Question

find the sum invested at 10% compounded annually on which the interest for the third year exceeds the interest of the first year by rupees 252​

Answer 1

Answer:

The invested sum is Rs.12000

Step-by-step explanation:

Let the investment be Rs.P

First year interest

$=\frac{Pr}{100}$

$=\frac{P*10}{100}$

$=\frac{P}{10}$

Second year investment

=First year invetment+ First year interest

$=P+\frac{P}{10}$

$=\frac{11P}{10}$

Second year interest

$=\frac{11P}{10}*\frac{10}{100}$

$=\frac{11P}{100}$

Third year investment

=Second year investment + Second year interest

$=\frac{11P}{10}+\frac{11P}{100}$

$=\frac{121P}{100}$

Third year interest

$=\frac{121P}{100}*\frac{10}{100}$

$=\frac{121P}{1000}$

But, given:

Third year interest-First year interest=Rs.252

$\implies\:\frac{121P}{1000}-\frac{P}{10}=252$

$\implies\:\frac{121P-100P}{1000}=252$

$\implies\:\frac{21P}{1000}=252$

$\implies\:\frac{P}{1000}=12$

$\implies\:P=12*1000$

$\implies\:P=Rs.12000$

Answer 2

Answer:

Rs 12000

Step-by-step explanation:

Let say Sum invested = 100P  Rs

Interest for 1st Year =  100 P * 10 * 1 /100  = Rs 10P

Investment Amount for  2nd Year = 100P + 10P = 110P

Interest for 2nd Year = 110P * 10 * 1/100 = 11P Rs

Investment amount for 3rd Year = 110P + 11P = 121P

Interest for 3rd Yeard = 121P * 10 * 1/100 = 12.1P Rs

Third year interest - 1st Year interest = Rs 252

=> 12.1 P - 10P = 252

=> 2.1P = 252

=> P = 120

Sum invested = 100P = 100 * 120 = Rs 12000