Chemical Engineering - Chemical Engineering Plant Economics - Discussion
Discussion Forum : Chemical Engineering Plant Economics - Section 2 (Q.No. 4)
4.
A reactor having a salvage value of Rs. 10000 is estimated to have a service life of 10 years. The annual interest rate is 10%. The original cost of the reactor was Rs. 80000. The book value of the reactor after 5 years using sinking fund depreciation method will be Rs.
Discussion:
4 comments Page 1 of 1.
Akshay Kore said:
4 years ago
The answer of this question should be C.
Solution:-
80000- (80000-10000){(1.1)^10-1)/(1.1)^5-1)} = 53185.31
Approx = 53196.
Solution:-
80000- (80000-10000){(1.1)^10-1)/(1.1)^5-1)} = 53185.31
Approx = 53196.
RAVI GUPTA said:
5 years ago
80000-((1.1)^5-1) * (80000-10000)/((1.1)^10-1) = 53196.
Ravi gupta said:
5 years ago
The answer is C.
Using the notation P-(P-S)*((1+m)^n----1))/((1+n)^n---1).
Using the notation P-(P-S)*((1+m)^n----1))/((1+n)^n---1).
Phahad said:
1 decade ago
Please I want the formula used in this case.
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