Compound Interest — Multiple Choice Questions

The compound interest on ₹1000 at 10% p.a. compounded annually for 2 years is

1. ₹190

2. ₹200

3. ₹210

4. ₹1210

Answer

C.I. = $P\Big[\Big(1 + \dfrac{r}{100}\Big)^n - 1\Big]$

Putting values in formula we get,

$C.I. = 1000 \times \Big[\Big(1 + \dfrac{10}{100}\Big)^2 - 1\Big] \\[1em] = 1000 \times \Big[\Big(\dfrac{110}{100}\Big)^2 - 1\Big] \\[1em] = 1000 \times \Big[\Big(\dfrac{11}{10}\Big)^2 - 1\Big] \\[1em] = 1000 \times \Big[\dfrac{121}{100} - 1\Big] \\[1em] = 1000 \times \dfrac{21}{100} \\[1em] = ₹210.$

Hence, Option 3 is the correct option.

If Sukriti borrows ₹8000 for two years at the rate of 10% per annum compound interest, then the amount to be paid by her at the end of two years to clear the debt is

1. ₹8800

2. ₹9600

3. ₹9680

4. ₹102400

Answer

A = $P\Big(1 + \dfrac{r}{100}\Big)^n$

Putting values in formula we get,

$A = ₹8000 \times \Big(1 + \dfrac{10}{100}\Big)^2 \\[1em] = ₹8000 \times \Big(\dfrac{110}{100}\Big)^2 \\[1em] = ₹8000 \times \Big(\dfrac{11}{10}\Big)^2 \\[1em] = ₹8000 \times \dfrac{11}{10} \times \dfrac{11}{10} \\[1em] = ₹8000 \times \dfrac{121}{100} \\[1em] = ₹9680.$

Hence, Option 3 is the correct option.

If a man invests ₹12000 for two years at the rate of 10% per annum compound interest, then the compound interest earned by him at the end of two years is

1. ₹2400

2. ₹2520

3. ₹2000

4. ₹1800

Answer

C.I. = $P\Big[\Big(1 + \dfrac{r}{100}\Big)^n - 1\Big]$

Putting values in formula we get,

$C.I. = ₹12000 \times \Big[\Big(1 + \dfrac{10}{100}\Big)^2 - 1\Big] \\[1em] = ₹12000 \times \Big[\Big(\dfrac{110}{100}\Big)^2 - 1\Big] \\[1em] = ₹12000 \times \Big[\Big(\dfrac{11}{10}\Big)^2 - 1\Big] \\[1em] = ₹12000 \times \Big[\dfrac{121}{100} - 1\Big] \\[1em] = ₹12000 \times \dfrac{21}{100} \\[1em] = ₹120 \times 21 \\[1em] = ₹2520.$

Hence, Option 2 is the correct option.

Mr. Rao bought 1-year, ₹10000 certificate of deposit that paid interest at an annual rate of 8% compounded semi-annually. The interest received by him on maturity is

1. ₹816

2. ₹864

3. ₹800

4. ₹10816

Answer

Rate = 8% i.e. $\dfrac{8}{2}$% = 4% when compounded semi-annually.

n (no. of conversion periods) = 2 half-years.

C.I. = $P\Big[\Big(1 + \dfrac{r}{100}\Big)^n - 1\Big]$

Putting values in formula we get,

$C.I. = ₹10000 \times \Big[\Big(1 + \dfrac{4}{100}\Big)^2 - 1\Big] \\[1em] = ₹10000 \times \Big[\Big(\dfrac{104}{100}\Big)^2 - 1\Big] \\[1em] = ₹10000 \times \Big[\Big(\dfrac{52}{50}\Big)^2 - 1\Big] \\[1em] = ₹10000 \times \Big[\dfrac{2704}{2500} - 1\Big] \\[1em] = ₹10000 \times \dfrac{2704 - 2500}{2500} \\[1em] = ₹10000 \times \dfrac{204}{2500} \\[1em] = ₹\dfrac{2040000}{2500} \\[1em] = ₹816.$

Hence, Option 1 is the correct option.

The compound interest on ₹5000 at 20% per annum for $1\dfrac{1}{2}$ years compounded half-yearly is

1. ₹6655

2. ₹1655

3. ₹1500

4. ₹1565

Answer

Rate = 20% i.e. $\dfrac{20}{2}$% = 10% when compounded semi-annually.

n (no. of conversion periods) = 3 half-years.

C.I. = $P\Big[\Big(1 + \dfrac{r}{100}\Big)^n - 1\Big]$

Putting values in formula we get,

$C.I. = ₹5000 \times \Big[\Big(1 + \dfrac{10}{100}\Big)^3 - 1\Big] \\[1em] = ₹5000 \times \Big[\Big(\dfrac{110}{100}\Big)^3 - 1\Big] \\[1em] = ₹5000 \times \Big[\Big(\dfrac{11}{10}\Big)^3 - 1\Big] \\[1em] = ₹5000 \times \Big[\dfrac{1331}{1000} - 1\Big] \\[1em] = ₹5000 \times \dfrac{1331 - 1000}{1000} \\[1em] = ₹5000 \times \dfrac{331}{1000} \\[1em] = ₹1655.$

Hence, Option 2 is the correct option.

If the number of conversion periods ≥ 2, then the compound interest is

1. less than simple interest

2. equal to simple interest

3. greater than or equal to simple interest

4. greater than simple interest

Answer

When the number of conversion periods ≥ 2, then the compound interest is greater than simple interest.

As, in compound interest the interest is always calculated on the compounded principal whereas in simple interest, the interest is calculated on the initial principle so for conversion periods ≥ 2 compound interest will be greater than simple interest.

Hence, Option 4 is the correct option.

The present population of a city is 12,00,000. If it increases at the rate of 8% every year then the population of the city after 2 years is

1. 199680

2. 1399680

3. 1500000

4. 1299680

Answer

By growth formula,

$V = V_0\Big(1 + \dfrac{r}{100}\Big)^n$

Substituting values in formula,

$V = 1200000\Big(1 + \dfrac{8}{100}\Big)^2 \\[1em] = 1200000\Big(\dfrac{108}{100}\Big)^2 \\[1em] = 1200000\Big(\dfrac{108}{100}\Big)^2 \\[1em] = 1200000\Big(\dfrac{27}{25}\Big)^2 \\[1em] = 1200000 \times \dfrac{27}{25} \times \dfrac{27}{25} \\[1em] = 1200000 \times \dfrac{729}{625} \\[1em] = 1920 \times 729 \\[1em] = 1399680.$

Hence, Option 2 is the correct option.

Consider the following two statements.

Statement 1: A sum of ₹ 1,000 invested at 10% p.a. rate of interest will earn ₹ 100 interest in first year.

Statement 2: Simple and compound interest is the same for first conversion period at the same rate of interest.

Which of the following is valid?

1. Both the statements are true.

2. Both the statements are false.

3. Statement 1 is true, and Statement 2 is false.

4. Statement 1 is false, and Statement 2 is true.

Answer

Given,

P = ₹ 1,000

R = 10%

T = 1 year

Using C.I. formula,

C.I. = $P\Big[\Big(1 + \dfrac{R}{100}\Big)^n - 1\Big]$

Substituting the values, we get

$C.I. = 1,000 \times \Big[\Big(1 + \dfrac{10}{100}\Big)^1 - 1\Big]\\[1em] = 1,000 \times \Big[\Big(1 + \dfrac{1}{10}\Big) - 1\Big]\\[1em] = 1,000 \times \Big(1 + \dfrac{1}{10} - 1\Big)\\[1em] = 1,000 \times \Big(\dfrac{1}{10}\Big)\\[1em] = 100.$

Calculating S.I.,

P = ₹ 1,000

R = 10%

T = 1 year

Using formula,

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

Substituting the values, we get

$\Rightarrow S.I. = \dfrac{1,000 \times 10 \times 1}{100}\\[1em] = 10 \times 10 \\[1em] = 100.$

Since,

C.I. = S.I. = ₹ 100.

Therefore, we can say that simple and compound interest is the same for first conversion period at the same rate of interest.

∴ Both statements are true.

Hence, option 1 is the correct option.