2.6 Milk-and-Water Ratio Problems (Adding / Removing a Pure Ingredient)

These do not always need the alligation diagram — often a direct ratio equation is faster, but alligation-based thinking helps you set them up without confusion.

Worked Example 2.6.1: A mixture of 60 L contains milk and water in the ratio 2 : 1. How much water must be added to make the ratio 1 : 2?

Solution: Milk =40 L, water =20 L (unchanged — we only add water). Let w L of water be added: (40)/(20+w) = (1)/(2) ⇒ 20+w = 80 ⇒ w = 60 L

Worked Example 2.6.2: 729 mL of a mixture contains milk and water in the ratio 7 : 2. How much water must be added to get a new ratio of milk to water of 7 : 3?

Solution: Milk = 729 × (7)/(9) = 567 mL, water =729×(2)/(9) = 162 mL. Milk stays constant; for the new ratio 7:3, water must become 567 × (3)/(7) = 243 mL. Water to be added = 243-162 = 81 mL.

Worked Example 2.6.3 (working backward from a target ratio to the original mixture): A mixture contains milk and water in the ratio 4 : 1. When 15 L of water is added, the ratio becomes 3 : 2. Find the total quantity of the original mixture.

Solution: Let milk =4x, water =x. After adding 15 L water: (4x)/(x+15)=(3)/(2) ⇒ 8x = 3x+45 ⇒ 5x=45 ⇒ x=9 Milk =36 L, water =9 L. Original mixture =36+9= 45 L. (Check: after adding 15 L water, water becomes 24 L, and 36:24=3:2 ✓.)

Worked Example 2.6.4 (edge case — removing part of the mixture and replacing with the PURE ingredient, not adding to a fixed component): A mixture of 35 L contains milk and water in the ratio 4 : 1. How many litres of the mixture must be removed and replaced with pure milk so that the resulting mixture has milk and water in the ratio 6 : 1?

Solution: Milk =35×(4)/(5)=28 L, water =35×(1)/(5)=7 L. Let y L of the mixture (which is (4)/(5) milk, (1)/(5) water) be removed and replaced with y L of pure milk. Milk after = 28-(4y)/(5)+y = 28+(y)/(5); Water after =7-(y)/(5). (28+(y)/(5))/(7-(y)/(5)) = (6)/(1) ⇒ 28+(y)/(5) = 42-(6y)/(5) ⇒ (7y)/(5)=14 ⇒ y=10 Check: Removing 10 L takes out 8 L milk and 2 L water, leaving milk =20, water =5; adding back 10 L pure milk gives milk =30, water =5, ratio =30:5=6:1 ✓. 10 L must be removed and replaced.

Worked Example 2.6.5 (two operations combined — remove mixture, then add pure water, in one problem): A vessel contains 60 L of a mixture of milk and water in the ratio 7 : 3. First, 10 L of this mixture is taken out, and then 10 L of pure water is added. Find the new ratio of milk to water.

Solution: Milk =60×(7)/(10)=42 L, water =18 L. Removing 10 L of mixture (in the same 7:3 ratio) removes 10×(7)/(10)=7 L milk and 10×(3)/(10)=3 L water, leaving milk =35 L, water =15 L. Adding 10 L pure water gives water =25 L (milk unchanged at 35 L). Milk : Water = 35:25 = 7:5 Check: Total volume =35+25=60 L, matching the original volume ✓.