A sum of money becomes 5 times at C.I. in 2 years in how many years will it become 55.9 times

We know the formula for compound interest-

CI = {P [1 + (R/100)t ] - P}

Where,

CI = Compound interest

P = Principal

R = Rate of interest

t = Time period

4P = {P [1 + (R/100)4 - P}

5 = [1 + (R/100)4]

Let ‘x’ be the number of years after which the sum becomes 25 times.

25P - P = {P [1 + (R/100)x - P}

25 = [1 + (R/100)x]

52 = [1 + (R/100)x]2

(? (am)n = am × n)

⇒ x = 4 × 2 = 8 years

∴ The number of years after which the sum becomes 25 times = 8 years

Let principal = P,

\( \Large \textbf{Case (1):} \) Time = 3 years,

Amount = 8P

\( \Large 8P=P \left(1+\frac{R}{100}\right)^{3}\)

\( \Large\left(2\right)^{3} = \left(1+\frac{R}{100}\right)^{3}\)

Taking cube root of both sides

2=\( \Large\left(1+\frac{R}{100}\right)\)

R=100 %

\( \Large \textbf{Case 2:} \) Let after t years it will be 16 times

\( \Large 16P=P \left(1+\frac{R}{100}\right)^{t}\)

\( \Large 16= \left(2\right)^{t}\)

\( \Large\left(2\right)^{4}= \left(2\right)^{t} \)

t = 4 years

Hence Required time (t) = 4 years

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A sum of money becomes 8 / 5 of itself in 5 years at a certain rate of simple interest .find the rate of interest.

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Solution

Given: Time, T=5 years

Let the principal be P

Rate of interest =R

Amount, A=8P5

Simple Interest, S.I= Amount − Principal
=8P5−P

=3P5

∴S.I=3P5
Again, S.I=P×R×T100

⇒3P5=P×R×5100

⇒35=5R100

⇒35=R20

⇒R=3×205

⇒R=3×4

∴R=12.

The required rate of interest is 12%.

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