Respuesta :
Answer:
Mark will have at the end of six years the amount of $25,865.74
Explanation:
According to the given data we have the following:
First investment = 2500
Investment increasing at rate of 10%
Interest rate = 13%
t=6 years
Present value is given by formula = C * [((1+g)^n/(1+i)^n) - 1 ] / (g-i)
C is first value = 2,500
g is increase in investment = 0.10
i is intrest rate = 0.13
n is no of years = 6
Putting values into the equation
P = 2500* [((1+ 0.10)^6/(1+0.13)^6) - 1 ] / (0.10-0.13) 1.771561 2.08195
P = 2500* [((1.10)^6/(1.13)^6) - 1 ] / (-0.03)
P = 2500* [0.8509142870866 - 1 ] / (-0.03)
P = 2500* (-0.14908571)/ (-0.03)
P = 2500* 4.9695236
P=$12,423.809
Future value = P*(1+i)^t
= $12,423.809 *(1+0.13)^6
= $25,865.74
Mark will have at the end of six years the amount of $25,865.74
Answer:
$29,228.47
Explanation:
year savings investment total
returns
1 $2,500 (1 + 13%)⁶ $5,204.88
2 $2,750 (1 + 13%)⁵ $5,066.70
3 $3,025 (1 + 13%)⁴ $4,932.18
4 $3,327.50 (1 + 13%)³ $4,801.24
5 $3,660.25 (1 + 13%)² $4,673.77
6 $4,026.28 (1 + 13%) $4,549.70
total $19,289.03 $29,228.47
Since Mark earns compound interest, then the returns will be:
- 1.13⁶ = 2.082
- 1.13⁵ = 1.8424
- 1.13⁴ = 1.6305
- 1.13³ = 1.443
- 1.13² = 1.2769
- 1.13¹ = 1.13