2.1 The Rule of Alligation — Statement and Diagrammatic (Criss-Cross) Method

Figure: The criss-cross alligation diagram — subtract diagonally to get the mixing ratio.

Definition (Mean Price): When two or more ingredients of different prices (or concentrations, or values) are mixed together, the price (or concentration) of the resulting mixture, calculated per unit quantity, is called the mean price. Intuitively, if you mix a cheap ingredient with an expensive one, the resulting mixture’s price per unit must land somewhere between the two original prices — and exactly where it lands depends on how much of each you used. Mix mostly the cheap one with a small amount of the expensive one, and the mean price sits close to the cheap end; mix them in equal amounts, and the mean price sits exactly at the midpoint. The rule of alligation formalises this intuition into an exact, reversible relationship, so that if you know the mean price you can recover the mixing ratio, and if you know the ratio you can recover the mean price.

Rule of Alligation: When two ingredients are mixed, the ratio of their quantities is inversely proportional to the difference of their prices from the mean price. In symbols, if a cheaper ingredient of price C_1 per unit and a dearer ingredient of price C_2 per unit (with C_1 < C_2) are mixed to give a mixture of mean price M (where C_1 ≤ M ≤ C_2), then:

(Quantity of cheaper)/(Quantity of dearer) = (Q_1)/(Q_2) = (C_2 - M)/(M - C_1)

Derivation: Let Q_1 units of the cheaper ingredient (price C_1) be mixed with Q_2 units of the dearer ingredient (price C_2). The total cost of the mixture must equal the cost of the two ingredients added separately:

Q_1 C_1 + Q_2 C_2 = (Q_1 + Q_2)M

Expand the right-hand side and rearrange:

Q_1 C_1 + Q_2 C_2 = Q_1 M + Q_2 M Q_1 C_1 - Q_1 M = Q_2 M - Q_2 C_2 Q_1 (M - C_1) = Q_2 (C_2 - M) (Q_1)/(Q_2) = (C_2 - M)/(M - C_1)

This is the complete derivation of the alligation rule. Notice that the mean price M must always lie between C_1 and C_2 — it is a weighted average, and a weighted average can never fall outside the range of the numbers being averaged. If a question ever gives you a mean price outside the two given prices, either the question is flawed or you have misread it — check immediately.

The Criss-Cross Diagram: This is the visual shortcut that lets you write down the ratio in five seconds without doing the algebra above every time. Draw it as:

   Cheaper (C1)         Dearer (C2)
          \                 /
           \               /
            \             /
             Mean Price (M)
            /             \
           /               \
          /                 \
   (C2 - M)              (M - C1)

Place the cheaper price on the top-left, the dearer price on the top-right, and the mean price in the middle. Then subtract diagonally and criss-cross: the difference (C_2 - M) written below the cheaper price gives the cheaper ingredient’s share of the ratio, and the difference (M - C_1) written below the dearer price gives the dearer ingredient’s share. So:

Cheaper : Dearer = (C_2 - M) : (M - C_1)

Always take the absolute (positive) difference — subtract the smaller from the larger in each diagonal so both numbers in your ratio come out positive.

Worked Example 2.1.1: In what ratio should tea worth Rs 60/kg be mixed with tea worth Rs 65/kg so that the mixture is worth Rs 62/kg?

Solution: Here C_1 = 60, C_2 = 65, M = 62. Q_1 : Q_2 = (65-62):(62-60) = 3:2 So tea at Rs 60/kg and tea at Rs 65/kg must be mixed in the ratio 3 : 2.

Worked Example 2.1.2: In what ratio should rice at Rs 9/kg be mixed with rice at Rs 7/kg so that the mixture costs Rs 8.20/kg?

Solution: C_1 = 7, C_2 = 9, M = 8.20. Q(Rs 7) : Q(Rs 9) = (9 - 8.20) : (8.20 - 7) = 0.8 : 1.2 = 2 : 3 Rice at Rs 7/kg and Rs 9/kg should be mixed in the ratio 2 : 3. (Tip: to avoid decimals, multiply both differences by 10 first — 8 : 12 = 2 : 3.)

Worked Example 2.1.3: A shopkeeper mixes two varieties of tea costing Rs 234/kg and Rs 130/kg in the ratio 2 : 3. Find the cost price of the mixture, and the selling price per kg if he wants a 10% profit.

Solution: This is the reverse use of the rule — quantities and prices given, find mean price. Simply take the weighted average: M = (234 × 2 + 130 × 3)/(2+3) = (468+390)/(5) = (858)/(5) = 171.60 CP of mixture = Rs 171.60/kg. For 10% profit: SP = 171.60 × 1.10 = Rs 188.76/kg.

Worked Example 2.1.4 (converting the ratio into actual quantities): Two qualities of sugar worth Rs 36/kg and Rs 44/kg are to be mixed to prepare 50 kg of a mixture worth Rs 40/kg. Find how many kg of each type are needed.

Solution: C_1=36, C_2=44, M=40. Ratio = (44-40):(40-36) = 4:4 = 1:1 Since the total mixture is 50 kg and the ratio is 1:1, each type contributes 25 kg. This step — converting a ratio into actual quantities using a given total — is used constantly once alligation is combined with real-world quantity questions, so always check whether the question asks for the ratio itself or for the actual quantities implied by a stated total.

Worked Example 2.1.5 (reverse — ratio and one price known, find the unknown price): A dealer mixes oil costing Rs 90/L with another variety of oil in the ratio 3 : 2 to obtain a mixture worth Rs 100/L. Find the price per litre of the second variety of oil.

Solution: Let the unknown price be C_2. Since the mean price (Rs 100) is higher than Rs 90, the second variety must be dearer, so C_2 > 100. Using the ratio form (first variety : second variety = 3:2): (Q_1)/(Q_2) = (C_2 - M)/(M - C_1) ⇒ (3)/(2) = (C_2-100)/(100-90) = (C_2-100)/(10) ⇒ C_2 - 100 = 15 ⇒ C_2 = 115 Check: (90(3)+115(2))/(5) = (270+230)/(5) = (500)/(5) = 100 ✓. The second variety costs Rs 115/L.

Worked Example 2.1.6 (ratio then actual quantities, decimal-free design): A dealer wants to prepare 45 L of a mixture worth Rs 60/L by mixing oil worth Rs 54/L with oil worth Rs 72/L. Find how many litres of each variety are required.

Solution: C_1=54, C_2=72, M=60. Ratio (cheaper : dearer) = (72-60):(60-54) = 12:6 = 2:1 Total parts =3, total mixture =45 L. Cheaper (Rs 54) = 45×(2)/(3) = 30 L; dearer (Rs 72) =45×(1)/(3)=15 L. Check: (54(30)+72(15))/(45) = (1620+1080)/(45) = (2700)/(45) = 60 ✓. Required: 30 L of Rs 54/L oil and 15 L of Rs 72/L oil.