A sum of money placed at compound interest doubles itself in 4 yr. In how many years will it amount to four times itself ?
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Answered by
3
8 years.
A = 2P in 4 years.
A = 4P in T years
2P = P[1 + r/100]^4....(a)
4P = P[1 + r/100]^n...(b)
Let [1 + r/100] = T
2P = P*T^4
2 = T^4....(i)
4P = P*T^n
4 = T^n...(ii)
we know that T^4 = 2
squaring both side: (T^4)^2 = 2^2
T^8 = 4...(iii)
Compare (ii) & (iii)
Hence T^n = T^8...n=8
[T= 1+ r/100]
So the sum will amount to 4 times in 8 years.
Hope it helps
A = 2P in 4 years.
A = 4P in T years
2P = P[1 + r/100]^4....(a)
4P = P[1 + r/100]^n...(b)
Let [1 + r/100] = T
2P = P*T^4
2 = T^4....(i)
4P = P*T^n
4 = T^n...(ii)
we know that T^4 = 2
squaring both side: (T^4)^2 = 2^2
T^8 = 4...(iii)
Compare (ii) & (iii)
Hence T^n = T^8...n=8
[T= 1+ r/100]
So the sum will amount to 4 times in 8 years.
Hope it helps
Answered by
0
Answer:
8 Years
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