Compound Interest Questions and Answers

Answer: D

The rate is 8% per annum.

For compound interest compounded annually, \(A = P(1 + r)^{2}\):

\[ \begin{aligned} (1 + r)^{2} &= \frac{5832}{5000} = 1.1664 \\ 1 + r &= 1.08 \\ r &= 8\% \end{aligned} \]

Check: 5000 × 1.08 = 5400 after one year, and 5400 × 1.08 = 5832 after two. At 10% the amount would be 6050, and at 5% or 7% it would be 5512.50 or 5724.50.

Answer: B

The tree was 375 cm tall when planted.

Growing by 20% a year for two years multiplies the height by \(1.2^{2} = 1.44\), so

\[ h = \frac{540}{1.44} = 375\ \text{cm} \]

Check: 375 rises to 450 after one year and to 540 after two. 324 cm comes from removing 40% of 540, two 20% cuts added together; undoing a rise needs division by 1.2 for each year, not a 20% cut. 432 cm is 540 less a single 20%, and 324 mm changes the unit.

Answer: B

The sum is ₹ 12,600.

For 2 years, the difference between compound and simple interest is the interest on the first year's interest:

\[ \begin{aligned} CI - SI &= P \left(\tfrac{R}{100}\right)^{2} \\ 126 &= P \times 0.01 \\ P &= 12600 \end{aligned} \]

Check: on ₹ 12,600 at 10%, simple interest for 2 years is ₹ 2,520 and compound interest is ₹ 2,646, a difference of ₹ 126.

Answer: A

C.I. when interest is compounded yearly 

= [5000 * (1 + 4/100) * (1 + (1/2 * 4)/100] 

= 5000 * 26/25 * 51/50 = Rs. 5304 

C.I. when interest is compounded half-yearly 

= [5000 * (1 + 2/100)²] 

= (5000 * 51/50 * 51/50 * 51/50) = Rs. 5306.04 Difference = (5306.04 - 5304) = Rs. 2.04

Answer: C

Let the rate of interest be R% p.a. 

4400{[1 + R/100]² - 1} = 11193.60 

[1 + R/100]² = (44000 + 11193.60)/44000

[1 + R/100]² = 1 + 2544/1000 = 1 + 159/625 

[1 + R/100]² = 784/625 = (28/25)2 

1 + R/100 = 28/25 

R/100 = 3/25 

Therefore R = 12 SI on Rs.44000 at 12% p.a. for two years = 44000(2)(12)/100 

=Rs.10560

Answer: A

Over two years the gap between compound and simple interest is

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

\[ 96 = 15000 \times \frac{R^{2}}{10000} = 1.5R^{2} \]

\[ R^{2} = 64 \quad\Longrightarrow\quad R = 8\% \]

The rate is fully determined, so "cannot be determined" is a trap: for a two-year period the difference, the principal and the rate are locked together by that single relation.

Answer: D

Compound growth is multiplicative, so doubling twice is the same as quadrupling.

\[ \left(1 + \frac{R}{100}\right)^{9} = 2 \]

\[ \left(1 + \frac{R}{100}\right)^{18} = \left[\left(1 + \frac{R}{100}\right)^{9}\right]^{2} = 2^{2} = 4 \]

So it takes 18 years. The rate never needs to be found.

This multiplicative behaviour is what separates compound from simple interest: under simple interest, doubling in 9 years means quadrupling in 27, since there the growth is additive.

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: A

An increase of \( \tfrac{1}{8} \) each year multiplies the value by \( \tfrac{9}{8} \).

\[ A = 64000 \times \left(\frac{9}{8}\right)^{2} = 64000 \times \frac{81}{64} \]

\[ = 1000 \times 81 = \text{Rs. } 81{,}000 \]

The 64 in the principal cancels the \( 8^{2} \) in the denominator exactly, which is why the answer is so clean. Adding \( \tfrac{1}{8} \) of the original twice would give only Rs. 80,000.

Answer: C

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

For two years the difference is

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

\[ 2166 = 60000 \times \frac{R^{2}}{10000} = 6R^{2} \]

\[ R^{2} = 361 \quad\Longrightarrow\quad R = 19\% \]

The rate appears squared, so a perfect square is the signal that the arithmetic is on track - here \( 19^{2} = 361 \).