SOLUTION: In testing a engine, various mixtures of gasoline and methanol are being tried. How much of a 90% gasoline mixture and a 75% gasoline mixture are needed for 1200L of an 85% gasolin

Question 491829: In testing a engine, various mixtures of gasoline and methanol are being tried. How much of a 90% gasoline mixture and a 75% gasoline mixture are needed for 1200L of an 85% gasoline mixture?
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
How much of a 90% gasoline mixture and a 75% gasoline mixture are needed for 1200L of an 85% gasoline mixture?
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Equation:
gas + gas = gas
0.90x + 0.75(1200-x) = 0.85*1200
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Multiply thru by 100 to get:
90x + 75*1200 - 75x = 85*1200
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15x = 10*1200
x = (2/3)1200
x = 800 L (amt. of 90% mixture needed)
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1200-x = 400 L (amt of 75% mixture needed)
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Cheers,
Stan H.
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