Sum
Find the compound interest on Rs.10000 for 2 years at 8% per annum compounded half-yearly.
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Solution
Principal P = ₹ 10,000
Rate of interest R = 8% p.a. compounded half-yearly
Duration T = 2 years
A = P`(1 + (("R"/2))/100)^"2T"`
= `10000(1 + (8/2)/100)^4`
= `10000(1 + 4/100)^4`
= 10000(1.04)4
= 11698.58
I = A – P
= 11648.58 – 10000
= 1698.58
∴ Compound interest is ₹ 1698.58.
Concept: Simple and Compound Interest (Entrance Exam)
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Chapter 9: Commercial Mathematics - Exercise 9.3 [Page 130]
Q 5Q 4Q 6
APPEARS IN
Balbharati Mathematics and Statistics 2 (Commerce) 11th Standard HSC Maharashtra State Board
Chapter 9 Commercial Mathematics
Exercise 9.3 | Q 5 | Page 130
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The compound interest on Rs $10000$ in $2$ years at $4%$ per annum being compounded half yearly is
A. $Rs\text{ 832}\text{.24}$
B. $Rs\text{ 828}\text{.82}$
C. $Rs\text{ 824}\text{.32}$
D. $Rs\text{ 912}\text{.86}$
Answer
Verified
Hint: First we recall the definition and formula of compound interest and then calculate the compound interest. The formula used to calculate the compound interest is
Compound interest = Amount – Principal
And \[\text{Amount =}P{{\left( 1+\dfrac{R}{100} \right)}^{T}}\]
Where, $P=$ Principal
\[R=\] Rate of interest
$T=$Time period
Complete step by step answer:
Now, we have given that Principal sum $=10,000$
Rate of interest $=4%$ per annum
Time
period \[=2\text{ years}\]
We have given that the compound interest being compounded half yearly, so the time period will be $4\text{ years}$and the rate of interest will be half i.e. $2%$ because when interest is compounded half yearly the rate of interest will be $\dfrac{R}{2}$.
Now, we have to calculate the Amount, so we put all values in the formula
\[\text{Amount =}P{{\left( 1+\dfrac{R}{100} \right)}^{T}}\]
$\Rightarrow 10000{{\left( 1+\dfrac{2}{100} \right)}^{4}}$
$\begin{align}
&
\Rightarrow 10000{{\left( 1+\dfrac{1}{50} \right)}^{4}} \\
& \Rightarrow 10000\times \left( \dfrac{51}{50} \right)\times \left( \dfrac{51}{50} \right)\times \left( \dfrac{51}{50} \right)\times \left( \dfrac{51}{50} \right) \\
& \Rightarrow 10.2\times 10.2\times 10.2\times 10.2 \\
& \Rightarrow 10824.32 \\
\end{align}$
The Amount will be Rs. $10824.32$
Now we have to calculate compound interest.
We know that Compound interest = Amount –
Principal
Putting the values, Compound interest will be
$\begin{align}
& =10824.32-10000 \\
& =824.32 \\
\end{align}$
So, the compound interest on Rs $10000$ in $2$ years at $4%$ per annum being compounded half yearly is $Rs.824.32$.
So, the correct answer is “Option C”.
Note: Compound interest is interest on interest; it means compound interest is additional amount of interest to the principal sum. Before calculating compound interest students have to calculate the amount by using the formula and then subtract principal from amount. Students must read questions carefully about the compounding frequency i.e. interest compounded yearly, half-yearly, quarterly, monthly or weekly. The time period will be changed accordingly.
Solution:
Here, Principal (P) = Rs. 10000, Rate of Interest (R) = 10% = 5% (compounded half yearly)
Time(n) = 1\ \frac{1}{2} years = 3 years (compounded half yearly)
Amount (A) = P\left(1+\frac{R}{100}\right)^n
= 10000\left(1+\frac{5}{100}\right)^3
= 10000\left(1+\frac{1}{20}\right)^3
= 10000\left(\frac{21}{20}\right)^3
= 10000\times\frac{21}{20}\times\frac{21}{20}\times\frac{21}{20}
= Rs. 11,576.25
Compound Interest (C.I.) = A – P
= Rs. 11,576.25 – Rs. 10,000 = Rs. 1,576.25
If it is compounded annually, then
Here, Principal (P) = Rs. 10000, Rate of Interest (R) = 10%, Time (n) = 1\ \frac{1}{2} years.
Amount (A) for 1 year = P\left(1+\frac{R}{100}\right)^n
= 10000\left(1+\frac{10}{100}\right)^1
= 10000\left(1+\frac{1}{10}\right)^1
= 10000\left(\frac{11}{10}\right)^1
= 10000\times\frac{11}{10}
= Rs. 11,000
Interest for \frac{1}{2} year = \frac{11000\times1\times10}{2\times100}=RS.\ 550
\therefore Total amount = Rs. 11,000 + Rs. 550 = Rs. 11,550
Now, C.I. = A – P = Rs. 11,550 – Rs. 10,000
= Rs. 1,550
Yes, interest Rs. 1,576.25 is more than Rs. 1,550.