Question

24. A sum of money at certain rate of compound interest, which compounds semi annually, becomes 60,500 and 73,205 after one year and two years respectively. Find the an
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Answer 1

Answer:

Difference between the C.I of two successive years=7410−5700=Rs.1710

Rate of interest=

C.I of preceding year×timeDifference between the C.I of two successive year×100

∴Rate of intereast= 5700×1

1710×100=30%

Answer 2

Given:

A sum of money at a certain rate of compound interest, which compounds semi-annually, becomes 60,500 and 73,205 after one year and two years respectively.

To Find:

The rate of interest

Solution:

Before solving this question we should know the formula for a sum that is being compounded at some intervals at a rate of interest annually, which is,

                                         $A=P(1+\frac{r}{n})^{nt}$

where,

          A= The amount at the end of the time period

          P= Initial principal amount

          r= rate of interest

          n= no of time sum is compounded

          t= time period

Now put the values for the 1st year in the formula that is,

$60500=P(1+\frac{r}{2})^2$       -(1)

Now put the values for the 2nd year in the formula that is,

$73205=P(1+\frac{r}{2} )^4$       -(2)

Now dividing equation 2 with equation 1, we have,

$\frac{73205}{60500} =\frac{P(1+\frac{r}{2})^4 }{P(1+\frac{r}{2})^2 } \\(1+\frac{r}{2} )^2=1.21$

Now square rooting both the sides, we will get

$1+\frac{r}{2} =1.1\\\frac{r}{2} =0.1\\r=0.2$

So if the r comes as 0.2 which is in fraction form, so 0.2 in fraction form considered in percentage form will be 20%.

Hence, the rate of interest will be 20%.