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# The compound interest on a sum of money for two years is ${\rm{Rs}}52$ and the simple interest for two years at the same rate is ${\rm{Rs}}50$. Then the rate of interest isA. $6\%$B. $8\%$C. $9\%$D. $10\%$

Last updated date: 20th Jun 2024
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Hint: Here, we will first use the formula of simple interest and find the principal in terms of rate of interest. Then we will use the formula of the compound interest and substitute the value of obtained principal in it. We will solve it further to get the required rate of interest.

Formula Used:
We will use the following formulas:
1. $S.I = \dfrac{{P \cdot R \cdot T}}{{100}}$ , where $S.I$ is the Simple Interest, $P$ is the Principal, $R$ is the rate of interest per annum and $T$ is the time period.
2. $C.I = P{\left( {1 + \dfrac{R}{{100}}} \right)^n} - P$ , where $C.I$ is the Compound Interest, $P$ is the Principal, $R$ is the rate of interest per annum and $n$ is the time period.

Let the sum of money invested, principal $= P$
Given time period, $T = 2{\rm{years}}$
Now, it is given that the simple interest, $S.I = {\rm{Rs}}50$
Substituting the given values in this formula $S.I = \dfrac{{P \cdot R \cdot T}}{{100}}$, we get,
$50 = \dfrac{{P \cdot R \cdot \left( 2 \right)}}{{100}}$
Multiplying both sides by 100, we get
$\Rightarrow 5000 = P \cdot R \cdot \left( 2 \right)$
Dividing both sides by 2, we get
$\Rightarrow P \cdot R = 2500$
$\Rightarrow P = \dfrac{{2500}}{R}$ ………………………………..$\left( 1 \right)$
Also, it is given that the Compound interest, $C.I = {\rm{Rs}}52$.
Hence, substituting the given values in this formula $C.I = P{\left( {1 + \dfrac{R}{{100}}} \right)^n} - P$, we get
$52 = P{\left( {1 + \dfrac{R}{{100}}} \right)^2} - P$
Taking $P$ common and opening the bracket using the formula ${\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}$, we get,
$\Rightarrow 52 = P\left( {1 + \dfrac{{2R}}{{100}} + \dfrac{{{R^2}}}{{10000}} - 1} \right)$
$\Rightarrow 52 = P\left( {\dfrac{{2R}}{{100}} + \dfrac{{{R^2}}}{{10000}}} \right)$
Taking LCM, we get
$\Rightarrow 52 = P\left( {\dfrac{{200R + {R^2}}}{{10000}}} \right)$
Multiplying both sides by 10000, we get
$\Rightarrow 520000 = P\left( {200R + {R^2}} \right)$
Dividing both sides by $P$, we get
$\Rightarrow 200R + {R^2} = \dfrac{{520000}}{P}$
Now, by substituting $P = \dfrac{{2500}}{R}$ in the above equation, we get
$\Rightarrow 200R + {R^2} = \dfrac{{520000R}}{{2500}}$
Dividing the terms on RHS, we get
$\Rightarrow 200R + {R^2} = 208R$
Subtracting the like terms, we get
$\Rightarrow {R^2} = \left( {208 - 200} \right)R = 8R$
Dividing both sides by $R$, we get
$\Rightarrow R = 8$

Therefore, the required rate of interest is $8\%$.
Hence, option B is the correct answer.

Note:
In this question, we have used the formula of Simple Interest as well as compound interest. Simple Interest is the interest earned on the Principal or the amount of loan. The second type of interest is Compound Interest. Compound Interest is calculated both on the Principal as well as on the accumulated interest of the previous year. Hence, this is also known as ‘interest on interest’.