SOLUTION: Angus invested $18,000, part at 3% and part at 5%. If the total interest at the end of the year is $660, how much did he invest at each rate?

Algebra ->  Absolute-value -> SOLUTION: Angus invested $18,000, part at 3% and part at 5%. If the total interest at the end of the year is $660, how much did he invest at each rate?      Log On


   



Question 69843: Angus invested $18,000, part at 3% and part at 5%. If the total interest at the end of the year is $660, how much did he invest at each rate?
Found 3 solutions by Edwin McCravy, checkley75, tutorcecilia:
Answer by Edwin McCravy(20056) About Me  (Show Source):
You can put this solution on YOUR website!
Angus (was he a bull? J) invested $18,000, part at 3%
and part at 5%.  If the total interest at the end of 
the year is $660, how much did he invest at each 
rate?

Let $x represent the first part, that is, the amount 
invested at 3%

Now use the formula for "the rest"

æThe rest of   ö    æ The  ö   æ the ö
ça total amount÷  = çtotal ÷ - çfirst÷
çbesides the   ÷    èamountø   è partø
èfirst part    ø	

So the rest of the $18000 besides the $x
is = ($18000 - $x)

This amount (the rest besides the $x) was invested
at 5%. 

So we have x dollars invested at 3%
and (18000 - x) dollars invested at 5%

The interest on the x dollars is .03x dollars.

The interest on the (18000 - x) dollars is
.05(18000 - x)

We read:
>>...the total interest at the end of the year 
is $660...<<

|Interest |     |Interest|
|   on    |     |   on   |
|  the    |  +  |  the   |  =  $660 
|x dollars|     |18000-x |
                |dollars |
       
     .03x    +  .05(18000-x) = 660

Can you solve that? If not post again asking how.

You get x = $12000, which is the amount invested
at 3%.

To get the rest, which was invested at 5%, subtract
that $12000 from the total, $18000, and get $6000, 
the amount invested at 5%.

Edwin

Answer by checkley75(3666) About Me  (Show Source):
You can put this solution on YOUR website!
.05X+.03(18000-X)=660
.05X+540-.03X=660
.02X=660-540
.02X=120
X=120/.02
X=6000 INVESTED AT 5% THUS
18000-6000=12000 INVESTED AT 3%
PROOF
.05*6000+.03*12000=660
300+360=660
660=660

Answer by tutorcecilia(2152) About Me  (Show Source):
You can put this solution on YOUR website!
$x+$y=$18,000
or x+(18,000-x)=18,000
.
.03x+.05(18,000-x)=660 because:
$x invested at 3% + $y invested at 5%=A total of $660 in interest.
.
.
Simplify
.03x+.05(18,000-x)=660
.03x+900-.05x=660
.03x+900-900-.05x=660-900
.03x-.05x=-240
-.02=-240
-.02x/-.02=-240/-.02
x=12,000
.
Plugging x=$12,000 back into the original equation:
.03x+.05(18,000-x)=660
.03(12,000)+.05(18,000-12,000)=660
360+ .05(6,000)=660
360+300=660
660=660 checks out