Question

The difference in simple interest and compound interest on a certain sum of money in 2 years at 18 % p.a. is Rs. 162. The sum is

Answer 1

Answer:

The difference in simple interest and compound interest on a certain sum of money in 2 years at 18 % p.a. is Rs. 162. The sum is.

Answer 2

Answer:

We can find the sum value using the given years (2), rate of interest (18%) and the difference of simple interest and compound interest (Rs.162).

The sum value is: $Rs.5000$

Step-by-step explanation:

Simple Interest  $=\frac{PRT}{100}$

Here $P$ → Principle, $R$ →Rate of Interest, $T$ → Number of years.

Compound Interest $=[P(1+\frac{R}{100} )^{T}]-P$

Here also $P$ → Principle, $R$ →Rate of Interest, $T$ → Number of years.

From the given question, difference in simple interest and compound interest is $Rs.162$. It should be written as,

compound interest - simple interest $= 162$

               $\left \{ [P(1+\frac{R}{100} )^{T}]-P \right\}-\frac{PRT}{100}=162$

Now apply $R=18$%, $T=2$ in the above equation.

          $\left \{ [P(1+\frac{18}{100} )^{2}]-P \right\}-(\frac{P*18*2}{100})=162$

Take L.C.M for $1$ and $\frac{18}{100}$. It can be written as,

           $\left \{ [P(\frac{100+18}{100} )^{2}]-P \right\}-(\frac{P*18*2}{100})=162$

                $\left \{ [P(\frac{118}{100} )^{2}]-P \right\}-(\frac{P*18*2}{100})=162$

                 $\left \{ [P(1.18 )^{2}]-P \right\}-(\frac{P*36}{100})=162$    

                         $1.3924P-P-0.36P=162$

                                $1.3924P-1.36P=162$

                                             $0.0324P=162$

                                                       $P=\frac{162}{0.0324}$

Multiply by $10000$ in both numerator and denominator.

                                                       $P=\frac{1620000}{324}$

                                                       $P=5000$

Final Answer: Sum Value or Principle $=Rs.5000$