# How To Solve Compound And Simple Interest Problems Quickly

## How To solve Compound Interest And Simple Interest Problems Quickly

The Basic Difference Between Simple Interest (SI) and Compound Interest(CI) is that the Simple Interest is the interest calculated on the Principal or sum of amount while Compound Interest is calculated on the principal amount and the previous earned interest also i.e.

### Definition of Simple and Compound Interest

• Compound Interest is the interest charged on the original principal and on the accumulated past interest of a deposit is known as Compound interest.
• Simple Interest is the interest When some money is borrowed by someone, then borrower is required to pay an additional amount of money other than the original sum. This additional amount of money is called interest.

### Type 2: Solve Simple Interest and Compound Interest Quickly.  Find the amount/time/rate of interest when CI or SI or their difference is given

Question 1. The difference between the CI and SI on a certain amount is at 10% p.a. for 3 years is Rs. 31. Find the principal?

Options

A. 1000

B. 3100

C. 310

D. 100

Solution    The difference between compound interest and simple interest for three years is 31.

Difference = $\frac{P × (R)^2}{(100)^2} × \frac{(300 + R) }{100}$

31 = $\frac{P × (10)^2}{(100)^2} × \frac{(300 + 10) }{100}$

31 = $P × \frac{31 }{1000}$

On solving further, we get

P = 1000

Correct option: A

Question 2. If the SI on a sum of money for 2 years at 5% p.a. is Rs. 500, what is the CI on the same sum at the same rate and for the same time?

Options

A. Rs. 512.5

B. Rs 521.5

C. Rs 515.2

D. Rs 215.5

Solution      Sum = $\frac{500 × 100 }{2 × 5}$

Sum = $\frac{50000 }{10}$

Sum = 5000

Amount =  $5000 (1+ \frac{5}{100})^{2}$

5000 × 1.05 × 1.05 = 5512.5

CI = 5512.5 – 5000

CI = Rs. 512.5

Correct option: A

Question 3. The difference between CI and SI on a principal of Rs. 15,000 for two years is Rs. 24. What is the annual rate of interest?

Options

A. 16%

B. 4%

C. 8%

D. 6%

Solution      CI – SI = $\frac{P × (R)^2}{(100)^2}$

On solving further we get,

24 × $(100)^{2}$= 15000 × R2

R2 = 16

R = 4%

Correct option: B

Question 4. The compound interest on Rs. 30,000 at 7% per annum is Rs. 4347. The period (in years) is:
A. 2
B.1/2
C. 3
D.4

Solution-  Amount = Rs. (30000 + 4347) = Rs. 34347.

Let the time be n years.

then,30000 $\left ( 1+\frac{7}{100} \right )^n$ = 34347

$\left ( \frac{107}{100} \right )^n$ = $\frac{34347}{30000}$

$\left ( \frac{107}{100} \right )^2$

Correct option: A

Question 5. There is 60% increase in an amount in 6 years at simple interest. What will be the compound interest of Rs. 12,000 after 3 years at the same rate?

A.Rs. 2160

B.Rs. 3120

C.Rs. 3972

D.Rs. 6240

Solution

Let P = Rs. 100. Then, S.I. Rs. 60 and T = 6 years.

$R= \left ( \frac{100\times60}{10\times6} \right )$= 10% p.a.
Now, P = Rs. 12000. T = 3 years and R = 10% p.a

C.I = Rs $\left [ 12000\times\left \{ \left ( 1+\frac{10}{100} \right )^3-1 \right \} \right ]$
=Rs$\left ( 12000\times\frac{331}{1000} \right )$

=3972

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