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

A sum of money becomes $\frac{41}{40}$ of itself in $\frac{1}{4}$ years at a certain rate of simple interest. The rate of interest per annum is

Answer:

Explanation:

Let the principal be Rs. $1$.

Hence, $S.I$ $=\frac{41}{40}-1$ $=\frac{1}{40}$

Rate % $=\frac{100\times S.I}{P\times N}$ $=\frac{\frac{1}{40}\times 100}{1\times\frac{1}{4}}$ = 10%

If in a certain number of years Rs. $3000$ amounts to Rs. $4320$ at compound interest, in half that time Rs. $3000$ will amount to

Answer:

Explanation:

Let rate of interest $=R\%$

Number of years $=N$

$4320$ $=3000\left(1+\frac{R}{100}\right)^{N}$

$\left(1+\frac{R}{100}\right)^{N/2}$ $=\sqrt{1.44}$ $=1.2$

In $\frac{n}{2}$ years Rs. $3000$ will amount to

$3000\left(1+\frac{R}{100}\right)^{N/2}$ $=3000\times 1.2$ = Rs. 3600.

A certain sum of money becomes three times of itself in 20 years at simple interest. In how many years does it become double of itself at the same rate of simple interest ?

Answer:

Explanation:

Let $P$ = Rs. $x$, $A=3x$, $I$ = Rs. $2x$

$SI$ $=\frac{PNR}{100}$ $\Rightarrow 2x$ $=\frac{x\times R\times 20}{100}$ $\Rightarrow R$ $= 10\%$

In second case :

$P$ = Rs. $x$, $A$ = Rs. $2x$, $I$ = Rs. $x$, $N$ = ?

$SI$ $=\frac{PNR}{100}$ $\Rightarrow 2x$ $=\frac{x\times N\times 10}{100}$ $\Rightarrow N$ = 10 years.

A person deposited Rs.400 for 2 years, Rs.550 for 4 years and Rs. 1,200 for 6 years. He received the total simple interest of Rs.1020. The rate of interest per annum is

Answer:

Explanation:

Let the rate of interest be $R\%$ per annum.

Given, $\frac{400\times 2\times R}{100}$ $+\frac{550\times 4\times R }{100}$ $+\frac{1200\times 6\times R}{100}$ $=1020$

$8R+22R+72R$ $=1020$ $\Rightarrow 102R =1020$ $\Rightarrow R$ = 10%.

The compound interest on a certain sum of money at 5% for 2 years is Rs. 328. The simple interest on that sum at the same rate and for the same period of time will be

Answer:

Explanation:

$C.I$ $=p\left[\left(1+ \frac{R}{100}\right)^{T}-1\right]$

$\Rightarrow p\left[\left(1+ \frac{5}{100}\right)^{2}-1\right]$

$\Rightarrow 328$ $=p\left(\frac{441}{400}-1\right)$

$\Rightarrow 328$ $=p\left(\frac{41}{100}\right)$

$\Rightarrow p$ $=\frac{328 \times 400}{41}$ $=3200$

$SI$ $=\frac{PRT}{100}$ $=\frac{3200\times 5\times 2}{100}$ = Rs. 320.

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