Question

A sum of money at simple interest amounts to Rs 600 in 2 years and to Rs 800 in 4 years. The sum is
A.100
B.200
c.300
D.400

Answer 1

Hint: We know that the accumulated value at simple interest is \[p\left( 1+\dfrac{tr}{100} \right)\]where p is called original principal amount or sum, t is called the number of years, r is the rate of interest. By this equation we will get two equations then we will divide those two equations we will get the rate of interest, by substituting the rate of interest in any one of the equations we will get principal amount or sum.

Complete step-by-step solution -
We know the formula to find the amount in case of simple interest is \[p+\dfrac{ptr}{100}\]
A sum of money at simple interest amount to Rs 600 in 2years
\[p\left( 1+\dfrac{2r}{100} \right)=600\]. . . . . . . . . . . . (1)
A sum of money at simple interest amount to Rs 800 in 4 years
\[p\left( 1+\dfrac{4r}{100} \right)=800\]. . . . . . . . . . . . (2)
Dividing equation (1) with equation (2)
\[\dfrac{1+\dfrac{2r}{100}}{1+\dfrac{4r}{100}}=\dfrac{3}{4}\]
\[4+\dfrac{8r}{100}=3+\dfrac{12r}{100}\]
\[\dfrac{4r}{100}=1\]
\[r=25\]
So we get the rate of interest is 25
Substitute the rate of interest in equation (1)
\[p\left( 1+\dfrac{2r}{100} \right)=600\]
\[p\left( 1+\dfrac{2\left( 25 \right)}{100} \right)=600\]
\[p\left( 1+\dfrac{1}{2} \right)=600\]
\[p=600\times \dfrac{2}{3}\]
\[p=400\]
So, the sum of money we get is Rs 400
The correct option is option (D).

Note: We can also solve the problem by finding the interest accumulated for each year by subtracting 600 from 800 we will get interest accumulated for 2 years and then we will get interest for one year. Now we have to subtract 200 from 600 because 600 sum is with 2 years so interest is 200 then the principal amount or sum obtained is 400.