SOLUTION: Hi I need help on this word problem. Could someone please help?
Susan invests 2 times as much money at 10% as she does at 6%. If her total interest after 1 year is $780, how muc
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Susan invests 2 times as much money at 10% as she does at 6%. If her total interest after 1 year is $780, how muc
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Question 1145188: Hi I need help on this word problem. Could someone please help?
Susan invests 2 times as much money at 10% as she does at 6%. If her total interest after 1 year is $780, how much does she have invested at each rate? Found 2 solutions by richwmiller, ikleyn:Answer by richwmiller(17219) (Show Source):
Let x be the amount invested at 6%, in dollars.
Then the amount invested at 10%, is 2x dollars.
The 6% investment produces the interest of 0.06x dollars.
The 10% investment produces the interest of 0.1*(2x) dollars.
The total interest is
0.06x + 0.1*(2x) = 780 dollars.
Simplify and solve for x
0.06x + 0.2x = 780
0.26x = 780
x = = 3000.
Answer. $3000 invested at 6%; 2*3000 = 6000 dollars invested at 10%.
CHECK. 0.06*3000 + 0.1*6000 = 780 dollars. ! Correct !.
Solved.
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It is a standard and typical problem on investments.
You will find there different approaches (using one equation or a system of two equations in two unknowns), as well as
different methods of solution to the equations (Substitution, Elimination).
