# Mathematics Questions and Answers – Applications of Compound Interest Formula – 2

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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%

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

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

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

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

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

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

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

Explanation: V0 = Rs. 5000000
Value after 3 years = V0 (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

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

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

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

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%

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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