SOLUTION: Didn't know exactly where to put this problem. It's a worksheet and in the text book for the certain type of problem it's under Investment Problems.
Solve and check:
An inve
Algebra.Com
Question 56757: Didn't know exactly where to put this problem. It's a worksheet and in the text book for the certain type of problem it's under Investment Problems.
Solve and check:
An investment of $1500 paid interest of $45 during a certain period. How much interest would an investment of $1200 invested at the same rate for the same length of time pay?
Answer by tutorcecilia(2152) (Show Source): You can put this solution on YOUR website!
Since two items are being compared, set it up as a ratio problem:
a/b=c/d and solve for the missing variable:
.
$1500/$45=$1200/x
1500x=(45)(1200) [Cross-miltiply]
1500x=54,000
x=54,000/1500
x=$36
.
Check by plugging (x=36) back into the original equation.
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