Question

Find the compound interest on R.s. 3200at 2.5p.a for 2 years,compounded annually

Answer 1

AnswEr :

• Principal = Rs. 3200
• Rate = 2.5% p.a.
• Time = 2 Years

$\Longrightarrow\mathsf{CI = P \bigg(1 + \dfrac{r}{100} \bigg)^{t} - 1}$

$\Longrightarrow\mathsf{CI = 3200 \bigg(1 + \dfrac{2.5}{100} \bigg)^{2} - 1}$

$\Longrightarrow\mathsf{CI = 3200 \bigg(1 + \dfrac{1}{40} \bigg)^{2} - 1}$

$\Longrightarrow\mathsf{CI = 3200 \bigg( \dfrac{41}{40} \bigg)^{2} - 1}$

$\Longrightarrow\mathsf{CI = 3200 \bigg( \dfrac{1681}{1600} - 1 \bigg) }$

$\Longrightarrow\mathsf{CI = \cancel{3200} \times \dfrac{81}{ \cancel{1600}} }$

$\Longrightarrow\mathsf{CI =Rs. \:(2 \times 81)}$

$\Longrightarrow\mathsf{CI =Rs. \: 168}$

⠀

$\therefore$ Compound Interest is Rs. 168

Answer 2

SOLUTION:-

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Given:

• Principal,[P]= Rs.3200
• Rate,[R]= 2.5% p.a.
• Time,[n]= 2 years.

To find:

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The compound Interest.

Explanation:

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Formula: Compound Interest

Formula: Amount- Principal

&

$A = P(1 + \frac{R}{100 } ) {}^{n}$

According to the question:

$A = 3200(1 + \frac{2 .5}{100} ) {}^{2} \\ \\ A=3200(1 + \frac{25}{1000} ) {}^{2} \\ \\ A= 3200( \frac{1000 + 25}{1000} ) {}^{2} \\ \\ A = 3200( \frac{1025}{1000} ) {}^{2} \\ \\ A = 3200 \times \frac{1025}{1000} \times \frac{1025}{1000} \\ \\ A = \frac{33620000}{10000} \\ \\ A = Rs.3362$

Now,

C.I= Rs.3362 - Rs.3200

C.I.= Rs.162

:)