This set of Class 8 Maths Chapter 8 Multiple Choice Questions & Answers (MCQs) 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 16704 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) 4230
b) 4320
c) 4240
d) 4420
View Answer
Explanation: Present population = 4000 (1 + \(\frac{2}{100}\))(1 + \(\frac{4}{100}\)) = 4000 (\(\frac{51}{50}\))(\(\frac{26}{25}\)) = 4243.20 ≈ 4240.
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 + \(\frac{10}{100})^3\).
⇒ P = \(\frac{532400\times10\times10\times10}{11\times11\times11}\)
⇒ P = 400000.
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: 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) 54675
c) 50460
d) 54000
View Answer
Explanation: Cost of the vehicle = Rs 90000
Rate of depreciation in the first two years = 10%
Rate of depreciation in the third 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. 54675.
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%
⇒ 6561 = 10000 (1 – \(\frac{10}{100})^n\)
⇒ \(\frac{6561}{10000} = (\frac{9}{10})^n\)
⇒ (\(\frac{9}{10})^4\) = (\(\frac{9}{10})^n\)
⇒ 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. 24000. If the cost of the gadget after 2 years depreciates to Rs. 20757.60, find the rate of depreciation.
a) 20%
b) 5%
c) 15%
d) 7%
View Answer
Explanation: Let R be the rate of depreciation
⇒ 20757.60 = 24000 (1 – \(\frac{R}{100})^2\)
⇒ \(\frac{20757.60}{24000}\) = (1 – \(\frac{R}{100})^2\)
⇒ \(\frac{93^2}{100^2}\) = (1 – \(\frac{R}{100})^2\)
⇒ \(\frac{93}{100}\) = 1 – \(\frac{R}{100}\)
⇒ R = 7%.
Sanfoundry Global Education & Learning Series – Mathematics – Class 8.
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