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

The effective annual rate of interest corresponding to nominal rate of 6% per annum payable half-yearly is

Answer:

Explanation:

Amount of Rs. $100$ for $1$ yrs when $CI$ half yearly $=\left[100\left(1+\frac{3}{100}\right)^{2}\right]$

$=106.09$

$\therefore$ Effective Rate $=106.09-100$ = 6.09 %

The sum of money at compound interest doubles itself in 5 years. It will amount to eight times itself at the same rate of interest in

Answer:

Explanation:

Suppose sum $=P$ ;Rate $=R\%$ ; $T=5$ yrs ; $A=2P$

$2P$ $=\left[P\left(1+\frac{R}{100}\right)^{5}\right]$

$1+\frac{R}{100}$ $=2\times\frac{1}{5}$

Now let amount becomes $8$ times in $x$ years

$8P$ $=\left[P\left(1+\frac{R}{100}\right)^{N}\right]$

$\Rightarrow 8$ $=\left(1+\frac{R}{100}\right)^{N}$

$\Rightarrow N$ = 15 Yrs

What annual payment will discharge a debt of Rs. 50,440 due in 3 years at 5% per annum compounded annually ?

Answer:

Explanation:

Let each installment = Rs. $a$

Debt Amount $=a[\left(\frac{100}{100+R}\right)$ $+\left(\frac{100}{100+R}\right)^{2}$ $+ ……………$ $+\left(\frac{100}{100+R}\right)^{T}]$

$\Rightarrow a[\left(\frac{100}{105}\right)$ $+\left(\frac{100}{105}\right)^{2}$ $+\left(\frac{100}{105}\right)^{3}]$ $=50,440$

$\Rightarrow a\times\frac{20}{21}\left[1+\frac{20}{21}+\frac{400}{441}\right]$ $=50,440$

$\Rightarrow a=50,440$ $\times\frac{441}{1,261}$ $\times\frac{21}{20}$ = Rs. 18,552.

$\therefore$ Annual payment = Rs. 18,552.

The compound interest on Rs. 1800 at a10% per annum, for certain period of time is Rs. 378. Find the same time in years.

Answer:

Explanation:

$P=1800$, $CI=378$

Time $=N$

$A$ $=1800+378$ $=2178$

$2178$ $=1800\left(1+\frac{10}{100}\right)^{N}$

N = 2 yrs

Simple interest on a sum of money for 3 years at $6\frac{1}{4}\%$ per annum is Rs. 2,400. What will be the compound interest on that sum at rate for the same period ?

Answer:

Explanation:

Let the sum be $P$

$\frac{(P\times\frac{25}{4}\times 3)}{100}$ $= 2400$

$P=12800$

Now $P$ $= 12800$, $N=3$, $R=6\frac{1}{4}\%$

$=12800\left[\left(1+\frac{25}{400}\right)^{3}\right]$

$=12800\times\frac{17}{16}\times\frac{17}{16}\times\frac{17}{16}$

$=15353.13$

$\therefore$ $CI$ $=15353.13-12800$ = 2553.13

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