Chapter 6.6, Problem 43SR

To determine

To find: the pair of system equations satisfying the given data and to solve the system

Answer to Problem 43SR

Hence 320 gallons of $25\%$ acid solution and 180 gallons of $50\%$ acid solution be mixed to get a 500 gallons of $34\%$ acid solution.

Explanation of Solution

Given:

500 gallons of $34\%$ acid solution

Available solution: $25\%$ acid solution and a $50\%$ acid solution.

Calculation:

Objective: To find the pair of system equations satisfying the given data and to solve the system.

Let $x$ gallons of $25\%$ acid solution and $y$ gallons of $50\%$ acid solution be mixed.

As Orion Labs wants to make 500 gallon of solution, thus,

  $x+y=500$

As Orion labs make 500 gallons of solution containing $34\%$ acid and $x$ gallons of solution contains $25\%$ acid, $y$ gallons of solution contains $50\%$ acid, therefore,

  $25\%(x)+50\%(y)=34\%(500)$

That is $\frac{25}{100}x+\frac{50}{100}y=\frac{34}{100}500$

Simplify $x+2y=680$

Therefore, the system of equation is

  $x+y=500$ First equation

  $x+2y=680$ Second equation

Subtract first equation from second equation,

  $x+2y-(x+y)=680-500$

Simplify $x+2y-x-y=180$

  $y=180$

Substitute $y=180$ in first equation,

  $x+180=500$

Subtract 180 from both sides,

  $x+180-180=500-180$

  $x=320$

Hence 320 gallons of $25\%$ acid solution and 180 gallons of $50\%$ acid solution be mixed to

get a 500 gallons of $34\%$ acid solution.

Conclusion:

Hence 320 gallons of $25\%$ acid solution and 180 gallons of $50\%$ acid solution be mixed to

get a 500 gallons of $34\%$ acid solution.