Answer:
its present worth is nearest to 483,566
Option d) 483,566 is the correct answer
Step-by-step explanation:
Given that;
gold mine is projected to produce $52,000 during its first year
produce $50,000 the second year
produce $48,000 the third year
the mine is expected to produce for 20yrs; i.e n = 20
annual interest rate = 4% = 0.04%
now let P represent the present worth
we determine the present worth ;
Present worth ⇒ Cashflow(Uniform series present worth) - (2000)(uniform gradient present worth)
⇒Cashflow(P | A,i%,n) - (2000)(P | G,i%,n)
= 52000[ ((1+i)ⁿ - 1) / (i(1+i)ⁿ) ] - (2000)[ {((1+i)ⁿ - 1) / (i²(1+i)ⁿ))} - (n/i(1 + i)ⁿ) ]
= 52000[ ((1+0.04)²⁰ - 1) / (0.04(1+0.04)²⁰) ] - (2000)[ {((1+0.04)20 - 1) / ((0.04)²(1+0.04)²⁰))} - (20/0.04(1 + 0.04)²⁰) ]
= 52000[ ((1.04)²⁰ - 1) / (0.04(1.04)²⁰) ] - (2000)[ {((1.04)²⁰ - 1) / ((0.04)²(1.04)²⁰))} - (20/0.04(1.04)²⁰) ]
= 52000[ ((2.191123 - 1) / (0.04(2.191123) ] - (2000)[ {((2.191123 - 1) / ((0.0016)(2.191123))} - (20/0.04(2.191123) ]
= 52000(1.191123/0.08764) - (2000){( 1.191123/0.003506) - (20/0.87645)}
= 52000(13.59033) - (2000)(339.7582 - 228.1935)
= 52000(13.59033) - (2000)(111.5647)
= 706695.6 - 223129.4
= 483,566.2 ≈ 483,566
Therefore its present worth is nearest to 483,566
Option d) 483,566 is the correct answer
, where B is the base and h is the height. We can rearrange this formula to solve for B.
.
.