Question

find the compound interest of 48000 for 2 years @12%per annum​

Answer 1

Interest on 1st year = (P*R*T)/100 = (48000*12*1)/100 = Rs 5760

Amount at end of 1st year = Rs (48000+5760) = Rs 53760

Principal for 2nd year = Rs 53760

Interest on 2nd year = (P*R*T)/100 = (53760*12*1)/100 = Rs 6451.20

Amount after 2nd year = Rs (53760+6451.2) = Rs 60211.20

Compound interest = Final amount - original principal

                                = Rs (60211.20-48000) = Rs 12211.20

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Answer 2

Given:

Principal (P) = 48000

Rate of Interest (R%) = 12%

Time (n) = 2 years

To find:

The Compound Interest (C.I)

• The formula to find the Compound Interest (C.I) is

$\bf C.I = P[(1+\frac{R}{100})^{n}-1]$

$\bf C.I = 48000[(1+\frac{12}{100})^{2}-1]$

$\bf C.I = 48000[(1+\frac{3}{25})^{n}-1]$

$\bf C.I = 48000[(\frac{25+3}{25})^{2}-1]$

$\bf C.I = 48000[(\frac{28}{25})^{2}-1]$

$\bf C.I = 48000[(\frac{784}{625})-1]$

$\bf C.I = 48000[\frac{784-625}{625}]$

$\bf C.I = 48000[\frac{159}{625}]$

$\bf C.I = [\frac{7632000}{625}]$

C.I = 12211.2

Hence,

• The Compound Interest is 12211.2