Simple and Compound Interest — Multiple Choice Questions

The compound interest on ₹1,000 at 10% p.a. for 2 years is

1. ₹190

2. ₹210

3. ₹1,210

4. ₹200

Answer

Given:

Principal (P) = ₹ 1,000

Rate of interest (R) = 10% p.a.

Number of conversion periods (n) = 2

As we know:

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

Substituting the values, we get

$\Rightarrow \text{A} = ₹ 1{,}000\Big(1 + \dfrac{10}{100}\Big)^2 \\[1em] \Rightarrow \text{A} = ₹ 1{,}000 \times \dfrac{11}{10} \times \dfrac{11}{10} \\[1em] \Rightarrow \text{A} = ₹ 1{,}210$

C.I. = A - P = ₹ 1,210 - ₹ 1,000 = ₹ 210

Hence, option 2 is the correct option.

The compound interest on ₹5,000 at 20% per annum for $1\dfrac{1}{2}$ years compounded half yearly is

1. ₹6,655

2. ₹1,655

3. ₹750

4. ₹1,000

Answer

Given:

Principal (P) = ₹ 5,000

Rate of interest = 20% p.a. compounded half yearly

As the interest is compounded half yearly,

Rate per conversion period (R) = $\dfrac{20}{2}$% = 10%

Number of conversion periods (n) = $1\dfrac{1}{2} \times 2$ = 3

As we know:

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

Substituting the values, we get

$\Rightarrow \text{A} = ₹ 5{,}000\Big(1 + \dfrac{10}{100}\Big)^3 \\[1em] \Rightarrow \text{A} = ₹ 5{,}000 \times \dfrac{11}{10} \times \dfrac{11}{10} \times \dfrac{11}{10} \\[1em] \Rightarrow \text{A} = ₹ 6{,}655$

C.I. = A - P = ₹ 6,655 - ₹ 5,000 = ₹ 1,655

Hence, option 2 is the correct option.

The compound interest on ₹10,000 at 8% per annum for 6 months compounded quarterly is

1. ₹408

2. ₹10,404

3. ₹404

4. ₹400

Answer

Given:

Principal (P) = ₹ 10,000

Rate of interest = 8% p.a. compounded quarterly

As the interest is compounded quarterly,

Rate per conversion period (R) = $\dfrac{8}{4}$% = 2%

Number of conversion periods (n) = $\dfrac{6}{3}$ = 2

As we know:

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

Substituting the values, we get

$\Rightarrow \text{A} = ₹ 10{,}000\Big(1 + \dfrac{2}{100}\Big)^2 \\[1em] \Rightarrow \text{A} = ₹ 10{,}000 \times \dfrac{51}{50} \times \dfrac{51}{50} \\[1em] \Rightarrow \text{A} = ₹ 10{,}404$

C.I. = A - P = ₹ 10,404 - ₹ 10,000 = ₹ 404

Hence, option 3 is the correct option.

The time periods and rate for a sum taken at 8% p.a. for $1\dfrac{1}{2}$ years compounded half yearly are

1. n = 3, R = 4%

2. n = 6, R = 2%

3. n = 3, R = 2%

4. n = 6, R = 4%

Answer

As the interest is compounded half yearly, the half yearly rate is half of the annual rate and there are two conversion periods in a year.

Rate per conversion period (R) = $\dfrac{8}{2}$% = 4%

Number of conversion periods (n) = $1\dfrac{1}{2} \times 2$ = 3

Hence, option 1 is the correct option.

If ₹12,000 taken for 2 years at 4% per annum compounded quarterly, then time period and rate is

1. n = 2, R = 16%

2. n = 4, R = 1%

3. n = 8, R = 1%

4. n = 8, R = 16%

Answer

As the interest is compounded quarterly, the quarterly rate is one fourth of the annual rate and there are four conversion periods in a year.

Rate per conversion period (R) = $\dfrac{4}{4}$% = 1%

Number of conversion periods (n) = 2 × 4 = 8

Hence, option 3 is the correct option.

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

1. less than or equal to simple interest

2. greater than or equal to simple interest

3. less than simple interest

4. greater than simple interest

Answer

In simple interest, the interest of every year is calculated on the original principal, whereas in compound interest, the interest of every conversion period is calculated on the amount of the previous conversion period.

For the first conversion period, the interest is the same in both the cases. From the second conversion period onwards, the principal in compound interest becomes greater than the original principal.

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

Hence, option 4 is the correct option.

The time in which ₹6,000 amounts to ₹7,986 at 10% p.a. compounded annually is

1. 2 years

2. 3 years

3. 4 years

4. 5 years

Answer

Given:

Principal (P) = ₹ 6,000

Amount (A) = ₹ 7,986

Rate of interest (R) = 10% p.a. compounded annually

As we know:

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

Substituting the values, we get

$\Rightarrow 7986 = 6000\Big(1 + \dfrac{10}{100}\Big)^n \\[1em] \Rightarrow \Big(\dfrac{11}{10}\Big)^n = \dfrac{7986}{6000} \\[1em] \Rightarrow \Big(\dfrac{11}{10}\Big)^n = \dfrac{1331}{1000} \\[1em] \Rightarrow \Big(\dfrac{11}{10}\Big)^n = \Big(\dfrac{11}{10}\Big)^3 \\[1em] \Rightarrow n = 3$

Hence, option 2 is the correct option.