Question 536220: Part of $11,000 was invested at 10% and the rest at 12%. If the annual income from these investments was $1260, how much was invested at each rate?
Please, Mr/Mrs Tutor i"ve tried countless times to evaluate this problem and several others similar to it. I need help. I need to learn how to answer any question like the stated one above by my-self, finals for me are approaching very quickly and I need to get the steps down and remembered. Thank you- Tray Tillmon
Answer by Earlsdon(6294)   (Show Source):
You can put this solution on YOUR website! First, let $x = the amount invested at 10%. The rest ($11,000-x) is the amount invested at 12%.
Now you need to express the amount earned in each case. Change the percentages to their decimal equivalents:
10% = 0.1
12% = 0.12
The amount earned on the $x can be expressed as: 0.1x and the amount earned on the rest can be expressed as 0.12($11,000-x). The sum of these two amounts is to equal $1260.00, so here's the equation that you'll need to solve for x.
0.1x+0.12(11000-x) = 1260 Simplify.
0.1x+1320-0.12x = 1260 Combine the x-terms.
-0.02x+1320 = 1260 Subtract 1320 from both sides.
-0.02x = -60 Finally, divide both sides by -0.02
x = $3,00 and $11,000-x = $8,000
$3,000 was invested at 10% and $8,000 was invested at 12%
Check:
0.1(3000)+0.12(8000) = 300+960 = 1260
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