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Maintenance cost of an equipment is P20,000 for 2 years, P40,000 at the end of 4 years, and P80,000 at the end of 8 years. Comp...Asked by Cathy
Maintenance cost of an equipment is P20,000 for 2 years, P40,000 at the end of 4 years, and P80,000 at the end of 8 years. Compute the semi-annual amount that will be set aside for this equipment. Money worth 10% compounded annually.
My answer is P152,166.38. Is it correct? If not, can anyone give me the final answer or the steps on how to correctly do it?
My answer is P152,166.38. Is it correct? If not, can anyone give me the final answer or the steps on how to correctly do it?
Answers
Answered by
Cathy
I think my answer was wrong. It should be P7,425. Am I right?
Answered by
Anonymous
Solution pls
Answered by
Ann
Hi! Can you show the solution to this problem? Thank you!
Answered by
Mari
Convert Annually to Semi-annually
i(annually)=i(semi-annually)
(1+0.10)^1=(1+X)^2
X=0.0488
P=A( (1-(1+x)^(-8(2)) )/x ) *note: times 2 at the exponent because of semi-annually
P=(20000/(1+x)^(2)(2)) + (40000/(1+x)^(4)(2)) + (80000/(1+x)^(8)(2))
P=81170.05
P=A( (1-(1+x)^(-8(2)) )/x )
81170.05 = A( (1-(1+x)^(-8(2)) )/x )
A=7426.19
i(annually)=i(semi-annually)
(1+0.10)^1=(1+X)^2
X=0.0488
P=A( (1-(1+x)^(-8(2)) )/x ) *note: times 2 at the exponent because of semi-annually
P=(20000/(1+x)^(2)(2)) + (40000/(1+x)^(4)(2)) + (80000/(1+x)^(8)(2))
P=81170.05
P=A( (1-(1+x)^(-8(2)) )/x )
81170.05 = A( (1-(1+x)^(-8(2)) )/x )
A=7426.19
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