SOLUTION: Joe wants to invest $30,000. He puts part of his money in company x with 0.4% interest return and part of his money in company y with 6% return. At the end of the year he received
Algebra ->
Customizable Word Problem Solvers
-> Finance
-> SOLUTION: Joe wants to invest $30,000. He puts part of his money in company x with 0.4% interest return and part of his money in company y with 6% return. At the end of the year he received
Log On
Question 912736: Joe wants to invest $30,000. He puts part of his money in company x with 0.4% interest return and part of his money in company y with 6% return. At the end of the year he received $1500 in interest. How much money did he invest in each company? Answer by richwmiller(17219) (Show Source):
You can put this solution on YOUR website! I will assume you really mean 4% and 6% other wise you earn about $21 at .4%
Total amount of money invested: $30000
x+y=30000,
Total yearly interest for the two accounts is: $1500
0.04*x+0.06*y=1500
x=30000-y
Substitute for x
0.04*(30000-y)+0.06*y=1500
Multiply out
1200-0.04*y+0.06*y=1500
Combine like terms.
0.02*y=300
Isolate y
y=$ 15000.00 at 6% earns $900 interest
x=30000-y
Calculate x
x=$ 15000.00 at 4% earns $600 interest
Check
0.04*15000+0.06*15000=1500
600+900=1500
1500=1500
If 1500=1500 is TRUE and neither x nor y is negative then all is well
codeint1