Question

A chemist needs to make 3 mL of a 10% methane solution from a 6% methane solution and a 40% methane solution.
If we let x be the number of mL of the 6% methane solution and y be the number of mL of the 40% methane
solution, how would you set up two equations using x and y to find how many mL of each solution the chemist
needs to mix to get the desired solution ?

Answer 1

Given:

A chemist needs to make 3 mL of a 10% methane solution from a 6% methane solution and a 40% methane solution

Aim:

We need to find equations to find the number of ml of each solution.

Explanation:

Let x be the number of mL of the 6% methane solution and y be the number of mL of the 40% methane

We know that the total number of ml of the 10% solution is 3ml.

The number of mL of the 6% methane solution + the number of mL of the 40% methane = the total number of ml of the 10% solution

$x+y=3$$6\text{ \% of x+40 6 of y =10 \% of 3}$

$\frac{6}{100}x+\frac{\text{40}}{100}\text{y =}\frac{10}{100}\times3$

Multiply both sides by 100.

$100\times\frac{6}{100}x+100\times\frac{\text{40}}{100}\text{y =100}\times\frac{10}{100}\times3$

$6x+40y=10\times3$

$6x+40y=30$

The required two equations are

$x+y=3$

$6x+40y=30$

$Consider\text{ the equation }x+y=3.$$x=3-y$

$Substitute\text{ x=3-y in the equation }6x+40y=30.$$6(y-3)+40y=30$

$6y-6\times3+40y=30$

$6y-18+40y=30$

Solve for y.

$6y+40y=30+18$

$46y=48$$y=\frac{48}{46}$$y=\frac{24}{23}$$y=1.04ml$

$Substitute\text{ }y=\frac{24}{23}\text{ in the equation x=3-y.}$$x=3-\frac{24}{23}$

$x=\frac{3\times23}{23}-\frac{24}{23}$

$x=\frac{69}{23}-\frac{24}{23}$

$x=\frac{69-24}{23}$

$x=\frac{45}{23}$$x=1.96ml$

Final answer:

The equations:

$x+y=3$

$6x+40y=30$

The number of mL of the 6% methane solution is 1.04 ml.

The number of mL of the 40% methane solution is 1.96 ml.