Arithmetic aptitude :: Compound Interest :: Data Sufficiency 3

• 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 at the end of $3$ years?

I. $S.I.$ earned on that amount at the same rate and for the same period is Rs. $4500$.

II. The rate of interest is $10$ p.c.p.a.

III. $C.I.$ for $3$ years is more than the simple interest for that period by Rs. $465$.

Answer:

Explanation:

I. gives, $S.I$ for $3$ years = Rs. $4500$.

II. gives, Rate $=10\%$ p.a.

III. gives, $(C.I.)-(S.I.)$ = Rs. $465$.

Clearly, using I and III we get $C.I.$ = Rs. $(465+4500)$.

Thus, II is redundant.

Also, from I and II, we get sum $=\left(\frac{100\times 4500}{10\times 3}\right)$ $=15000$.

Now $C.I.$ on Rs. $15000$ at $10\%$ p.a. for $3$ years may be obtained.

Thus, III is redundant.

$\therefore$ Either II or III is redundant.

What is the rate of interest per annum ?

I. The amount becomes Rs. $11,025$ with compound interest after $2$ years.

II. The same amount with simple interest becomes Rs. $11,000$ after $2$ years.

III. The amount invested is Rs. $10,000$.

Answer:

Explanation:

I gives, Amount after $2$ years = Rs. $11025$, when compounded.

II gives, Amount after $2$ years at $S.I.$ = Rs. $11000$.

III gives, Principal = Rs. $10000$.

From II and III, we have :

Principal = Rs. $10000$, $S.I.$ = Rs. $(11000-10000)$ = Rs. $1000$ and Time = $2$ years.

Hence, Rate can be calculated.

$\therefore$ I is redundant.

From I and III, we get $11025$ $=10000\times \left(1+\frac{R}{100}\right)^{2}$. This gives $R$.

$\therefore$ II is redundant.

From I and II, we have

$P\left(1+\frac{R}{100}\right)^{2}$ $=11025$ ---(i)

and $P\left[1+\frac{R\times 2}{100}\right]$ $=11000$ ---(ii)

On dividing (i) by (ii), we get $\frac{\left(1+\frac{R}{100}\right)^{2}}{(50+R)}$ $=\frac{11025}{550000}$.

This gives $R$.

Thus, III is redundant.

Hence I or II or II is redundant.

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