q14: Solve for the variables

Sol:

Q15: Simplify the following

Sol: