Question

A fuel oil distributor has 120,000 gallons of fuel with 0.9% sulfur content, which exceeds pollution control standards of 0.8% sulfur content. How many gallons of fuel oil with a 0.3% sulfur content must be added to the 120,000 gallons to obtain fuel oil that will comply with the pollution control standards?​

Answer 1

Answer:

Here you have three amounts of oil: the original 120,000 at 0.9%, the amount added at 0.3% and the total amount at 0.8%.

 

Let X = the amount of fuel added.

Now 120,000 + X is the total amount of fuel you end up with.

 

You're actually interested in the amount of sulfur in each, so we can set up an equation that tracks it.

 

The first amount has 120,000 * 0.009 gallons of sulfur.

The amount added has X * 0.003 gallons of sulfur.

The total has (120,000+X) * 0.008 gallons of sulfur.

 

120,000 * 0.009 + X * 0.003 = (120,000+X) * 0.008

 

Distribute on the right and you should be able to solve for X. Good luck!

Answer 2

Given,

The fuel distributor has $120,000$ gallons of fuel with $0.9\%$ sulfur content.

The sulfur content in fuel that complies with the pollution control standards is $0.8\%$.

Let the fuel oil with $0.3\%$ sulfur content to be added to $120,000$ gallons to make the sulfur content in total fuel volume as $0.8\%$ is ‘$x$’.

The final volume of fuel oil is $(120,000+x)$.

Let us make the equation from the quantities of sulfur content in first, second and final volumes of fuel oil:

$(0.9\%\times120,000)+(0.3\%\times x)=(120,000+x)\times0.8\%$

$(0.9/100\times120,000)+(0.3/100\times x)=(120,000+x)\times0.8/100$

$(0.9\times1200)+0.003x=(120,000\times0.008)+0.008x$

$1080+0.003x=960+0.008x$

$1080-960=0.008x-0.003x$

$120=0.005x$

$x= 120/0.005$

$x=24000$.

Hence, $24,000$ gallons of fuel oil with $0.3\%$ sulfur is to be added to $120,000$ gallons of fuel oil with $0.9\%$ sulfur to comply with the pollution control standards.