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S Chand Class 9 ICSE Maths Solutions Chapter 2 Compound Interest Ex 2(D)

Question 1.
The cost of a machine depreciates by 10% every year. If its present value is ₹ 18,000; what will be its value after three years.
Solution:
Present value of machine (P) = ₹ 18000
Rate of depreciation (R) = 10% p.a.
Period (n) = 3 years
∴ Value of after 3 years = P( 1 – $$\frac { R }{ 100 }$$)n
= ₹ 18000 $$\left(1-\frac{10}{100}\right)^3$$
= ₹ 18000 ($$\frac { 9 }{ 10 }$$)³
= ₹ 18000 x $$\frac { 9 }{ 10 }$$ x $$\frac { 9 }{ 10 }$$ x $$\frac { 9 }{ 10 }$$ = ₹ 13122

Question 2.
The population of a town increased by 20% every year. If its present population is 2,16,000, find it population (i) after 2 years (ii) 2 years ago.
Solution:
Present population = 216000
Rate of increase = 20% p.a.
Period (n) = 2 years
(i) ∴ Population after 2 years = P ( 1 + $$\frac { R }{ 100 }$$)n
= 216000 ( 1 + $$\frac { 20 }{ 100 }$$)2
= 216000 x ($$\frac { 6 }{ 5 }$$)²
= 216000 x $$\frac { 6 }{ 5 }$$ x $$\frac { 6 }{ 5 }$$
= 8640 x 36 = 3,11,040

(ii) Population 2 years ago
= A ÷ ( 1 + $$\frac { R }{ 100 }$$)n
= 216000 ÷ ( 1 + $$\frac { 20 }{ 100 }$$)2
= 216000 ÷ ($$\frac { 6 }{ 5 }$$)2
= 216000 ÷ $$\frac { 5 }{ 6 }$$ x $$\frac { 5 }{ 6 }$$ = 1,50,000

Question 3.
The machinery of a certain factory is valued at ₹ 18,400 at the end of 1980. If it is supposed to depreciate each year at 8% of the value at the beginning of the year, calculate the value of the machine at the end of 1979 and 1981. (ICSE)
Solution:
Value machinery at the end of 1980 = ₹ 18400
P,ate of depreciation (R) = 8% p.a.
(a) Period = 1 year ago i.e., in 1979
∴ P = A ÷ ( 1 – $$\frac { R }{ 100 }$$)
= 18400 ÷ ( 1 – $$\frac { 8 }{ 100 }$$)
= 18400 ÷ ($$\frac { 23 }{ 25 }$$) = $$\frac{18400 \times 25}{23}$$ = 20, 000

(b) Population after one year i.e., in 1981
= P ( 1 – $$\frac { R }{ 100 }$$) = 18400( 1 – $$\frac { 8 }{ 100 }$$)
= 18400 x $$\frac { 23 }{ 25 }$$ = 16928

Question 4.
The present value of a scooter is ₹ 15,360.
If its value depreciates 12 $$\frac { 1 }{ 2 }$$ % every’ year, find its value after 3 years.
Solution:
Present value of scooter (P) = ₹ 15360
Rate of depreciation (R) = 12$$\frac { 1 }{ 2 }$$% = $$\frac { 25 }{ 2 }$$% p.a.
Period (n) = 3 years
∴ Value after 3 years = P ( 1 – $$\frac { R }{ 100 }$$)n
= ₹ 15360 x ( 1 – $$\frac { 25 }{ 20×100 }$$)3
= ₹ 15360 x ($$\frac { 7 }{ 8 }$$)3
= ₹ 15360 x $$\frac { 7 }{ 8 }$$ x $$\frac { 7 }{ 8 }$$ x $$\frac { 7 }{ 8 }$$ = ₹ 10290

Question 5.
A new car is purchased for ₹ 12,50,000. Its value depreciates at the rate of 10% in the first year, 8% in the 2nd year and then 6% every year. Find its value after 4 years.
Solution:
Value of car (P) = ₹ 2,50,000
Rate of depreciation (R) = 10% in first year, 8% in second year and then 6% every year Value of car after 4 years
= P$$\left(1-\frac{\mathrm{R}_1}{100}\right)\left(1-\frac{\mathrm{R}_2}{100}\right)\left(1-\frac{\mathrm{R}_3}{100}\right)\left(1-\frac{\mathrm{R}_4}{100}\right)$$
= ₹ 2,50,000$$\left(1-\frac{10}{100}\right)\left(1-\frac{8}{100}\right)\left(1-\frac{6}{100}\right)\left(1-\frac{6}{100}\right)$$
= ₹ 2,50,000 x $$\frac { 9 }{ 10 }$$ x $$\frac { 23 }{ 25 }$$ x $$\frac { 47 }{ 50 }$$ x $$\frac { 47 }{ 50 }$$
= ₹ 182905.20

Question 6.
The bacteria in a culture grows by 10% in the first hour, decreases by 10% in the second hour and again increases by 10% in the third hour. If the original count of the bacteria in a sample is 10000, find the bacteria count at the end of 3 hours.
Solution:
Original number of bacteria (P) = 10000
Increase in first hour (R1) = 10%
Decrease in second hour (R2) = 10%
Increase in third hour (R3) = 10%
∴ Number of bacteria after 3 hours
= P$$\left(1+\frac{\mathrm{R}_1}{100}\right)\left(1-\frac{\mathrm{R}_2}{100}\right)\left(1+\frac{\mathrm{R}_3}{100}\right)$$
= 1000 $$\left(1+\frac{10}{100}\right)\left(1-\frac{10}{100}\right)\left(1+\frac{10}{100}\right)$$
= 1000 $$\frac{11}{10} \times \frac{9}{10} \times \frac{11}{10}$$ = 10890

Question 7.
The production of refrigerators in factory rose from 40000 to 48400 in 2 years. Find the rate of growth p.a.
Solution:
Present production (P) = 40000
Production after 2 years (A) = 48400
Period (n) = 2 years
Let rate of increase = R% p.a.

∴ Rate of growth = 10% p.a.

Question 8.
The value of a flat worth ₹ 5,00,000 is depreciating at the rate of 10% p.a. In how many years will its value be reduced to ₹ 364500?
Solution:
Present value of plot (P) = ₹ 5,00,000
Value after depreciation (A) = ₹ 3,64,500
Rate of depreciation (R) = 10% p.a.
Let period = n years

Comparing both sides, n = 3
Period = 3 years

Question 9.
Rachit bought a flat for ₹ 10 lakh and a car for ₹ 3,20,000 at the same time. The price of the flat appreciates uniformly at the rate of 20% p.a.; while the price of the car depreciates at the rate of 15% p.a. If Rachit sells the flat and car after 3 years, what will be his profit or loss?
Solution:
Price of flat = ₹ 10,00,000
and price of car = ₹ 3,20,000
Rate of appreciation of flat (R1) = 20%
Rate of depreciation of car (R2) = 15%
Period (n) = 3 years
(a) Value of flat after 3 years = $$\mathrm{P}\left(1+\frac{\mathrm{R}}{100}\right)^n$$

Total cost of flat and car = ₹ 10,00,000 + ₹ 3,20,000 = ₹ 3,20,000
and selling price = ₹ 17,28,000 + ₹ 1,96,520 = ₹ 19,24,520
Gain = ₹ 19,24,520 – ₹ 13,20,000
= ₹ 16,04,520

Question 10.
8000 workers were employed by a company to complete a job in 4 years. At the end of first year, 5% of the workers were retrenched. At the end of second year, 5% of those working at that time were retrenched. However to complete the job in time, the number of workers was increased by 10% of those working at the end of third year. How many workers were working during the fouth year ?
Solution:
Number of workers in the beginning = 8000
Rate of retrenchment after first year = 5%
Rate of retrenchment after second year = 5%
Rate of increase after the third year = 10%
∴ Number of workers after third year or in the beginning of fourth year
= 8000 $$\left(1-\frac{5}{100}\right)\left(1-\frac{5}{100}\right)\left(1+\frac{10}{100}\right)$$
= 8000 x $$\frac { 19 }{ 20 }$$ x $$\frac { 19 }{ 20 }$$ x $$\frac { 11 }{ 10 }$$
= 7942