Ms. Wilson invested $11.000 in two accounts, one yielding 8% interest and the other yielding 12%. If she received a total of $1,080 in interest at the end of the year how much did she invest in each account?

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Answer 1 · 2016-08-05T17:22:03

8% account - $6000
12% account - $5000

Let's call the money invested in the 8% account $a$ and the money in the 12% account $b$ .

We know that $a + b = 11000$ .

To work out the interest, let's convert the percentages to decimals.

$8% = 0.08 and 12% = 0.12$

So $0.08a + 0.12b = 1080$

We now have a system of simultaneous equations, I'm going to solve via substitution.

$a = (1080-0.12b)/(0.08)$

$(1080-0.12b)/(0.08) + b = 11000$

Multiply both sides by $0.08$

$1080 - 0.12b + 0.08b = 11000*0.08$

$0.04b = 1080 - 11000*0.08$

$b = (1080-11000*0.08)/(0.04) = 5000$

$a+b = 11000 implies a = 6000$