SOLUTION: Harry invested $4,000.00 in two investments: one investment was a CD that paid 3.4% annual interest, and the other investment was a bond that paid 2.5% annual interest. If he rec

Algebra ->  Equations -> SOLUTION: Harry invested $4,000.00 in two investments: one investment was a CD that paid 3.4% annual interest, and the other investment was a bond that paid 2.5% annual interest. If he rec      Log On


   

Question 1200438: Harry invested $4,000.00 in two investments: one investment was a CD that paid 3.4% annual
interest, and the other investment was a bond that paid 2.5% annual interest. If he received a check in
the mail at the end of the year for $114.40 for both investments, how much did he invest in the CD?
Found 2 solutions by josgarithmetic, greenestamps:
Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
v invested at 3.4% rate CD
4000-v at the 2.5% rate bond

0.034v%2B0.025%284000-v%29=114.4
-
0.034v-0.025v=114.4-0.025%2A4000

v=%28114.4-0.025%2A4000%29%2F%280.034-0.025%29--------compute this.

Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


The response from the other tutor sets the problem up using the standard formal algebraic method.

If formal algebra is not required, here is a way to solve any 2-part mixture problem like this with less work.

A return of $114.40 on an investment of $4000 is an interest rate of 114.40/4000 = 0.0286 or 2.86%.

Compare that to the two separate interest rates of 2.50% and 3.40% and observe/calculate that 2.86% is 0.36/0.90 = 2/5 of the way from 2.5% to 3.4%. That means 2/5 of the total was invested at the higher rate.

ANSWERS: 2/5 of $4000, or $1600, was invested at 3.4%; the other $2400 at 2.5%.

CHECK: .034($1600)+.025($2400) = $54.40+$60.00 = $144.40