Quick Revision

Reading a Mixture

Milk and water in the ratio a : b means a parts of milk in every a + b parts.
Share of milk = $\frac{a}{a + b}$ . Times 100, it is the percentage, or strength.
Litres of a part = total × its share. 54 litres in the ratio 5 : 4 hold 30 litres of milk and 24 of water.
20% acid has 20 parts acid in every 100, so 45 litres of it hold 9 litres of acid.
Write down the litres of each part before changing anything.

Only the part you add changes. The other stays fixed, so build the answer on it.
Ratio target: 42 litres in 5 : 1 has milk 35 and water 7. For 5 : 2 the water must be 14, so add 7 litres.
Percentage target: 30 litres with 20% alcohol has 24 litres of water. For 40% alcohol that water must be 60%, so the total is 40: add 10 litres of alcohol.
Trap: never take the new percentage of the old total. The total grows.

Mixture taken out carries both parts in the same ratio as the whole mixture.
So find what is left of each part first, then add the new liquid to its own part only.
64 litres in the ratio 5 : 3 has milk 40 and water 24. Take out 16 litres, a quarter, and milk 30 and water 18 are left. Add 12 litres of water: 30 : 30 = 1 : 1.
Taking out 16 litres of mixture is not the same as taking out 16 litres of water.

Find the litres of each part in every vessel, then add part by part.
Don't add ratio terms unless amounts and ratio totals match.
Equal vessels: add the shares. 3 : 1 and 3 : 2 give milk $\frac{3}{4} + \frac{3}{5} = \frac{27}{20}$ out of 2, so 27 : 13.
LCM trick: totals 4 and 5 become 20, so 15 : 5 and 12 : 8 add to 27 : 13.
In a ratio p : q: weight each share by p or q.

Between vessels, track both parts in both vessels after every pour.
A has 30 litres of milk, B 10 litres of water. Pour 10 litres from A into B: B is 10 milk, 10 water. Pour 5 litres back: 2.5 of each, so A is 22.5 milk, 2.5 water, 9 : 1.
Alloys are mixtures by weight. 12 kg of copper and zinc in 2 : 1 plus 18 kg in 1 : 2: copper 8 + 6 = 14, zinc 4 + 12 = 16, so 7 : 8.

Price of the mix = total cost ÷ total quantity.
30 kg at ₹38 and 10 kg at ₹54: (1,140 + 540) ÷ 40 = ₹42 per kg.
Equal quantities: the plain average of the prices works.
Equal money: find how much each sum buys. ₹480 each at ₹12 and ₹24 a kg buys 40 + 20 = 60 kg for ₹960, which is ₹16 a kg, not ₹18.
Free water costs nothing: count its litres, not its cost.

Cheaper (c) and dearer (d) varieties give a mean price m when cheaper : dearer = (d − m) : (m − c).
Prices on top, mean in the middle, differences taken crosswise.
₹32 and ₹50 for ₹44: (50 − 44) : (44 − 32) = 6 : 12 = 1 : 2.
The mean must lie between the two prices.
One quantity known: scale the ratio. Ratio and mean known: find the missing price.
Check by mixing forward again.

Percentages work just like prices. 12% and 42% acid for 30%: (42 − 30) : (30 − 12) = 12 : 18 = 2 : 3.
Water is a 0% solution; the pure liquid is 100%.
For ratios, first turn each mixture into the share of one part. 3 : 2 is milk $\frac{3}{5}$ and 1 : 3 is milk $\frac{1}{4}$ . For 1 : 1, the target is $\frac{1}{2}$ : $\left(\frac{1}{2} - \frac{1}{4}\right) : \left(\frac{3}{5} - \frac{1}{2}\right)$ = 5 : 2.
Never allige the ratio numbers themselves.

Any weighted average of two groups can be alliged.
Interest: ₹20,000 lent partly at 6% and partly at 11% earns ₹1,600 a year, an average of 8%. Parts: (11 − 8) : (8 − 6) = 3 : 2, so ₹12,000 at 6% and ₹8,000 at 11%.
Two profit percents on parts of a stock work the same way.
Speeds: allige the times, not the distances.

Pair each variety cheaper than the mean with one dearer than it, and allige each pair.
₹20, ₹28 and ₹40 for ₹32: 20 with 40 gives 2 : 3, and 28 with 40 gives 2 : 1. Add the shares of ₹40: 2 : 2 : 4 = 1 : 1 : 2.
Many ratios can work, so in an MCQ test each option for the mean.
Two varieties in a fixed ratio: blend them into one price first, then allige with the third.

Turn the selling price into the mix's cost price: CP = SP × 100 ÷ (100 + gain%), or ÷ (100 − loss%).
Then allige with that CP, never the selling price.
Sold at ₹52.80 with a 10% gain: CP = ₹48.
Water costs ₹0. Sold at the cost of milk, water : milk = gain% : 100, so a 15% gain means 3 : 20.
Sold above the cost of milk: find the mix's CP, then allige ₹0 water with milk.

V litres of pure liquid, x litres taken out and replaced with water, n times: liquid left = $V\left(1 - \frac{x}{V}\right)^n$ .
50 litres, 5 litres replaced three times: $50 \times (0.9)^3$ = 36.45 litres.
Water = V − liquid left.
Not pure at the start: use the same factor on the part being drawn down.
Different amounts: multiply by each step's own factor, $\frac{V - x_1}{V}$ , then $\frac{V - x_2}{V}$ .
Never just subtract x once for each round.

Write the final share of liquid left and match it with $\left(1 - \frac{x}{V}\right)^n$ .
Two operations leave 36%: $\left(1 - \frac{x}{V}\right)^2 = \frac{9}{25}$ , so $\frac{x}{V} = \frac{2}{5}$ .
For n, count the factors: $\left(\frac{4}{5}\right)^n = \frac{64}{125}$ gives n = 3.
Fraction f to replace once with water: old share × (1 − f) = new share. Milk 80% down to 60%: $1 - f = \frac{3}{4}$ , so $f = \frac{1}{4}$ .
Replacing with the other part instead: work on the part that only goes down.

Adding the ratio terms of two mixtures whose sizes or ratio totals differ.
Taking the new percentage of the old total when something is added.
Alliging with the selling price instead of the cost price.
Taking the plain average of prices when equal money, not equal quantity, is spent.
Subtracting the removed amount n times instead of multiplying by $\left(1 - \frac{x}{V}\right)$ n times.
Forgetting that mixture taken out takes both parts with it.
Alliging speeds on distance instead of time.