If you have 20 gallons of 10%, how many gallons of 70% do you need to mix to have a total mixture of 50%
2006-06-27 11:48:14 · 4 answers · asked by boo 1 in Science & Mathematics ➔ Mathematics
20(.10) + .70(x) = .50(x+20)
2 + .7x = .5x+10
.2x=8
x=40
40 gallons
2006-06-27 11:54:09 · answer #1 · answered by jenh42002 7 · 0⤊ 0⤋
20*10/100+x*70/100=(20+x)50/100
Which means:
we have 20 gallons of 10%(10/100=10%) and you need an unknown (x) number of gallons of 70%(70/100) to mix and produce a "y" number of gallons where y=20+x - considering you keep the total number of gallons constant (20+x : which means the total number of gallons used in the mixing process)
Steps:
20*10/100+x*70/100=(20+x)50/100 (*10)multiply the ecuation by 3 and divide the procentual coefficient to it's numitor =>
20+x*7=(20+x)*5 =>
20+7*x=100+5*x =>
2*x=80 =>
x=40
So, it means you need 40 gallons of 70% to mix with 20 gallons of 10% to result a normal distribution of 50% per gallon in 20+40=60 total gallons used in the process
2006-06-27 20:06:49 · answer #2 · answered by theslimdim 1 · 0⤊ 0⤋
40
2006-06-27 20:33:36 · answer #3 · answered by becca 3 · 0⤊ 0⤋
who knows who cares?
2006-06-27 18:53:33 · answer #4 · answered by btcldm 1 · 0⤊ 0⤋
how many total gallons of the 50% mixture do you need?
2006-06-27 18:56:21 · answer #5 · answered by lobster17 2 · 0⤊ 0⤋