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A person invested $6100 for 1 year, part at 4%, part at 9%, and the remainder at 15%. The total annual income from these investments was $681. The amount of money invested

at 15% was $300 more than the amounts invested at 4% and 9% combined. Find the amount invested at each rate.

User Chenna V
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The amount invested at 4% was $81, and the amount invested at 9% was $825. The amount invested at 15% was $300 more than the combined amount invested at 4% and 9%, which is $81 + $825 + $300 = $1206.

Let x be the amount of money invested at 4%, and y be the amount of money invested at 9%. We know that x + y + (x + y + 300) = 6100, so the amount invested at 15% is x + y + 300.

We also know that 0.04x + 0.09y + 0.15(x + y + 300) = 681, which is the total annual income from these investments.

Substituting the value of x + y + 300, we get:

0.04x + 0.09y + 0.15(x + y + 300) = 681

0.04x + 0.09y + 0.15x + 0.15y + 0.15*300 = 681

0.19x + 0.24y + 45 = 681

0.19x + 0.24y = 636

We can now solve this system of equations using elimination method. First, we can multiply the first equation by 4 and the second equation by 9:

4x + 4y = 6100

9x + 9y = 5676

Then, we can add the two equations to eliminate the y-term:

(4x + 4y) + (9x + 9y) = 6100 + 5676

13x + 13y = 11776

x + y = 906

Substituting this value back into either of the original equations, we can solve for x and y:

x + y = 906

x = 906 - y

4(906 - y) + y = 6100

4*906 - 4y + y = 6100

3624 - 3y = 6100

-3y = -2476

y = 825

Substituting this value back into the equation x + y = 906, we can solve for x:

x + y = 906

x + 825 = 906

x = 81

Therefore, the amount invested at 4% was $81, and the amount invested at 9% was $825. The amount invested at 15% was $300 more than the combined amount invested at 4% and 9%, which is $81 + $825 + $300 = $1206.

We can check our solution by substituting the values of x and y into the original equations to verify that they are both satisfied.

User Manish Chhabra
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