The difference between the simple interest and the compound interest on a sum of money for 3 years at 12% per annum is ₹216. Find the sum.


Given :

The difference between the simple interest the compound interest on a sum of money for 3 years at 12% per annum $=$ Rs. 216.


To find :

We have to find the sum.


Solution :

Let the sum be P.

Compound interest $= P[(1+\frac{r}{100})^n - 1]$

                           $= P [(1+\frac{12}{100})^3 - 1]$

                            $= P [(1+\frac{3}{25})^3 - 1]$

                           $= P [(\frac{28}{25})^3 - 1]$

                           $= P [\frac{21952}{15625} - 1]$

                           $= P[\frac{21952-15625}{15625}]$

                          $= \frac{6327}{15625}P$

Simple interest $= \frac{Pnr}{100}$

                           $= \frac{P \times 3 \times 12}{100}$

                           $= \frac{36P}{100} = \frac{9}{25}P$

$CI - SI = 216$

$ \frac{6327}{15625}P - \frac{9}{25}P = 216$

$ \frac{6327}{15625}P - \frac{9 \times 625}{25 \times 625}P = 216$

$\frac{6327}{15625}P -\frac{5625}{15625}P=216 $

$ \frac{6327 - 5625}{15625}P=216$

$\frac{702}{15625}P = 216$

$ P = \frac{216 \times 15625}{702}$

$P = \frac{337500}{702} = 4807.69$

Therefore, the sum is ₹4807.69.

   

Compoundinterest=x[(1+r100)n1]=x[(1+12100)31]=x[(1+325)31]=x[(2825)31]=x(21952156251)=x(219521562515625)=x(632715625)Compound interest=x\left[\left( 1+\frac{r}{100}\right)^{n} -1\right] =x\left[\left( 1+\frac{12}{100}\right)^{3} -1\right] =x\left[\left( 1+\frac{3}{25}\right)^{3} -1\right] =x\left[\left(\frac{28}{25}\right)^{3} -1\right] =x\left(\frac{21952}{15625} -1\right) =x\left(\frac{21952-15625}{15625}\right) =x\left(\frac{6327}{15625}\right)

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Updated on: 10-Oct-2022

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