Question

Mrs. Ming invested an amount of money in two accounts for one year. She invested some at 8% interest and the rest at 6% interest. Her total amount invested was $1,500. At the end of the year, she had earned $106.40 in interest. How much had Mrs. Ming invested in the account paying 6%? $117 $680 $760 $820

Answer 1

0.06x+0.08 (1500-x)=106.4
Solve for x
0.06+120-0.08x=106.4
0.06x-0.08x=106.4-120
-0.02x=-13.6
X=13.6÷0.02
X=680 invested at 6%

Answer 2

Answer:

The correct option is B) $680.

Step-by-step explanation:

Consider the provided information.

Her total amount invested was $1,500. At the end of the year, she had earned $106.40 in interest.

Let she invested x amount with 8% interest rate.

Total amount she invested was $1,500, thus the amount she invested with 6% interest rate was 1500-x.

Total interest she earn was $106.40

Write this information into mathematical form.

$x\times \frac{8}{100}+\frac{6}{100}(1500-x)=106.40$

$x\times 0.08+0.06(1500-x)=106.40$

$0.08x+90-0.06x=106.40$

$0.08x-0.06x=106.40-90$

$0.02x=16.4$

$x=820$

Hence, she invested $820 with 8% interest rate.

The amount she invested with 6% interest rate was 1500-x.

Substitute the value of x in above.

1500-820=680

Hence, she invested $680 with 6% interest rate.

The correct option is B) $680.