2.3 The Repeated Dilution / Replacement Formula

Figure: Each replacement removes some of the current mixture and adds pure water, diluting it further.

This is the single most tested advanced idea in the chapter: a vessel contains a pure liquid (or a mixture); a certain quantity is withdrawn and replaced by water (or another liquid); this operation is repeated n times. We need a formula for how much of the original liquid remains.

Setup: Let the vessel hold P units of pure liquid initially. At each step, x units of the current mixture (not necessarily pure liquid, after the first step) are withdrawn and replaced with x units of water, keeping total volume constant at P.

Derivation: - Before any operation: quantity of original liquid = P. - After operation 1: We withdraw x units of pure liquid (since before the first operation, the vessel contains only the original liquid). Liquid remaining = P - x = P(1 - (x)/(P)). - After operation 2: The vessel now has volume P, of which (P-x) is liquid and x is water — the concentration of liquid is (P-x)/(P). When we withdraw x units of this mixture, the liquid removed is x × (P-x)/(P). So liquid remaining: (P-x) - x·(P-x)/(P) = (P-x)(1-(x)/(P)) = P(1-(x)/(P))^2 - By induction, after each operation the remaining liquid is multiplied by the same factor (1-(x)/(P)), because the volume withdrawn is always the same fraction (x)/(P) of the total (constant) volume P. Therefore, after n such operations:

[Liquid remaining = P(1-(x)/(P))^n]

This formula is valid only when: (i) the total volume P stays constant throughout (equal volume of liquid is removed and the same volume of water is added back each time), and (ii) the quantity removed, x, is the same at every single step. If either condition is broken (e.g., different quantities removed each time), you cannot use this shortcut and must work step-by-step.

Worked Example 2.3.1: A vessel contains 40 L of milk. 4 L of milk is withdrawn and replaced with water. This is repeated two more times (three times total). Find the final quantity of milk left.

Solution: P = 40, x = 4, n = 3. Milk left = 40(1-(4)/(40))^3 = 40(0.9)^3 = 40 × 0.729 = 29.16 L

Worked Example 2.3.2: From a container full of wine, a third part is drawn out and replaced with water. This operation is repeated 4 times. Find the final ratio of wine to water.

Solution: Here (x)/(P) = (1)/(3) at each step, n=4. Taking P = 1 (fraction of the whole): Wine remaining = (1-(1)/(3))^4 = ((2)/(3))^4 = (16)/(81) So wine : total = 16:81, which gives wine : water = 16 : 65 (since water = 81-16=65).

Worked Example 2.3.3: A container has 80 L of pure milk. 8 L is replaced with water, and this is repeated 2 more times (3 operations total). What percentage of milk remains?

Solution: P=80, x=8, n=3, and (x)/(P) = 0.1. Milk left = 80(0.9)^3 = 80 × 0.729 = 58.32 L Percentage = (58.32)/(80)× 100 = 72.9%

Worked Example 2.3.4 (finding the number of operations, n): A container has 128 L of pure spirit. Each time, 16 L is withdrawn and replaced with water. After how many such operations will the quantity of spirit first fall below 50 L?

Solution: (x)/(P)=(16)/(128)=(1)/(8), so the multiplying factor per operation is (7)/(8). We compute 128((7)/(8))^n for successive n until it drops below 50: n=6: 128(0.875)^6 ≈ 57.44 L (still above 50) n=7: 128(0.875)^7 ≈ 50.26 L (still above 50) n=8: 128(0.875)^8 ≈ 43.98 L (below 50) So the spirit content first falls below 50 L after the 8th operation. When a question asks for n itself rather than giving n, this trial-based approach (checking successive powers) is usually faster than trying to solve the exponential equation algebraically, especially with the reduced fraction ((7)/(8))^n.

Worked Example 2.3.5 (reverse — given the final quantity, find the number of operations n, exact power): A vessel contains 81 L of pure milk. Each time, one-third of the mixture currently present is withdrawn and replaced with water. After how many operations will the quantity of milk in the vessel become exactly 24 L?

Solution: Removing one-third each time means the multiplying factor per operation is 1-(1)/(3)=(2)/(3). We need: 81((2)/(3))^n = 24 ⇒ ((2)/(3))^n = (24)/(81)=(8)/(27)=((2)/(3))^3 So n=3. Check: 81×(2)/(3)=54; 54×(2)/(3)=36; 36×(2)/(3)=24 ✓. It takes 3 operations.

Worked Example 2.3.6 (reverse — given P, n, and the final quantity, find x): A container has 64 L of pure alcohol. A certain quantity is withdrawn and replaced with water, and this operation is carried out 3 times in total. If 27 L of alcohol remains at the end, find the quantity withdrawn each time.

Solution: Using P(1-(x)/(P))^n = final quantity, with P=64, n=3: 64(1-(x)/(64))^3 = 27 ⇒ (1-(x)/(64))^3 = (27)/(64) = ((3)/(4))^3 ⇒ 1-(x)/(64) = (3)/(4) ⇒ (x)/(64)=(1)/(4) ⇒ x=16 Check: 64(3/4)^3 = 64×(27)/(64)=27 ✓. The quantity withdrawn each time is 16 L.