# Compound Interest Questions

**FACTS AND FORMULAE FOR COMPOUND INTEREST QUESTIONS**

Let Principal = P, Rate = R% per annum, Time = n years.

**I.**

**1. When interest is compound Annually:**

Amount =$P{\left(1+\frac{R}{100}\right)}^{n}$

**2. When interest is compounded Half-yearly:**

Amount = $P{\left[1+\frac{\left({\displaystyle \raisebox{1ex}{$R$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right)}{100}\right]}^{2n}$

**3. When interest is compounded Quarterly:**

Amount = $P{\left[1+\frac{\left({\displaystyle \raisebox{1ex}{$R$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}\right)}{100}\right]}^{4n}$

**II.**

**1. When interest is compounded Annually but time is in fraction, say $3\frac{2}{5}$ years.**

Amount = $P{\left(1+\frac{R}{100}\right)}^{3}\times \left(1+\frac{{\displaystyle \frac{2}{5}}R}{100}\right)$

**2. When Rates are different for different years, say R1%, R2%, R3% for 1st, 2nd and 3rd year respectively.**

Then, Amount = $P\left(1+\frac{{R}_{1}}{100}\right)\left(1+\frac{{R}_{2}}{100}\right)\left(1+\frac{{R}_{3}}{100}\right)$

**III. Present worth of Rs. x due n years hence is given by:**

Present Worth = $\frac{x}{{\left(1+{\displaystyle \frac{R}{100}}\right)}^{n}}$

A) 2 years | B) 2.5 years |

C) 3 years | D) 4 years |

Explanation:

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

let the time be n years

Then,30000(1+7/100)^n = 34347

(107/100)^n = 34347/30000 = 11449/10000 = (107/100)^2

n = 2years

A) 4360 | B) 4460 |

C) 4560 | D) 4660 |

Explanation:

Let the sum be Rs.P.then

P(1+R/100)^3=6690…(i) and P(1+R/100)^6=10035…(ii)

On dividing,we get (1+R/100)^3=10025/6690=3/2.

Substituting this value in (i),we get:

P*(3/2)=6690 or P=(6690*2/3)=4460

Hence,the sum is rs.4460.

A) I only | B) II and III only |

C) All I, II and III | D) I only or II and III only |

Explanation:

$I.\frac{P*R*5}{100}=P\iff R=20$

$II.P{\left(1+\frac{R}{100}\right)}^{2}-P-\frac{P*R*2}{100}=400=p{R}^{2}=4000000$

$III.\frac{P*R*1}{100}=2000=PR=200000$

$\frac{P{R}^{2}}{PR}=\frac{4000000}{200000}\iff R=20$

Thus I only or (II and III) give answer.

Correct answer is (D)

A) 2109 | B) 3109 |

C) 4109 | D) 6109 |

Explanation:

Time = 2 years 4 months = 2(4/12) years = 2(1/3) years.

Amount = Rs'. [8000 X (1+(15/100))^2 X (1+((1/3)*15)/100)]

=Rs. [8000 * (23/20) * (23/20) * (21/20)]

= Rs. 11109. .

:. C.I. = Rs. (11109 - 8000) = Rs. 3109.

A) 2422 | B) 2522 |

C) 2622 | D) 2722 |

Explanation:

Principal = Rs. 16000; Time = 9 months =3 quarters;

Rate = 20% per annum = 5% per quarter.

Amount = Rs. [16000 x (1+(5/100))3] = Rs. 18522.

CJ. = Rs. (18522 - 16000) = Rs. 2522

A) 2.04 | B) 3.04 |

C) 4.04 | D) 5.04 |

Explanation:

C.I. when interest

compounded yearly=rs.[5000*(1+4/100)(1+1/2*4/100)]

= Rs. 5304.

C.I. when interest is

compounded half-yearly=rs.5000(1+2/100)^3

= Rs. 5306.04

Difference = Rs. (5306.04 - 5304) = Rs. 2.04

A) 612 | B) 712 |

C) 812 | D) 912 |

Explanation:

Amount = Rs [7500*(1+(4/100)2] = Rs (7500 * (26/25) * (26/25) ) = Rs. 8112.

therefore, C.I. = Rs. (8112 - 7500) = Rs. 612.

A) 10000 | B) 20000 |

C) 40000 | D) 50000 |

Explanation:

Let sum=Rs.x

C.I. when compounded half yearly = $\left[x{\left(1+\frac{10}{100}\right)}^{4}-x\right]=\frac{4641}{10000}$

C.I. when compounded annually =$\left[x{\left(\frac{20}{100}\right)}^{2}-x\right]=\frac{11}{25}$

$\frac{4641}{10000}x-\frac{11}{25}x=482$

=> x=20000