DAV Class 8 Maths Chapter 6 Worksheet 4 Solutions: Compound Interest
Worksheet 4 is where compound interest steps out of the bank and into real life. There are no loans or deposits here. Instead, you deal with towns that grow, firms whose profits go up and down, a house that loses value, and a workforce that keeps changing. The good news is that the same formula you already know does all the work. This article solves all 7 questions of the worksheet, one by one, with the reasoning explained so you can follow it and not just copy it.
The idea behind this worksheet
Compound interest is just repeated percentage change. If something grows by R% every year, you multiply it by (1 + R/100) each year. If something falls by R% every year, you multiply it by (1 − R/100) each year.
For growth at a constant rate: Final value = Original value × (1 + R/100)n
For depreciation at a constant rate: Final value = Original value × (1 − R/100)n
When the rate changes from year to year, you do not use a power. You simply multiply the original value by one factor for each year, using that year’s own rate.
Question 1. Population growth at 8% per annum
The population of a town increases at 8% per annum. What will the population be after 2 years if it is 12,500 now?
The growth rate is constant, so we use the compound growth formula.
Population after 2 years = 12,500 × (1 + 8/100)2
= 12,500 × (27/25)2
= 12,500 × 729/625
= 20 × 729
= 14,580
Answer: 14,580
Question 2. Population with changing rates
Three years ago a town had 50,000 people. The annual increase in the three successive years was 4%, 5% and 4%. Find the present population.
The rate is different each year, so we multiply by a separate factor for each year, in order.
Present population = 50,000 × (26/25) × (21/20) × (26/25)
Take it step by step.
50,000 × 26/25 = 52,000
52,000 × 21/20 = 54,600
54,600 × 26/25 = 56,784
Answer: 56,784
Question 3. Profit that rises and then falls
A firm’s profit was ₹72,000 in 2005. The next year it increased by 7%, and in the following year it decreased by 5%. What is the profit after 2 years?
This question is a good lesson in why you cannot just add or subtract percentages. A 7% rise followed by a 5% fall is not a net 2% rise, because the fall applies to the new, larger profit.
Profit after the first year = 72,000 × 1.07 = ₹77,040
Profit after the second year = 77,040 × 0.95 = ₹73,188
Answer: ₹73,188
As a quick check, a straight 2% rise would give ₹73,440. The real figure is slightly lower, as expected.
Question 4. Population with birth and death rates
The population of a town is 64,000. The annual birth rate is 10.7% and the annual death rate is 3.2%. Find the population after 3 years.
Births add to the population and deaths take away from it, so the net growth rate is the difference.
Net growth rate = 10.7% − 3.2% = 7.5% per annum
Population after 3 years = 64,000 × (1 + 7.5/100)3
To keep the arithmetic clean, write 1.075 as 43/40. Then (43/40)3 = 79,507/64,000, so the population = 64,000 × 79,507/64,000 = 79,507.
Answer: 79,507
Notice how the 64,000 cancels completely. That is a useful clue whenever the starting number is a perfect cube like 403.
Question 5. Finding the rate of growth
A factory made 40,000 motor bikes in a year, and production rose to 48,400 in 2 years. Find the uniform rate of growth per annum.
This time we know the start and the end, and we need the rate.
48,400 = 40,000 × (1 + R/100)2
(1 + R/100)2 = 48,400/40,000 = 1.21
1 + R/100 = 1.1, since 1.1 × 1.1 = 1.21
R = 10
Answer: 10% per annum
Question 6. Depreciation of a house
Madhu bought a house for ₹1,31,25,000. Its value depreciates at 10% per annum. What is its sale price after 3 years?
Depreciation means the value falls every year, so we use (1 − R/100).
Sale price = 1,31,25,000 × (9/10)3
= 1,31,25,000 × 729/1,000
= 13,125 × 729
= 95,68,125
Answer: ₹95,68,125
Question 7. Changing number of workers
A builder employed 4,000 workers on a project. At the end of the first year 10% were removed. At the end of the second year 5% of those then working were retrenched. At the end of the third year the number was increased by 15%. How many workers worked in the fourth year?
Each change applies to the number of workers at that time, so we go year by year.
After the first year: 4,000 × 0.90 = 3,600
After the second year: 3,600 × 0.95 = 3,420
After the third year: 3,420 × 1.15 = 3,933
Answer: 3,933 workers in the fourth year
Quick answer table
| Question | Answer |
|---|---|
| 1 | 14,580 |
| 2 | 56,784 |
| 3 | ₹73,188 |
| 4 | 79,507 |
| 5 | 10% per annum |
| 6 | ₹95,68,125 |
| 7 | 3,933 workers |
Common mistakes in this worksheet
The biggest trap is adding up percentages when the rate changes. In Questions 2, 3 and 7, each year’s change works on the value left by the previous year, so you multiply the factors. You never add the percentages.
Another slip is using (1 + R/100) for depreciation. When something loses value, the factor is (1 − R/100), as in Question 6.
In Question 4, some students use the birth rate alone and forget to subtract the death rate. Always read the hint: net growth is births minus deaths.
Finally, in Question 5, you are finding the rate, not the amount. Divide first, take the square root, and only then convert back to a percentage.
Frequently asked questions
Is compound interest used only for money?
No. Any quantity that changes by a fixed percentage each period follows the same rule, including population, profit, the value of a machine, and the number of workers.
When do I multiply separate factors instead of using a power?
When the rate is different in each year. A power like (1.08)2 only works when the same rate repeats.
What is the difference between growth and depreciation in the formula?
Growth uses (1 + R/100), because the value rises. Depreciation uses (1 − R/100), because the value falls.
How can I check my answer quickly?
Estimate first. In Question 1, 8% of 12,500 is 1,000, so after one year you expect about 13,500 and after two years a bit above 14,500. The answer 14,580 fits.
Written for students following the DAV Secondary Mathematics book for Class 8. All answers were calculated and double-checked.