This set of Mathematics Objective Questions and Answers for Class 8 focuses on “Applications of Compound Interest Formula – 2”.

1. A company increased the production of cars from 15625 in 2002 to 27000 in 2005. Find the annual rate of growth of production of the cars.

a) 15%

b) 2%

c) 10%

d) 20%

View Answer

Explanation: Let the annual rate of growth be R% per annum. Then,

⇒ 27000 = 15625 (1 + \(\frac{R}{100})^3\)

⇒ \(\frac{27000}{15625}\) = (1 + \(\frac{R}{100})^3\)

⇒ (\(\frac{30}{25})^3\) = (1 + \(\frac{R}{100})^3\)

⇒ \(\frac{30}{25}\) = 1 + \(\frac{R}{100}\)

⇒ \(\frac{1}{5}\) = \(\frac{R}{100}\)

⇒ R = 20%.

2. Fe-57 decays at the constant rate in such a way that it reduces to 50% in 5568 years. Find the age of an old rod which the iron (Fe) is only 12.5% of the original.

a) 16700

b) 16705

c) 16704

d) 16000

View Answer

Explanation: Let the rate of decay be R% per annum and the age of rod be n years. Let the original amount of iron in the rod be P. Then in 5568 years the amount left is \(\frac{P}{2}\)

⇒ \(\frac{P}{2}\) = P (1 – \(\frac{R}{100})^{5568}\)

⇒ \(\frac{1}{2}\) = (1 – \(\frac{R}{100})^{5568}\)

After n years, the iron left in the rod is 12.5% of P

⇒ \(\frac{P}{8}\) = P (1 – \(\frac{R}{100})^n\)

⇒ \(\frac{1}{8}\) = (1 – \(\frac{R}{100})^n\)

⇒ (\(\frac{1}{2})^3\) = (1 – \(\frac{R}{100})^n\)

⇒ (1 – \(\frac{R}{100})^{16704}\) = (1 – \(\frac{R}{100})^n\)

⇒ n = 5568 × 3 = 16704 years.

Hence the age of the rod is 16707 years.

3. The present population of a village is 10000. If it increases at the rate of 2% per annum, find the population after 2 years.

a) 10404

b) 10004

c) 10402

d) 10400

View Answer

Explanation: Initial population = 10000

⇒ Let after 2 years be P.

⇒ P = 10000 (1 + \(\frac{2}{100})^2\) = 10000 (\(\frac{51}{50})^2\) = 10404.

4. The initial population of a country is 1000000. If the birth rate and the death rate is 10% and 4% respectively, then find the population of the country after 2 years.

a) 1126000

b) 1023600

c) 1123000

d) 1123600

View Answer

Explanation: Initial population = 1000000

Increment in population = 10%

Decrement in population = 4%

Net growth in population = 10-4 = 6%

Population after 2 years = 1000000 (1 + \(\frac{6}{100})^2\) = 1000000 (\(\frac{53}{50})^2\) = 1123600.

5. Two years ago the population of a village was 4000. If the annual increase during the two successive years be at the rate of 2% & 4%, find the present population.

a) 7074

b) 7080

c) 7072

d) 7000

View Answer

Explanation: Present population = 4000 (1 + \(\frac{2}{100}\))(1 + \(\frac{4}{100}\)) = 4000 (\(\frac{51}{50}\))(\(\frac{26}{25}\)) = 7072.

6. The annual rate of growth in population of a certain city is 10%. If its present population is 532400, find the population 3 years ago.

a) 404000

b) 300000

c) 400400

d) 400000

View Answer

Explanation: Population 3 years ago = P (1 + \(\frac{R

}{100})^3\)

⇒ 532400 = P (1 + \(\frac{10}{100})^3\).

7. Vidhya started a business with initial investment of 500000. She incurred a loss of 2% in the first year and in the second year she gained the profit of 10%. Calculate her net profit earned.

a) 39000

b) 38000

c) 41000

d) 40000

View Answer

Explanation: Initial investment = 500000

Profit = 500000 (1 – \(\frac{2}{100}\))(1 + \(\frac{10}{100}\)) = 500000 (\(\frac{49}{50}\))(\(\frac{11}{10}\)) = 539000

⇒ Net profit earned = 500000 – 539000 = 39000.

8. An apartment of two floors is constructed at the cost of Rs. 5000000. It is depreciating at the rate of 20% per annum. Find its value 3 years after construction.

a) 2400000

b) 2460000

c) 2500000

d) 2560000

View Answer

Explanation: V

_{0}= Rs. 5000000

Value after 3 years = V

_{0}(1 – \(\frac{R}{100})^3\) = 5000000 (1 – \(\frac{20}{100})^3\) = 5000000 (\(\frac{4}{5})^3\) = Rs. 2560000.

9. A new two-wheeler costs Rs. 90000. Its price depreciates at the rate of 10% a year during first two years and the rate of 25% a year thereafter. Find the cost of the two-wheeler after 3 years.

a) 55000

b) 546750

c) 50460

d) 540000

View Answer

Explanation: Cost of the vehicle = Rs 90000

Rate of depreciation in the first two years = 10%

Rate of depreciation in the second year = 25%

⇒ Price of the vehicle after 3 years = 90000 (1 – \(\frac{10}{100})^2\) (1 – \(\frac{25}{100}\)) = 90000 (\(\frac{9}{10})^2\)(\(\frac{3}{4}\)) = Rs. 546750.

10. The present price of a cycle is Rs 8000. If its value is decreasing every year by 10% then find the price before 3 years.

a) 10977

b) 10976

c) 10975

d) 10974

View Answer

Explanation: Let the price of the cycle be Rs. P before 3 years. Then its present value is Rs P (1 – \(\frac{10}{100})^3\)

But the present price is Rs. 8000

⇒ 8000 = P (1 – \(\frac{10}{100})^3\)

⇒ 8000 = P (\(\frac{9}{10})^3\)

⇒ P = 10974.

11. The value of a gadget worth Rs.10000 is depreciating at the rate of 10% per annum. In how many years will its value be reduced to Rs 6561?

a) 5

b) 3

c) 4

d) 2

View Answer

Explanation: Present value = Rs. 10000

Depreciated value = Rs. 6561

Rate of depreciation = 10%

⇒ 6562 = 10000 (1 – \(\frac{10}{100})^n\)

⇒ \(\frac{6562}{10000} = (\frac{9}{10})^n\)

⇒ (\(\frac{9}{10})^n\) = (\(\frac{9}{10})^n\)

⇒ n = 4.

12. The value of a land increases every year at the rate of 8%. If its value at the end of 3 years be Rs 6000000, find its original value at the beginning of these years.

a) 4762994

b) 3762994

c) 3000000

d) 4000000

View Answer

Explanation: R = 8% and n = 3

⇒ 6000000 = P (1 + \(\frac{8}{100})^3\)

⇒ 6000000 = P (\(\frac{27}{25})^3\)

⇒ P = 4762994.

13. Max bought a gadget for Rs. 21000. If the cost of the gadget after 2 years depreciates to Rs. 18162, find the rate of depreciation.

a) 20%

b) 5%

c) 15%

d) 7%

View Answer

Explanation: Let R be the rate of depreciation

⇒ 18162 = 21000 (1 + \(\frac{R}{100})^2\)

⇒ \(\frac{18162}{21000}\) = (1 + \(\frac{R}{100})^2\)

⇒ (0.8648)

^{2}= (1 + \(\frac{R}{100})^2\)

⇒ (0.929) = (1 + \(\frac{R}{100}\))

⇒ R = 7%.

**Sanfoundry Global Education & Learning Series – Mathematics – Class 8**.

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