# 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) 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) 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

A) Rs.24.64 | B) Rs.21.85 |

C) Rs.16 | D) Rs.16.80 |

Explanation:

For 1st year S.I =C.I.

Thus, Rs.16 is the S.I. on S.I. for 1 year, which at 8% is thus Rs.200

i.e S.I on the principal for 1 year is Rs.200

Principle = $Rs.\frac{100*200}{8*1}$ = Rs.2500

Amount for 2 years, compounded half-yearly

$Rs.\left[2500*{\left(1+\frac{4}{100}\right)}^{4}\right]=Rs.2924.4$

C.I = Rs.424.64

Also, $S.I=Rs.\left(\frac{2500*8*2}{100}\right)=Rs.400$

Hence, [(C.I) - (S.I)] = Rs. (424.64 - 400) = Rs.24.64

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) 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) Rs.6000 | B) Rs.8000 |

C) Rs.12000 | D) none of these |

Explanation:

Compound Interest on P at 10% for 2 years when interest is compounded half-yearly

=$P{\left(1+\raisebox{1ex}{$\frac{R}{2}$}\!\left/ \!\raisebox{-1ex}{$100$}\right.\right)}^{2T}-P=P{\left(1+\frac{1}{20}\right)}^{4}-P=P{\left(\frac{21}{20}\right)}^{4}-P$

Simple Interest on P at 10% for 2 years = $\frac{PRT}{100}=\frac{P\times 10\times 2}{100}=\frac{P}{5}$

Given that difference between compound interest and simple interest = 124.05

$P*{\left(\frac{21}{20}\right)}^{4}-P-\frac{P}{5}=124.05$

=>$P\left[{\left(\frac{21}{20}\right)}^{4}-1-\frac{1}{5}\right]=124.05$

P=8000

A) 5430 | B) 4430 |

C) 3430 | D) 2430 |

Explanation:

Let each installment be Rs.x. Then,

(P.W. of Rs.x due 1 year hence) + (P.W of Rs.x due 2 years hence) + (P.W of Rs. X due 3 years hence) = 7620.

$\left[\frac{{\displaystyle x}}{{\displaystyle \left(1+\frac{50}{3x100}\right)}}\right]+\left[\frac{x}{1+{\left({\displaystyle \frac{50}{3x100}}\right)}^{2}}\right]+\left[\frac{{\displaystyle x}}{{\displaystyle 1+{\left(\frac{50}{3x100}\right)}^{3}}}\right]=7620$

$\frac{6x}{7}+\frac{36x}{49}+\frac{216x}{343}=7620$

$294x+252x+216x=7620x343$

=> x = 3430

Amount of each installment = Rs.3430