Question

The compound interest is RS. 6.40 more than the simple interest, if a sum is lent for 2 yr at 8% compound interest. Find the sum.
A). RS.1800
B). RS.10000
C). RS.800
D). RS.1000​

Answer 1

Answer:

Correct Answer:

D) RS.1000

Description for Correct answer:

Given, CI - SI = 6.40

T=2yr and R=8%

=> \( \Large P \left[(1+ \frac{R}{100}\right)^{T}-1] \)-\( \Large \frac{P\times R\times T}{100} \)=6.40

=> \( \Large P \left[(1+ \frac{8}{100}\right)^{2}-1] \)-\( \Large \frac{P\times 8\times 2}{100} \)=6.40

=> \( \Large P \left[(\frac{27}{25}\right)^{2}-1] \)-\( \Large \frac{16P}{100} \)=6.40

=> \( \Large P \left[\frac{27^{2}-25^{2}}{625}\right] \)-\( \Large \frac{16P}{100} \)=6.40

=> \( \Large \frac{104 P}{625} \)-\( \Large \frac{16P}{100} \)=6.40

=> \( \Large \frac{416P-400P}{2500} \)=6.40

=> 16P = \( \Large 6.40\times 2500 \)

P = \( \Large \frac{6.40\times 2500}{16} \)= RS.1000

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Answer 2

Given,

CI - SI = 6.40

T=2yr and R=8%

=> P[(1+R100)T−1]-P×R×T100=6.40

=> P[(1+8100)2−1]-P×8×2100=6.40

=> P[(2725)2−1]-16P100=6.40

=> P[272−252625]-16P100=6.40

=> 104P625-16P100=6.40

=> 416P−400P2500=6.40

=> 16P = 6.40×2500

P = 6.40×250016= RS.1000

option d is a right answer