A sum of $3500 is invested in two parts. One part brings a return of 5% and the other a return of 8%. The total annual return is $250. Find the amount invested at each rate.
2006-11-14 03:05:19 · 7 answers · asked by TIM H 1 in Science & Mathematics ➔ Mathematics
Solve a system of two equations
x + y=3500 --Originaly investeed
x(.05) + y(.08)=250 -- Annual return
I got y = $2500
and x = $1000
I hope it was helpful
2006-11-14 03:08:52 · answer #1 · answered by Edward 7 · 0⤊ 0⤋
Let the sum of money invested with 5% return be $a.
Thus sum of money invested with 8% return will be $(3500-a).
(5/100) x (a) + (8/100) x (3500-a) = 250
0.05a + 280 - 0.08a = 250
280 - 250 = 0.08a - 0.05a
30 = 0.03a
a = 30/0.03 = 1000
Thus $1000 is invested at 5% per annum while $2500 is invested at 8% per annum.
Hope this helps :)
2006-11-14 11:10:51 · answer #2 · answered by chyrellos 2 · 1⤊ 0⤋
Let investment in first part=x
then investment in second part=3500-x
total return=5*x/100+8*(3500-x)/100
5*x/100+8*(3500-x)/100=250
solve this for x
2006-11-14 11:13:16 · answer #3 · answered by Dupinder jeet kaur k 2 · 0⤊ 0⤋
i= prt, int = prin*rate*time
i1+i2 = 250, the sum of the two interests is 250
.05*x+.08*(3500-x) = 250,
x invested at 5%, (3500-x) at 8%...
2006-11-14 11:09:41 · answer #4 · answered by modulo_function 7 · 0⤊ 0⤋
Let the first part=x $
Other part=3500-x $
.05x+.08[3500-x]=250
.05x-.08x+280=250
.03x=30
x=1000 $
3500-x=2500 $
The parts were
1000$,2500$
verify
1000*.05+2500*.08
=50+200
=250
ok
2006-11-14 11:14:05 · answer #5 · answered by openpsychy 6 · 0⤊ 0⤋
One part is $1000 and the other part is $2500
2006-11-14 11:21:06 · answer #6 · answered by Anonymous · 0⤊ 0⤋
one amount is $1346.15
next amount is $2153.85
$3500.00
2006-11-14 11:22:38 · answer #7 · answered by colinhughes333 3 · 0⤊ 0⤋