Power Play Class 8 Extra Questions Maths Chapter 2 - #NCSOLVE 📚

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Class 8 Power Play Extra Questions

Class 8 Maths Chapter 2 Power Play Extra Questions

Class 8 Maths Chapter 2 Extra Questions – Power Play Extra Questions Class 8

Question 1.
Simplify: \(\left(4^0+2^0\right)^0 \div\left(\frac{1}{2}\right)^{-2}\)
Solution:

Question 2.
Simplify: \(\left(4^2 \times 2^{-3}\right) \div\left(\frac{1}{10}\right)^{-2}\)
Solution:

Question 3.
Write the following in expanded form using exponents of 10.
(a) 425.2394
(b) 76.359898
Solution:

Question 4.
Express the numbers used in the following facts in scientific notation.
(a) The speed of light is 300000000 m/s.
(b) The distance from the Earth to the Moon is 384400000 m.
Solution:
(a) The speed of light = 300000000 m/s = 3 × 10⁸ m/s.
(b) The distance from the Earth to the Moon = 384400000 m = 3.844 × 10⁸ m.

Question 5.
If the mass of the Earth is 5.97 × 10²⁴ kg and the mass of the Moon is 7.35 × 10²² kg, which is heavier and by how much?
Solution:
Here, we have to compare the mass of the Earth and the Moon.
We will convert them into numbers with the same exponents 22, as 22 < 24.
Mass of Earth = 5.97 × 10²⁴ kg
= 5.97 × 100 × 10²²
= 597 × 10²² kg
Mass of Moon = 7.35 × 10²² kg
Since 597 × 10²² kg > 7.35 × 10²² kg.
So, Earth is heavier than the Moon.
Further, the difference of masses = (597 × 10²² kg) – (7.35 × 10²² kg)
= (597 – 7.35) × 10²² kg
= 589.65 × 10²² kg
= 5.8965 × 10²⁴ kg
Hence, the Earth is approximately 5.8965 × 10^24 kg heavier than the Moon.

Question 6.
Express the height of a bundle of 500 papers placed on each other if the thickness of one paper is 0.0016 cm, in standard form.
Solution:
Number of papers = 500
Thickness of one paper = 0.0016 cm
Total height = Number of papers × Thickness of one paper
= 500 × 0.0016 cm
= 0.8 cm
= 8 × 10^-1 cm
Therefore, the height of the bundle of 500 papers is 8 × 10^-1 cm.

Question 7.
Express the following in exponential form,
(i) 5 × 5 × 5 × 5
Solution:
We have, 5 × 5 × 5 × 5
Here, 5 is multiplied 4 times.
So, 5 × 5 × 5 × 5 = 5⁴

(ii) x × x × x
Solution:
We have, x × x × x
Here, x is multiplied 3 times.
So, x × x × x = x³

(iii) 3 × 3 × 5 × 5 × 5
Solution:
We have, 3 × 3 × 5 × 5 × 5
Here, 3 is multiplied 2 times and 5 is multiplied 3 times.
So, 3 × 3 × 5 × 5 × 5 = 3² × 5³

(iv) x × x × y × y × z × z × z
Solution:
We have, x × x × y × y × z × z × z
Here, x is multiplied 2 times, y is multiplied 2 times and z is multiplied 3 times.
So, x × x × y × y × z × z × z = x²y²z³

Question 8.
Find the numerical value of the following.
(i) 7⁴ × 3²
Solution:
We have, 7⁴ × 3² = 7 × 7 × 7 × 7 × 3 × 3
= 2401 × 9
= 21609

(ii) (-3)³ × (-4)²
Solution:
We have, (-3)³ × (-4)²
= (-3) × (-3) × (-3) × (-4) × (-4)
= (-27) × 16
= -432

(iii) 2² × 10⁴
Solution:
We have, 2² × 10⁴ =2 × 2 × 10 × 10 × 10 × 10
= 4 × 10000
= 40000

(iv) (-2)⁴ × (4)²
Solution:
We have, (-2)⁴ × (4)²
= (-2) × (-2) × (-2) × (-2) × 4 × 4
= 16 × 16
= 256

Question 9.
Write (81)^-2 as a power with the base 3.
Solution:
We have, 81 = 3 × 3 × 3 × 3 = 3⁴
(81)^-2 = (3⁴)^-2
= 3^-8 [∵ (a^m)ⁿ =a^mn]

Question 10.
Simplify and write in exponential form.
(i) (-8)^-5 × (-8)^-2
Solution:
We have. (-8)^-5 × (-8)^-2
= [-8]^{(-5) + (-2)} [∵ a^m × aⁿ = a^m+n]
= (-8)“s-2 =(-8)-7,
which is the required exponential form.

(ii)(-7)⁵ ÷ (-7)^-10
Solution:
We have, (-7)⁵ ÷ (-7)^-10
= (-7)^{5 – (-10)} [∵ a^m ÷ aⁿ = a^m-n]
= (-7)⁵⁺¹⁰
=(-7)¹⁵
which is the required exponential form.

(iii) a^-5 × a^-2 × a⁹
Solution:
We have, a^-5 × a^-2 × a⁹

which is the required exponential form.

Question 11.
Find the unit digit in the value of 3100.
Solution:
We have, 3¹⁰⁰
We know that 3¹ = 3
3² = 9
3³ = 27 [unit digit is 7]
3⁴ = 81 [unit digit is 1]
3⁵ = 243 [unit digit is 3]

Here, the unit digits repeat is a cycle of 4. (3 9 7 1).
On dividing the exponent 100 by the length of cycle 4, we get
100 ÷ 4 = 25, remainder = 0
A remainder of 0 means the unit digit is the last digit in the cycle, which is 1.
So, the unit digit in the value of 3¹⁰⁰ is 1.

Question 12.
Write the exponential form of 2500.
Solution:
The prime factors of 2500
= 2 × 2 × 5 × 5 × 5 × 5
So, the exponential form = 2² × 5⁴

Question 13.
Expand the number 3462 in power of 10.
Solution:
We have, 3462 = 3 × 1000 + 4 × 100 + 6 × 10 + 2
= 3 × 10³ + 4 × 10² + 6 × 10¹ + 2 × 10⁰ [∵ a° = 1]

Question 14.
Express 0.00000532 in scientific notation.
Solution:
To express 0.00000532 in scientific notation.
Firstly, count the number of decimal places you need to move.
Here, we need to move the decimal point by 6 places.

Decimal is moved 6 places to the right.
We would write it as 5.32 i.e. 1 < 5.32 < 10
Now, determine the power of 10 required.
Decimal is moved to 6 places right which means that we need to multiply by 10 “6.
So, in scientific notation, 0.00000532 can be expressed as 5.32 × 10^-6.

Question 15.
Express 5329800000 in standard form.
Solution:
To express 5329800000 in standard form.
First, count the number of decimal places you need to move.

Write it as 1 < 53298 < 10.
Secondly, decimal is moved to 9 places left which means that we need to multiply by 109.
So, the standard form will be 5.3298 × 10⁹.

Question 16.
Write the following numbers in standard form.
(i) 0.0000000635
Solution:
We have, 0.0000000635 = \(\frac{635}{10000000000}=\frac{635}{10^{10}}\)
= \(\frac{6.35 \times 100}{10^{10}}=\frac{6.35 \times 10^2}{10^{10}}\)
= 6.35 × 10² × 10^-10 [∵ a^-m = \(\frac{1}{a^m}\)]
= 6.35 × 10^2-10
= 6.35 × 10^-8

(ii) 31820000
Solution:
We have, 31820000 = 3182 × 10⁴
= \(\frac{3182 \times 10^4 \times 10^3}{10^3}\)
= \(\frac{3182 \times 10^{4+3}}{1000}\)
= 3.182 × 10⁷

Question 17.
Express the following numbers in usual form,
(i) 6.671 × 10⁵
Solution:
We have, 6.671 × 10⁵ = 6.67 × 100000
= 667100

(ii) 5.332 × 10^-6
Solution:
We have, 5.332 × 10^-6 = \(\frac{5.332}{10^6}\)
= \(\frac{5.332}{10^6}\) [∵ a^-m = \(\frac{1}{a^m}\)]
= 0.000005332

(iii) 5 × 10^-4
Solution:
We have, 5 × 10^-4
= \(\frac{5}{10^4}=\frac{5}{10000}\)
= 0.0005 [∵ a^-m = \(\frac{1}{a^m}\)]

Question 18.
Find the standard form of 0.00006 + 0.00132.
Solution:
We have, 0.00006 + 0.00132 = 0.00138 = 1.38 × 10^-3

Question 19.
Express the product of 1.5 × 10³ and 2.5 × 10^-5 in the standard form.
Solution:
We have, 1.5 × 10³ and 2.5 × 10^-5
Product = 1.5 x 2.5 × 10³ × 10^-5
= 3.75 × 10^3-5
= 3.75 × 10^-2 [∵ a^m × aⁿ = a^m+n]

Question 20.
Express the number appearing in the following statements in standard form.
(i) The population of a city is 650000.
(ii) The distance between Earth and Sun is about 93000000 miles.
(iii) A petri dish contains 0.000045 g of bacteria.
(iv) The mass of a hydrogen atom is approximately 0.00000000000000000000000167 g.
Solution:
(i) The population of a city is 650000.
∴ Standard form = 6.5 × 10⁵
(ii) 93000000 miles = 9.3 × 10⁷ miles
(iii) 0.000045 g= 4.5 × 10^-5 g
(iv) 0.00000000000000000000000167 g= 1.67 × 10^-24 g

Question 21.
The diameter of Earth is 1.2756 × 10⁷ m and diameter of Sun is 1.4 × 10⁹ m. How many times is the diameter of the Sun to the diameter of the Earth?
Solution:
The diameter of Earth is 12756 × 10⁷ m and diameter of Sun is 14 × 10⁹ m.
Suppose we want to compare the diameter of Sun with the diameter of Earth.

which is approximately 100.
So, the diameter of th£Sa»4s 100 times to the diameter of the Earth.

Practice Questions

Question 1.
Convert the following to exponential form:
(a) 8 × 8 × 8 × 8
(b) z × z
(c) 4 × 4 × 6 × 6 × 6
(d) m × m × m × m × m × g × g × g
(e) 3 × 3 × p × p
Answer:
(a) 8⁴
(b) z²
(c) 4² × 6³
(d) m⁵ × g³
(e) 3² × p²

Question 2.
Write the following as a product of their prime factors using powers.
(a) 924
(b) 875
(c) 1080
(d) 242
(e) 735
Answer:
(a) 2² × 3 × 7 × 11
(b) 5³ × 7
(c) 2³ × 3³ × 5
(d) 2 × 11²
(e) 3 × 5 × 7²

Question 3.
Find the actual value of these expressions:
(a) 5 × 10²
(b) 6² × 2³
(c) 4 × 3⁴
(d) (-4)² × (-6)²
(e) (-3)³ × (-2)⁴
Answer:
(a) 500
(b) 288
(c) 324
(d) 576
(e) -432

Question 4.
What is the units digit of the result of this expression: 3²⁴⁵ ÷ 9⁸⁰?
Answer:
3

Question 5.
Ajar contains 6 marbles. Each day, one more jar with 6 marbles is added. How many marbles will there be after 30 days?
Answer:
180 marbles

Question 6.
Express the number as a product of powers in three different ways.
(You can use positive or negative exponents.)
(a) 81³
(b) 256⁴
(c) 16^-6
Answer:
(a) 3¹², 9⁶, 27³ × 3³
(b) 2³², 16⁸, 4¹⁶
(c) 2^-24, 4^-12, 8^-16 × 2^-6

Question 7.
For each statement below, say whether it is Always True, Only Sometimes True, or Never True. Give a reason for your answer:
(i) Every square number is also a cube number.
(ii) The square of any even number is divisible by 4.
(iii) The sixth power of a number is divisible by its square.
(iv) The product of a cube and a square number is always a cube.
(v) If m²⁴, then mmm must be a square and cube root of that number.
Answer:
(i) Never True
(ii) Always True
(iii) Always True
(iv) Only Sometimes True
(v) Always True

Question 8.
Evaluate the expressions using power laws:
(a) 10^-3 × 10^-6
(b) 6⁸ ÷ 6⁵
(c) 8^-4 ÷ 8^-2
(d) (15^-3)⁴
(e) a⁴ b³ × (ab²)⁵
Answer:
(a) 10^-9
(b) 6³
(c) 8^-2
(d) 15^-12
(e) a⁹ b¹³

Question 9.
Use known squares to find values. If 11² = 121, then find the value of:
(a) (1.1)²
(b) (0.11)²
(c) (0.011)²
(d) 110²
(e) 0.0011²
Answer:
(a) 1.21
(b) 0.0121
(c) 0.000121
(d) 12,100
(e) 0.00000121

The post Power Play Class 8 Extra Questions Maths Chapter 2 appeared first on Learn CBSE.

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