Question

Find the amount and compound interest on 36000 for 2 years at the rate in 10% compound
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Answer 1

Answer:

For the first year

P=Rs36,000

N=1year

R=10 %

We have S.I.=

100

PNR

=

100

36,000×1×10

=Rs3,600

And Amount at the end of first year P+S.I.=Rs36,000+Rs3,600=Rs39,600

Now, for the second year

P=Rs39,600

N=1year

R=12 %

We have S.I.=

100

PNR

=

100

39,600×1×12

=Rs4,752

And Amount at the end of second year P+S.I.=Rs39,600+Rs4,752=Rs44,352

Step-by-step explanation:

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Answer 2

Step-by-step explanation:

P = 36,000 ( currency not given )

R = 10% pa

T = 2 yrs

$a = p(1 + { \frac{r}{100} })^{n} \\ = 36000(1 + { \frac{10}{100} })^{2} \\ = 36000( \frac{11}{10} \times \frac{11}{10} ) \\ = 36000 \times \frac{121}{100} \\ = 360 \times 121 \\ = 43560$

Now , CI = A - P

= 43560 - 36000

= 7560