Arithmetic aptitude :: Compound Interest :: General Questions

• Compound Interest - General Questions
• Compound Interest - Data Sufficiency 1
• Compound Interest - Data Sufficiency 2
• Compound Interest - Data Sufficiency 3
• Compound Interest - Important Formulas and Facts

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

2. Rate compounded Annually :

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

3. Rate compounded Half-yearly:

Amount $=P\left[1+\frac{R/2}{100}\right]^{2n}$

4. Rate compounded Quarterly:

Amount $=P\left[1+\frac{R/4}{100}\right]^{4n}$

5. Rate compounded Annually but time is in fraction, say $3\frac{2}{5}$ years.

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

6. When Rates are different for different years, say $R_{1}\%$, $R_{2}\%$, $R_{3}%$ for $1^{st}$, $2^{nd}$ and $3^{rd}$ year respectively.

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

7. Present worth of Rs. $x$ due $n$ years hence is given by:

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

Sathish invests Rs. 35,500 in a scheme which earns him simple interest at 15% p.a for 2 years. At the end of 2 years he reinvests the total amount at 20% p.a. What will be the total interest earned by Satish over the principal at the end of 5 years ?

Answer:

Explanation:

$SI=$ $\frac{35,500\times 15}{100\times 2}$ $=10,650$

$A$ $=35,500+10,650$ $=46,150$

$CI$  for $3$ yrs  $= 46,150\left[\left(1+\frac{20}{100}\right)^{3}\right]$ $=33,597.20$

$\therefore$ Total interest $=10,650 + 33,597.20$ = 44,247.20

Population of a city decreases by 3% of the initial at the beginning of every year. If the present population is 1,25,000. What will be the population after 2 years ?

Answer:

Explanation:

$A$ $=P\left[\left(1-\frac{R}{100}\right)^{n}\right]$

=$125000\left(1-\frac{3}{100}\right)^{2}$

$=125000\times\frac{97}{100}\times\frac{97}{100}$ = 1,17,613

A sum of money is lent out at compound interest for 2 years at 20% per annum. It would fetch Rs. 482 more if the interest is compounded half yearly, than if it were compounded yearly, the sum of money is

Answer:

Explanation:

$P\left[\left(1+\frac{10}{100}\right)^{4}\right]$ $-P\left[\left(1+\frac{20}{100}\right)^{2}\right]$ $= 482$

$P[\left(1+\frac{10}{100}\right)^{4}$ $-\left(1+\frac{20}{100}\right)^{2}]$ $= 482$

$\Leftrightarrow P[\left(\frac{11}{10}\times\frac{11}{10}\times\frac{11}{10}\times\frac{11}{10}\right)$ $-\left(\frac{6}{5}\times\frac{6}{5}\right)]$ $=482$

$\therefore P$ = 20,000

A sum of money on compound interest amounts to Rs. 10,648 in 3 years and Rs. 9680 in 2 years. The rate of interest per annum is

Answer:

Explanation:

Let  Sum be $P$ and rate be $R\%$ per annum.

Then $P\left(1+\frac{R}{100}\right)^{2}$ $=9680$    ---------------------- (1)

$P\left(1+\frac{R}{100}\right)^{3}$ $=10648$   -----------------------------(2)

Divide equ. (2) by (1)

$\Rightarrow 1+\frac{R}{100}$ $=\frac{10648}{9680}$

$\frac{R}{100}$ $=\frac{1}{10}$ = 10%

The difference between simple and compound interest on a certain sum of money for 3 years at 10% p.a. is Rs. 15.50. Find the sum.

Answer:

Explanation:

The difference in C.I. and S.I. for 3 years is given by

$C.I.-S.I$ $=P\{\left(\frac{R}{100}\right)^{3}$ $+3 \left(\frac{R}{100}\right)^{2}\}$

$\Rightarrow 15 \frac{1}{2}$ $=P\{ \left(\frac{10}{100}\right)^{3}$ $+3 \left(\frac{10}{100}\right)^{2}\}$

$\Rightarrow \frac{31}{2}$ $=P\left[\frac{1}{1000}+\frac{3}{100}\right]$

$P$ $=\frac{31}{2}\times\frac{1000}{31}$ = Rs. 500.

• Compound Interest - General Questions
• Compound Interest - Data Sufficiency 1
• Compound Interest - Data Sufficiency 2
• Compound Interest - Data Sufficiency 3