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05 Nov 2019, 17:30
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A certain sum of money becomes 8 times of itself in 20 years at simple interest. In how many years does it becomes 22 times of itself at the same rate of simple interest?
(a) 60 years
(b) 50 years
(c) 65 years
(d) 66.66 years
(e) 52 years
(a) 60 years
(b) 50 years
(c) 65 years
(d) 66.66 years
(e) 52 years
09 Nov 2019, 02:11
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TheNightKing
A certain sum of money becomes 8 times of itself in 20 years at simple interest. In how many years does it becomes 22 times of itself at the same rate of simple interest?
(a) 60 years
(b) 50 years
(c) 65 years
(d) 66.66 years
(e) 52 years
(a) 60 years
(b) 50 years
(c) 65 years
(d) 66.66 years
(e) 52 years
Answer : 60 years.
Explanation :-
Let the principal be P
Amount will be 8P
Interest = Amount - Principal
=> 8P - P = 7P
Simple Interest = PTR/100
According to the question,
=> 7P = P*R*20/100
=> Rate = 35%
Let the time = t years
After t years it will be 22 times of itself
=> Interest = 22P - P = 21P
=> 21P = P*35*t/100
=> t =100*21P/P*35
= 60 years.
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09 Nov 2019, 20:15
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Let the sum of money be x.
Since the amount in the 1st case is 8x. so, SI = 7x
SI = PRT/100
R=(100*SI) / (P*T)
=(100*7x) / (x*20)
=35%
Applying this rate on the 2nd case:
SI= 21x
So time = (SI * 100) / (P * R)
= (21x * 100) / (x * 35)
= 20 years (A)
We can also equate both the cases for R and solve.
Since the amount in the 1st case is 8x. so, SI = 7x
SI = PRT/100
R=(100*SI) / (P*T)
=(100*7x) / (x*20)
=35%
Applying this rate on the 2nd case:
SI= 21x
So time = (SI * 100) / (P * R)
= (21x * 100) / (x * 35)
= 20 years (A)
We can also equate both the cases for R and solve.
