
A sum was put at simple interest at a certain rate for 4 years. Had it been put at 2% higher rate, it would have fetched Rs. 112 more. Find the sum.
A. 1400
B. 1200
C. 1000
D. 1600
Answer
519.3k+ views
Hint: Consider a variable for the sum and a variable for rate of interest (say r %). Calculate simple interest at r % and at (r +2)%. Using the formula: $\text{simple interest }=\dfrac{P\times r\times t}{100}$ where, P = principal amount (the sum given for interest), r = rate of interest and t = time.
After calculating the two simple interests equate their difference to Rs. 112 and solve the obtained equation.
Complete step-by-step answer:
Let the sum be $Rs.x$. So, our principal amount will be $'x'$.
Let, the rate of interest for 4 years = r %
So, Principal amount = $Rs.x$
Rate of interest = r %
Time = 4 years
So, the simple interest in this case $\text{ }=\dfrac{P\times r\times t}{100}$
$\text{simple interest }=\dfrac{x\times r\times 4}{100}............\left( 1 \right)$
According to the question, if the rate of interest would be 2 % higher than the current rate of interest, simple interest would have been Rs. 112 more than current simple interest.
Let us calculate the simple interest at a rate of (r + 2)%,
Now, Principal amount = $Rs.x$
Rate of interest = (r + 2)%
Time = 4 years
New simple interest $\text{ }=\dfrac{x\times \left( r+2 \right)\times 4}{100}............\left( 2
\right)$ (using the formula for simple interest)
According to question,
(New simple interest) – (Initial simple interest) = Rs.112
$\Rightarrow \left( \dfrac{x\times \left( r+2 \right)\times 4}{100} \right)-\left( \dfrac{x\times r\times 4}{100} \right)=Rs.112$
Taking $\left( \dfrac{x\times 4}{100} \right)$ common from the LHS,
$\begin{align}
& \Rightarrow \dfrac{x\times 4}{100}\left[ \left( r+2 \right)-r \right]=112 \\
& \Rightarrow \dfrac{x\times 4\times 2}{100}=112 \\
& \Rightarrow \dfrac{x\times 8}{100}=112 \\
\end{align}$
Multiplying both sides by 100, we get,
$\Rightarrow 8x=112\times 100$
Dividing both sides by 8, we will get,
$\begin{align}
& \Rightarrow x=\dfrac{112\times 100}{8} \\
& \Rightarrow x=1400 \\
\end{align}$
So, the required sum = Rs. 1400.
Note: Short method,
According to the question, for an increase in 2 % of rate of interest, simple interest will increase by Rs. 112. In the formula of simple interest, put the increase in rate of interest in the place of rate of interest. i.e. put $'\Delta r'$in place of ‘r’ and in the place of simple interest, put change in simple interest.
i.e. put $\Delta $(simple interest) in place of simple interest.
$'\Delta '$ is a symbol used for ‘change’.
Formula: $\text{simple interest }=\dfrac{P\times r\times t}{100}$
Let P = $Rs.x$
On putting $\Delta r$in place of r and $\Delta $ simple interest in place of simple interest,
$\begin{align}
& \Rightarrow \Delta \left( \text{simple interest} \right)\text{ }=\dfrac{x\times \Delta r\times t}{100} \\
& \Rightarrow +112=\dfrac{x\times \left( 2 \right)\times 4}{100} \\
& \Rightarrow 112\times 100=8x \\
& \Rightarrow x=\dfrac{112\times 100}{8} \\
& \Rightarrow x=Rs.1400 \\
\end{align}$
After calculating the two simple interests equate their difference to Rs. 112 and solve the obtained equation.
Complete step-by-step answer:
Let the sum be $Rs.x$. So, our principal amount will be $'x'$.
Let, the rate of interest for 4 years = r %
So, Principal amount = $Rs.x$
Rate of interest = r %
Time = 4 years
So, the simple interest in this case $\text{ }=\dfrac{P\times r\times t}{100}$
$\text{simple interest }=\dfrac{x\times r\times 4}{100}............\left( 1 \right)$
According to the question, if the rate of interest would be 2 % higher than the current rate of interest, simple interest would have been Rs. 112 more than current simple interest.
Let us calculate the simple interest at a rate of (r + 2)%,
Now, Principal amount = $Rs.x$
Rate of interest = (r + 2)%
Time = 4 years
New simple interest $\text{ }=\dfrac{x\times \left( r+2 \right)\times 4}{100}............\left( 2
\right)$ (using the formula for simple interest)
According to question,
(New simple interest) – (Initial simple interest) = Rs.112
$\Rightarrow \left( \dfrac{x\times \left( r+2 \right)\times 4}{100} \right)-\left( \dfrac{x\times r\times 4}{100} \right)=Rs.112$
Taking $\left( \dfrac{x\times 4}{100} \right)$ common from the LHS,
$\begin{align}
& \Rightarrow \dfrac{x\times 4}{100}\left[ \left( r+2 \right)-r \right]=112 \\
& \Rightarrow \dfrac{x\times 4\times 2}{100}=112 \\
& \Rightarrow \dfrac{x\times 8}{100}=112 \\
\end{align}$
Multiplying both sides by 100, we get,
$\Rightarrow 8x=112\times 100$
Dividing both sides by 8, we will get,
$\begin{align}
& \Rightarrow x=\dfrac{112\times 100}{8} \\
& \Rightarrow x=1400 \\
\end{align}$
So, the required sum = Rs. 1400.
Note: Short method,
According to the question, for an increase in 2 % of rate of interest, simple interest will increase by Rs. 112. In the formula of simple interest, put the increase in rate of interest in the place of rate of interest. i.e. put $'\Delta r'$in place of ‘r’ and in the place of simple interest, put change in simple interest.
i.e. put $\Delta $(simple interest) in place of simple interest.
$'\Delta '$ is a symbol used for ‘change’.
Formula: $\text{simple interest }=\dfrac{P\times r\times t}{100}$
Let P = $Rs.x$
On putting $\Delta r$in place of r and $\Delta $ simple interest in place of simple interest,
$\begin{align}
& \Rightarrow \Delta \left( \text{simple interest} \right)\text{ }=\dfrac{x\times \Delta r\times t}{100} \\
& \Rightarrow +112=\dfrac{x\times \left( 2 \right)\times 4}{100} \\
& \Rightarrow 112\times 100=8x \\
& \Rightarrow x=\dfrac{112\times 100}{8} \\
& \Rightarrow x=Rs.1400 \\
\end{align}$
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