
A sum of Rs. 5000 was lent partly at 6% and partly at 9% simple interest. If the total interest received after 1 year was Rs. 390, what was the ratio in which the money was lent at 6% and 9%?
(A) \[1:5\]
(B) \[2:5\]
(C) \[2:3\]
(D) \[1:3\]
Answer
580.8k+ views
Hint: To solve this question we need to first find the simple interest respective to both the interests for complete year. Then compare the interest which came as a result to the total interest given in the question. When we will get the final equation after comparing the interests, further solve it. The result which will come will be sum at one of the interests, then we will find the sum due to another interest. The ratio of these final sums will be the final answer. The formula for finding the Simple Interest is \[S.I.=\dfrac{PRT}{100}\]
Where, P means the principle amount, R means the rate of interest and T means the time interval.
Complete step by step answer:
To start solving this question, first of all let us assume the sum lent at 6% per annum be x.
Then, the sum at 9% per annum \[=5000-x\]
Simple Interest at 6% \[=\dfrac{x\times 6\times 1}{100}\]
Simple Interest at 9% \[=\dfrac{(5000-x)\times 9\times 1}{100}\]
According to the question,
S.I. at 6% + S.I at 9% = 390
\[\begin{align}
& \dfrac{x\times 6}{100}+\dfrac{(5000-x)\times 9}{100}=390 \\
&\Rightarrow \dfrac{6x}{100}+\left( \dfrac{5000\times 9}{100}- \dfrac{9x}{100} \right)=390 \\
& \Rightarrow \dfrac{6x}{100}+\left( 450- \dfrac{9x}{100} \right)=390 \\
&\Rightarrow \dfrac{9x-6x}{100}=450-390 \\
&\Rightarrow \dfrac{3x}{100}=60 \\
&\Rightarrow x=2000 \\
\end{align}\]
The sum lent at 6% per annum is Rs. 2000
Now, the sum lent at 9% \[=5000-2000=3000\]
The sum lent at 9% per annum is Rs. 3000
The ratio in which the money was lent at 6% and 9%,
\[\begin{align}
& =\dfrac{\text{money lent at }6\%}{\text{money lent at }9\%} \\
& =\dfrac{2000}{3000} \\
& =\dfrac{2}{3} \\
& =2:3 \\
\end{align}\]
Therefore, the ratio at which the money was lent at 6% and 9% is \[2:3\]
Note: In this it is mentioned partially, which does not mean that the simple interest which is to be found out for one year will be found in any other time, this will make the solution wrong. Make the equation of the sum of interest carefully, as a small mistake will make the complete solution wrong.
Where, P means the principle amount, R means the rate of interest and T means the time interval.
Complete step by step answer:
To start solving this question, first of all let us assume the sum lent at 6% per annum be x.
Then, the sum at 9% per annum \[=5000-x\]
Simple Interest at 6% \[=\dfrac{x\times 6\times 1}{100}\]
Simple Interest at 9% \[=\dfrac{(5000-x)\times 9\times 1}{100}\]
According to the question,
S.I. at 6% + S.I at 9% = 390
\[\begin{align}
& \dfrac{x\times 6}{100}+\dfrac{(5000-x)\times 9}{100}=390 \\
&\Rightarrow \dfrac{6x}{100}+\left( \dfrac{5000\times 9}{100}- \dfrac{9x}{100} \right)=390 \\
& \Rightarrow \dfrac{6x}{100}+\left( 450- \dfrac{9x}{100} \right)=390 \\
&\Rightarrow \dfrac{9x-6x}{100}=450-390 \\
&\Rightarrow \dfrac{3x}{100}=60 \\
&\Rightarrow x=2000 \\
\end{align}\]
The sum lent at 6% per annum is Rs. 2000
Now, the sum lent at 9% \[=5000-2000=3000\]
The sum lent at 9% per annum is Rs. 3000
The ratio in which the money was lent at 6% and 9%,
\[\begin{align}
& =\dfrac{\text{money lent at }6\%}{\text{money lent at }9\%} \\
& =\dfrac{2000}{3000} \\
& =\dfrac{2}{3} \\
& =2:3 \\
\end{align}\]
Therefore, the ratio at which the money was lent at 6% and 9% is \[2:3\]
Note: In this it is mentioned partially, which does not mean that the simple interest which is to be found out for one year will be found in any other time, this will make the solution wrong. Make the equation of the sum of interest carefully, as a small mistake will make the complete solution wrong.
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