2.2 Alligation for More Than Two Ingredients (Mixture of Mixtures)

Figure: With three or more ingredients, pair them up around the target mean and solve two at a time.

When three or more ingredients are combined, there are two distinct question types you will encounter, and it is important to tell them apart quickly.

Type A — Quantities and individual prices given, find the mean price of the combined mixture. This is simply the weighted average formula extended to n terms — no criss-cross needed: M = (Q_1C_1 + Q_2C_2 + Q_3C_3 + …)/(Q_1+Q_2+Q_3+…)

Worked Example 2.2.1: Three qualities of wheat costing Rs 8, Rs 10 and Rs 12 per kg are mixed in the ratio 1 : 1 : 2. Find the price of the mixture per kg.

Solution: M = (8(1)+10(1)+12(2))/(1+1+2) = (8+10+24)/(4) = (42)/(4) = 10.50 The mixture costs Rs 10.50/kg.

Type B — Mean price given, find the ratio of three (or more) ingredients. Here a single criss-cross cannot directly give you three unknowns from one equation — you need either an extra condition (such as a fixed ratio between two of the three quantities) or you solve it by pairing ingredients against the mean price.

Worked Example 2.2.2: In what ratio should wheat costing Rs 8/kg be mixed with wheat costing Rs 12/kg to get a mixture worth Rs 10.50/kg?

Solution: Straight two-ingredient alligation with C_1=8, C_2=12, M=10.5: Q_1 : Q_2 = (12-10.5):(10.5-8) = 1.5:2.5 = 3:5 So wheat at Rs 8/kg and Rs 12/kg should be mixed in the ratio 3 : 5 (note this is the exact reverse-check of Example 2.2.1’s data, confirming consistency).

Worked Example 2.2.3 (three ingredients, unequal quantities, direct weighted average): Three grades of pulses costing Rs 60/kg, Rs 80/kg and Rs 100/kg are mixed in the ratio 2 : 3 : 5. Find the price per kg of the resulting mixture.

Solution: This is again Type A — quantities and prices given, so simply take the weighted average: M = (60(2)+80(3)+100(5))/(2+3+5) = (120+240+500)/(10) = (860)/(10) = 86 The mixture costs Rs 86/kg. Note how, with three or more ingredients, as long as the actual quantity (or ratio) of each is known, you never need the criss-cross diagram at all — a plain weighted average always works and is faster.

Worked Example 2.2.4 (Mixture of Mixtures — combining ready-made mixtures): Container A holds 20 L of a milk-water mixture in which milk : water = 3 : 1. Container B holds 30 L of a milk-water mixture in which milk : water = 2 : 3. Both are poured into a larger container. Find the ratio of milk to water in the final mixture.

Solution: Find the actual milk and water quantities in each container first. Container A: milk = 20 × (3)/(4) = 15 L, water = 20 × (1)/(4) = 5 L. Container B: milk = 30 × (2)/(5) = 12 L, water = 30 × (3)/(5) = 18 L. Total milk = 15+12 = 27 L. Total water = 5+18 = 23 L. Milk : Water = 27 : 23

This “break into actual quantities, then add” method is the safest way to handle any “mixture of mixtures” problem — resist the temptation to average the ratios directly (that is a common and costly mistake, covered in Section 4).

Worked Example 2.2.5 (genuine three-unknown alligation, extra condition supplied): In what ratio must three varieties of tea costing Rs 50, Rs 60 and Rs 80 per kg be mixed to get a mixture costing Rs 65 per kg, given that the quantities of the Rs 50 and Rs 60 varieties used are equal?

Solution: A single criss-cross diagram cannot resolve three unknowns, so we use the given extra condition. Let the quantity of the Rs 50 and Rs 60 varieties each be x, and the quantity of the Rs 80 variety be y. The weighted average must equal 65: (50x+60x+80y)/(x+x+y) = 65 ⇒ 110x+80y = 65(2x+y) = 130x+65y ⇒ 15y = 20x ⇒ y = (4)/(3)x Taking x=3 gives y=4, so the ratio is x:x:y = 3:3:4. Check: Quantities 3, 3, 4 of Rs 50, Rs 60, Rs 80 → total cost =150+180+320=650; total quantity =10; mean =65 ✓. Required ratio: 3 : 3 : 4.

Worked Example 2.2.6 (mixture combined with another mixture of an unknown quantity — reverse Type A): A shopkeeper mixes 20 kg of rice costing Rs 40/kg with an unknown quantity of rice costing Rs 60/kg. He then combines this lot with 15 kg of a third variety costing Rs 70/kg, and finds that the overall mixture costs Rs 58/kg. Find the quantity of the Rs 60/kg rice used.

Solution: Let the unknown quantity be q kg. Total cost =20(40)+60q+15(70) = 800+60q+1050 = 1850+60q. Total quantity =20+q+15=35+q. (1850+60q)/(35+q) = 58 ⇒ 1850+60q = 2030+58q ⇒ 2q=180 ⇒ q=90 Check: Total cost =1850+60(90)=1850+5400=7250; total quantity =35+90=125; mean =(7250)/(125)=58 ✓. The Rs 60/kg rice used was 90 kg.