Question 694590
Jack has $10,000 to invest in three accounts. He invests part of the money at 7% annual interest and twice that amount in an account paying 5.5% annual interest. The remaining money is invested at 4% annual interest. How much was invested in Each account if the total annual income from the three investments was $490?

I tried the question and got it wrong. I did:

.07x+5.5(2x)=490
.07x+11x=490
11.07x=490
x=44.26

4y=10000
y=2500

$44.26=7%
$88.52=5.5%
$2500=4%


Sorry to say but you're way off!!


Let amount invested in the 7% and 4% interest-accounts be S, and F, respectively
Amount he invested in account paying 5.5% = 2S
Therefore, S + F + 2S = 10,000 ---- 3S + F = 10,000 ----- eq (i)
Also, .07S + .055(2S) + .04(F) = 490 --- .07S + .11S + .04F = 490 --- .18S + .04F = 490 --- eq (ii) 
- .12S - .04F = - 400 ------ Multiplying eq (i) by - .04
.06S = 90 ---- Adding eqs (ii) & (i)
S, or amount invested in the {{{highlight(7)}}}% interest-account = {{{90/.06}}}, or ${{{highlight_green(1500)}}}


Amount invested in the {{{highlight(5.5)}}}% interest-account = 2(1,500), or ${{{highlight_green(3000)}}}


3(1,500) + F = 10,000 ------ Substituting 1,500 for S in eq (i)
4,500 + F = 10,000
F, or amount invested in the {{{highlight(4)}}}% interest-account = 10,000 – 4,500, or ${{{highlight_green(5500)}}}


You can do the check!!


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