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A total of $6000 is invested: part at 7% and the remainder at 11 %. How much is invested at each rate if the annual interest is $460?

A total of $6000 is invested: part at 7% and the remainder at 11 %. How much is invested-example-1
User Yvon
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If a total of $6000 is invested: part at 7% and the remainder at 11 %. $5000 is invested at 7%, and $1000 is invested at 11%.

What is the amount invested?

Let's assume that the amount invested at 7% is x dollars.

Amount invested at 11% can be represented as (6000 - x) dollars.

Use the formula for simple interest to set up an equation:

Interest = Principal * Rate * Time

Set up two equations using the amounts invested at each rate:

0.07x + 0.11(6000 - x) = 460

0.07x + 660 - 0.11x = 460

Combine like terms:

-0.04x + 660 = 460

Subtract 660 from both sides:

-0.04x = -200

Divide both sides by -0.04:

x = 5000

Therefore $5000 is invested at 7% and ($6000 - $5000 = $1000) is invested at 11%.

User Steve Saporta
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Let x = the amount invested at 11%.

Then 6000-x = the amount invested at 7%

The total interest earned is $460.

So we can write

0.11x + 0.07(6000-x) = 460

Solve for x:

0.11x - 0.07x + 420 = 460

x = 40/0.04

x=1000

Therefore the amount invested at 7% interest = 6000 - 1000 = $5000

User Rohit Nandi
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