Aptitude · Mixtures and Alligations · Model 2

Draw off, top up, repeat — and why the milk never runs out

Q: State the formula and say what each symbol is.

Milk after n operations is V(1 − x/V)^n , where V is the vessel's volume, x is the amount drawn off and replaced each time, and n is the number of operations. The water is simply V minus that.

Q: Give the final ratio of milk to water directly, without computing litres.

It is (V−x)^n to V^n − (V−x)^n . For 40 litres drawing 8 twice that is 4² : 5² − 4² = 16 : 9. This is the fastest route when the question only wants a ratio.

A vessel of milk has some drawn off and is refilled with water, and then it happens again. Students subtract the same number of litres each time. The process actually multiplies by the same fraction each time, and that difference is the entire model.

Set the vessel and the draw, and watch each round multiply →

Each operation removes the same fraction of what is left, not the same amount. So the milk is multiplied by (1 − x/V) every round.

01 The idea

The volume never changes, so think in fractions

Take 40 litres of pure milk. Draw off 8 litres and replace them with 8 litres of water. Now draw off 8 litres of the mixture and replace with water again. How much milk is left?

The tempting answer is 40 − 8 − 8 = 24 litres, and it is wrong. The second withdrawal takes 8 litres of a mixture that is already part water, so it removes less than 8 litres of milk. Treating each round as a flat subtraction overstates the loss every time after the first.

What is actually constant is the proportion removed. You take 8 out of 40, which is a fifth of everything in the vessel — and because the mixture is uniform, a fifth of the milk goes with it. So four fifths of the milk survives each round, whatever the milk happens to be at the time.

That is why the answer is a power rather than a subtraction. Four fifths of four fifths is sixteen twenty-fifths, so after two rounds the milk is 40 × 16/25 = 25.6 litres — noticeably more than the 24 the wrong method predicts. It also explains something students find odd: the milk never reaches zero, because a fraction of a positive quantity is always positive.

The vessel is refilled each time, so the total is constant and each round removes the same fraction. Multiply by (1 − x/V) once per operation.

Replacement operationDraw off x litres of the mixture and top the vessel back up to V with water. The volume ends where it started; only the composition moved.

Surviving fraction(V − x)/V, the share of the milk left after one operation. Constant across rounds, which is exactly why the answer is a power.

Uniform mixtureThe assumption that what you draw off has the same milk-to-water ratio as the vessel. Without it the model does not hold, and every exam question assumes it.

02 Worked example

40 litres, 8 drawn off, twice

This vessel carries the lesson and the console. A vessel holds 40 litres of pure milk. 8 litres are drawn off and replaced with water. From the mixture, 8 litres are again drawn off and replaced with water. Find the final quantity of milk.

Find the fraction that survives one roundEight litres out of forty is a fifth of the vessel. The mixture is uniform, so a fifth of the milk leaves and four fifths stay.(40 − 8) / 40 = 32/40 = 4/5 survives

Apply it onceThe vessel started as pure milk, so all 40 litres are milk before the first round.40 × 4/5 = 32 litres of milk (and 8 of water)

Apply it again — to 32, not to 40The second withdrawal takes a fifth of what is now present. A fifth of 32 is 6.4, so only 6.4 litres of milk leave, not 8.32 × 4/5 = 25.6 litres of milk

Read off the waterThe volume is back to 40, so the water is whatever is not milk.water = 40 − 25.6 = 14.4 litres ⇒ milk : water = 25.6 : 14.4 = 16 : 9

The formula, now earnedTwo rounds of the same multiplication is that fraction squared. For n rounds it is the nth power.milk = 40 × (4/5)² = 40 × 16/25 = 25.6 litres

Compare 25.6 against the 24 you get by subtracting 8 twice. The gap of 1.6 litres is exactly the milk that the second withdrawal did not remove because water had already taken its place. Notice also that the final ratio 16 : 9 is 4² : (5² − 4²) — the powers appear in the ratio too, which is often the fastest way to answer a ratio-only question.

03 The method

The formula, and the three things it is not

One formula covers the model. Most errors come from applying it to the wrong quantity rather than from misremembering it.

After n operations of drawing off x from a vessel of V and refilling: milk = V × (1 − x/V)ⁿ, and the water is V − milk. The ratio of milk to water is (V−x)ⁿ : Vⁿ − (V−x)ⁿ.

Three things this is not. It is not V − nx — that double-counts. The power applies to the original volume, not to whatever is currently in the vessel, so do not multiply by V again each round. And if the vessel is not refilled, the total does change and this formula does not apply at all.

Round	Milk (from 40 L, 8 drawn)	Naive V − nx	Error
0	40	40	0
1	32	32	0
2	25.6	24	1.6
3	20.48	16	4.48
4	16.384	8	8.384
5	13.107	0	13.107

05 Cheat sheet

Model 2 on one page

One formula and the traps around it. The last two rows are the ones that decide whether the formula may be used at all.

What you want	Use	On 40 L, draw 8, n = 2
Milk after n rounds	`V(1 − x/V)^n`	40 × (4/5)² = 25.6
Water after n rounds	`V − milk`	40 − 25.6 = 14.4
Milk : water	`(V−x)^n : V^n − (V−x)^n`	16 : 9
Surviving fraction	`(V − x)/V`	4/5
Find n from a final ratio	`solve the power`	16/25 → n = 2
Flat subtraction	`V − nx — WRONG`	would give 24
No refill	`formula does not apply`	the total changes

The milk never hits zeroEach round keeps a positive fraction of a positive quantity. The milk approaches zero and never arrives, which is why “after how many rounds is the milk gone” has no answer.

The power is on the original volumeV(1−x/V)^n uses the starting volume once. Multiplying by V again each round is the second most common error after flat subtraction.

Refilling is what makes it workThe formula depends on the volume returning to V each time. If liquid is removed and not replaced, the fraction changes every round and you must track it by hand.

06 Where & why

Where this model shows up

The replacement model is set often because the wrong method gives a clean-looking number that sits in the options.

Bank PO · SSC CGL

Milk and water, two or three rounds

The standard form. Two rounds is most common; three appears in mains papers because the gap from the naive answer is larger and more tempting.

TCS NQT · Infosys

Ratio-only versions

“Find the final ratio of milk to water.” The powers form (V−x)^n : V^n − (V−x)^n answers it without computing any litres.

Reverse questions

Find n, or find x

Given that the milk is now 16/25 of the original, the power tells you two rounds happened at a fifth each. These are harder and reward knowing the structure rather than the formula.

Compound interest, structurally

The same repeated multiplication

This is compound decay: a fixed fraction applied repeatedly. Recognising it as the mirror of compound interest makes both chapters easier.

The tell for this model is the word replaced or refilled. If liquid comes out and nothing goes back in, you are not in this model and the formula will mislead you.

07 Interview questions

What gets asked

Ten, and the first two are the ones that decide whether the rest goes right.

40 litres of milk, 8 drawn off and replaced with water, twice. How much milk is left?

25.6 litres. Eight out of forty is a fifth, so four fifths of the milk survives each round: 40 × (4/5)² = 40 × 16/25 = 25.6. The water is 14.4 litres, giving a ratio of 16 : 9.

Why is the answer not 24 litres?

Because the second withdrawal takes 8 litres of a mixture that is already one-fifth water, so it removes only 6.4 litres of milk rather than 8. Subtracting 8 twice assumes you are always drawing pure milk, which stops being true after the first round.

State the formula and say what each symbol is.

Milk after n operations is V(1 − x/V)^n, where V is the vessel's volume, x is the amount drawn off and replaced each time, and n is the number of operations. The water is simply V minus that.

Why is it a power rather than a product of different fractions?

Because the vessel is refilled to V every time, so the proportion you remove is x/V on every single round. The same fraction applied n times is that fraction to the nth power. If the volume changed between rounds the fractions would differ and no power would appear.

Does the milk ever become zero?

No. Each operation keeps a fixed positive fraction of whatever milk remains, and a positive fraction of a positive number is positive. The milk tends to zero as the rounds increase but never reaches it, so a question asking when the milk is fully gone is ill-posed.

Give the final ratio of milk to water directly, without computing litres.

It is (V−x)^n to V^n − (V−x)^n. For 40 litres drawing 8 twice that is 4² : 5² − 4² = 16 : 9. This is the fastest route when the question only wants a ratio.

The milk is now 16/25 of what it was, and a fifth is drawn each time. How many operations happened?

Two. The surviving fraction per round is 4/5, and (4/5)^n = 16/25 means n = 2. Reverse questions like this are why it pays to see the formula as a repeated fraction rather than as a black box.

What changes if the liquid is removed but not replaced?

Everything. The volume no longer returns to V, so the fraction removed differs each round and the power formula does not apply. You have to track the composition round by round. The word “replaced” in the question is what licenses the formula.

A vessel has milk and water in the ratio 3 : 1, and some is replaced by water. Does the formula still work?

Hard

A vessel holds 50 litres of pure milk. 10 litres are drawn off and replaced with water. Then 5 litres of the mixture are drawn off and replaced with water. (a) Find the final quantity of milk. (b) Explain why the single-power formula cannot be used here and what replaces it. (c) Returning to equal draws of 10 litres each round, show that no number of operations can reduce the milk below 5 litres... or, if that is false, find the smallest number of rounds that takes it below 5.

Follow-up

Part (b) breaks the formula deliberately: unequal draws mean unequal fractions, so you get a product of different factors rather than a power. Part (c) makes you decide whether “never reaches zero” also means “never gets below a given positive number” — and it does not, which is the distinction worth having.

Show the hint

For (c), each round multiplies by 4/5; ask how many multiplications take 50 below 5, and remember that tending to zero means every positive bound is eventually crossed.