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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 (R_{1}) = 10%

Decrease in second hour (R_{2}) = 10%

Increase in third hour (R_{3}) = 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 (R_{1}) = 20%

Rate of depreciation of car (R_{2}) = 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