A certain sum was borrowed at 15% per annum. If at the end of 2 years, Rs 1,290 was compounded as C.I., then find the sum borrowed.

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Solution

Amount = $P {(1 + \frac{r}{100})}^{n}$ $= P {(1 + \frac{15}{100})}^{2} = P {(1 + \frac{3}{20})}^{2}$ $= P {(\frac{23}{20})}^{2} = P \times \frac{23}{20} \times \frac{23}{20}$

$∴ C . I . = (P \times \frac{23}{20} \times \frac{23}{20}) - P = P (\frac{529}{400} - 1)$
But, $C . I . = R s 1290$

$∴ P (\frac{529}{400} - 1) = 1290$

or $P (\frac{529 - 400}{400}) = 1290$

or $P (\frac{129}{400}) = 1290$

or $P = \frac{1290 \times 400}{129}$
or $P = 10 \times 400 = 4000$
Hence, Principal = Rs $4, 000.$

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A certain sum was borrowed at 15% per annum. If at the end of 2 years, Rs 1,290 was compounded as C.I., then find the sum borrowed.

Amount = P(1+r100)n=P(1+15100)2=P(1+320)2=P(2320)2=P×2320×2320 ∴C.I.=(P×2320×2320)−P=P(529400−1) But, C.I.=Rs1290 ∴P(529400−1)=1290 or P(529−400400)=1290 or P(129400)=1290 or P=1290×400129 or P=10×400=4000 Hence, Principal = Rs 4,000.