Source: ENGR 216 Practice Challenges, Texas A&M
Tags: mixture problems, blending, systems of equations, mass balance, concentration, ENGR 216, experimental physics
Mixture and blending problems require setting up two simultaneous equations: one for total quantity (mass, volume, barrels) and one for the component of interest (caffeine content, cost per barrel, concentration). Solve the system to find unknown amounts. These are algebra problems dressed in engineering context.
Mass balance
The principle that mass in equals mass out. The total mass of the mixture must equal the sum of the masses of the individual components.
Component balance
The amount of a specific substance (caffeine, a chemical, cost contribution) in the mixture equals the sum of that substance's contribution from each source.
System of linear equations
Two or more equations with two or more unknowns, solved simultaneously by substitution or elimination.
Let A and B be the unknown quantities from two sources. You always write two equations:
Total equation: A + B = Total
Component equation: (concentration_A × A) + (concentration_B × B) = (concentration_mix × Total)
Substitute the first equation into the second and solve.
From the total equation: A = Total − B
Substitute into the component equation
Solve for B
Back-substitute to find A
Total: A + B = T
Component: c_A · A + c_B · B = c_T · T
Where c_A, c_B, c_T are the concentrations (or costs, or percentages) of each source and the mixture.
Three containers combined into 450 g of solution at 3.4% caffeine.
Sources:
Pure caffeine powder: 2.7 g
Solution A: 2.7% caffeine, unknown mass (call it A)
Solution B: 6.2% caffeine, unknown mass (call it B)
Total equation:
2.7 + A + B = 450
So: A + B = 447.3
Caffeine balance:
Total caffeine in the mix = 3.4% × 450 = 15.3 g
Caffeine from powder + caffeine from A + caffeine from B = 15.3
2.7 + 0.027A + 0.062B = 15.3
0.027A + 0.062B = 12.6
Solve by substitution:
A = 447.3 − B
0.027(447.3 − B) + 0.062B = 12.6
12.0771 − 0.027B + 0.062B = 12.6
0.035B = 0.5229
B = 14.94 g
A = 447.3 − 14.94 = 432.36 g
Solution A: 432.36 g, Solution B: 14.94 g
Blend 5,000 barrels of crude at ≤ $67.00/barrel. Light crude: $82.00/barrel. Heavy crude: $54.00/barrel. Find barrels of light crude.
Let L = barrels of light crude, H = barrels of heavy crude.
Total equation:
L + H = 5000
Cost equation:
82L + 54H = 67 × 5000 = 335,000
Solve by substitution:
H = 5000 − L
82L + 54(5000 − L) = 335,000
82L + 270,000 − 54L = 335,000
28L = 65,000
L = 2,321.43 barrels (approximately)
Light crude needed: about 2,321 barrels
⚠️ Always write two equations, even if it feels obvious. One for total quantity, one for the component of interest. This keeps you from losing track.
⚠️ Be careful with percentage vs. decimal. If caffeine is 2.7%, use 0.027 in the equation, not 2.7.
⚠️ For three-source problems (like the caffeine example), one quantity may be given outright (the 2.7 g of powder), reducing the system to two unknowns.
⚠️ Check your answer by plugging back into both equations. Does the total add up? Does the component balance check out?
Q: A mixture of 100 kg contains 30% salt. It is made from solution X (20% salt) and solution Y (50% salt). How much of each?
A: X + Y = 100. 0.20X + 0.50Y = 30. Substituting: 0.20(100 − Y) + 0.50Y = 30 → 20 + 0.30Y = 30 → Y = 33.33 kg, X = 66.67 kg.
Q: Why do mixture problems always need two equations?
A: Because there are two unknowns. One equation alone has infinitely many solutions. The total constraint and the component constraint together pin down a unique answer.
Q: What if the problem has three unknowns and only two equations?
A: You need additional information (a third equation or a fixed value for one unknown). In the caffeine problem, the powder mass was given, which effectively reduced the system to two unknowns.
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