In testing an 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 1200 L of an 85% gasoline mixture?

Dolly Robinson 2021-03-08 Answered
In testing an 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 1200 L of an 85% gasoline mixture?
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Ezra Herbert
Answered 2021-03-09 Author has 99 answers
Let x be the amount of 90% gasoline mixture so that 1200-x is the amount of 75% gasoline mixture, both in liters.
In terms of percentage, we write: 0.90(x)+0.75(1200-x)=0.85(1200)
Solve for x: 0.90x+900-0.75x=1020
0.15x+900=1020
0.15x=120
x=1200.15
x=800
Hence, we need 1800 liters of the 90% gasoline mixture and 400 liters of the 75% gasoline mixture.
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Jeffrey Jordon
Answered 2021-08-10 Author has 2495 answers

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