Class 8 Maths Ganita Prakash Chapter 2 Power Play NCERT Solutions
Textbook Page 20
The following table lists the thickness after each fold. Observe that the thickness doubles after each fold.
(We use the sign ‘≈’ to indicate ‘approximately equal to’.)
After 10 folds, the thickness is just above 1 cm (1.024 cm). After 17 folds, the thickness is about 131 cm (a little more than 4 feet).
Q. Now, what do you think the thickness would be after 30 folds? 45 folds? Make a guess.
Solution:
I think that the thickness after 30 folds would be 10 km, and after 45 folds it would be 20,000 km.
Q. Fill the table below.
Textbook Page 22
Q. Which expression describes the thickness of a sheet of paper after it is folded 10 times? The initial thickness is represented by the letter-number v.
(i) 10v (ii) 10 + v (iii) 2 × 10 × v
(iv) 210 (v) 210v (vi) 102v
Solution:
Initial thickness of sheet = v
Since the thickness of the sheet doubles after every fold.
Therefore, the thickness of a sheet of paper after 10 folds = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × v = 210v.
Therefore, (v) 210v is the correct answer.
Q. What is (-1)5? Is it positive or negative? What about (-1)56?
Solution:
(-1)odd = −1 → Negative
(-1)even = 1 → Positive
(i) (-1)5
Since 5 is an odd number, (-1)5 = -1, which is negative.
(ii) (-1)56
Since 56 is an even number, (-1)56 = 1, which is positive.
Q. Is (-2)4 = 16? Verify.
Solution:
(-2)4 = (-2) × (-2) × (-2) × (-2)
(-2)4 = (-2) × (-2) × (-2) × (-2)
(-2)4 = 4 × 4
(-2)4 = 16.
So yes, (-2)4 = 16 is correct.
Q. What is 02, 05? What is 0n?
Solution:
02 = 0 × 0 = 0.
05 = 0 × 0 × 0 × 0 × 0 = 0.
Similarly, 0n = 0.
Textbook Page 22 – 23
Figure it Out
1. Express the following in exponential form:
(i) 6 × 6 × 6 × 6
(ii) y × y
(iii) b × b × b × b
(iv) 5 × 5 × 7 × 7 × 7
(v) 2 × 2 × a × a
(vi) a × a × a × c × c × c × c × d
Solution:
(i) 6 × 6 × 6 × 6 = 64
(ii) y × y = y2
(iii) b × b × b × b = b4
(iv) 5 × 5 × 7 × 7 × 7 = 52 × 73.
(v) 2 × 2 × a × a = 22 × a2.
(vi) a × a × a × c × c × c × c × d = a3 × c4 × 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 × 2 × 2 × 3 × 3 × 3 × 3 = 23 × 34.
(ii) 405 = 3 × 3 × 3 × 3 × 5 = 34 × 5.
(iii) 540 = 2 × 2 × 3 × 3 × 3 × 5 = 22 × 33 × 5.
(iv) 3600 = 2 × 2 × 2 × 2 × 3 × 3 × 5 × 5 = 24 × 32 × 52.
3. Write the numerical value of each of the following:
(i) 2 × 103
(ii) 72 × 23
(iii) 3 × 44
(iv) (-3)2 × (-5)2
(v) 32 × 104
(vi) (-2)5 × (-10)6
Solution:
(i) 2 × 103
= 2 × (10 × 10 × 10)
= 2 × 1000
= 2000.
(ii) 72 × 23
= (7 × 7) × (2 × 2 × 2)
= 49 × 8
= 392.
(iii) 3 × 44
= 3 × (4 × 4 × 4 × 4)
= 3 × (16 × 16)
= 3 × 256
= 768.
(iv) 32 × 104
= {(-3) × (-3)} × {(-5) × (-5)}
= 9 × 25
= 225.
(v) 32 × 104
= (3 × 3) × (10 × 10 × 10 × 10)
= 9 × 10000
= 90000.
(vi) (-2)5 × (-10)6
= {(-2) × (-2) × (-2) × (-2) × (-2)} × {(-10) × (-10) × (-10) × (-10) × (-10) × (-10)}
= {4 × 4 × (-2)} × {100 × 100 × 100}
= (-32) × (1000000)
= -32000000.
Textbook Page 24
Q. 37 can also be written as 32 × 35. Can you reason out why?
Solution:
Yes, 37 can also be written as 32 × 35 because 37 = (3 × 3) × (3 × 3 × 3 × 3 × 3), which becomes 32 × 35.
Q. na × nb = n(a+b), where a and b are counting numbers.
Use this observation to compute the following.
(i) 29
(ii) 57
(iii) 46
Solution:
(i) 29 = (2 × 2 × 2) × (2 × 2 × 2) × (2 × 2 × 2) = 23 × 23 × 23 = 8 × 8 × 8 = 512.
(ii) 57 = (5 × 5) × (5 × 5) × (5 × 5) × 5 = 52 × 52 × 52 × 5 = 25 × 25 × 25 × 5 = 78,125.
(iii) 46 = (4 × 4) × (4 × 4) × (4 × 4) = 42 × 42 × 42 = 16 × 16 × 16 = 4096.
Q. Write the following expressions as a power of a power in at least two different ways:
(i) 86
(ii) 715
(iii) 914
(iv) 58
Solution:
(i) 86
Two possible ways:
86 = (83)2
86 = (82)3
(ii) 715
Two possible ways:
715 = (73)5
715 = (75)3
(iii) 914
Two possible ways:
914 = (92)7
914 = (97)2
(iv) 58
Two possible ways:
58 = (52)4
58 = (54)2
Textbook Page 25
In the middle of a beautiful, magical pond lies a bright pink lotus. The number of lotuses doubles every day in this pond. After 30 days, the pond is completely covered with lotuses. On which day was the pond half full?
If the pond is completely covered by lotuses on the 30th day, how much of it is covered by lotuses on the 29th day?
Since the number of lotuses doubles every day, the pond should be half covered on the 29th day.
Q. Write the number of lotuses (in exponential form) when the pond was —
(i) fully covered (ii) half covered
Solution
The number of lotuses doubles every day.
On the 30th day, the pond is fully covered.
So, on the 29th day, the pond must be half full.
Number of lotuses:
Day 1 → 1 = 20
Day 2 → 21
Day 3 → 22
Day 4 → 23
…….
Day 29 → 228
Day 30 → 229
(i) The number of lotuses when the pond was fully covered (Day 30) = 229.
(ii) The number of lotuses when the pond was fully covered (Day 29) = 228.
Q. ma × na = (mn)a, where a is a counting number.
Use this observation to compute the value of 25 × 55.
Solution:
25 × 55 = (2 × 5)5 = (10)5.
Q. Simplify 104/54 and write it in exponential form.
Solution: