Arithmetic aptitude :: Compound Interest :: Data Sufficiency 2

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

Find out the compound interest earned on an amount of Rs. $5000$ in $2$ years?

I. The simple interest on the same amount at the same rate of interest in $5$ years is Rs. $2000$.

II. The compound interest and the simple interest earned in $1$ year is the same.

III. The amount becomed more than double on compound interest in $10$ years.

Answer:

Explanation:

$P$ = Rs. $5000$ and $T=2$ years.

I. $S.I.$ on Rs. $5000$ in $5$ years is Rs. $2000$.

$\frac{5000\times R\times 5}{100}$ $=2000$ $\Rightarrow R=8$.

Thus I only gives the answer .

$\therefore$ Correct answer is (A).

A sum of money is placed at compound interest. In how many years will it amount to sixteen times of itself ?

I. The sum doubles itself in $4$ years.

II. The sum amounts to eight times of itself in $12$ years.

III. The sum amounts to four times of itself in $8$ years.

Answer:

Explanation:

I. $P\left(1+\frac{R}{100}\right)^{4}$ $=2P$ $\Rightarrow \left(1+\frac{R}{100}\right)^{4}$ $=2$ ---(i)

II. $P\left(1+\frac{R}{100}\right)^{12}$ $=8P$ $\Rightarrow \left(1+\frac{R}{100}\right)^{12}$ $=8$ ---(ii)

III. $P\left(1+\frac{R}{100}\right)^{8}$ $=4P$ $\Rightarrow \left(1+\frac{R}{100}\right)^{8}$ $=4$ ---(iii)

Let the given sum become $16$ times in $n$ years. Then,

$P\left(1+\frac{R}{100}\right)^{n}$ $=16P$

$\Rightarrow \left(1+\frac{R}{100}\right)^{n}$ $=16$ ---(iv)

$\therefore$ Any one of (i), (ii) and (iii) with (iv) will give the value of $n$.

$\therefore$ Correct answer is (E).

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