Question 4 (d)
Diminishing balance depreciation: average rate and the rate for the last two years
6 marks
- Time to write
- About 12 minutes
Question
Question 4 (d) · 6 marksA machinery installed at a cost of ₹1,00,000 on a diminishing balance method stood at ₹34,020 after five years.
Compute the average rate at which the machinery was depreciated over the five years.
The rates of depreciation applied in the first three years were 20%, 30% and 25% respectively. Find 'k' if depreciation at k % per year was applied during the last two years. (6×3)
Reworded from the university paper: names and figures differ, the task is the same. The original paper belongs to the University of Delhi. See DU's question bank.
Before you write
Read off the question and its marks
Read from the question
- Compute…Find…2 asks, each with its own marks below. Answer both.
- ComputeFormula first, then the working, then the answer with its unit (₹, %, units).
What the paper leaves open
Part of "Attempt any three"; the marker printed here, (6×3), is read as 6 marks for each of the three parts attempted.
Write it in this order
- Read the data; note exactly what is asked 1 min
- 1 Average rate of depreciation 3 marks 5 min
- 2 Value after three years and k 3 marks 5 min
- Check: arithmetic, units (₹, %), final answer underlined 1 min
Try it first. About 12 minutes for 6 marks. Write it out, then open the solution.
Solution
Hover or tap a marked figure, like 72,000, to see how it is made and where else it appears.
Under the diminishing (reducing) balance method the book value falls by a fixed percentage of the opening value each year, so after n years at r a year the value is C(1 − r)ⁿ: the compound-interest relation A = P(1 + i)ⁿ with a negative rate.
(i) Average rate over the five years
1,00,000(1 − r)⁵ = 34,020 ⟹ (1 − r)⁵ = 0.3402 ⟹ 1 − r = (0.3402)^(1/5).
log 0.3402 = −0.468266, so log(1 − r) = −0.093653 and 1 − r = 0.806022.
r = 0.193978, i.e. about 19.40% a year.
(ii) The rate k for the last two years
After the first three years: 1,00,000 × 0.80 × 0.70 × 0.75 = ₹42,000.
| Year | Rate | Opening value (₹) | Depreciation (₹) | Closing value (₹) |
|---|---|---|---|---|
| 1 | 20% | 1,00,000 | 20,000 | 80,000 |
| 2 | 30% | 80,000 | 24,000 | 56,000 |
| 3 | 25% | 56,000 | 14,000 | 42,000 |
| 4 | 10% | 42,000 | 4,200 | 37,800 |
| 5 | 10% | 37,800 | 3,780 | 34,020 |
| Total | 65,980 |
42,000(1 − k/100)² = 34,020 ⟹ (1 − k/100)² = 0.81 ⟹ 1 − k/100 = 0.9 ⟹ k = 10.
Depreciation at 10% a year in years 4 and 5 brings the value to ₹34,020, as the table shows.
Where the 6 marks sit
| # | Component | Marks |
|---|---|---|
| 1 | Average rate of depreciation | 3 |
| 2 | Value after three years and k | 3 |
| Total | 6 |
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