Question

Mahesh borrowed a certain sum for two years at a simple interest from Bhim. Mahesh lent this sum to Vishnu at the same rate for two years of compound interest. At the end of two years, Mahesh received Rs. 410 as compound interest but paid Rs. 400 as simple interest. Find the sum and rate of interest.
(a) 400, 15 %
(b) 5000, 10 %
(c) 4000, 5 %
(d) 7000, 15 %

Answer 1

Hint:First of all, take the value of the sum pas P and rate of interest as R %. Now, write the simple interest and compound interest using the formulas $\dfrac{P\times R\times T}{100}$ and $P\left[ {{\left( 1+\dfrac{R}{100} \right)}^{T}}-1 \right]$ respectively. Now equate S.I and C.I to 400 and 410 respectively and get the value of P and R from these equations.

Complete step-by-step answer:
Here, we are given that Mahesh borrowed a certain sum for two years at a simple interest from Bhim. Mahesh lent this sum to Vishnu at the same rate for two years of compound interest. At the end of two years, Mahesh received Rs. 410 as compound interest but paid Rs. 400 as simple interest. We have to find the sum and rate of interest.
Let us consider the sum that Mahesh borrowed from Bhim = P
Now, we are given that Mahesh borrowed this amount at simple interest for 2 years. Let the interest at which he borrowed this amount = R %
We know that, simple interest, $S.I=\dfrac{P\times R\times T}{100}$
where P = principal amount, R = Rate of interest, and T = time for which the money is borrowed or given.
By substituting the values of P = P, R = R % and T = 2 years, we get,
$S.I=\dfrac{P.R.2}{100}$
We are given that the simple interest is equal to 400, so by equation the above value to 400, we get,
$S.I=\dfrac{P.R.2}{100}=400$
$P.R=\dfrac{400\times 100}{2}$
$\Rightarrow P=\dfrac{20000}{R}.....\left( i \right)$
Now, we are given that Mahesh lent the sum that is P to Vishnu at the same rate R % for two years at compound interest. We know that compound interest, $C.I=P\left[ {{\left( 1+\dfrac{R}{100} \right)}^{T}}-1 \right]$
where P is the principal amount, R is the rate of interest and T is the time for which the money is borrowed or lent.
By substituting the values of P = P, R = R % and T = 2 years, we get,
$C.I=P\left[ {{\left( 1+\dfrac{R}{100} \right)}^{2}}-1 \right]$
We are given that the compound interest is equal to 410. So by equation the above value to 410, we get,
$C.I=P\left[ {{\left( 1+\dfrac{R}{100} \right)}^{2}}-1 \right]=410$.
By substituting the value of P in the above equation from equation (i), we get,
$\left( \dfrac{20000}{r} \right)\left[ {{\left( 1+\dfrac{R}{100} \right)}^{2}}-1 \right]=410$
$\left( \dfrac{20000}{r} \right)\left[ {{\left( \dfrac{100+R}{100} \right)}^{2}}-1 \right]=410$
By using the identity ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$ in the above equation, we get,
$\Rightarrow \left( \dfrac{20000}{r} \right)\left[ \dfrac{10000+{{R}^{2}}+200R}{10000}-1 \right]=410$
$\Rightarrow \left( \dfrac{20000}{r} \right)\left[ \dfrac{10000+{{R}^{2}}+200R-10000}{10000} \right]=410$
$\Rightarrow \left( \dfrac{20000}{r} \right)\left[ \dfrac{{{R}^{2}}+200R}{10000} \right]=410$
$\Rightarrow \left( \dfrac{2}{R} \right)\left( {{R}^{2}}+200R \right)=410$
$\Rightarrow 2\left[ \dfrac{{{R}^{2}}}{R}+\dfrac{200R}{R} \right]=410$
$\Rightarrow 2\left( R+200 \right)=410$
$\Rightarrow 2R=410-400$
$2R=10$
$R=\dfrac{10}{2}$
$R = 5\%$
Hence, we get the rate of interest as 5 %. Now, by substituting the value of R in equation (i), we get,
$P=\dfrac{20000}{R}$
$\Rightarrow P=\dfrac{20000}{5}$
$P = Rs. 4000$
Hence, we get the value of the sum as Rs. 4000. So, option (c) is the right answer.

Note: First of all, students must remember the formulas of simple interest and compound interest, $\dfrac{P\times R\times T}{100}$ and $P\left[ {{\left( 1+\dfrac{R}{100} \right)}^{T}}-1 \right]$ respectively. For this question, students make this mistake of using compound interest as $P{{\left( 1+\dfrac{R}{100} \right)}^{T}}$ which is wrong as $P{{\left( 1+\dfrac{R}{100} \right)}^{T}}$ is the total amount after T years and not the value of the interest.