Question

Find the amount and the compound interest on Rs. $10,000$ for $1\dfrac{1}{2}$ years at $10\% $ per annum, compounded half-yearly. Would this interest be more than the interest he would get if it was compounded annually?

Answer 1

Hint: Here we will use the formula to find the amount A is $A = P{\left( {1 + \dfrac{R}{{100}}} \right)^n}$ where A is the amount, P is principal, R is the rate of interest and n is the time period. Here we will take $n = 3$ for the amount after one and half year, since the rate of interest is compounded half yearly. Also, in the second case where the interest is calculated annually, use both the formulas for the compound interest and simple interest $I = \dfrac{{PRT}}{{100}}$ .

Complete step-by-step answer:
Given that-
Principal, $P = \,{\text{Rs 1}}0,000$
Rate of interest, $R = 10\% $ per annum
$ \Rightarrow R = 5\% {\text{ }}\;per\,{\text{6 months}}$
Case (i) Interest is compounded half-yearly

Here, the interest is calculated half-yearly, $\therefore n = 18\;{\text{months = 3}}$
Place values in the formula –
$A = P{\left( {1 + \dfrac{R}{{100}}} \right)^n}$
$\therefore A = 10000 \times {\left( {1 + \dfrac{5}{{100}}} \right)^3}$
Take LCM (Least common multiple) and simplify the above equation –
$
   \Rightarrow A = 10000 \times {\left( {\dfrac{{100 + 5}}{{100}}} \right)^3} \\
   \Rightarrow A = 10000 \times {\left( {\dfrac{{105}}{{100}}} \right)^3} \\
 $
Convert the above fraction in the decimal form-
$ \Rightarrow A = 10000 \times {(10.5)^3}$
Simplify the above equation-
$ \Rightarrow A = 11,576.25{\text{ Rs}}{\text{.}}$
Now, the compound Interest $ = Amount{\text{ - Principal (Sum)}}$
Place the values
$
  C.I. = 11576.25 - 10000 \\
  C.I. = Rs.{\text{ 1576}}{\text{.25}}\,{\text{ }}.....{\text{ (A)}} \\
 $
Case (ii) Interest is compounded annually-
Rate of interest, $R = 10\% $ per annum
Term period, $T = 1\dfrac{1}{2}years,{\text{ }} \Rightarrow {\text{n = 1}}$
Amount, $A = P{\left( {1 + \dfrac{R}{{100}}} \right)^n}$
Place values in the above equation –
$ \Rightarrow A = 10000{\left( {1 + \dfrac{{10}}{{100}}} \right)^1}$
Simplify the above equation –
$ \Rightarrow A = 10000{\left( {1 + \dfrac{1}{{10}}} \right)^1}$
Take LCM (Least common multiple) and simplify
$ \Rightarrow A = 10000{\left( {\dfrac{{11}}{{10}}} \right)^1}$
Convert the fraction in the decimal form –
$ \Rightarrow A = 10000{\left( {1.1} \right)^1}$
Simplify the above equation –
$ \Rightarrow A = 11000\;{\text{ }}.....{\text{ (B)}}$
Now, Simple interest for last half year is –
$I = \dfrac{{PRT}}{{100}}$
Place values in the above equation, where $T = \dfrac{1}{2}year$ and $P = 11000\;{\text{Rs}}{\text{.}}$
$I = \dfrac{{11000 \times 10 \times \dfrac{1}{2}}}{{100}}$
Numerator’s denominator goes to denominator-
$I = \dfrac{{11000 \times 10 \times 1}}{{100 \times 2}}$
Simplify the above equation –
$ \Rightarrow I = 550{\text{ Rs}}{\text{.}}\;{\text{ }}....{\text{ (C)}}$
By using equations (B) and (C)-
The Net amount after \[1\dfrac{1}{2}{\text{year is}}\]
$
  {A_N} = 11000 + 550 \\
  {A_N} = 11550{\text{ Rs}}{\text{.}} \\
 $
Now, the compound Interest $ = Amoun{t_{Net}}{\text{ - Principal (Sum)}}$
Place the values
$
  C.I. = 11550 - 10000 \\
  C.I. = Rs.{\text{ 1550}}{\text{ }}....{\text{ (D)}} \\
 $
By comparing the equations (A) and (D), we can say that interest would be more in case it is compounded half yearly.

Note: Interest is the amount of money paid for using someone else’s money. There are two types of interest. $1)$ Simple Interest and $2)$ Compound interest. Know the difference between both the types of interest and apply them accordingly.