A2ZLY

Chapter 2:Β Power Play

by

This chapter focuses on powers, exponents, and their applications. Using these Class 8 Maths Ganita Prakash NCERT Solutions, students can quickly grasp the rules of exponents and solve problems accurately.

Textbook Page 22 – 23

Figure it Out

1. Express the following in exponential form:
(i) 6 Γ— 6 Γ— 6 Γ— 6
(ii) y Γ— y
(iii) b Γ— b Γ— b Γ— b
(iv) 5 Γ— 5 Γ— 7 Γ— 7 Γ— 7
(v) 2 Γ— 2 Γ— a Γ— a
(vi) a Γ— a Γ— a Γ— c Γ— c Γ— c Γ— c Γ— d
Solution:
(i) 6 Γ— 6 Γ— 6 Γ— 6 = 64             (ii) y Γ— y = y2  (iii) b Γ— b Γ— b Γ— b = b4             (iv) 5 Γ— 5 Γ— 7 Γ— 7 Γ— 7 = 52 Γ— 73 (v) 2 Γ— 2 Γ— a Γ— a = 22 Γ— a2                (vi) a Γ— a Γ— a Γ— c Γ— c Γ— c Γ— c Γ— d = a3 Γ— c4 Γ— d

2. Express each of the following as a product of powers of their prime factors in exponential form.
(i) 648 (ii) 405 (iii) 540 (iv) 3600
Solution:
(i) 648 = 2 Γ— 2 Γ— 2 Γ— 3 Γ— 3 Γ— 3 Γ— 3 = 23 Γ— 34

(ii) 405 = 3 Γ— 3 Γ— 3 Γ— 3 Γ— 5 = 34 Γ— 5

(iii) 540 = 2 Γ— 2 Γ— 3 Γ— 3 Γ— 3 Γ— 5 = 22 Γ— 33 Γ— 5

(iv) 3600 = 2 Γ— 2 Γ— 2 Γ— 2 Γ— 3 Γ— 3 Γ— 5 Γ— 5 = 24 Γ— 32 Γ— 52

3. Write the numerical value of each of the following:
(i) 2 Γ— 103
(ii) 72 Γ— 23
(iii) 3 Γ— 44
(iv) (-3)2 Γ— (-5)2
(v) 32 Γ— 104
(vi) (-2)5 Γ— (-10)6
Solution:
(i) 2 Γ— 103
= 2 Γ— (10 Γ— 10 Γ— 10) = 2 Γ— 1000 = 2000

(ii) 72 Γ— 23
= (7 Γ— 7) Γ— (2 Γ— 2 Γ— 2)  = 49 Γ— 8  = 392

(iii) 3 Γ— 44
= 3 Γ— (4 Γ— 4 Γ— 4 Γ— 4)  = 3 Γ— (16 Γ— 16)  = 3 Γ— 256  = 768

(iv) 32 Γ— 104
= {(-3) Γ— (-3)} Γ— {(-5) Γ— (-5)}   = 9 Γ— 25   = 225

(v) 32 Γ— 104
= (3 Γ— 3) Γ— (10 Γ— 10 Γ— 10 Γ— 10)  = 9 Γ— 10000  = 90000

(vi) (-2)5 Γ— (-10)6
= {(-2) Γ— (-2) Γ— (-2) Γ— (-2) Γ— (-2)} Γ— {(-10) Γ— (-10) Γ— (-10) Γ— (-10) Γ— (-10) Γ— (-10)}
= {4 Γ— 4 Γ— (-2)} Γ— {100 Γ— 100 Γ— 100}   = (-32) Γ— (1000000)  = -32000000

Textbook Page 44 – 45

Figure it Out

1. Find out the units digit in the value of 2224 Γ· 432 ?
Solution:
2224 Γ· 432
= 2224 Γ· (22)32  = 2224 Γ· 22 Γ— 32    = 2224 Γ· 264  = 2224 – 64 = 2160 

21 = 2 (units digit 2)
22 = 4 (units digit 4)
23 = 8 (units digit 8)
24 = 16 (units digit 6)
25 = 32 (units digit 2)
26 = 64 (units digit 4)
27 = 128 (units digit 8)

The units digit repeats in a cycle of 4: 2, 4, 8, 6.

Since 2160Γ·4=540 (exact multiple of 4), the units digit of 22160 will be the same as that of 24, which is 6.

2. There are 5 bottles in a container. Every day, a new container is brought in. How many bottles would be there after 40 days?
Solution:

Containers added per day = 1
Containers after 40 days = 40
Bottles in each container = 5
Total bottles after 40 days = 40Γ—5=200

∴ After 40 days, there will be 200 bottles.

3. Write the given number as the product of two or more powers in three different ways. The powers can be any integers.
(i) 643
(ii) 1928
(iii) 32-5
Solution: (i) 643
64 = 2 Γ— 2 Γ— 2 Γ— 2 Γ— 2 Γ— 2 = 26
643 = (26)3 = 218
Three different ways are-
1. 643 = 218 = 29 Γ— 29
2. 643 = 218 = 210 Γ— 28
3. 643 = 218 = 26 Γ— 26 Γ— 26

(ii) 1928
192 = 2 Γ— 2 Γ— 2 Γ— 2 Γ— 2 Γ— 2 Γ— 3 = 26 Γ— 3
1928 = (26 Γ— 3)8 = 26 Γ— 8 Γ— 38 = 248 Γ— 38
Three different ways are-

1. 1928 = 248 Γ— 38 = 240 Γ— 28 Γ— 38 = 240 Γ— (2Γ—3)8 = 240 Γ— 68
2. 1928 = 248 Γ— 38 = (224Γ—34)Γ—(224Γ—34)
3. 1928 = 248 Γ— 38 = (26)8 Γ— 38

(iii) 32-5
32 = 2 Γ— 2 Γ— 2 Γ— 2 Γ— 2 = 25
32-5 = (25)-5 = 2-25
Three different ways are-
1. 32-5 = 2-25 = 2-15 Γ— 2-10
2. 32-5 = 2-25 = 2βˆ’5 Γ— 2βˆ’5 Γ— 2βˆ’5 Γ— 2βˆ’5 Γ— 2βˆ’5
3. 32-5 = 2-25 = 2-5 Γ— 2-20

4. Examine each statement below and find out if it is β€˜Always True’, β€˜Only Sometimes True’, or β€˜Never True’. Explain your reasoning.

(i)  Cube numbers are also square numbers.

(ii)  Fourth powers are also square numbers.

(iii)  The fifth power of a number is divisible by the cube of that number.

(iv)  The product of two cube numbers is a cube number.

(v)  q46 is both a 4th power and a 6th power (q is a prime number).

Solution:

(i) Cube numbers are also square numbers.

Answer: Only sometimes true.
Reason: A number n3 is a square exactly when n itself is a square, i.e. n=k2. Then n3=(k2)3=k6, a sixth power. So only those cubes that are sixth powers are also squares.
Example: 64=43=82 (true). 27=33 is not a square (false).

(ii) Fourth powers are also square numbers.
Answer: Always true.
Reason: n4=(n2)2, so every fourth power is a perfect square.
Example: 16=24=(22)2=42.

(iii) The fifth power of a number is divisible by the cube of that number.
Answer: Always true (for every nonzero integer).
Reason: n5=n3β‹…n2, so n5 is divisible by n3. (Note: division by 0 is undefined, so the usual interpretation excludes n=0.)
Example: 55=53β‹…52, so 55 is divisible by 53.

(iv) The product of two cube numbers is a cube number.
Answer: Always true.
example- 8 = 23, 27 = 33; 8 Γ— 27 = 216, which is 63.

(v) q46 is both a 4th power and a 6th power ( q is prime).
Answer: Never true.
Reason: For q46 to be a 4th power, 4 must divide 46; for it to be a 6th power, 6 must divide 46. Neither divides 46. More generally, a number is both a 4th and a 6th power only if its exponent is a multiple of lcm(4,6)=12; 46 is not a multiple of 12. (Also q being prime rules out trivial exceptions like 1.)

5. Simplify and write these in the exponential form.
(i) 10–2 Γ— 10–5
(ii) 57 Γ· 54
(iii) 9–7 Γ· 94
(iv) (13–2)–3
(v) m5n12(mn)9
Solution:
(i) 10–2 Γ— 10–5 = 10-2 – 5 = 10-7  (ii) 57 Γ· 54 = 57 – 4 = 53  (iii) 9–7 Γ· 94 = 9-7 – 4 = 9–11  (iv) (13–2)–3 = 13-2 Γ— -3 = 136  (v) m5n12(mn)9 = m5n12 Γ— m9n9 = m5+9n12+9 = m14n21

6. If 122 = 144 what is
(i) (1.2)2
(ii) (0.12)2
(iii) (0.012)2
(iv) 1202
Solution:

7. Circle the numbers that are the same β€”
24 Γ— 36, 64 Γ— 32, 610, 182 Γ— 62, 624

Solution:

(i) 24 Γ— 36

(ii) 64 Γ— 32 = (2 Γ— 3)4 Γ— 32 = 24 Γ— 34 Γ— 32 = 24 Γ— 36

(iii) 610 = (2 Γ— 3)10 = 210 Γ— 310

(iv) 182 Γ— 62 = (2 Γ— 3 Γ— 3)2 Γ— (2 Γ— 3)2 = 22 Γ— 32 Γ— 32 Γ— 22 Γ— 32 = 24 Γ— 36

(v) 624 = (2 Γ— 3)24 = 224 Γ— 324

Same numbers are: 24 Γ— 36 ,64 Γ— 32 , 182 Γ— 62

8. Identify the greater number in each of the following β€”
(i) 43 or 34 (ii) 28 or 82 (iii) 1002 or 2100
Solution: Let’s compare carefully step by step:

(i) 43 or 34
43 = 64 , 34 = 81 ; So, 34 is greater.

(ii) 28 or 82
28 = 256, 82 = 64; So, 28 is greater.

(iii) 1002 or 2100
1002 = 10,000; 2100 = (210)10 = 102410, which is far greater than 10,000.
So, 2100 is greater than 1002.

9. A dairy plans to produce 8.5 billion packets of milk in a year. They want a unique ID (identifier) code for each packet. If they choose to use the digits 0–9, how many digits should the code consist of?
Solution:

8.5 billion = 8,500,000,000

With digits 0–9 there are 10n possible codes of length n.
We need at least 8.5Γ—10^9 codes.
10^9=1 000 000 000 (too few) and 1010=10 000 000 000 (enough).
So the minimum length is n=10.

10. 64 is a square number (82) and a cube number (43). Are there other numbers that are both squares and cubes? Is there a way to describe such numbers in general?
Solution: Yes. Any number that is both a perfect square and a perfect cube must be a sixth power.

Reason:

  • Squares require every prime’s exponent to be a multiple of 2.
  • Cubes require every prime’s exponent to be a multiple of 3.
  • For both, exponents must be multiples of lcm(2,3) = 6.

So such numbers are exactly n^6.

General Rule: The sixth power of any number (i.e., n6) is both a square and a cube. i.e.
16 = 1, 26 = 64, 36 = 729, 46 = 4096, 56 = 15,625

11. A digital locker has an alphanumeric (it can have both digits and letters) passcode of length 5. Some example codes are G89P0, 38098, BRJKW, and 003AZ. How many such codes are possible?

Solution:

Passcode length = 5

Choices for each position = 26 (letters) + 10 (digits) = 36
Total number of possible passcodes =

36Γ—36Γ—36Γ—36Γ—36=365

So, the total possible codes = 36⁡.

12. The worldwide population of sheep (2024) is about 109, and that of goats is also about the same. What is the total population of sheep and goats?
(i) 209 (ii) 1011 (iii) 1010
(iv) 1018 (v) 2 Γ— 109 (vi) 109 + 109
Solution:

Population of sheep = 10⁹

Population of goats = 10⁹

Total population = 10⁹ + 10⁹ = 2 Γ— 10⁹

Hence, both (v) 2 Γ— 10⁹ and (vi) 10⁹ + 10⁹ represent the correct total population.

13. Calculate and write the answer in scientific notation:
(i)  If each person in the world had 30 pieces of clothing, find the total number of pieces of clothing.
(ii)  There are about 100 million bee colonies in the world. Find the number of honeybees if each colony has about 50,000 bees.
(iii)  The human body has about 38 trillion bacterial cells. Find the bacterial population residing in all humans in the world.
(iv) Total time spent eating in a lifetime in seconds.
Solution:

(i) Clothing pieces

World population β‰ˆ 8Γ—109 people
Each person has 30 pieces

Total pieces=30Γ—8Γ—109=240Γ—109=2.4Γ—1011

Answer: 2.4Γ—1011 pieces of clothing

(ii) Honeybees
Number of colonies β‰ˆ 1Γ—108
Bees per colony = 50,000 = 5Γ—104

Total bees=(1Γ—108)(5Γ—104)=5Γ—1012

Answer: 5Γ—1012 honeybees

(iii) Bacteria in all humans
Bacteria per person β‰ˆ 3.8Γ—1013
World population β‰ˆ 8Γ—109

Total bacteria=(3.8Γ—1013)(8Γ—109)=30.4Γ—1022=3.04Γ—1023

Answer: 3.04Γ—1023 bacterial cells

(iv) Lifetime eating time in seconds
Average eating time β‰ˆ 1.5 hours/day = 1.5 Γ— 3600 = 5400 seconds/day
Average lifespan β‰ˆ 80 years = 80Γ—365Γ—24Γ—3600β‰ˆ2.52Γ—109 seconds

Eating fraction = 540086400=116 of the day

Eating lifetime seconds=116Γ—2.52Γ—109β‰ˆ1.575Γ—108

Answer: 1.58Γ—108 seconds

14. What was the date 1 arab/1 billion seconds ago?
Solution:

Step 1: Convert 1 billion seconds into years

1 billion seconds = 1,000,000,000 seconds.

  • Seconds in a minute: 60
  • Seconds in an hour: 60Γ—60=3600
  • Seconds in a day: 3600Γ—24=86400
  • Seconds in a year (approx): 86400Γ—365.25β‰ˆ31,557,600

Now divide:

1,000,000,00031,557,600β‰ˆ31.69 years

So 1 billion seconds β‰ˆ 31 years and 8 months.

Step 2: Subtract from today

Today is 24 August 2025.

  • Subtract 31 years β†’ 24 August 1994
  • Subtract 8 months β†’ 24 December 1993

Answer: Approximately 24 December 1993.

error: Content is protected !!