Compound Interest Questions and Answers

Answer: D

Apply each year's change as a multiplier in turn; a decrease multiplies by less than 1.

\[ 50000 \times 1.25 \times 0.70 \times 1.40 \]

\[ 1.25 \times 0.70 = 0.875, \qquad 0.875 \times 1.40 = 1.225 \]

\[ 50000 \times 1.225 = 61{,}250 \]

Adding the percentages (\( +25 - 30 + 40 = +35\% \)) would give 67,500. Successive percentage changes never simply add, because each acts on a different base.

Answer: A

Let the rate of interest be R% p.a. 

(49500)(3)(R)/100 = 17820 

495R = 5940 

R = 12 

CI on Rs.49500 at 12% p.a. for three years 

= 49500{[1 + 12/100]³ - 1} = 69543.936 - 49500 = Rs.20043.94 

when rounded off to two decimal places.

Answer: D

The sum more than doubles once \( (1.25)^{n} > 2 \).

\[ (1.25)^{2} = 1.5625, \qquad (1.25)^{3} = 1.953125 \]

\[ (1.25)^{4} = 2.4414 \]

The third year still falls just short of doubling, so the answer is 4 years.

The near miss at year 3 - 1.953, only 2.3% below 2 - is exactly what the question is testing.

Answer: B

\[ A = P\left(1 + \frac{R}{100}\right)^{n}, \qquad \text{CI} = A - P, \qquad \text{SI} = \frac{PRn}{100} \]

Half-yearly compounding at 5% per annum means 2.5% per half-year. The January deposit earns for two half-years and the July deposit for one.

\[ A = 1600(1.025)^{2} + 1600(1.025) \]

\[ = 1600(1.050625) + 1640 = 1681 + 1640 = \text{Rs. } 3321 \]

\[ \text{interest} = 3321 - 3200 = \text{Rs. } 121 \]

Each deposit must be tracked from its own date; treating the full Rs. 3200 as present all year would give Rs. 162.

Answer: B

The simple interest for the first year is 10200/2 is Rs.5100 and compound interest for first year also is Rs.5100. The compound interest for second year on Rs.5100 for one year 

So rate of the interest = (100 * 1530)/ (5100 * 1) = 30% p.a. 

So P = (100 * 10200)/ (30 * 2) = Rs.17000

Answer: A

\[ A = P\left(1 + \frac{R}{100}\right)^{n}, \qquad \text{CI} = A - P, \qquad \text{SI} = \frac{PRn}{100} \]

Recover the principal from the simple interest.

\[ 50 = \frac{P \times 5 \times 2}{100} = 0.1P \quad\Longrightarrow\quad P = \text{Rs. } 500 \]

\[ \text{CI} = 500\left[(1.05)^{2} - 1\right] = 500(1.1025 - 1) \]

\[ = \text{Rs. } 51.25 \]

The Rs. 1.25 excess is 5% of the first year's Rs. 25 of interest - the only difference compounding makes over two years.

Answer: A

Over two years the gap between compound and simple interest is

\[ \text{CI} - \text{SI} = P\left(\frac{R}{100}\right)^{2} \]

\[ 48 = P(0.08)^{2} = 0.0064P \]

\[ P = \frac{48}{0.0064} = \text{Rs. } 7500 \]

Checking: SI = Rs. 1200 and CI = \( 7500(1.1664 - 1) = 1248 \) - exactly Rs. 48 more.

Answer: C

\[ A = P\left(1 + \frac{R}{100}\right)^{n}, \qquad \text{CI} = A - P, \qquad \text{SI} = \frac{PRn}{100} \]

Find the rate from the simple-interest statement first.

\[ \frac{R \times 6}{100} = 0.60 \quad\Longrightarrow\quad R = 10\% \]

Now compound Rs. 12,000 for three years at that rate:

\[ \text{CI} = 12000\left[(1.1)^{3} - 1\right] = 12000(1.331 - 1) \]

\[ = 12000 \times 0.331 = \text{Rs. } 3972 \]

Simple interest for the same three years would be Rs. 3600; the extra Rs. 372 is the compounding.

Answer: B

Let A be the amount received at the end of the three years. 

A = 4800[1 + 10/100][1 + 20/100][1 + 25/100] 

A = (4800 * 11 * 6 * 5)/(10 * 5 * 4) 

A = Rs.7920 

So the interest = 7920 - 4800 = Rs.3120

Answer: D

\[ A = P\left(1 + \frac{R}{100}\right)^{n}, \qquad \text{CI} = A - P, \qquad \text{SI} = \frac{PRn}{100} \]

Prakash's money compounds for 17 years and Prashant's for 19, and both end equal:

\[ x(1.04)^{17} = y(1.04)^{19} \quad\Longrightarrow\quad x = y(1.04)^{2} = 1.0816\,y \]

Prashant's money has two extra years to grow, so he needs the smaller share.

\[ 1.0816y + y = 1301 \quad\Longrightarrow\quad 2.0816y = 1301 \]

\[ y = \text{Rs. } 625 \]

Prakash therefore gets Rs. 676, and indeed \( 625 \times 1.0816 = 676 \).