Stan invested $5,000, part at 8% and part at 17%. If the total interest at the end of the year is $490, how much did he invest at 8%?
A) $5,000 B) $4,000 C) $3,000 D) $1,000
2006-09-20 02:06:27 · 3 answers · asked by BlueEyes4172004 1 in Science & Mathematics ➔ Mathematics
let stan invest $x in first part.
stan invest $5000-x in second part
Interest earned in 1st part =xX1X8%
=$8x%
Interest earned in 2nd part=(5000-x)X1X17%
Total interest earned = 8x%+(5000-x)X17% which is equal to 490
that is 8x%+(5000-x)X17% = 490
Solving this simple equation, you get
8x+85000-17x =49000
-9x=49000-85000
-x= -36000
x= 4000
Therefore he invested $4000 at 8%
2006-09-20 02:51:24 · answer #1 · answered by dudul 2 · 0⤊ 0⤋
Let x=investment at 8%.
Therefore, 5000-x was invested at 17%.
Given: x(8%)+(5000-x)(17%)=490
Simplify and solve for x:
8x/100+5000(17)/100-17x/100=490.
Sorry, something wrong with my computer.
The 49..shown above should read
490.
Note: 8% =8/100 and 17%=17/100.
Multiply each term by 100:
8x+5000(17)-17x=490(100)
8x+85000-17x=49000
8x-17x=49000-85000
-9x=-36000
The negative sign will cancel out. Divide each term by 9:
x=36000/9
=4000
He invested $4000 at 8%.
2006-09-20 11:54:57 · answer #2 · answered by tul b 3 · 0⤊ 0⤋
4,000 @ 8%
and 1,000 @ 17%
2006-09-20 09:10:47 · answer #3 · answered by Jazz 4 · 0⤊ 0⤋