|
Question 1185556: A woman has a total of $7,000 to invest. She invests part of the money in an account that pays 8% per year and the rest in an account that pays 10% per year. If the interest earned in the first year is $620, how much did she invest in each account?
Answer by greenestamps(13198)   (Show Source):
You can put this solution on YOUR website!
(1) Using a traditional formal algebraic approach (good practice in solving problems using algebra)....
x = amount invested at 8%
7000=x = amount invested at 10%
The total interest was $620:
.08(x)+.10(7000-x)=620
Solve using basic algebra -- probably start by multiplying everything by 100 to clear the decimals....
(2) A quick and easy way to solve any 2-part "mixture" problem like this, if a formal algebraic solution is not required....
$7000 all invested at 8% would yield $560 interest; all invested at 10% would yield $700 interest; the actual interest was $620.
The amount invested at each rate is exactly determined by where the actual interest of $620 lies between the two extremes $560 and $700.
(1) Picture the three interest amounts $560, $620, and $700 on a number line and observe/calculate that $620 is 60/140 = 3/7 of the way from $560 to $700.
(2) That means 3/7 of the total was invested at the higher rate.
ANSWER: 3/7 of the $7000, or $3000, was invested at 10%; the other $4000 at 8%.
CHECK: .10(3000)+.08(4000)=300+320=620
|
|
|
| |
