Courses
Courses for Kids
Free study material
Offline Centres
More
Store Icon
Store

A sum of money is invested at 10% per annum compounded half yearly. If the difference of amount at the end of 6 months and 12 months is Rs.189, find the sum of money invested.

seo-qna
Last updated date: 24th Jul 2024
Total views: 424.5k
Views today: 6.24k
Answer
VerifiedVerified
424.5k+ views
Hint: Assuming the sum of money be Rs.y and at rate 10% per annum compounded half yearly. Calculate amount for the first six months, which is the equation of variable y and then calculating amount for first 12 months i.e. 1 year, which again is an equation of variable y. Substituting these two equations would give the value of y, actual sum of money.

Complete step-by-step answer:
Before solving the question, we must first understand the meaning of interest and compound interest. Interest is the money paid regularly at a particular rate for the use of money lent, or for delaying the repayment of a debt.
Compound interest is the addition of interest to the principal sum of a loan or deposit or in other words, interest on interest.
In compound interest, P = Principal, initial amount of money taken.
R = Rate % per annum
n = time in years.
When interest is compounded annually,
\[\text{Amount=P}{{\left( 1+\dfrac{R}{100} \right)}^{n}}\]
When interest is compounded half yearly,
\[\text{Amount=P}{{\left( 1+\dfrac{\dfrac{R}{2}}{100} \right)}^{2n}}\]
Now, according to the question:
Let the sum of money be Rs. y and given rate = 10% per annum compounded half yearly.
For first six months (half year, $ n=\dfrac{1}{2} $ )
\[\text{A=P}{{\left( 1+\dfrac{\dfrac{R}{2}}{100} \right)}^{2n}}\]
Here, P = y
\[A=y{{\left( 1+\dfrac{\left( \dfrac{10}{2} \right)}{100} \right)}^{2\times \dfrac{1}{2}}}=y\left( \dfrac{21}{20} \right)\]
For first 12 months (1 year, n=1)
\[\begin{align}
  & \text{A=P}{{\left( 1+\dfrac{\dfrac{R}{2}}{100} \right)}^{2n}}\text{ (Considering P=y)} \\
 & \Rightarrow \text{y}{{\left( 1+\dfrac{10}{100} \right)}^{2\times 1}}=y{{\left( \dfrac{21}{20} \right)}^{2}}=y\left( \dfrac{441}{400} \right) \\
\end{align}\]
Given the difference between the above amounts is Rs.189
\[\begin{align}
  & \Rightarrow y\left( \dfrac{441}{400} \right)-y\left( \dfrac{21}{20} \right)=189 \\
 & \Rightarrow \left( \dfrac{21}{400} \right)y=189 \\
 & \Rightarrow y=\dfrac{189\times 400}{21} \\
 & \Rightarrow y=3600 \\
\end{align}\]
Therefore, the sum of money invested at 10% per annum compounded half yearly is Rs.3600

Note: Students might get confused while keeping the values of n time, which is in year. For half year n should be $ \dfrac{1}{2} $ and for 1 year n should be 1. Since, students are much more familiar with simple interest formula i.e. $ \text{SI=}\dfrac{\text{P}\times I\times R}{100} $ might use this formula but this would lead to wrong results. Students must use the CI formula as stated here.