Diagram first, then the mixing formula.
The problem
How many litres of a 30% alcohol solution must be mixed with 12 litres of a 70% alcohol solution to make a 45% alcohol mixture?
Track the alcohol, not the liquid: 0.30x + 0.70(12) = 0.45(x + 12). That gives 0.3x + 8.4 = 0.45x + 5.4, so 0.15x = 3 and x = 20 litres. The total on the right is x + 12 — forgetting to add the 12 there gives 0.3x + 8.4 = 0.45x, so x = 56, which is C. D (32 litres) is the whole mixture, 20 + 12, not the 30% part. B (12) assumes equal amounts, but equal amounts of 30% and 70% make 50%, not 45%.
Every mixture item has three containers. Draw them before any algebra.
Solution A
30%
? L
Solution B
70%
12 L
Mixture
45%
? L
C = concentration (the %) · V = volume (litres)
Two unknowns show up as “?” — the volume of A and the volume of the mixture. Step 2 links them so only one letter is left.
Nothing leaks out, so the two volumes simply add up.
Solution A
30%
x L
Solution B
70%
12 L
Mixture
45%
(x + 12) L
x + 12
Writing the mixture as its own letter (say y) gives two unknowns and one equation. Writing x + 12 keeps it solvable.
The alcohol from A plus the alcohol from B lands in the mixture.
Solution A
water
0.30 × x
x L in all
Solution B
water
0.70 × 12
12 L in all
Mixture
water
0.45 × (x + 12)
(x + 12) L in all
Gold band = litres of pure alcohol.
Percents cannot be added: 30% + 70% is not 45%. Litres of pure alcohol can be added — that is the whole trick.
C1V1 + C2V2 = CnewVtotal
| Symbol | Means | Value |
|---|---|---|
| C1V1 | solution A | 0.30 and x |
| C2V2 | solution B | 0.70 and 12 |
| CV | the mixture | 0.45 and x + 12 |
45% is written 0.45. Percent means “per hundred”, so divide by 100 before multiplying.
The unknown can be a percent instead of a volume. Mixing 5 L of 60% with 15 L of 40%: the pure parts are 3 + 6 = 9 L in 20 L, so the mixture is 9 ÷ 20 = 45%. And if 4 L of 25% plus 6 L of an unknown c make a 37% mix: 0.25(4) + 6c = 0.37(10), so 1 + 6c = 3.7, 6c = 2.7 and c = 0.45, a 45% solution.
0.30x + 0.70(12) = 0.45(x + 12)
the Step 4 equation
0.30x + 8.4 = 0.45x + 5.4
brackets multiplied out
pure alcohol already sitting in solution B
the alcohol the 12 L contributes at 45%
Both bracket terms came from the same 12 litres — once at its own strength, once at the mixture’s strength.
0.30x + 8.4 = 0.45x + 5.4
from Step 5
8.4 = 0.15x + 5.4
− 0.30x both sides
3 = 0.15x
− 5.4 both sides
20 = x
÷ 0.15 both sides
Dividing by 0.15 is the same as dividing by 15 then × 100. 3 ÷ 0.15 = 20, not 0.05.
Call one x; the other is whatever is left of the total.
x of 60%, 500 − x of 20%
one letter
0.60x + 0.20(500 − x) = 0.30(500)
the pure part balances
0.40x + 100 = 150
x = 125 mL of the 60%
and 375 mL of the 20%
Check: 75 + 75 = 150 mL of pure substance, 30% of 500 ✓. The mix sits closer to 20%, so more of the 20% is used. Price mixtures work the same way, with ₱ per kg in place of the percent.
The pure part still balances; the volumes move in a new way.
Evaporation: how many litres of water must evaporate from 40 L of a 5% salt solution to make it 8%?
salt: 0.05 × 40 = 2 L
evaporating water leaves the salt
0.08(40 − x) = 2
the volume shrinks by x
40 − x = 25, so x = 15 L
2 ÷ 0.08 = 25
Removing water is adding water in reverse: the total is 40 − x, and the solute is unchanged.
Adding x to the total, as when diluting, gives a negative x. Taking water away makes the solution smaller and stronger.
Drain and replace: a radiator holds 24 L of 25% antifreeze. How much must be drained and replaced with pure antifreeze to make it 50%?
0.25(24 − x) + 1.00x = 0.50(24)
x L out at 25%, x L in at 100%
6 − 0.25x + x = 12
the total stays 24 L
0.75x = 6, so x = 8 L
check: 4 + 8 = 12 L of 24 ✓
The drained liquid takes antifreeze with it, so the 25% applies to 24 − x, not to 24.
No solution: can 20% and 35% solutions make 10 L of a 40% solution?
0.20x + 0.35(10 − x) = 0.40(10)
fixed total of 10 L
3.5 − 0.15x = 4, so x = −103
a negative volume
impossible
40% is above both ingredients
A mixture always lands between its ingredients. A target outside that range gives a negative amount; that is the signal, not an arithmetic slip.
Give the added liquid its percent and use the same equation.
40 L of 25% plus x L of water → 20%
dilute it
0.25(40) + 0(x) = 0.20(40 + x)
water adds no solute
10 = 8 + 0.2x, so x = 10 L
20 L of 10% acid plus x L of pure acid → 25%
strengthen it
0.10(20) + 1.00x = 0.25(20 + x)
pure acid is 100%
2 + x = 5 + 0.25x, so x = 4 L
0.75x = 3
Added water or acid enlarges the total: the right side is 0.20(40 + x), not 0.20(40).
Mixture items balance the pure part: C₁V₁ + C₂V₂ = CV. Watch for a fixed total (use x and total − x), water at 0% and pure substance at 100%. Check that the final percent sits between the two you mixed, nearer the one used more.
The problem
A tank holds 40 litres of a 15% salt solution. How many litres of water must evaporate to raise the concentration to 25%?
Only water leaves, so the salt stays at 0.15 × 40 = 6 litres. After x litres evaporate: 6 = 0.25(40 − x), so 24 = 40 − x and x = 16 litres. Check: 6 ÷ 24 = 25%. A (24) is the volume left, not the amount removed. B (4) multiplies only the 40 by 0.25, writing 6 = 10 − x. D (10) reads the 10-point rise in percent as 10 litres.
Hint: salt does not evaporate. Fix the 6 litres of salt, and let the volume that remains be 40 − x.
