Colin invested $16,000, part at 7% and part at 5%. If the total interest at the end of the year is $920?

Question

Colin invested $16,000, part at 7% and part at 5%. If the total interest at the end of the year is $920, how much did he invest at each rate?

Answer 1

x=.07
y=.05

.07x+.05y=920
x+y=16,000

Multiply the second equation by -.05:
.07x+.05y=920
-.05x-.05y=-800
=============
.02x=120
x=$6,000

.07(6,000)+.05y=920
420+.05y=920
.05y=500
y=$10,000

$6,000 was invested at 7% and $10,000 was invested at 5%.

Answer 2

Let x represent the amount invested at 7%,
let y represent the amount invested at 5%.
x + y = 16000
.07x + .05y = 920
x = 16000-y
.07(16000-y) + .05y = 920
7(16000-y) + 5y = 92000
112000-7y +5y = 92000
2y = 20000
y = $10,000 invested at 5% = $500
$6,000 invested at 7% = $420

Answer 3

7%x + (16000 -- x)*5% = 920
=> 7x + 80000 -- 5x = 92000
=> 2x = 12000
=> x = 6000
$6000 @ 7% and $10000 @ 5%
= $420 + $ 500
= $920

Answer 4

howdy Nicole! a million. enable's say that he invested $a at sixteen% and $b at 3%. we've a gadget: a+b=37000 .16a +.03b=3320 fixing it, we've a=17000 and b=20000. answer: $17000 2. enable's say that the dimensions of the rectangle is l, and the width is w. we've a gadget: l=2w+a million 2l+2w=32 fixing it, we've l=11 and a couple of=5. answer: 5 in. 3. enable's call the bigger sort p and the smaller sort q. we've a gadget: p=q+12 q+2p=39 fixing it, we've p=17 and q=5. answer: 17 wish this helps! in case you will like greater help on truthfully fixing the structures, incredibly than merely a thank you to set them up and the solutions, positioned it in greater information and that i'll gladly complicated! -M?TH ?V?NG?R