An excavator costs Rs. 20,00,000 and has an estimated life of 8 years. It has no salvage value at the end of 8 years. The book value of the excavator at the end of 3 years using general double declining balance method is
A. Rs. 8,43,750
B. Rs. 8,75,000
C. Rs. 10,50,000
D. Rs. 11,56,250
Answer: Option A
Solution (By Examveda Team)
The double declining balance (DDB) method is an accelerated depreciation method that depreciates the asset faster in the earlier years.The formula for DDB depreciation is:
Depreciation for the year = 2 × (1 / useful life) × book value at beginning of year
Given:
Cost of excavator = Rs. 20,00,000
Estimated life = 8 years
Salvage value = Rs. 0
Depreciation rate = 2 × (1 / 8) = 25%
Year 1:
Depreciation = 25% of 20,00,000 = Rs. 5,00,000
Book value at end of Year 1 = 20,00,000 - 5,00,000 = Rs. 15,00,000
Year 2:
Depreciation = 25% of 15,00,000 = Rs. 3,75,000
Book value at end of Year 2 = 15,00,000 - 3,75,000 = Rs. 11,25,000
Year 3:
Depreciation = 25% of 11,25,000 = Rs. 2,81,250
Book value at end of Year 3 = 11,25,000 - 2,81,250 = Rs. 8,43,750
Therefore, the correct answer is: Rs. 8,43,750 (Option A).
This Question Belongs to Civil Engineering >> Construction Planning And Management
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Comments (4)
The normal time required for the completion of project in the above problem is
A. 9 days
B. 13 days
C. 14 days
D. 19 days
A. $$\frac{{{{\text{t}}_{\text{o}}} + 3{{\text{t}}_{\text{m}}} + {{\text{t}}_{\text{p}}}}}{2}$$
B. $$\frac{{{{\text{t}}_{\text{o}}} + 3{{\text{t}}_{\text{m}}} + {{\text{t}}_{\text{p}}}}}{3}$$
C. $$\frac{{{{\text{t}}_{\text{o}}} + 4{{\text{t}}_{\text{m}}} + {{\text{t}}_{\text{p}}}}}{4}$$
D. $$\frac{{{{\text{t}}_{\text{o}}} + 4{{\text{t}}_{\text{m}}} + {{\text{t}}_{\text{p}}}}}{5}$$
E. $$\frac{{{{\text{t}}_{\text{o}}} + 4{{\text{t}}_{\text{m}}} + {{\text{t}}_{\text{p}}}}}{6}$$
A construction schedule is prepared after collecting
A. Number of operations
B. Output of labour
C. Output of machinery
D. All the above
A. 3.5 and $$\frac{5}{6}$$
B. 5 and $$\frac{{25}}{{36}}$$
C. 3.5 and $$\frac{{25}}{{36}}$$
D. 4 and $$\frac{5}{6}$$
FDBB=2/n = 2/8 = 0.25
Book value after 3 year = Ci (1-FDDB)^3. = 20000000(0.75)^3 = 843750
Cost of asset=20, 00,000
Useful life= 8years
Depreciation rate=1/8*100=12.5%
Double declining balance formula=2*cost of asset*depreciation rate
1st year=2*2000000*12.5%=500000
So book value at the end of 1st year= 20L-5L=15L
2nd year=2*1500000*12.5%=375000
Book value at the end of 2nd year=15L-3.75L=11.25L
3rd year=2*1125000*12.5%=281250
Book value at the end of 3rd year= 11.25L-281250=843750
Solve?
DEPRECIATION (D) = {(2/N)*Initial Cost}
Book Value = Initial Cost - D
D1 = (2/8)*2000000 =500000
BV1= 2000000-500000 =1500000
D2 = (2/8)*1500000=375000
BV2=1500000-375000=1125000
D3=(2/8)*1125000=281250
BV3=1125000-281250=843750
ANS: A