Arithmetic aptitude :: Compound Interest :: Data Sufficiency 1

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

What was the total compound interest on a sum after $3$ years ?

I. The interest after one year was Rs. $100$ and the sum was Rs. $1000$.
II. The difference between simple and compound interest on a sum of Rs. $1000$ at the end of $2$ years was Rs. $10$.

Answer:

Explanation:

I. gives : $P$ = Rs. $1000$ and $S.I$ for $1$ year = Rs. $100$.

$\therefore$ Rate $=\frac{100\times S.I.}{P\times T}$

$=\frac{100\times 100}{1000\times 1}$ $=10\%$ p.a.

Thus, $P$ = Rs. $1000$,$T=3$ years and $R=10%$ p.a.

$\therefore$ $C.I.$ may be obtained.

II. Sum = Rs. $1000$,$(C.I.)-(S.I.)$ for $2$ years = Rs. $10$.

Let the rate be $R\%$ p.a.

$1000\times\left[\left(1+\frac{R}{100}\right)^{2}-1\right]$ $-\left(\frac{1000\times R\times 2}{100}\right)$ $=10$.

From this, we can find $R$.

Thus $P$, $T$ and $R$ are given and therefore, $C.I$ may be calculated.

Thus, I alone as well as II alone is sufficient to get the answer.

$\therefore$ Correct answer is (C).

What is the compund interest earned by James at the end of $2$ years ?

I. Simple interest at the same rate for one year is Rs. $1020$ and the rate of interest is $12$ p.c.p.a

II. The amount invested is Rs. $8500$.

Answer:

Explanation:

I. $S.I.$ = Rs. $1020$, $R$ $=12\%$ p.a and $T$ = $1$ year.

$\therefore P$ $=\frac{100\times S.I.}{R\times T}$

$\Rightarrow P$ = Rs. $\left(\frac{100\times 1020}{12\times 1}\right)$ = Rs. $8500$.

$\therefore$ $C.I$ for $2$ years = Rs. $\left[8500\times \left\{\left(1+\frac{12}{100}\right)^{2}-1\right\}\right]$

II. gives : only $P$ and $T$

$\therefore$ II alone does not give the answer.

$\therefore$ Correct answer is (A).

What was the rate of interest on a sum of money ?

I. The sum fetched a total of Rs. $2522$ as compound interest at the end of $3$ years.

II. The difference between the simple interest and the compound interest at the end of $2$ years at the same rate was Rs. $40$

Answer:

Explanation:

I gives : $C.I$ for $3$ years = Rs. $2522$.

II gives : $(C.I)-(S.I)$ for $2$ years at same rate is Rs. $40$.

$\left[\left(1+\frac{R}{100}\right)^{3}-1\right]$ $=2522$ ---(i)

$\left[\left(1+\frac{R}{100}\right)^{2}-1\right]$ $-\frac{P\times R\times 2}{100}$ $=40$ ---(ii)

On dividing (i) by (ii) we get :

$\frac{\left(1+\frac{R}{100}\right)^{3}-1}{\left(1+\frac{R}{100}\right)^{2}-1-\frac{R}{50}}$ $=\frac{2522}{40}$

$\Rightarrow \frac{\frac{R^{3}}{1000000}+\frac{3R}{100}+\frac{3R^{2}}{10000}}{\frac{R{2}}{10000}}$ $=\frac{1261}{20}$

$\Rightarrow \frac{R}{100}$ $+\frac{300}{R}$ $=\frac{1201}{20}$ $\Rightarrow R^{2}$ $-6005R$ $+30000$ $=0$

$\Rightarrow R^{2}-6000R$ $-5R+30000$ $=0$

$\Rightarrow R(R-6000)$ $-5(R-6000)$ $=0$

$\Rightarrow (R-5)(R-6000)$ $=0$

$\Rightarrow R=5$

$\therefore$ Both I and II are needed to get $R$.

$\therefore$ Correct answer is (E).

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