In what time will a sum of money double itself at 10% compound interest payable half yearly?

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Bihar Police SI Prelims 2020: Full Mock Test

100 Questions 200 Marks 120 Mins

Let, the sum be Rs x and Rate of interest be r%

According to the question,

2x = x (1 + r/100)15

⇒ (1 + r/100)15 = 2      ----(i)

Now, Required amount = 8x

Let time period be ‘t’ years

8x = x(1 + r/100)t

⇒ (1 + r/100)t = 8 = 23

⇒ (1 + r/100)t = [(1 + r/100)15]3                      [from (i)]

⇒ (1 + r/100)t = (1 + r/100)45

⇒ t = 45

∴ In 45 years, it will become 8 times.

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Solution

Let sum of money be P=100Interest per annum =10% amount = 2× 100 = 200 simple interest = 200 - 100 = 100Since, SI=PTR100⇒T=100×SIPR=100×100100×10∴t=10 years.

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At what rate of percent a sum will double itself in 10 years?

Hence the required rate in which the sum becomes double itself in 10 years is 10%.

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Therefore, the number of years it will take to double the money at 5% per annum when compounded annually is 12.5 years.

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=(x×4100×x)years = 25 years.

In what time a sum will double itself at 12% annual rate of simple interest?

In this problem, it is given that the rate is 12 % per annum and we need to find the time in which the principal amount doubles. The total amount at the end of N years is the sum of simple interest and the principal amount. Hence, the required time is 8 years and 4 months.