Achievable logoAchievable logo
GRE
Sign in
Sign up
Purchase
Textbook
Practice exams
Support
How it works
Resources
Exam catalog
Mountain with a flag at the peak
Textbook
Welcome
1. Vocabulary approach
2. Quantitative reasoning
2.1 Quant intro
2.2 Arithmetic & algebra
2.2.1 Positive negative problems
2.2.2 Defined & undefined
2.2.3 GRE vocabulary list 01 (alacrity)
2.2.4 Odd even problems
2.2.5 GRE vocabulary list 02 (adulterate)
2.2.6 Algebra
2.2.7 Fraction math
2.2.8 GRE vocabulary list 03 (abstain)
2.2.9 Percent change
2.2.10 GRE vocabulary list 04 (anachronism)
2.2.11 Function problems
2.2.12 GRE vocabulary list 05 (ameliorate)
2.2.13 Divisors, prime factors, multiples
2.2.14 Greatest common factor (GCF) & Least common multiple (LCM)
2.2.15 GRE vocabulary list 06 (acumen)
2.2.16 Permutations and combinations
2.2.17 GRE vocabulary list 07 (aesthetic)
2.2.18 Decimals
2.2.19 GRE vocabulary list 08 (aggrandize)
2.2.20 FOIL and quadratic equations
2.2.21 GRE vocabulary list 09 (anodyne)
2.2.22 Exponent rules
2.2.23 GRE vocabulary list 10 (aberrant)
2.2.24 Square roots and radicals
2.2.25 Sequences
2.2.26 Venn diagrams & tables
2.2.27 Ratios
2.2.28 Mixtures
2.2.29 Probability
2.2.30 Algebra word problems
2.2.31 Number line, absolute value, inequalities
2.2.32 Simple and compound interest
2.2.33 System of linear equations (SOLE)
2.3 Statistics and data interpretation
2.4 Geometry
2.5 Strategies
3. Verbal reasoning
4. Analytical writing
Wrapping up
Achievable logoAchievable logo
2.2.28 Mixtures
Achievable GRE
2. Quantitative reasoning
2.2. Arithmetic & algebra

Mixtures

4 min read
Font
Discuss
Share
Feedback

Mixture problems involve combining parts, solutions, or liquids in given ratios or concentrations. The problem usually describes a change to the mixture - for example, adding a new liquid or adding more of one component already present.

Your job is to:

  • Decide what quantity you’re solving for
  • Translate the situation into an algebraic equation
  • Solve for the unknown

As you set up your equations, be consistent about what each variable represents (total amount of drink, amount of milk, amount of coffee, etc.).

Mixture problems can vary a lot, so the best preparation is practice. Try these examples.

If Rhea likes her coffee-to-milk ratio at 9 to 1, and Asia likes her coffee-to-milk ratio at 3 to 1, what ratio of Rhea’s to Asia’s drinks is required to make a mixture with a 4 to 1 ratio of coffee to milk?

A. 1 to 3
B. 1 to 2
C. 1 to 1
D. 2 to 1
E. 3 to 1

(spoiler)

Answer: B. 1 to 2

To get a 4:1 coffee-to-milk ratio, you need to mix some of Asia’s drink with some of Rhea’s drink, because 4:1 lies between their preferred ratios of 3:1 and 9:1. The question asks for the ratio of the amounts you should mix.

Begin by writing an equation. One convenient approach is to write everything in terms of the amount of milk (you could choose coffee instead, but stick to one choice).

Let:

  • a = total amount of Asia’s drink used
  • r = total amount of Rhea’s drink used

Step 1: Write the milk fraction in each drink

Asia’s ratio is 3 parts coffee to 1 part milk. That’s 4 total parts, so milk is 41​=0.25 of Asia’s drink.

a(.25)

Rhea’s ratio is 9 parts coffee to 1 part milk. That’s 10 total parts, so milk is 101​=0.1 of Rhea’s drink.

r(.1)

Step 2: Add the milk amounts

The total milk in the combined mixture is:

a(.25)+r(.1)

Step 3: Set this equal to the milk in the final mixture

The target ratio is 4 parts coffee to 1 part milk. That’s 5 total parts, so milk is 51​=0.2 of the final mixture. The total amount of mixture is (a+r), so the milk in the final mixture is:

.2(a+r)

Set the two expressions for milk equal and solve:

a(.25)+r(.1).25a+.1r.05a.05/11/2​=.2(a+r)=.2a+.2r=.1r=r/a=r/a​

So the ratio of Rhea to Asia is 1 to 2.

Using weighted averages

A common method for mixture problems is the weighted average equation. (If you want a refresher, revisit the “mean, median, mode, and range” chapter.) Use weighted averages to solve the problem below.

A very hot mixture consists of three solutions, each containing a specific concentration of a melted rare metal.

Solution Rare metal concentration level
A 5%
B 15%
C 50%

If 15% of the mixture is made up of Solution A, 25% is made up of Solution B, and 60% is made up of Solution C, approximately what is the concentration of the rare metal in the mixture? Round to the nearest percent.

(spoiler)

Answer: 35%

Set up a weighted average. Remember:

  • The mixture percentages (15%, 25%, 60%) should add to 100%.
  • These are not the same as the concentrations (5%, 15%, 50%).

Key information (as decimals):

  • Percent of mixture that’s A = .15
  • Concentration of A = .05
  • Percent of mixture that’s B = .25
  • Concentration of B = .15
  • Percent of mixture that’s C = .6
  • Concentration of C = .5

Compute the mixture’s concentration:

mixture’s cocentration​=(.15)(.05)+(.25)(.15)+(.6)(.5)=.0075+.0375+.3=.345​

So the concentration of the rare metal in the mixture is approximately 35%.

As a quick check, notice that 35% is between the smallest and largest concentrations (between 5% and 50%). This is a useful fail-safe: a weighted average must fall between the minimum and maximum of the values being averaged. If your result is outside that range, something went wrong in your setup or arithmetic.

Sign up for free to take 2 quiz questions on this topic

All rights reserved ©2016 - 2026 Achievable, Inc.