Answer:
The amount of money he should invest in a 1-year deposit is $3,333.333
The amount of money he should invest in a 2-year deposit is $4,333.333
The amount of money he should invest in a 3-year deposit is $8666.67
Step-by-step explanation:
The given parameters are;
Total investment = $12,000
Total interest = $600
Interest rate = r
Amount invested in a 2-year deposit, B = $1,000 + Amount invested in a 1-year deposit, A
Amount invested in a 3-year deposit, C = $12,000 - (A + B)
C = 12000 - 2·B + 1000 = 13000 - 2·B
∴ A + B + C = $12,000
A×r + 2×r×B + 3×r×C = $600
∴ r×(A + 2·B + 3·C) = $600
r×(12000 + B + 2·C) = $600
B = A + 1000
∴ r×(B - 1000 + 2·B + 3·(13000 - 2·B) = $600
r×(B - 1000 + 2·B + 3·(13000 - 2·B) = $600
r×(38000 - 3·B) = $600
∴B + 2·C = -3·B
4·B = 2·C
C = 2·B
Hence;
B - 1000 + B + 2·B = $12,000
3·B = 13,000
∴ B = 13000/3 = [tex]\$4,333\tfrac{1}{3}[/tex] = $4,333.333
B = A + 1000
∴ A = B - 1000 = [tex]\$3,333\tfrac{1}{3}[/tex] = $3,333.33
C = 2×B = [tex]\$8,666\tfrac{2}{3}[/tex] = $8666.67
Therefore;
The amount of money he should invest in a 1-year deposit = $3,333.333
The amount of money he should invest in a 2-year deposit = $4,333.333
The amount of money he should invest in a 3-year deposit = $8666.67.