Mathematics Objective Questions and Answers for Class 8 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%
Answer: d
Clarification: 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
Answer: c
Clarification: 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
Answer: a
Clarification: 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
Answer: d
Clarification: 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
Answer: c
Clarification: 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
Answer: d
Clarification: 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
Answer: a
Clarification: 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
Answer: d
Clarification: 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
Answer: b
Clarification: 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
Answer: d
Clarification: 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