Aptitude - Compound interest online test

While calculating compound interest, the simple interest which is calculated for the principal of first year will add to the principal which we taken already. The addition of this amount will be the principle for the next year. For this amount simple is calculated for second year and added to principal for second year. It will go on increase like this. Finally principal is subtracted from last year amount. The result of this subtraction gives compound interest. It is denoted by CI.

1. If the interest is compounded annually, then

Amount = P [(1 + (R/100))] power n.

Compound interest = P [(1+ (R/100)) power n – 1]

Or CI = Amount – Principal

2. If interest is compounded half yearly, then

Amount = P[1 + (R/(2*100))] power 2n.

3. If interest is compounded quarterly, then

Amount = P[1 + (R/(4*100))]power 4n.

4. If interest is compounded annually but time is in fraction

Amount =P{1 + (R/100)}whole power t * {1 + [(p/q)R]/100]}

P = Principal; R% = Rate per annum; Time = n yr.

Previous year solved Questions:

1. What total amount will Ravali get in 2 yr. if she invests Rs. 5000 to obtain compound interest at the rate of 5 pcpa?

A) Rs. 5511.50 B) Rs. 5312.50 C) Rs. 5542.50 D) Rs. 5512.50 E) None of these

Ans: D

Solution: Given data P = 5000, R = 5%, n = 2yr.

Formulae: Amount = P{1 + (R/100)} power n

=> 5000{1 + (5/100)} power 2

=> 5000 * (21/20) power 2

=> Rs.5512.50

2. The population of a city is 126800. It increases by 15% in the first year and decreases by 20% in the second year. What is the population of the town at the end of 2 yr?

A) 174984 B) 135996 C) 116656 D) 145820 E) None of these

Ans: C

Solution: As per given data, P = 126800, R1 = 15% increase, R2 = 20% decrease

Population after 2 yrs. P {1 + (R1/100)}{1 – (R2/100)}

=> 126800 {1 + (15/100)} {1 – (20/100)}

= > 116656