Question

The compound interest on a sum in 8 years is 400 and compound interest. Find ci on the same sum in 20 years on the same sum in 16 years is 1300

Answer 1

Can you please explain it a bit more.. I couldn't get the 5===400 part

Answer 2

Answer:

$Rs.2115.91$

Step-by-step explanation:

Given:

Compound interest for $8$ years is $Rs.400$

Compound interest for $16$ years is $Rs.1300$

Let the sum be $Rs.P$ and rate of compound interest be $r$‰

For $8$ years,

$P(1+\frac{r}{100})^8-P=400$   or  $P[(1+\frac{r}{100})^8-1]=400...(i)$

For $16$ years,

$P(1+\frac{r}{100})^1^6-P=1300$  or  $P[(1+\frac{r}{100})^1^6-1]=1300$

or  $P[(1+\frac{r}{100})^8+1][(1+\frac{r}{100})^8-1] =1300...(ii)$

Divide $(ii)$ by $(i),$ we get

$(1+\frac{r}{100} )^8+1=\frac{13}{4}$

     $(1+\frac{r}{100} )^8=\frac{9}{4}$

         $1+\frac{r}{100}=(\frac{9}{4} )^(^\frac{1}{8} ^)$

                $\frac{r}{100}=(\frac{9}{4} )^(^\frac{1}{8} ^)-1$

                   $r=100*[(\frac{9}{4} )^(^\frac{1}{8} ^)-1]$

                   $r=10.7$

Putting $r=10.7$ in $(i),$  we get

$P[(1+\frac{10.7}{100} )^8-1]=400$

                        $P=318.78$

We have found:

Sum, $P=Rs.318.78$

Rate of interest, $r=10.7$‰

Therefore the compound interest in $20$ years will be

$=P(1+\frac{r}{100})^2^0-P$

$=318.78(1+\frac{10.7}{100})^2^0-1$

$=Rs.2115.91$