Math, asked by sashreeksom, 5 months ago

The difference between the compound interest and the simple interest accrued on an amount of Rs. 18,000 in 2 years was Rs. 405. What was the rate of interest p.c.p.a.

Answers

Answered by parijaini
6

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  • Let the rate be R% p.a. Then,

  • [1800(1+R100)2−1800]−(1800*R*2100)=405
  • 1800[(100+R)210000−1−2R100]=405
  • 1800[(100+R)2−10000−200R10000]=405

  • 95*R2=405 ⇔R2=(405*59)=225⇔R=15
  • Rate = 15%.

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sashreeksom: Thank You
parijaini: wlcm!!
Answered by Anonymous
5

Answer : 15% ✔✔

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(p {(1 +  \frac{r}{100}) }^{t}  - p) -  \frac{p \times r \times t}{100}  = 405 \\  \\  =  > 18000( {(1 +  \frac{r}{100}) }^{2}  - 1 -  \frac{r \times 2}{100} ) = 405 \\  \\  =  >  {(1 +  \frac{r}{100}) }^{2}  - 1 -  \frac{2r}{100}  =  \frac{405}{18000}  \\  \\  =  > 1 +  \frac{ {r}^{2} }{10000}  +  \frac{2r}{100}   - 1 -  \frac{2r}{100}  =  \frac{405}{18000}  \\  \\  =  >  \frac{ {r}^{2} }{10000}  =  \frac{405}{18000}  =  >  {r}^{2}  =  \frac{4050}{18}  \\  \\  =  >  {r}^{2}  = 225 \\  \\  =  > r =  \sqrt{225}  = 15

So, the rate of interest p.c.p.a will be 15%

_____________________________

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