A sum triples in ten years under compound interest at a certain rate of interest, the interest is being compounded annually. In how many years, it would become nine times ? ,

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Q: 116 (CAPF/2020)
A sum triples in ten years under compound interest at a certain rate of interest, the interest is being compounded annually. In how many years, it would become nine times ? ,

question_subject: 

Maths

question_exam: 

CAPF

stats: 

0,5,7,5,6,1,0

keywords: 

{'compound interest': [0, 0, 0, 2], 'sum triples': [0, 0, 0, 1], 'certain rate': [0, 0, 0, 1], 'interest': [1, 3, 3, 15], 'years': [1, 0, 0, 2], 'many years': [3, 0, 0, 4]}

In this question, we are given that a sum triples in ten years under compound interest at a certain rate. We need to find out in how many years the sum will become nine times.

To solve this problem, we can use the formula for compound interest:

A = P(1 + r/n)^(nt)

Where:

A = the final amount

P = the principal amount (initial sum)

r = the annual interest rate (in decimal form)

n = the number of times that interest is compounded per year

t = the number of years

Given that the sum triples in ten years, we can write the equation:

3P = P(1 + r/1)^(1*10)

Simplifying this equation, we get:

3 = (1 + r)^10

To find out when the sum becomes nine times, we can set up the equation:

9P = P(1 + r/1)^(1*t)

Simplifying this equation, we get:

9 = (1 + r)^t

Comparing the two equations, we can see that we need to find the value of t such that (1 + r)^t = 9.

Solving this equation, we find that t = 20.

Therefore,

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