Question

You invested $21,000 in two accounts paying 2% and 4% annual interest, respectively.
If the total interest earned for the year was $660, how much was invested at each rate?
The amount invested at 2% is $​

Answer 1

Answer:

• $12,000 at 4%
• $9,000 at 2%

Step-by-step explanation:

If x represents the amount invested at 4%, then the total interest earned is ...

  0.04x + 0.02(21000 -x) = 660

  0.02x = 240 . . . . . . . . . . . simplify, subtract 420

  x= 240/0.02 = 12,000 . . . . amount invested at 4%

  21,000 -x = 9,000 . . . . . . . . amount invested at 2%

The amount invested at 2% is $9,000.

The amount invested at 4% is $12,000.

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Comment on method of solution

These problems can usually be solved using one variable for the amount associated with the highest contribution to the mix. Here, that is the amount getting 4% interest. This keeps the numbers in the solution positive, helping to minimize errors. As you can see, the solution develops in about 2 or 3 steps. Other methods can take longer.

Answer 2

Answer:

• $12,000 at 4%

• $9,000 at 2%

Step-by-step explanation:

If x represents the amount invested at 4%, then the total interest earned is ...

 0.04x + 0.02(21000 -x) = 660

 0.02x = 240 . . . . . . . . . . . simplify, subtract 420

 x= 240/0.02 = 12,000 . . . . amount invested at 4%

 21,000 -x = 9,000 . . . . . . . . amount invested at 2%

The amount invested at 2% is $9,000.

The amount invested at 4% is $12,000.

_____

Comment on method of solution

These problems can usually be solved using one variable for the amount associated with the highest contribution to the mix. Here, that is the amount getting 4% interest. This keeps the numbers in the solution positive, helping to minimize errors. As you can see, the solution develops in about 2 or 3 steps. Other methods can take longer.