Compound Interest — Competency Focused Questions

What annual instalment will discharge a debt of ₹ 4,600 due in 4 years at 10% simple interest?

1. ₹ 1,000

2. ₹ 1,150

3. ₹ 1,500

4. ₹ 1,610

Answer

Let annual installment be ₹ x.

Annual installment 1 (Paid at the end of 1st year) : This installment will earn interest for 3 years.

$\Rightarrow x + x \times \dfrac{10}{100} \times 3 \\[1em] \Rightarrow x + \dfrac{3x}{10} \\[1em] \Rightarrow \dfrac{13x}{10} \\[1em] \Rightarrow 1.3x$

Annual installment 2 (Paid at the end of 2nd year) : This installment will earn interest for 2 years.

$\Rightarrow x + x \times \dfrac{10}{100} \times 2 \\[1em] \Rightarrow x + \dfrac{2x}{10} \\[1em] \Rightarrow \dfrac{12x}{10} \\[1em] \Rightarrow 1.2x$

Annual installment 3 (Paid at the end of 3rd year) : This installment will earn interest for 1 year.

$\Rightarrow x + x \times \dfrac{10}{100} \times 1 \\[1em] \Rightarrow x + \dfrac{x}{10} \\[1em] \Rightarrow \dfrac{11x}{10} \\[1em] \Rightarrow 1.1x$

Annual installment 4 (Paid at the end of 4th year) : This installment will not earn any interest.

⇒ x

The sum of these values must be equal to debt value of ₹ 4,600.

⇒ 1.3x + 1.2x + 1.1x + x = 4,600

⇒ 4.6x = 4,600

⇒ x = $\dfrac{4,600}{4.6}$

⇒ x = ₹ 1,000.

Hence, Option 1 is correct option.

The simple interest on a sum of money at 8% per annum for 6 years is half the sum. The sum is:

1. ₹ 4,000

2. ₹ 6,000

3. ₹ 10,000

4. Data inadequate

Answer

Let the sum be ₹ P.

Given,

I = $\dfrac{P}{2}$

T = 6 years

R = 8%

By formula,

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

Substituting values we get :

$\Rightarrow \dfrac{P}{2} = \dfrac{P \times 8 \times 6}{100}$

By doing this calculation we will not obtain the sum, we need to have the interest to calculate the sum.

Hence, Option 4 is correct option.

What is the principal amount which earns ₹ 132 as compound interest for the second year at 10% per annum?

1. ₹ 800

2. ₹ 1,050

3. ₹ 1,200

4. Data inadequate

Answer

Let the principal amount be ₹ P.

For 1 year :

R = 10%

T = 1 year

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

Substituting values we get :

$\Rightarrow I = \dfrac{P \times 10\times 1}{100} = \dfrac{P}{10}$

A = $P + \dfrac{P}{10} = \dfrac{11P}{10}$

For 2nd year :

Principal = $\dfrac{11P}{10}$

R = 10%

T = 1 year

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

$I = \dfrac{\dfrac{11P}{10} \times 10 \times 1}{100} \\[1em] = \dfrac{11P \times 10}{1000} \\[1em] = \dfrac{11P}{100}.$

Given,

Interest for 2nd year = ₹ 132

$\therefore \dfrac{11P}{100} = 132 \\[1em] \Rightarrow P = \dfrac{132 \times 100}{11} \\[1em] \Rightarrow P = 1,200.$

Hence, Option 3 is correct option.

In an industrial area, the price of a plot of area 1 acre is ₹ 5 crores. If the price of land increases by 2% quarterly, then what will be the price of the plot after 1 year?

1. $5 \times \Big(1 + \dfrac{2}{100}\Big)^4$ crores

2. $5 \times \Big(1 + \dfrac{8}{100}\Big)^4$ crores

3. $5 \times \Big(1 + \dfrac{2}{100}\Big)$ crores

4. $5 \times \Big(1 + \dfrac{8}{100}\Big)$ crores

Answer

Given,

V = ₹ 5 crores

R = 2%

Price of land increases by 2% quarterly.

n = There are 4 Quarters in a year = 4

By Formula,

Value after n years = $V \times \Big(1 + \dfrac{R}{100}\Big)^n$

By substituting values,

$\text{Value after 1 year}= 5 \times \Big(1 + \dfrac{2}{100}\Big)^4$ crores.

Hence, Option 1 is correct option.

If the compound interest on a certain sum for 2 years at 10% p.a. is ₹ 2,100, the simple interest on it at the same rate for 2 years will be:

1. ₹ 1,500

2. ₹ 1,800

3. ₹ 2,000

4. ₹ 2,050

Answer

Given,

C.I = ₹ 2,100

n = 2 years

R = 10%

By Formula,

C.I = Amount - Principal

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

Substituting the values in formula

$\Rightarrow 2100 = P \times \Big(1 + \dfrac{10}{100}\Big)^2 - P\\[1em] \Rightarrow 2100 = P \times \Big(\dfrac{100 + 10}{100}\Big)^2 - P\\[1em] \Rightarrow 2100 = P \times \Big(\dfrac{110}{100}\Big)^2 - P\\[1em] \Rightarrow 2100 = P \times (1.10)^2 - P\\[1em] \Rightarrow 2100 = 1.21P - P \\[1em] \Rightarrow 2100 = 0.21P \\[1em] P = \dfrac{2100}{0.21} \\[1em] P = 10,000$

To calculate the simple interest on the same principal.

T = 2 years

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

Substituting the Values,

I = $\dfrac{10000 \times 10 \times 2}{100}$ = ₹ 2,000

Hence, Option 3 is correct option.

Mr. Goyal bought a car for ₹ 6,25,000. Its value depreciates at 20% p.a. How many years it would take for the price of the car to go down by ₹ 3,69,000?

Answer

Given,

Present value of machine (V) = ₹ 6,25,000

r = 20%

Depreciated value = ₹ 6,25,000 - ₹ 3,69,000 = ₹ 2,56,000

By formula,

Value of machine after n years = ₹ $V \times \Big(1 - \dfrac{r}{100}\Big)^n$

Substituting the values in formula,

$\Rightarrow 256000 = 625000 \times \Big(1 - \dfrac{20}{100}\Big)^n\\[1em] \Rightarrow \Big(1 - \dfrac{20}{100}\Big)^n=\dfrac{256000}{625000} \\[1em] \Rightarrow \Big(1 - \dfrac{1}{5}\Big)^n=\dfrac{256}{625} \\[1em] \Rightarrow \Big(\dfrac{4}{5}\Big)^n=\Big(\dfrac{4}{5}\Big)^4 \\[1em] \Rightarrow n = 4$

Hence, It will take 4 years for the price of the car to depreciate.

The population of a town was decreasing every year due to migration. The present population of the town is 6,31,680. Last year the migration was 4% and the year before last, it was 6%. What was the population 2 years ago?

Answer

Given,

P = 6,31,680

r₁ = 4%

r₂ = 6%

n = 1 year

By formula,

Population n years ago = $\dfrac{P}{\Big(1 - \dfrac{r_1}{100}\Big) \times \Big(1 - \dfrac{r_2}{100}\Big)}$

Substituting the values in formula,

$\text{Population 2 years ago}= \dfrac{631680}{\Big(1 - \dfrac{4}{100}\Big) \times \Big(1 - \dfrac{6}{100}\Big)} \\[1em] = \dfrac{631680}{\Big(\dfrac{100 - 4}{100}\Big) \times \Big(\dfrac{100 - 6}{100}\Big)} \\[1em] = \dfrac{631680}{\Big(\dfrac{96}{100}\Big) \times \Big(\dfrac{94}{100}\Big)} \\[1em] = \dfrac{631680 \times 100 \times 100}{96 \times 94} \\[1em] = \dfrac{6316800000}{9024} \\[1em] = 700000.$

Hence, the population of the town 2 years ago was 700000.