2.8 Composite Problems — Combining Alligation with Ratio, Average and Replacement

Exam-setters frequently chain two or three of the ideas above into a single multi-step question. The key skill here is not new mathematics — it is correctly identifying which sub-tool (alligation, ratio, weighted average, or the replacement formula) applies at each stage, and carrying the output of one stage cleanly into the next.

Composite Example 2.8.1 (Alligation for mean price → Average for profit → Ratio for total profit): A trader buys wheat of type A at Rs 20/kg and type B at Rs 30/kg, and mixes them in the ratio 3 : 2. He sells the entire mixture at a 10% profit. If the total mixture weighs 100 kg, find his total profit in rupees.

Solution (three stages): Stage 1 (Average / weighted mean price): Mean CP = (20(3)+30(2))/(3+2) = (60+60)/(5) = (120)/(5) = 24 per kg. Stage 2 (Profit-Loss link): SP per kg with 10% profit = 24 × 1.10 = 26.40. Stage 3 (Scale up using given total quantity): Profit per kg =26.40-24=2.40. For 100 kg: Total profit = 100 × 2.40 = Rs 240

Composite Example 2.8.2 (Alligation to find the initial ratio → Replacement formula on the result): A container holds 50 L of a milk-water mixture. The mixture was originally prepared by mixing pure milk (worth Rs 50/L) with water (free) so that the mean cost of the mixture works out to Rs 40/L per litre. Now, 10 L of this mixture is withdrawn and replaced with water, and this is repeated once more (2 operations total). Find the final quantity of milk left in the container.

Solution (two stages): Stage 1 (Alligation to recover the milk:water ratio): Using milk (Rs 50) and water (Rs 0) with mean Rs 40: Q_(milk):Q_(water) = (0-40):(40-50) → magnitudes 40:10 = 4:1 Since the total is 50 L, milk =50×(4)/(5)=40 L and water =10 L. Stage 2 (Replacement formula): Now apply P(1-(x)/(P))^n with P=50 (total volume, constant throughout), x=10 (volume withdrawn each time), n=2: Milk left = 50(1-(10)/(50))^2 = 50(0.8)^2 = 50 × 0.64 = 32 L

This two-stage pattern — “use alligation to reconstruct an unknown ratio, then feed that ratio’s actual quantities into the replacement formula” — is exactly the kind of layered question that separates moderate from hard-level papers, and it is entirely manageable once you treat each stage as a separate, familiar sub-problem rather than trying to solve everything in one giant equation.

Composite Example 2.8.3 (percentage → actual quantity → repeated replacement → back to percentage): A vessel holds 125 L of a mixture in which milk constitutes 64% of the total. If 25 L of the mixture is withdrawn and replaced with water, and this is repeated two more times (3 operations in total), find the quantity of milk left, and its final percentage concentration.

Solution (three stages): Stage 1 (percentage → actual quantity): Milk =125×0.64=80 L. Stage 2 (replacement formula): P=125, x=25, so the factor per operation is 1-(25)/(125)=0.8; with n=3: Milk left = 80(0.8)^3 = 80×0.512 = 40.96 L Stage 3 (back to percentage): (40.96)/(125)×100 = 32.768%. Note the shortcut this reveals: since the total volume P stays fixed, the percentage itself also simply gets multiplied by the same factor each time: 64%×(0.8)^3 = 64×0.512=32.768%, confirming the two routes agree.

Composite Example 2.8.4 (alligation for mean CP → split-lot sale at different profit/loss% → overall profit%): A shopkeeper prepares 90 kg of a mixture of two varieties of rice costing Rs 20/kg and Rs 30/kg, mixed in the ratio 2 : 1. He then sells 60 kg of this mixture at a profit of 10%, and the remaining 30 kg at a loss of 5%. Find his overall profit or loss percentage on the entire 90 kg.

Solution (three stages): Stage 1 (alligation → mean CP): Mean CP =(20(2)+30(1))/(3)=(40+30)/(3)=(70)/(3) per kg. Total CP of 90 kg =90×(70)/(3)=2100. Stage 2 (split sale at different profit/loss%): CP of the 60 kg portion =60×(70)/(3)=1400; SP =1400×1.10=1540. CP of the 30 kg portion =30×(70)/(3)=700; SP =700×0.95=665. Stage 3 (overall profit%): Total SP =1540+665=2205. Overall profit =2205-2100=105. Overall Profit% = (105)/(2100)×100 = 5% The shopkeeper makes an overall profit of 5%.