ML Aggarwal Solutions Class 9 Math 2nd Chapter Compound Interest Exercise 2.3
ML Aggarwal Understanding ICSE Mathematics Class 9 Solutions Second Chapter Compound Interest Exercise 2.3. APC Solution Class 9 Exercise 2.3.
(1) The present population of a town is 200000. Its population increases by 10% in the first year and 15% in the second year. Find the population of the town at the end of the two years.
Solution:
Given, Present population = 200000
1st year = 10% increase, 2nd year 15% increase
∴ Population in 2 years = 200000 × (1 + 10/100) (1 + 15/100)
= 200000 × 11/10 × 23/20
= 253000
(2) The present population of a town is 15625. If the population increases at the rate of 4% every year, what will be the increase in the population in next 3 years?
Solution:
Given, present population = 15625
% increase = 4% p.a.
n = 3 years
∴ Population in 3 years = 15625 (1 + 6/100)3
= 15625 × 17576/15625
= 17576
∴ Increase in population in 3 years = 17576 – 15625
= 1951
(3) The population of a city increases each year by 4% of what it had been at the beginning of each year. If its present population is 6760000, find:
(I) its population 2 years hence
(ii) Its population 2 years ago
Solution:
Given, present population = 6760000
Rate of increase = 4%/year
(i) ∴ Population in 2 years = 6760000 (1 + 4/100)2
= 6760000 × 676/625
= 10816 × 676
= 7311616
(ii) Let, the population 2 years ago be x.
∴ Present population = x (1 + 4/100)2
Or, 6760000 = x × (26/25)2
Or, x = 6760000 × 625/670
∴ Population 2 years back was 6250000
(4) The cost of a refrigerator is ₹9000. Its value depreciates at the rate 5% every year. Find the total depreciation in its value at the end of 2 years.
Solution:
Given, cost of refrigerator = ₹9000
% of depreciation = 5%/year
∴ Cost of refrigerator in 2 years = 9000 (1 – 8/100)2
= 9000 × (19/20)2
= 9000 × 361/400
= 16245/2
= ₹8122.5
∴ Depreciation value = 9000 – 8122.5
= ₹877.5
(5) Dinesh purchased a scooter for ₹24000. The value of the scooter is depreciating at the rate 5% per annum. Calculate its value after 3 years.
Solution:
Given, value of scooter at present = ₹24000
% of depreciation = 5%/year
∴ Value after 3 years = 24000 (1 – 5/100)3
= 24000 × (19/20)3
= 24000 × 6859/8000 = ₹20577
(6) A farmer increases his output of wheat in his farm every year by 8%. This year he produced 2187 quintals of wheat. What was the yearly produce of wheat two years ago?
Solution:
Given, % of in erase in production = 8%/year
Present production quantity = 2187. Quintals
Let, the production quantity 2 years ago be x.
∴ 2187 = x (1+ 8/100)2
Or, 2187 = x (27/25)2
Or, x = 2187×625/729 = 1875 quintals
∴ Production quantity 2 years ago was 1875 quintals.
(7) The value of property decreases every year at the rate of 5%. If its present value is ₹41154, what was its value three years ago?
Solution:
Given, % of decrease in value = 5%/year
Present value = ₹411540
Let, the value of property 3 years ago be ₹ x.
∴ 411540 = x (1 – 5/100)3
Or, 411540 = x (14/20)3
Or, x = 411540×8000/6859
Or, x = ₹420000
∴ The value of property 3 years ago was ₹420000
(8) Ahmed purchased an old scooter for ₹16000. If the cost of the scooter after 2 years depreciates to ₹14440, find the rate of depreciation.
Solution:
Given, present value of vertex = ₹16000
Cost of sector = 2 years from now = ₹14440
Let, the % depreciation be x%
∴ 14440 = 16000 (1 – x/100)2
Or, 14440/16000 = (100-x)4/10000
Or, (100 – x)2 = 361×25
Or, 100 – x = √9025
Or, x = 100 – 95
Or, x = 5%
∴ The rate of depreciation is 5% per year.
(9) A factory increased its production of cars from 80000 in the year 2011-2012 to 92610 in 2014-2015. Find the annual rate of growth of production of cars.
Solution:
Production in year 2011 – 2012 = 80000
Production in year 2014 – 2015 = 92610
∴ Increase in production occurred in 2014 – 2011
= 3 years
Let, the % in growth per year be x %
∴ 92610 = 8000 (1 + x/100)3
Or, 92610/80000 = (100+x/100)3
Or, 100+x/100 = 3√9261/8000
Or, 100+x/100 = 21/20
Or, x = 105 – 100 = 5%
∴ % of production growth per year is 5%.
(10) The value of a machine worth ₹500000 is depreciating at the rate of 10% every year. in how many years will its value be reduced to ₹364500?
Solution:
Given, present value = ₹500000
% of depreciation on = 10%/Year
Reduced value = ₹364500
Let, the time taken for this values depreciation be n years
∴ 364500 = 500000 (1 – 10/100)n
Or, (9/10)n = 5000/3645 3645/5000
Or, (9/10)n = 729/1000
Or, (9/10)n = (9/10)3
Equating both sides we get,
n = 3 years
∴ Time taken for this value to depreciate is 3 years.
(11) Afzal purchased an old motorbike for ₹16000. If the value of the motorbike after 2 years is ₹14440, find the rate of depreciation.
Solution:
Given, present of motorbike = ₹16000
Value of motorbike after 2 years = ₹14440.
∴ Let, the % of depreciation be x %/ year
∴ 14440 = 16000 (1 – x/100)2
Or, 1444/1600 = (100 – x/100)2
Or, 100-x/100 = √361/400
Or, 100-x/100 = 19/20
Or, x = 100 – 95 = 5%
∴ The rate of depreciation is 5%
(12) Mahindra setup a factory by investing ₹2500000. During the first two years, his profits were 5% and 10% respectively. If each year the profit was on previous year’s capital calculate his total profit.
Solution:
Many invested = ₹2500000
Profit rate 5%, 10% for 2 successive years
∴ Total many after 2 years.
= 2500000 (1 + 8/100) (1 + 10/100)
= 2500000 × 21/20 × 11/20
= ₹2887500
∴ Total profit made in 2 years = 2887500 – 2500000
= ₹387500
(13) The value of a property is increasing at the rate of 25% every year. By what percent will the value of the property increase after 3 years?
Solution:
Given, % of inverse every year = 25%
Let, the percent value be ₹100
∴ Value in 3 years = 100 (1 + 25/100)3
= 100 × (5/4)3
= 100 × 125/64
= ₹ 3125/16
∴ Increase in value = 3125/16 – 100
= 3125-1600/16
= 1525/16
∴ Increase in value =
(14) Mr.Durani bought a plot of land for ₹180000 and car for ₹320000 at the same time. The value of the plot of land grows uniformly at the rate of 30% p.a., while the value of the car depreciates by 20% in the first year and by 15% p.a. thereafter. If he sells the plot of land as well as the car after 3 years, what will be his profit or loss?
Solution:
Given, present value of land = ₹18000
Present value of car = ₹320000
% increase in value of land p.a. = 30%
% decrease in value car p.a. = 20% 1st year 15% thereafter.
∴ Value of land in 3 years = 180000 (1 + 30/100)3
= 180000 × (13/10)3
= 180000 × 2197/1000
= ₹ 395460
Value of car in 3 years = 320000 (1 – 20/100) (1 – 15/100) (1 – 15/100)
= 320000 × 4/5 × 17/20 × 17/20
= 160 × 4 × 17 × 17
= 184960
Total value of land and car presently = 180000 + 320000
= ₹ 500000
Total value of land & car after 3 years = 395460 + 184960
= ₹580420
∴ Total profit made after 3 years = 580420 – 50000
= ₹ 80420
Multiple Choice Questions:
(1) The compound interest on ₹1000 at 10% p.a. compounded annually for 2 years is
(a) ₹190
(b) ₹200
(c) ₹210
(d) ₹1210
Solution:
1000 [(1 + 10/100)2 – 1]
= 1000 (121-100/100) = 1000 × 21/100 = ₹210
Ans – (c) ₹210
(2) If Sukriti borrows ₹8000 for two years at the rate of 10% per annum compound interest, then the amount to be paid by her at the end of two years to clear the debt is
(a) ₹8800
(b) ₹9600
(c) ₹9680
(d) ₹102400
Solution:
8000 (1 + 10/100)2 = 8000 × 121/100 = ₹96980
Ans – (c) ₹9680
(3) If a man invests ₹12000 for two years at the rate of 10% per annum compound interest, then the compound interest earned by him at the end of two years is
(a) ₹ 2400
(b) ₹2520
(c) ₹2000
(d) ₹1800
Solution:
12000 [(1 + 10/100)2 – 1] . 1200 (121-100/100) = 12000 × 21/100
= ₹2520
Ans – (b) ₹2520
(4) Mr. Rao bought 1-year, ₹10000 certificate of deposit that paid interest at an annual rate of 8% compounded semi-annually. The interest received by him on maturity is
(a) ₹816
(b) ₹864
(c) ₹800
(d) ₹10816
Solution:
10000 [(1 + 4/100)2 – 1] = 10000 (676-625/625)
= 10000 × 51/625 = ₹816
Ans – (a) ₹816
(5) The compound interest on ₹5000 at 20% per annum for 1 1/2 years compounded half-yearly is
(a) ₹6655
(b) ₹1655
(c) ₹1500
(d) ₹1565
Solution:
(6) If the number of conversion periods ≥ 2, then the compound interest is
(a) Less than simple interest
(b) Equal to simple interest
(c) Greater than or equal to simple interest
(d) Greater than simple interest
Solution:
Greater than simple interest.
(7) The present population of a city is 12,00,000. If it increases at the rate of 8% every year, then the population of the city after 2 years is
(a) 199680
(b) 1399680
(c) 1500000
(d) 1299680
Solution:
1200000 (1 + 8/100) = 1200000 × 729/625 = 1399680
Ans – (b) 1399680.