An earth moving equipment costs Rs. 5,00,000 and has an estimated life of 10 years and a salvage value of Rs. 50,000. What uniform annual amount must be set aside at the end of each of the 10 years for replacement if the interest rate is 8% per annum and if the sinking fund factor at 8% per annum interest rate for 10 years is 0.069?
A. Rs. 31050
B. Rs. 34500
C. Rs. 37950
D. Rs. 50000
Answer: Option A
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Comments ( 6 )
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 $$\frac{5}{6}$$
F=500000-50000=450000
F=P/sinking factor
Sinking factor=i/[{(1+i)^n}-1]
P=F*SF=450000*0.069=31050
F=500000-50000=450000
F=P/sinking factor
Sinking factor=1/(1+I)^n
P=F*SF=450000*0.069=31050
Ans a
Solve with method
Future value=((R X (1+i)^n -1)/i)
we have to find R=?
Future value =500000-50000=450000
i=0.08( as 8% rate of interest)
n=10 yrs
apply all and get the answer.
Please give the method of calcution
A