A gold mine is projected to produce $52,000 during its first year of operation, $50,000 the second year, $48,000 the third year, and so on. If the mine is expected to produce for a total of 20 years, and the effective annual interest rate is 4%, its present worth is nearest to: a. $679,515
b. $1,127,810
c. $456,385
d. $483,566

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

t= half of the arc subtended by t.

r= half of the arc subtended by r.  

t+r= half of the arc subtended by t + half of the arc subtended by r= half of 360=180

step-by-step explanation: