Bilbo asked in Science & MathematicsMathematics · 1 month ago

# Help with a mixture problem?

How many gallons of a 30% acid solution must be added to how many gallons of a 70% acid solution to make 50 gallons of a 40% acid solution?

I have no idea how to solve this.  Most helpful explanation will be awarded best answer.

Relevance
• 1 month ago

Start by defining a variable.

Let x be the amount of 30% acid solution (in gallons)

The rest of the mixture will add up to 50 gallons, so that means you have 50 - x of the second solution.

Let 50 - x represent the amount of 70% acid solution (in gallons).

Now write an expression for the amount of acid that results from the mixture.

0.30x + 0.70(50 - x)

That must be equal to 0.40 of a 50 gallon mix.

0.30x + 0.70(50 - x) = 0.40 * 50

Now we can expand and simplify:

0.3x + 35 - 0.7x = 20

-0.4x + 35 = 20

-0.4x = -15

x = -15 / -0.4

x = 15 / (2/5)

x = 15 * 5/2

x = 75/2

x = 37.5 gallons <-- amount of 30% acid solution

And the remaining amount (50 - x) would be the other solution

50 - x = 12.5 gallons <-- amount of 70% acid solution

• Ian H
Lv 7
1 month ago

0.3n + 0.7(50 – n) = 0.4*50

35 - 20 = 0.4n

n = (5/2)*15 = 37.5 gallons (of the 30%)

• 1 month ago

50 gallons of a 40% acid solution means 20 gal of acid and 30 gal of water

30% has 0.3 gal per gal, call quan. of solution x

70% has 0.7 gal per gal, call quan. of solution y

0.3x + 0.7y = 20

x + y = 50

multiply by 10 and –3 and add

3x + 7y = 200

–3x – 3y = –150

4y = 50

y = 12.5 gal

x = 50–12.5 = 37.5 gal

check

37.5 gal of 30% = 11.25 gal acid

12.5 gal of 70% =   8.75 gal acid

total                       20 gal acid

• 1 month ago

50 gallons of a 40% solution is 50(.4) = 20 gallons of pure acid.

Let x be the gallons of 30% solution to be added.

Let y be the gallons of 70% solution to be added.

x + y = 50

.3x + .7y = 20

x = 75/2 = 37.5 gallons

y = 25/2 = 12.5 gallons