SOLUTION: Question:
Jane invested some amount at the rate of 12% simple interest and some other amount at the rate of 10% simple interest. She received yearly interest of $130. Randy also
Question 69242: Question:
Jane invested some amount at the rate of 12% simple interest and some other amount at the rate of 10% simple interest. She received yearly interest of $130. Randy also invested in the same scheme, but he interchanged the amounts invested, and received $4 more as interest. How much amount did each of them invest a different rates?
So far, I have:
x(.12) + y(.10)= $130
x + y = $134 Found 2 solutions by stanbon, checkley75:Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website! Jane invested some amount at the rate of 12% simple interest and some other amount at the rate of 10% simple interest. She received yearly interest of $130. Randy also invested in the same scheme, but he interchanged the amounts invested, and received $4 more as interest. How much amount did each of them invest a different rates?
So far, I have:
1st: x(.12) + y(.10)= $130
2nd: x(0.10) + y(0.12) = $134
--------------------------------
Multiply each equation by 100 to get:
3rd: 12x + 10y = 13000
4th: 10x + 12y = 13400
-----------
Multiply 3rd by 10 and 4th by 12 to get:
5th: 120x + 100y = 130000
6th: 120x + 144y = 160800
---------------------
Subtract 5th from 6th to get:
44y = 30800
y=$700
Substitute into 4th to find "x"
10x+12*700 = 13400
10x + 8400 = 13400
10x = 5000
x=$500
-------------
Amounts invested were $700 and $500
Cheers,
Stan H.
You can put this solution on YOUR website! .12X+.10Y=130
.10X+.12Y=134 OR
.12X+.10Y-(.10X+.12Y)=-4
.12X-.10X+.10Y-.12Y=-4
.02X-.02Y=-4
.02X=.02Y-4
X=.02Y/.02-4/.02
X=Y-200 NOW SUBSTITUTE (Y-200) FOR X IN EITHER EQUATION. THUS:
.12(Y-200)+.10Y=130
.12Y-24+.10Y=130
.22Y=130+24
.22Y=154
Y=154/.22
Y=$700 INVESTED @ 10% TO GET THE $130 INTEREST.
X=7-200
X=500 INVESTED @ 12% TO GET $130 INTEREST.
PROOF
.10*500+.12*700=134
50+84=134
134=134
(