How many gallons of a 30% salt solution must be mixed with 10 gallons of a 10% salt solution to create a 20% salt solution?

Question

Mixture problems like this show up in SAT and other standardized tests, and they’re great for modeling real-world math. You're trying to create a 20% salt solution by mixing a stronger (30%) solution with a weaker (10%) one. How many gallons of the stronger solution do you need to add to get exactly 20% concentration overall?

Give it a shot — and make sure to set up and solve the equation based on the amount of salt in each part!

Answer 1

During the SAT, a timed test, you cannot spend a whole lot of time on this. Think!

Answer 2

10 gallons

20% is midway between 30 and 10%

so you need an equal amount of each to get a mixture of 20%

10 of 30% & 10 of 10% gives you 20 gallons of 20%

or

10(.30) + x(.10) = (10+x)(.20)

3 + .1x = 2 +.2x

3-2 = .2x-.1x

1 = .1x

.1x = 1

x = 10 gallons of 10%

Answer 3

Mixing a 10% solution with a 30% solution will form something with a concentration between 10% and 30%.

20% is right in the middle, halfway between 10% and 30%.

So, an even-steven, 50-50, half-and-half mixture, of equal parts 10% and 30%, will make 20%.

Hence, mix 10 gallons of 30% with 10 gallons of 10% to make (20 gallons of) 20%.

This intuitive approach could save time on an exam for you!

Answer 4

To solve this problem, let x represent the number of gallons of the 30% salt solution we need to add.

We’ll set up an equation based on the total amount of salt in the mixture:

From the 10 gallons of 10% solution:

0.10 × 10 = 1 gallon of salt

From the x gallons of 30% solution:

0.30 × x = 0.3x gallons of salt

The final mixture will be (10 + x) gallons of a 20% solution:

0.20 × (10 + x) = 2 + 0.2x gallons of salt

Now, set up the equation:

1 + 0.3x = 0.2(10 + x)

Distribute on the right:

1 + 0.3x = 2 + 0.2x

Subtract 0.2x from both sides:

1 + 0.1x = 2

Subtract 1 from both sides:

0.1x = 1

Divide both sides by 0.1:

x = 10

Final Answer: x = 10

So, you need to add 10 gallons of the 30% salt solution to make a 20% solution.

Answer 5

Time is of the essence whenever you are solving a question on a standardized exam so sometimes you can tell just based on the way the question is phrased … so like one this one you are trying to get half way in between 10% and 30% to get to 20%, meaning that the amounts have to be equal for each concentration. That means that both have 10 gallons.

If you want to set up a formal equation though you can do

10(10) + x(30) = 20(x+10) and then solve that for x!

That ends up being

100 + 30x = 20x + 200

10x = 100

x = 10

so the end value here would be 10 gallons using either approach

Answer 6

You can do this problem algebraically, or you can reason it out with logic.

Algebra:

0.30x + 0.10(10 gallons) = 0.20(x+10 gallons)

0.30x +1 gallon = 0.20x + 2 gallons

x = 10 gallons

(You can also observe that the above is simply the dilution expression ΣMV = M*V*; where ΣMV represents the summation of the product of the initial concentrations and volumes, and M*V* represent the product of the final concentration and volume.)

Logic: Since the target solution's concentration is the average/mean of the two starting solutions, you can conclude that you need an equivalent amount of each solution from a basic understanding of dilution. It helps to look at problems not just mathematically, but realistically as well. Reasoning problems out using logic can help save time on the SAT, particularly for problems like this.

Answer 7

Let’s call the number of gallons of the 30% salt solution x.

You’re mixing:

1. 10 gallons of 10% solution → contributes 0.10 × 10 = 1 gallon of salt
2. x gallons of 30% solution → contributes 0.30 × x gallons of salt

You want a 20% solution, and the total volume will be (10 + x) gallons.

So, total salt = 0.20 × (10 + x)

Now set up the equation:

0.30x + 1 = 0.20(10 + x)

Distribute on the right:

0.30x + 1 = 2 + 0.20x

Subtract 0.20x from both sides:

0.10x + 1 = 2

Subtract 1:

0.10x = 1 ⇒ x = 10

Answer: You need to mix 10 gallons of the 30% solution.

Let me know if you want help with more mixture or SAT-style algebra problems!

Best regards,

Sid Bhatia