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# Rajan lent Rs. 1200 to Rakesh for 3 years at a certain rate of interest and Rs. 1000 to Mukesh for the same time at the same rate. If he gets Rs. 50 more from Rakesh than from Mukesh, then find the rate percent.A. $8\dfrac{1}{3}$%B. $6\dfrac{2}{3}$%C. $10\dfrac{1}{3}$%D. $9\dfrac{2}{3}$%

Last updated date: 20th Jun 2024
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Hint: We first assume the rate of interest for the principal amount has been under interest. We use the theorem of $a=\dfrac{pnr}{100}$ to relate the principal and the interest. We find the multiplied values of the variable from the first condition. We use that to find a solution to the problem.

Complete step-by-step solution
Rajan gave two people money of different value for 3 years at a certain rate of interest. For both cases, the rate of interest and time is the same.
Let’s assume the rate is r% annually. We term the time as n. So, $n=3$.
The amount of money for Rakesh and Mukesh is 1200 and 1000 respectively.
He got back Rs. 50 more from Rakesh than from Mukesh.
We have to calculate the interesting part.
We know that if the principal amount is p and the amount of interest is a, then the formula of finding the interest is defined by ${{a}_{i}}=\dfrac{pnr}{100}$.
For our given problem Rajan gave Rakesh 1200 rupees at a rate of r for 3 years.
We put the values $p=1200,n=3$ in the equation ${{a}_{1}}=\dfrac{pnr}{100}$. ${{a}_{1}}$ is the interest for Rakesh.
We get ${{a}_{1}}=\dfrac{1200\times 3\times r}{100}=36r$.
Rajan gave Mukesh 1000 rupees at a rate of r for 3 years.
We put the values $p=1000,n=3$ in the equation ${{a}_{2}}=\dfrac{pnr}{100}$. ${{a}_{2}}$ is the interest for Rakesh.
We get ${{a}_{2}}=\dfrac{1000\times 3\times r}{100}=30r$.
The difference of those two terms is $36r-30r=6r$ which is equal to 50.
Solving the equation, we get $6r=50\Rightarrow r=\dfrac{50}{6}=8\dfrac{1}{3}$.
The rating percentage is $8\dfrac{1}{3}$%. The correct option is A.

Note: We need to use simple interest if otherwise mentioned. In this problem, the principal was responsible for the change of interest value as the time span and the rate both were constant. In case of compound interest, the formula for interest would have been $a=p{{\left( 1+\dfrac{r}{100} \right)}^{n}}-p$.