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

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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

$x=\frac{120}{0.15}$

x=800

Hence, we need 1800 liters of the 90% gasoline mixture and 400 liters of the 75% gasoline mixture.

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=800

Hence, we need 1800 liters of the 90% gasoline mixture and 400 liters of the 75% gasoline mixture.

Jeffrey Jordon

Answered 2021-08-10
Author has **2495** answers

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