SOLUTION: One type of gasohol is 10% alcohol and another type is 35% alcohol. How much of each type must be mixed to obtain 400 gallons of a 25% alcohol mixture?
Question 62496: One type of gasohol is 10% alcohol and another type is 35% alcohol. How much of each type must be mixed to obtain 400 gallons of a 25% alcohol mixture? Found 4 solutions by checkley71, josmiceli, stanbon, venugopalramana:Answer by checkley71(8403) (Show Source):
You can put this solution on YOUR website! 400*.25=.10X+.35(400-X)
100=.10X+140-.35X
.25X=40
X=40/.25
X=160 GALLONS OF 10 % GASOHOL
400-160=240 GALLONS OF 35 % ALCOHOL
PROOF
400*.25=.10*160+.35*240
100=16+84
100=100
You can put this solution on YOUR website! a = gasohol 10% alcohol
b = gasohol 35% alcohol
multiply 2nd equation by .1
subtract 2nd from 1st
240 gal of the .1 mixture and
160 gal of the .35 misture
You can put this solution on YOUR website! One type of gasohol is 10% alcohol and another type is 35% alcohol. How much of each type must be mixed to obtain 400 gallons of a 25% alcohol mixture?
-------------
Let amount of 10% solution be "x" gallons; this has 0.10x gals. of alcohol
----------
Amount of 35% solution is "400-x" gallons; this has 0.35(400-x)=140-0.35x gals
------------------
EQUATION:
alcohol + alcohol = 25%(400) gallons
0.10x+140-0.35x=100
-0.25x=-40
x=160 gallons (amount of 10% solution)
400-x=240 gallons (amount of 35% solution)
Cheers,
Stan H.
You can put this solution on YOUR website! One type of gasohol is 10% alcohol( T SAY) and another type is 35% alcohol(F SAY). How much of each type must be mixed to obtain 400 gallons of a 25% alcohol mixture?
T+F = 400 G......F = 400 - T G....................1
ALCOHOL IN T = T*10/100=0.1T G.....................2
ALCOHOL IN F = F*35/100=0.35F=0.35(400-T)=140-0.35T G.............3
TOTAL ALCOHOL IN THE MIX = 400*25/100=100 G.................4
HENCE
0.1T+140-0.35T=100
0.35T-0.1T=140-100=40
0.25T = 40
T=40/0.25=160 G............IS 10% ALCOHOL IN THE MIX
F=400-160=240 G............IS 35% ALCOHOL IN THE MIX