How many gallons of a 50​% antifreeze solution must be mixed with 100 gallons of 15​% antifreeze to get a mixture that is 40​% ​antifreeze? Use the​ six-step method.

Question

How many gallons of a 50​% antifreeze solution must be mixed with 100 gallons of 15​% antifreeze to get a mixture that is 40​% ​antifreeze? Use the​ six-step method.

Answer 1

Hi Angela W.

I hope this matches your method if it doesn't please let me know

Step 1 Deftine the problem: Make a 40% solution of antifreeze by combining an unknown amount of a 50% solution of antifreeze with 100 gals of 15% antifreeze

Step 2 Declare a variable Let x = the unknown amount of the 50% antifreeze

Step 3 Set up an equation

Equation requires your starting amount of 100 gal be increased by x

Equation also requires that the concentration of the final total amount is moved from 15% to 40%

Equation is basically Total amount of Antifreeze in New Solution /Total Amount of New Solution = 0.4

100(0.15) + 0.50x = Total Amount of Antifreeze in the New Solution

100 + x = Total Amount of the New Solution

Final Equation

100(0.15) + 0.50x/(100 + x) = 0.40

Step 4 Solve for x

Multiply both sides by (100 + x)

100(0.15) + 0.50x = 0.40(100 + x)

Mulitiply to remove the parentheses on both sides

15 + 0.50x = 40 + 0.40x

Move the variables to one side of the equation and the constants to the other by adding the opposites

0.50x - 0.40x = 40 - 15

0.10x = 25

x = 250

Step 5 answer the question

You need to mix 100 gal of 15% antifreeze solution with 250 gal of 50% antifreeze solution to obtain a final solution that is 40% antifreeze

Step 6 Check it in your original equation

100(0.15) + 0.50x/(100 + x) = 0.40

100(0.15) + 0.50(250)/(100 + 250) = 0.4

15 + 125/350 = 0.40

140/350 = 0.40

I hope this helps and please let me know if you have any questions. Let me know if these are your 6 steps some texts have 5 steps