Answer to Question #237442 in Algebra for Mr. Lee

Question

Two liters of a 20% alcohol solution is mixed with x liters of

a 50% alcohol solution. if the new mixture contains 38%

alcohol, what is the value of x?

Answer 1

Explanations & Calculations

• The amount of alcohol provided to the final mixture by the 20% solution,

$\qquad\qquad \begin{aligned} \small q_1&=\small 2\times\frac{20}{100}\,l \end{aligned}$

• That from the 50% solution,

$\qquad\qquad \begin{aligned} \small q_2&=\small x\times\frac{50}{100}\,l \end{aligned}$

• These 2 amounts are included in the final mixture. The volume of the final mixture is,

$\qquad\qquad \begin{aligned} \small V&=\small (2+x)\,l \end{aligned}$

• Then from the final mixture,

$\qquad\qquad \begin{aligned} \small 38&=\small \Bigg[\frac{(2\times\frac{20}{100})+(x\times\frac{50}{100})}{2+x}\Bigg]\times100\\ \small x&=\small 3\,l \end{aligned}$