Simple Interest and Compound Interest — Exercise 8(C)

Mr Dubey borrows ₹ 1,00,000 from a bank at 10% per annum Compound interest. What amount will he have to pay to the bank after 3 years?

Answer

Clearly, we have

P = ₹ 1,00,000, R = 10% p.a. and $n$ = 3 years.

Using the formula, $\text{A} = \text{P}\left(1 + \dfrac{\text{R}}{100}\right)^n$, we get :

$\text{Amount after 3 years} = 100000 \times \left(1 + \dfrac{10}{100}\right)^3\\[1em] = 100000 \times \dfrac{11}{10} \times \dfrac{11}{10} \times \dfrac{11}{10}\\[1em] = 133100$

Hence, Mr Dubey will have to pay ₹ 1,33,100 to the bank after 3 years.

Find the amount and compound interest on ₹ 10,240 for 3 years at $12\dfrac{1}{2}$% per annum compounded annually.

Answer

Clearly, we have

P = ₹ 10,240, R = $12\dfrac{1}{2}$% p.a. = $\dfrac{25}{2}$% p.a. and n = 3 years.

$\text{Amount after 3 years} = 10240 \times \left(1 + \dfrac{25}{2 \times 100}\right)^3\\[1em] = 10240 \times \dfrac{9}{8} \times \dfrac{9}{8} \times \dfrac{9}{8}\\[1em] = 14580$

∴ C.I. after 3 years = Amount − Principal = ₹ (14,580 − 10,240) = ₹ 4,340

Hence, amount = ₹ 14,580 and compound interest = ₹ 4,340.

The present population of a town is 176400. If it increases at the rate of 5% per annum, what will be its population after 2 years?

Answer

Present population = 176400 and rate of increase = 5% p.a.

$\text{Population after 2 years} = 176400 \times \left(1 + \dfrac{5}{100}\right)^2\\[1em] = 176400 \times \dfrac{21}{20} \times \dfrac{21}{20}\\[1em] = 194481$

Hence, the population after 2 years will be 194481.

Amit started a shop by investing ₹ 50,000. He gained 4% during first year, 5% during second year and 10% during 3rd year. What will be his capital amount after these 3 years?

Answer

Here,

P = ₹ 50,000, r₁ = 4% p.a., r₂ = 5% p.a. and r₃ = 10% p.a.

$\text{Capital amount after 3 years} = 50000 \times \left(1 + \dfrac{4}{100}\right) \times \left(1 + \dfrac{5}{100}\right) \times \left(1 + \dfrac{10}{100}\right)\\[1em] = 50000 \times \dfrac{26}{25} \times \dfrac{21}{20} \times \dfrac{11}{10}\\[1em] = 60060$

Hence, Amit's capital amount after 3 years will be ₹ 60,060.

A mango tree of height 125 cm was planted 3 years ago. If it increases at the rate of 20% per annum, what is its present height?

Answer

Here,

height 3 years ago = 125 cm and rate of increase = 20% p.a.

$\text{Present height} = 125 \times \left(1 + \dfrac{20}{100}\right)^3\\[1em] = 125 \times \dfrac{6}{5} \times \dfrac{6}{5} \times \dfrac{6}{5}\\[1em] = 216$

Hence, the present height of the mango tree is 216 cm.

Two years ago, the population of a town was 10000. During first year, it increased at the rate of 5% per annum and during second year, it increased at the rate of 6% per annum. What is its present population?

Answer

Here,

population 2 years ago = 10000, r₁ = 5% p.a. and r₂ = 6% p.a.

$\text{Present population} = 10000 \times \left(1 + \dfrac{5}{100}\right) \times \left(1 + \dfrac{6}{100}\right)\\[1em] = 10000 \times \dfrac{21}{20} \times \dfrac{53}{50}\\[1em] = 11130$

Hence, the present population of the town is 11130.

Mahesh borrowed ₹ 16,000 at $7\dfrac{1}{2}$% per annum simple interest. On the same day, he lent it to Gagan at the same rate but compounded annually. What does he gain at the end of 2 years?

Answer

Here,

P = ₹ 16,000, R = $7\dfrac{1}{2}$% p.a. = $\dfrac{15}{2}$% p.a. and T = 2 years.

Interest paid by Mahesh (simple interest) :

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

$= ₹ \left(16000 \times \dfrac{15}{2} \times 2 \times \dfrac{1}{100}\right)$ = ₹ 2,400

Interest received by Mahesh (compound interest) :

$\text{Amount after 2 years} = 16000 \times \left(1 + \dfrac{15}{2 \times 100}\right)^2\\[1em] = 16000 \times \dfrac{43}{40} \times \dfrac{43}{40}\\[1em] = 18490$

C.I. = Amount − Principal = ₹ (18,490 − 16,000) = ₹ 2,490

∴ Gain = C.I. − S.I. = ₹ (2,490 − 2,400) = ₹ 90

Hence, Mahesh gains ₹ 90 at the end of 2 years.

A machine is purchased for ₹ 6,25,000. Its value depreciates at the rate of 8% per annum. What will be its value after 2 years?

Answer

Here,

present value = ₹ 6,25,000 and rate of depreciation = 8% p.a.

$\text{Value of the machine after 2 years} = 625000 \times \left(1 - \dfrac{8}{100}\right)^2\\[1em] = 625000 \times \dfrac{23}{25} \times \dfrac{23}{25}\\[1em] = 529000$

Hence, the value of the machine after 2 years will be ₹ 5,29,000.

A car is purchased for ₹ 3,48,000. Its value depreciates at 10% per annum during the first year and at 20% per annum during the second year. What will be its value after 2 years?

Answer

Here,

present value = ₹ 3,48,000, r₁ = 10% p.a. and r₂ = 20% p.a.

$\text{Value of the car after 2 years} = 348000 \times \left(1 - \dfrac{10}{100}\right) \times \left(1 - \dfrac{20}{100}\right)\\[1em] = 348000 \times \dfrac{9}{10} \times \dfrac{4}{5}\\[1em] = 250560$

Hence, the value of the car after 2 years will be ₹ 2,50,560.