Compound Interest Questions and Answers

Introduction

With the help of Nithra Jobs solve Compound interest problems. Keep on practicing compound interest questions and answers regularly by using the compound interest formula to be selected easily in competitive exams.

Formulas to use:

Interest is Compounded Annually

Amount = P (1+ (r/100)) ⁿ

Compound Interest = Total amount - Principal

Interest is Compounded Half-Yearly

Amount = P (1 + ((r/2) / 100)) ²ⁿ

Compound Interest = Total amount - Principal

Interest is Compounded Quarterly

Amount = P (1 + ((r/4) / 100)) ⁴ⁿ

Compound Interest = Total amount - Principal

Interest is Compound Monthly

Amount = P (1 + (r / 12 / 100) ¹²ⁿ

Interest is Compounded Annually but Time is in Fraction, say 2(3/2) years Amount = P (1 + r/100) ² × (1 + (3/2)r/100)

CI when Rates are Different for Different Years Amount = P (1 + r1/100) (1 + r2/100) (1 + r3/100)

1. Rohit invested an amount of Rs. 40000 for 2 years at compound interest at the rate of 6 % per annum. Find the amount he receives at the end of 2 years?

Answer: Rs. 44944

Explanation:

Principal P = Rs. 40000 Rate of Interest r = 6 %

Number of years n = 2 Amount = P ( 1 + ( r / 100 ) )ⁿ

Amount = 40000 × ( 1 + ( 6 / 100 ))²

= 40000 × ( 1 + ( 3 / 50 ))²

= 40000 × ( 53 / 50 )²

= 40000 × ( 53 / 50 ) × ( 53 / 50 )

= ( 400 × 53 × 53 ) / 25

= 1123600 / 25

= 44944

The amount that Rohit receives at the end of 2 years = Rs. 44944

2. If the compound interest on a certain sum for 2 years at 15% per annum is 258. Find the Principal?

Answer: Rs. 800

Explanation:

Let the Principal is P

Compound interest = Amount - Principal

So compound interest = P [ 1 + ( 15 / 100 )]² - P 258 = P [ ( 100 + 15 ) / 100 ]² - P

258 = P ( 115 / 100) × ( 115 / 100 ) - P

258 = P ( 23 / 20) × ( 23 / 20) - P

258 = P ( 529 / 400 ) - P

258 = P [ ( 529 - 400 ) / 400 ]

258 = P ( 129 / 400 )

( 258 × 400 ) / 129 = P

2 × 400 = P

P = Rs. 800

3. A sum amounts to Rs. 2704 in 2 years at 4% compound interest. The sum is?

Answer: Rs. 2500

Explanation:

Let the sum be P. R = 4 %

n = 2

Amount = Rs. 2704

Amount = P [ 1 + ( R / 100 )]n 2704 = P ( 1 + ( 4 / 100) )²

2704 = P [ ( 100 + 4 ) / 100 ]²

2704 = P ( 104 / 100) × ( 104 / 100 )

2704 = P ( 26 / 25 ) × ( 26 / 25 )

P = ( 2704 × 25 × 25 ) / ( 26 × 26 )

= ( 67600 × 25 ) / 676

= 1690000 / 676

= 2500

Principal = Rs. 2500

4. What compound interest will be obtained on an amount of Rs. 5000 at the rate of 12 % p.a. in 2 years?

Answer: Rs. 1272

Explanation:

Principal = Rs. 5000 R = 12 %

n = 2 years

Compound interest = Amount - Principal

Compound Interest = Principal × [ 1 + ( R / 100 )]ⁿ - Principal

= [ 5000 × [ 1 + ( 12 / 100 )]² ] - 5000

= [ 5000 × [ ( 100 + 12 ) / 100 ]² ] - 5000

= [ 5000 × ( 112 / 100) × ( 112 / 100 ) ] - 5000

= [ 5000 × ( 56 / 50 ) × ( 56 / 50 ) ] - 5000

= ( 2 × 56 × 56 ) - 5000

= Rs. 1272

Compound Interest = Rs. 1272

5. Simple interest on a certain sum of money for 3 years at 4% per annum is 1200. Compound interest on the same amount of money for 2 years is?

Answer: Rs. 816

Explanation:

Given, Time = 3 Rate = 4%

Simple interest = 1200

Formula, Simple interest = ( Principal × Time × Rate ) / 100 So, 1200 = ( Principal × 3 × 4 ) / 100

P = ( 1200 × 100 )/12

= 100 × 100

P = 10000

For compound interest, Rate = 4, Principal = 10000 Compound interest = Amount - Principal

Formula, Compound interest = P × [ 1 + ( R/100 ) ]ⁿ - P

= 10000 × [ 1 + ( 4 / 100 ) ]² - 10000

= 10000 × (104 / 100 ) × ( 104 / 100 ) - 10000

= 104 × 104 - 10000

= 10816 - 10000

= 816

Compound interest = Rs. 816

Conclusion

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