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?
4 answers
I think my answer was wrong. It should be P7,425. Am I right?
Solution pls
Hi! Can you show the solution to this problem? Thank you!
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
1 answer