A machine costing Rs. one lakh depreciates at constant rate 10%. Estimated useful life of machine is 8 years.
Match List I with List II
| LIST I | LIST II | ||
| A. | Total depreciation in 2nd and 3rd year is | I. | Rs. 81,000 |
| B. | Value of machine after one year is | II. | Rs. 17,100 |
| C. | Value of machine after 2 years is: | III. | Rs. 43050 |
| D. | Scrap value of machine is ; given \((1.1)^{8} = 2.144\) & \((0.9)^{8} = 0.4305\) | IV. | Rs. 90,000 |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → A-II, B-IV, C-I, D-III
Depreciation at a constant rate of 10% is charged on the value at the start of each year (the reducing balance method), so each year the value is multiplied by \(1 - 0.10 = 0.9\):
\(\text{value after } n \text{ years} = 1{,}00{,}000 \times (0.9)^{n}\)
(B) Value of machine after one year
This matches (IV) Rs. 90,000.
\(1{,}00{,}000 \times 0.9 = 90{,}000\)
(C) Value of machine after 2 years
This matches (I) Rs. 81,000.
\(90{,}000 \times 0.9 = 81{,}000\)
(A) Total depreciation in 2nd and 3rd year
This matches (II) Rs. 17,100.
2nd year: 10% of 90,000 \(= 9000\)
3rd year: 10% of 81,000 \(= 8100\)
total \(= 9000 + 8100 = 17{,}100\)
(D) Scrap value of machine (after 8 years)
This matches (III) Rs. 43050.
\(1{,}00{,}000 \times (0.9)^{8} = 1{,}00{,}000 \times 0.4305 = 43{,}050\)
(The given value of \((1.1)^{8}\) is not needed.)
Note: This topic was not in the CUET 2026 syllabus.