SSC CHSL 15 Oct 2020 Shift 1 Paper

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Question : 61 of 100

Marks: +1, -0

The compound interest on

₹ 4000

after

3 {\yr}

is ₹

630.50

. Then, the rate of interest compounded yearly is

$7 \%$
$5 \%$
$8 \%$
$6 \%$

Solution:

Given that, principal

(\P)=₹ 4000

time

(n)=3 {\yr}

and

{\Cl}=₹ 630.50

According to the formula,
Compound interest

=\P[(1+{\R}/{100})^{n}-1]

⇒

630.50=4000[(1+{\R}/{100})^{3}-1]

⇒

1+{6305}/{40000}=(1+{\R}/{100})^{3}

⇒

(1+{\R}/{100})^{3}={46305}/{40000}

⇒

(1+{\R}/{100})^{3}={9261}/{8000}

⇒

(1+{\R}/{100})=({9261}/{8000})^{1 / 3}={21}/{20}

⇒

{\R}/{100}={21}/{20}-1

={21-20}/{20}={1}/{20}

∴

\R={100}/{20}=5 \%

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