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Class 8 RD Sharma Solutions – Chapter 14 Compound Interest – Exercise 14.4 | Set 2

  • Last Updated : 08 Apr, 2021

Chapter 14 Compound Interest – Exercise 14.4 | Set 1

Question 11. 6400 workers were employed to construct a river bridge in four years. At the end of the first year, 25% workers were retrenched. At the end of the second year, 25% of those working at that time were retrenched. However, to complete the project in time, the number of workers was increased by 25% at the end of the third year. How many workers were working during the fourth year?

Solution:

We have,

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Initial number of workers are = 6400



At the end of first year = 25% retrenched

At the end of second year = 25% retrenched

At the end of third year = 25% increased

By using the formula,

A = P (1 + \frac{R}{100}  )

Substituting the values, we have

x = 6400 (1 – \frac{25}{100}  ) (1 – \frac{25}{100}  ) (1 + \frac{25}{100}  )

= 6400 (\frac{75}{100}  ) (\frac{75}{100}  ) (\frac{125}{100}  )



= 6400 (0.75) (0.75) (1.25)

= 4500

Therefore,

Number of workers working during the fourth year is 4500.

Question 12. Aman started a factory with an initial investment of Rs 100000. In the first year, he incurred a loss of 5%. However, during the second year, he earned a profit of 10% which in the third year rose to 12%. Calculate his net profit for the entire period of three years.

Solution:

We have,

Initial investment by Aman = Rs.100000

In first year = incurred a loss of 5%

In second year = earned a profit of 10%

In third year = earned a profit of 12 %

By using the formula,

A = P (1 + \frac{R}{100}  )

Substituting the values, we have

x = 100000 (1 – \frac{5}{100}  ) (1 + \frac{10}{100}  ) (1 + \frac{12}{100}  )

= 100000 (\frac{95}{100}  ) (\frac{110}{100}  ) (\frac{112}{100}  )

= 100000 (0.95) (1.1) (1.12)

= 117040

Therefore,

Aman’s net profit for entire three years is 117040 – 100000 = Rs 17040.

Question 13. The population of a town increases at the rate of 40 per thousand annually. If the present population be 175760, what was the population three years ago.

Solution:



We have,

Annul increase rate of population of town = 40/1000× 100 = 4%

Present population of town = 175760

So, let the population of town 3 years ago be = x

By using the formula,

A = P (1 + \frac{R}{100}  )

Substituting the values, we have

175760 = x (1 + \frac{4}{100}  ) (1 + \frac{4}{100}  ) (1 + \frac{4}{100}  )

175760 = x (\frac{104}{100}  ) (\frac{104}{100}   (\frac{104}{100}

175760 = x (1.04) (1.04) (1.04)



175760 = 1.124864x

x = 175760/1.124864

= 156250

Therefore,

Population 3 years ago was 156250.

Question 14. The population of a mixi company in 1996 was 8000 mixies. Due to increase in demand it increases its production by 15% in the next two years and after two years its demand decreases by 5%. What will its production after 3 years?

Solution:

We have,

Population of mixi company in 1996 was = 8000 mixies

Production growth rate in next 2 years is = 15 %

Decrease rate in 3rd year is = 5%

By using the formula,

A = P (1 + \frac{R}{100}  )

Substituting the values, we have

x = 8000 (1 + \frac{15}{100}  ) (1 + \frac{15}{100}  ) (1 – \frac{5}{100}  )

= 8000 (\frac{115}{100}  0) (\frac{115}{100}  ) (\frac{95}{100}  )

= 8000 (1.15) (1.15) (0.95)

= 10051

Therefore,

Production after three years will be 10051 mixies.

Question 15. The population of a city increases each year by 4% of what it had been at the beginning of each year. If the population in 1999 had been 6760000, find the population of the city in (1) 2001 (ii) 1997.

Solution:



We have,

Annual increase rate of population of city is = 4%

Population in 1999 was = 6760000

(i) Population of the city in 2001 (2 years after)

By using the formula,

A = P (1 + \frac{R}{100}  )

Substituting the values, we have

x = 6760000 (1 + \frac{4}{100}  ) (1 + \frac{4}{100}  )

= 6760000 (\frac{104}{100}  ) (\frac{104}{100}  )

= 6760000 (1.04) (1.04)

= 7311616

Therefore,

Population in the year 2001 is 7311616

(ii) Population of city in 1997 (2 years ago)

By using the formula,

A = P (1 + \frac{R}{100}  )

Substituting the values, we have

x = 6760000 (1 – \frac{4}{100}  ) (1 – \frac{4}{100}  )

= 6760000 (\frac{96}{100}  ) (\frac{96}{100}  )

= 6760000 (0.96) (0.96)



= 6230016

Therefore,

Population in the year 1997 was 6230016.

Question 16. Jitendra set up a factory by investing Rs. 2500000. During the first two successive years his profits were 5% and 10% respectively. If each year the profit was on previous year’s capital, compute his total profit.

Solution:

We have,

Initial investment by Jitendra was = Rs 2500000

Profit in first 2 successive years were = 5% and 10%

Final investment after two successive profits = 2500000 (1 + \frac{5}{100}  ) (1 + \frac{10}{100}  )

= 2500000 (\frac{105}{100}  ) (\frac{110}{100}  )

= 2500000 (1.05) (1.1)

= 2887500

Therefore,

Jitendra total profit is = 2887500 – 2500000 = Rs 387500.




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