Question

a certain sum p amounts to 29400 in two years and to 34300 in three years on compound interest .then the sum p is ( in ruppes)???

Answer 1

Answer:

The sum p in rupees is 21,600

Explanation:

Let the sum be p ( in rupees)

For a certain sum p amounts to 29400 in two years and to 34300 in three years on compound interest are:

Using the formula of Amounts(A) :

$Amount(A)=p\left (1+ \frac{r}{100}\right )^n$ , where r is the rate of interest and n is the number of years.

First find the rate of interest r% :

from the given condition and using above formula, we have

$29400=p\left ( 1+\frac{r}{100} \right )^2$             ......(1)

$34300=p\left ( 1+\frac{r}{100} \right )^3$             .......(2)

On dividing 2 by 1 equation we get:

$\frac{34300}{29400}=\frac{p\left ( 1+\frac{r}{100} \right )^3}{p\left ( 1+\frac{r}{100} \right )^2}$

On simplifying we get,

$\frac{343}{294}=\left ( 1+\frac{r}{100} \right)$

$\frac{7}{6}=\left ( 1+\frac{r}{100} \right)$

$\frac{7}{6}-1=\frac{r}{100}$

$\frac{1}{6}=\frac{r}{100}$

∴ $r=\frac{100}{6}$ $=\frac{50}{6}$$=16\frac{2}{3}$ %

Therefore, the rate of interest is ,$r=16\frac{2}{3}$%

Now, to find the sum p in rupees put the value of r in equation (1),

$29400=p\left ( 1+\frac{50}{3\cdot 100} \right )^2$

$29400=p\left ( 1+\frac{1}{6} \right )^2$

$29400=p\left ( \frac{7}{6} \right )^2$

$29400=p\cdot \frac{49}{36}$

∵ $p=29400\cdot\frac{36}{49}$

On solving  we get, the value of sum p=Rs. 21,600