Removal and Replacement

One Removal and Replacement: Track Both Parts

Removal and replacement means taking some liquid out of a vessel and pouring the same amount of something else, usually water, back in. The vessel stays full, but the milk inside drops every time. This type comes up often in CGL papers.

Once. A vessel holds 72 litres of pure milk. 12 litres is taken out and replaced with water.

• Taking out 12 L of pure milk leaves 60 L of milk.
• Adding 12 L of water gives 60 L milk and 12 L water, which is 5 : 1. The vessel still holds 72 L.

Twice. Now 12 litres of this mixture is taken out and replaced with water again. This jugful is no longer pure milk; it is the 5 : 1 mixture. So it carries 10 L of milk and 2 L of water.

• Left after the removal: 50 L milk and 10 L water.
• After adding 12 L of water: 50 L milk and 22 L water, which is 25 : 11.

The key idea. Each removal takes away the same fraction of whatever milk is there. Here 12 L is $\frac{1}{6}$ of 72 L, so every removal takes $\frac{1}{6}$ of the milk and leaves $\frac{5}{6}$ of it:

72 × $\frac{5}{6}$ = 60, and then 60 × $\frac{5}{6}$ = 50.

The water is whatever is left of the 72 litres: 72 − 50 = 22 L.

The trap. "12 litres was taken out twice, so 24 litres of milk is gone and 48 litres remain." Wrong: the second jug carried only 10 L of milk, not 12. Milk falls by multiplying, not by subtracting. The subtraction answer (48 L here) is usually one of the options.

Percentage form. "20% of the milk in a can is taken out and replaced with water" means 80% of the milk stays. Do it again and 80% of that stays: 0.8 × 0.8 = 0.64, so the can is 64% milk. Two 20% steps take out 36% of the milk, not 40%.

Two rules to hold on to.

1. When equal amounts go out and come in, the total stays the same.
2. Only the part that is not poured back in shrinks by the fixed fraction. The other part is simply the total minus that.

Taking mixture out and then adding a different amount of one part (not a replacement) is covered in Mixture Basics: Ratios, Concentration and Average Price.

[SSC Tip] Before you use any formula, check that the amount poured in equals the amount taken out. If it does not, the total changes and you must track the litres step by step, as in Different Amounts, Percentage Steps and Leaks below.

Repeating It n Times: The Power Formula

Each operation multiplies the milk by the same fraction $\left(1 - \frac{x}{V}\right)$, where V is the volume in the vessel and x is the amount taken out and replaced each time. After n operations:

$$\text{Milk left} = V\left(1 - \frac{x}{V}\right)^n$$

and the water is V minus the milk left.

Three times. A can holds 108 litres of pure juice. 36 litres is taken out and replaced with water, three times in all. Each time $\frac{36}{108} = \frac{1}{3}$ goes, so $\frac{2}{3}$ stays:

Juice left = 108 × $\left(\frac{2}{3}\right)^3$ = 108 × $\frac{8}{27}$ = 32 litres, and water = 108 − 32 = 76 L.

So juice : water = 32 : 76 = 8 : 19.

The ratio shortcut. If $1 - \frac{x}{V} = \frac{a}{b}$ in lowest terms, then after n operations

$$\text{Milk} : \text{Water} = a^n : \left(b^n - a^n\right)$$

Here a = 2, b = 3 and n = 3, so the ratio is 8 : (27 − 8) = 8 : 19 without any litres at all. With $\frac{7}{8}$ three times the ratio is 343 : 169; with $\frac{4}{5}$ twice it is 16 : 9.

As a percentage. The milk share after n operations is $\left(\frac{a}{b}\right)^n$. The juice above is $\frac{8}{27}$ of the can, about 29.6%.

Concentration. If the vessel starts as a 60% acid solution instead of pure acid, the acid share follows the same rule: 60% × $\left(\frac{2}{3}\right)^2$ = $26\frac{2}{3}\%$ after two operations with $\frac{1}{3}$ replaced each time. Only the starting value changes.

The trap of adding the changes. Taking out $\frac{1}{3}$ three times does not take out everything; it leaves $\frac{8}{27}$. With small fractions the subtraction answer looks believable: 100 litres with 10 litres replaced twice leaves 81 litres of milk, not 80.

Why it works. After each removal, every litre in the vessel has the same make-up. Taking out x of the V litres takes out $\frac{x}{V}$ of every part in it. Water poured back adds to the water only. So the milk is cut by the same fraction every time, like a price that falls by the same percentage every year.

Put the fraction that stays, $1 - \frac{x}{V}$, in lowest terms before raising it to a power. $\frac{60}{72}$ becomes $\frac{5}{6}$, and $\left(\frac{5}{6}\right)^2 = \frac{25}{36}$ is far easier than squaring $\frac{60}{72}$.

Working Backwards: Finding x, V or n

Many questions give the end result and ask for the amount taken out each time (x), the size of the vessel (V) or the number of operations (n). Write the fraction of milk left, then take the root.

Powers worth knowing.

Finding x. "A vessel holds 125 litres of pure milk. x litres is taken out and replaced with water, three times. 64 litres of milk is left. Find x." Fraction of milk left = $\frac{64}{125} = \left(\frac{4}{5}\right)^3$. So each time $\frac{4}{5}$ stays and $\frac{1}{5}$ goes: x = $\frac{1}{5}$ × 125 = 25 litres.

Finding V. "From a can of oil, 9 litres is taken out and replaced with kerosene. This is done once more. Oil : kerosene is now 16 : 33. How much oil did the can hold?" Oil share = $\frac{16}{16 + 33} = \frac{16}{49} = \left(\frac{4}{7}\right)^2$. So $\frac{3}{7}$ goes each time, and 9 litres is $\frac{3}{7}$ of the can. V = 9 × $\frac{7}{3}$ = 21 litres.

Finding n. "A vessel has 40 litres of pure milk, and 10 litres is replaced with water each time. After how many operations is milk : water = 81 : 175?" Each time $\frac{3}{4}$ stays. The milk share is $\frac{81}{81 + 175} = \frac{81}{256} = \left(\frac{3}{4}\right)^4$, so it takes 4 operations.

"First time below" questions work the same way: keep multiplying by the fraction left until you cross the target, and count the steps. With $\frac{3}{4}$ staying each time, milk is 75%, 56.25%, 42.2% … of the vessel, so it first falls below half after the third operation.

[SSC Tip] Add the two terms of the final ratio. After 2 operations the sum is a perfect square, after 3 a perfect cube, and so on (16 + 33 = 49 = $7^2$, 81 + 175 = 256 = $4^4$). The root of the sum is b and the root of the milk term is a, which hands you the fraction left at once. If the sum is not a perfect power, recheck the reading of the question.

Starting with a Mixture, Not a Pure Liquid

Often the vessel already holds a mixture, say milk and water in the ratio 5 : 3, and some of it is replaced with water. The same rule works for the part that is not poured back in: the milk is multiplied by $\left(1 - \frac{x}{V}\right)$ each time, and the water is the rest.

$$\text{Milk after } n \text{ operations} = \text{milk at the start} \times \left(1 - \frac{x}{V}\right)^n$$

Worked through. An 80-litre can holds milk and water in the ratio 5 : 3, so milk = 50 L and water = 30 L. 16 litres is replaced with water, twice. Each time $\frac{16}{80} = \frac{1}{5}$ of the mixture goes.

• After the first: milk = 50 × $\frac{4}{5}$ = 40 L and water = 80 − 40 = 40 L.
• After the second: milk = 40 × $\frac{4}{5}$ = 32 L and water = 48 L.
• Milk : water = 32 : 48 = 2 : 3.

A step-by-step check of the first operation: the 16 litres taken out is 5 : 3, so it carries 10 L milk and 6 L water. That leaves 40 L milk and 24 L water, and adding 16 L of water makes it 40 : 40. Same answer.

Strength form. A 60-litre solution is 45% acid. 20 litres is taken out and replaced with water, twice. Each time $\frac{1}{3}$ goes, so the strength becomes 45% × $\frac{2}{3}$ × $\frac{2}{3}$ = 20%. You never need the litres of acid.

Finding the vessel size from a mixture. "A jar has acid and water in the ratio 4 : 1. When 10 litres of the mixture is replaced with water, the ratio becomes 2 : 1. Find the capacity of the jar." The acid share goes from $\frac{4}{5}$ to $\frac{2}{3}$. The fraction that stays is $\frac{2}{3} \div \frac{4}{5} = \frac{5}{6}$, so $\frac{1}{6}$ was taken out. 10 litres is $\frac{1}{6}$ of the jar, so the jar holds 60 litres. Check: 48 L acid becomes 40 L, and water becomes 20 L.

Why the ratio changes at all. Taking mixture out never changes the ratio on its own (see Mixture Basics: Ratios, Concentration and Average Price). It is the water poured back that changes it. So the first half of each operation shrinks both parts equally; the second half adds only to the water.

Never apply $\left(1 - \frac{x}{V}\right)^n$ to the water when water is what you are pouring back; the water grows, it does not shrink.

Replacing with Milk or with Another Mixture

Replacing with the other part. If the mixture is topped up with milk instead of water, the water is now the part that is not poured back, so the water is the one that shrinks.

"A 45-litre can has milk and water in the ratio 2 : 3. 15 litres is taken out and replaced with milk, twice." Water at the start = 27 L. Each time $\frac{15}{45} = \frac{1}{3}$ goes, so water = 27 × $\left(\frac{2}{3}\right)^2$ = 12 L and milk = 45 − 12 = 33 L. Milk : water = 11 : 4.

Replacing with another mixture. Now both parts come in, so neither part follows a clean power. Go step by step: take out, then add what comes in.

"36 litres of a mixture has milk and water in the ratio 3 : 1. 12 litres is taken out and replaced with 12 litres of a mixture with milk : water = 1 : 3."

• Before: milk 27 L, water 9 L.
• Take out 12 L, which is $\frac{1}{3}$ of the vessel: milk 18 L, water 6 L.
• Pour in 12 L of the 1 : 3 mixture: 3 L milk and 9 L water.
• After: milk 21 L, water 15 L, which is 7 : 5.

One step is a weighted average. Replacing x litres of a V-litre vessel is the same as mixing (V − x) litres of the old mixture with x litres of the new one. Milk share above: $\frac{24 \times \frac{3}{4} + 12 \times \frac{1}{4}}{36} = \frac{21}{36}$, the same 7 : 5.

[Tier 2] Many steps with another solution. If the vessel starts at strength $C_0$ and each time x of V litres is replaced with a solution of strength s, then after n operations

$$C_n = s + \left(C_0 - s\right)\left(1 - \frac{x}{V}\right)^n$$

40 litres of 60% acid, with 10 litres replaced by 20% acid twice: $C_2 = 20 + 40 \times \left(\frac{3}{4}\right)^2 = 20 + 22.5 = 42.5$, so the strength is 42.5%. Step by step: the acid goes 24 L → 18 + 2 = 20 L (50%) → 15 + 2 = 17 L (42.5%). Pouring back water is just the case s = 0, which gives the power formula from earlier in this section.

The idea behind the formula: only the gap between the vessel's strength and the incoming strength shrinks by the factor. Here the gap starts at 40 points and becomes 40 × $\frac{9}{16}$ = 22.5 points.

With another mixture coming in, the strength moves towards the strength of what is poured in but never reaches or passes it. Here it falls from 60% towards 20%, so any option at or below 20% is impossible.

What Fraction to Replace for a Target Ratio or Strength

Here the final ratio is given and you find how much of the mixture to take out and replace. Let f be the fraction of the mixture replaced.

Replaced with water. The milk share is multiplied by (1 − f), so

$$1 - f = \frac{\text{target milk share}}{\text{present milk share}}$$

"A vessel is full of milk and water in the ratio 7 : 3. What fraction of the mixture must be taken out and replaced with water so that the vessel holds equal amounts of milk and water?" The milk share goes from $\frac{7}{10}$ to $\frac{1}{2}$. So $1 - f = \frac{1}{2} \div \frac{7}{10} = \frac{5}{7}$ and $f = \frac{2}{7}$. In a 70-litre vessel that is 20 litres.

The tempting answer is the drop in share, $\frac{7}{10} - \frac{1}{2} = \frac{1}{5}$. It is wrong because the litres you take out are only 70% milk.

Replaced with the other part. "Milk and water are in the ratio 3 : 2. What fraction must be replaced with milk to make the ratio 4 : 1?" Now the water shrinks. The water share goes from $\frac{2}{5}$ to $\frac{1}{5}$, so $1 - f = \frac{1}{2}$ and half the mixture must be replaced.

Replaced with another solution. "Some of a 45% alcohol solution is replaced with a 15% solution, and the result is 25% alcohol. What fraction was replaced?" One replacement mixes (1 − f) of the old with f of the new, so

$$f = \frac{C_0 - C_{\text{target}}}{C_0 - s} = \frac{45 - 25}{45 - 15} = \frac{2}{3}$$

This is the alligation rule in another form; see The Alligation Rule and Mixing for Profit.

Doing it twice to reach a target. "What fraction must be replaced with water each time, twice, so that pure milk becomes 64% milk?" $(1 - f)^2 = 0.64$, so 1 − f = 0.8 and $f = \frac{1}{5}$, which is 20% each time.

In litres. Multiply f by the volume of the vessel. A question that asks "how many litres must be drawn off" is the same question with one more step.

Check the answer by running it forward. With $f = \frac{2}{7}$ of 70 litres: milk 49 L × $\frac{5}{7}$ = 35 L and water 35 L. It takes ten seconds and catches the "difference of shares" trap every time.

Different Amounts, Percentage Steps and Leaks

Different amounts each time. The power formula needs the same fraction every time. When the amounts differ, use each step's own fraction, worked out on the volume at that moment.

"A tank holds 100 litres of pure milk. 10 litres is replaced with water, then 20 litres of the mixture is replaced with water, then 25 litres." The fractions left are $\frac{9}{10}$, $\frac{4}{5}$ and $\frac{3}{4}$. Milk = 100 × $\frac{9}{10}$ × $\frac{4}{5}$ × $\frac{3}{4}$ = 54 litres.

[Tier 2] Taking out one amount and putting in another. Now the volume changes, so track both parts in litres.

"A can holds 60 litres of pure milk. 12 litres is taken out and 18 litres of water is poured in. Then 11 litres of the mixture is taken out."

• After step 1: milk 48 L and water 18 L, 66 L in all.
• 11 L is $\frac{1}{6}$ of 66 L, so it takes 8 L milk and 3 L water, leaving milk 40 L and water 15 L.
• Milk : water = 8 : 3.

Percentage steps. "20% of a can of pure milk is taken out and replaced with water, and then 25% of the mixture is taken out and replaced with water." Milk = 100% × 0.8 × 0.75 = 60%. It is not 100 − 20 − 25 = 55%. Percentage steps multiply, just like successive discounts.

Leaks. "A can of pure milk leaks $\frac{1}{10}$ of its contents every hour, and at the end of each hour it is filled up with water." Each hour is one removal and replacement with $\frac{x}{V} = \frac{1}{10}$, so after 2 hours the milk is $\left(\frac{9}{10}\right)^2$ = 81% of the can. A milkman who sells a fifth of his milk each day and tops the can up with water is the same story with $\frac{4}{5}$.

Does the volume stay fixed? Read the wording. "Taken out and replaced" keeps the volume fixed. "Taken out, and then 18 litres added" does not. A leak that is topped up "to the brim" keeps it fixed; one that is not topped up is just a removal, and the ratio does not change.

[SSC Tip] The trap is to add the amounts taken out. In the 100-litre tank above, 10 + 20 + 25 = 55 litres were drawn, but only 46 litres of milk left the tank, because the later jugs were part water.

Pouring Back and Forth Between Two Vessels

[Tier 2] In this puzzle two vessels swap equal amounts.

"Vessel A holds 30 litres of milk and vessel B holds 30 litres of water. 6 litres of milk is poured from A into B and mixed well. Then 6 litres of the mixture is poured from B back into A."

• After the first pour: A has 24 L milk. B has 6 L milk and 30 L water, 36 L in all, with milk share $\frac{1}{6}$.
• The 6 litres poured back carries 1 L milk and 5 L water.
• A: 25 L milk and 5 L water, so 5 : 1. B: 5 L milk and 25 L water, so 1 : 5.

The symmetry. Both vessels end with 30 litres. Every litre of water that went into A pushed one litre of milk out to B. So the water in A always equals the milk in B (5 litres each here), however many times equal amounts go back and forth.

Shortcut for one round. When A starts with V litres of milk and B with V litres of water, and x litres goes across and back, A ends with milk : water = V : x. Here 30 : 6 = 5 : 1.

Quick checks for any replacement answer.

• Milk never reaches zero by replacement; it only shrinks by a fraction each time.
• After two or more operations, the milk left is more than the subtraction answer (V − nx) and less than the one-step answer.
• The two terms of the final ratio add up to a perfect power when the same fraction is used every time.
• If the volume stayed fixed, milk + water must equal the volume of the vessel.

All the formulas in one place.

Solved Examples

Question 1

A vessel contains 36 litres of pure milk. 6 litres of milk is taken out and replaced with water. Then 6 litres of the mixture is taken out and replaced with water. How much milk is left in the vessel? (a) 24 litres (b) 30 litres (c) 25 litres (d) 27 litres

Answer. Each time 6 L out of 36 L goes, which is $\frac{1}{6}$ of whatever milk is in the vessel. So milk = 36 × $\frac{5}{6}$ × $\frac{5}{6}$ = 25 litres. Step by step: after the first round there is 30 L milk and 6 L water; the second 6-litre jug is 5 : 1, so it takes only 5 L of milk, leaving 25 L. The answer is (c) 25 litres. Subtracting 6 litres twice would give 24 litres.

Question 2

A can holds 32 litres of pure juice. 8 litres is drawn out and replaced with water, and this is done three times in all. How much juice is left in the can? (a) 13.5 litres (b) 8 litres (c) 18 litres (d) 16 litres

Answer. $\frac{8}{32} = \frac{1}{4}$ goes each time, so $\frac{3}{4}$ of the juice stays. Juice = 32 × $\left(\frac{3}{4}\right)^3$ = 32 × $\frac{27}{64}$ = (a) 13.5 litres.

Question 3

A container holds 49 litres of pure alcohol. 14 litres is taken out and replaced with water, and this is done once more. What is the ratio of alcohol to water in the container now? (a) 3 : 4 (b) 25 : 24 (c) 24 : 25 (d) 5 : 2

Answer. $\frac{14}{49} = \frac{2}{7}$ goes each time, so $\frac{5}{7}$ stays. After two rounds the alcohol share is $\left(\frac{5}{7}\right)^2 = \frac{25}{49}$, so alcohol : water = 25 : (49 − 25) = (b) 25 : 24. In litres that is 25 L and 24 L.

Question 4

A tank contains 98 litres of pure milk. x litres of milk is taken out and replaced with water. Then x litres of the mixture is taken out and replaced with water. The tank now has 72 litres of milk. What is the value of x? (a) 13 (b) 12 (c) 16 (d) 14

Answer. Milk left = 98 × $\left(1 - \frac{x}{98}\right)^2$ = 72, so $\left(1 - \frac{x}{98}\right)^2 = \frac{72}{98} = \frac{36}{49}$. Taking the square root, $1 - \frac{x}{98} = \frac{6}{7}$, so $\frac{x}{98} = \frac{1}{7}$ and x = 14. The answer is (d) 14. Check: 98 L becomes 84 L and then 72 L.

Question 5

A thief draws 10 litres of wine from a full cask and replaces it with water. He does this two more times. The ratio of wine to water in the cask is now 343 : 169. How much wine did the cask hold at first? (a) 70 litres (b) 80 litres (c) 90 litres (d) 120 litres

Answer. Three operations, so I look for cubes. 343 + 169 = 512 = $8^3$ and 343 = $7^3$, so the wine share is $\left(\frac{7}{8}\right)^3$. Each time $\frac{7}{8}$ stays and $\frac{1}{8}$ is taken. The 10 litres is $\frac{1}{8}$ of the cask, so the cask held (b) 80 litres. Check: out of every 512 parts of the cask, 343 are wine and 512 − 343 = 169 are water, which is the ratio given.

Question 6

A drum holds 90 litres of pure juice. Each time, 30 litres is taken out and replaced with water. After how many such operations does the juice in the drum first fall below 20 litres? (a) 3 (b) 5 (c) 4 (d) 2

Answer. $\frac{30}{90} = \frac{1}{3}$ goes each time, so the juice is multiplied by $\frac{2}{3}$. It goes 90 → 60 → 40 → $26\frac{2}{3}$ → $17\frac{7}{9}$ litres. It is still above 20 after the third operation and first drops below 20 after the fourth. The answer is (c) 4. Subtracting 30 each time would wrongly say the drum is empty of juice after 3 operations.

Question 7

A 105-litre mixture contains milk and water in the ratio 6 : 1. 35 litres of the mixture is taken out and replaced with water, and the process is repeated once more. What is the ratio of milk to water now? (a) 13 : 8 (b) 4 : 17 (c) 4 : 3 (d) 8 : 13

Answer. Milk at the start = 90 L. Each time $\frac{35}{105} = \frac{1}{3}$ of the mixture goes, and water is what comes back, so only the milk follows the power rule: milk = 90 × $\left(\frac{2}{3}\right)^2$ = 90 × $\frac{4}{9}$ = 40 L. Water = 105 − 40 = 65 L. Milk : water = (d) 8 : 13.

Question 8

A 30-litre mixture has milk and water in the ratio 7 : 3. 10 litres of the mixture is taken out and replaced with 10 litres of pure milk. What is the new ratio of milk to water? (a) 31 : 9 (b) 3 : 1 (c) 4 : 1 (d) 7 : 3

Answer. Milk = 21 L and water = 9 L. The 10 litres taken out is $\frac{1}{3}$ of the vessel, so it carries 7 L milk and 3 L water, leaving 14 L and 6 L. Adding 10 L of milk gives 24 : 6 = (c) 4 : 1. Quicker: water is the part not poured in, so water = 9 × $\frac{2}{3}$ = 6 L and milk = 30 − 6 = 24 L.

Question 9

A vessel holds 72 litres of milk and water in the ratio 5 : 1. 18 litres of the mixture is taken out and replaced with 18 litres of a mixture of milk and water in the ratio 1 : 2. What is the ratio of milk to water in the vessel now? (a) 17 : 7 (b) 11 : 4 (c) 2 : 1 (d) 5 : 2

Answer. At the start: milk 60 L and water 12 L. Taking out 18 L, which is $\frac{1}{4}$ of the vessel, leaves 45 L milk and 9 L water. The 18 litres poured in has 6 L milk and 12 L water. Now milk = 51 L and water = 21 L, so the ratio is (a) 17 : 7.

Question 10

[Tier 2] A 50-litre solution contains 80% alcohol. 10 litres of it is taken out and replaced with 10 litres of a 30% alcohol solution, and this is done once more. What is the strength of the final solution? (a) 70% (b) 51.2% (c) 60% (d) 62%

Answer. First round: alcohol is 40 L. Taking out $\frac{1}{5}$ leaves 32 L, and the 10 litres poured in brings 3 L, so 35 L (70%). Second round: $\frac{1}{5}$ of 35 leaves 28 L, plus 3 L makes 31 L out of 50, which is (d) 62%. By the formula: 30 + (80 − 30) × $\left(\frac{4}{5}\right)^2$ = 30 + 32 = 62. Treating the incoming liquid as plain water would give 51.2%.

Question 11

A vessel is full of milk and water in the ratio 7 : 5. What fraction of the mixture must be drawn off and replaced with water so that the vessel contains equal amounts of milk and water? (a) $\frac{1}{12}$ (b) $\frac{1}{7}$ (c) $\frac{1}{6}$ (d) $\frac{2}{7}$

Answer. The milk share has to fall from $\frac{7}{12}$ to $\frac{1}{2}$. Replacing a fraction f with water multiplies the milk share by (1 − f), so $1 - f = \frac{1}{2} \div \frac{7}{12} = \frac{6}{7}$ and f = (b) $\frac{1}{7}$. Check with an 84-litre vessel: milk 49 L; drawing off 12 L leaves 49 × $\frac{6}{7}$ = 42 L of milk, exactly half. Don't just subtract the shares: $\frac{7}{12} - \frac{1}{2}$ gives $\frac{1}{12}$, but the litres drawn off are partly water.

Question 12

A solution has acid and water in the ratio 2 : 3. What fraction of the solution must be taken out and replaced with pure acid so that the ratio of acid to water becomes 3 : 2? (a) $\frac{1}{3}$ (b) $\frac{1}{5}$ (c) $\frac{2}{5}$ (d) $\frac{1}{2}$

Answer. Acid is poured in, so the water is the part that shrinks. Its share must fall from $\frac{3}{5}$ to $\frac{2}{5}$, so $1 - f = \frac{2}{5} \div \frac{3}{5} = \frac{2}{3}$ and f = (a) $\frac{1}{3}$. Check with 15 litres: 6 L acid and 9 L water; take out 5 L (2 L acid, 3 L water) and add 5 L acid to get 9 L acid and 6 L water, which is 3 : 2.

Question 13

Some of a 70% sugar solution is replaced with an equal quantity of a 30% sugar solution. The new solution is 46% sugar. What fraction of the original solution was replaced? (a) $\frac{2}{5}$ (b) $\frac{2}{3}$ (c) $\frac{3}{5}$ (d) $\frac{3}{4}$

Answer. If a fraction f is replaced, the new solution is (1 − f) of the 70% solution and f of the 30% one. So 70(1 − f) + 30f = 46, which gives 40f = 24 and f = (c) $\frac{3}{5}$. Check with 5 litres: 2 L of 70% and 3 L of 30% hold 1.4 + 0.9 = 2.3 L of sugar, which is 46% of 5 L.

Question 14

A 90-litre mixture has milk and water in the ratio 5 : 1. How many litres of the mixture must be taken out and replaced with water so that the ratio of milk to water becomes 2 : 1? (a) 15 litres (b) 30 litres (c) 20 litres (d) 18 litres

Answer. The milk share must fall from $\frac{5}{6}$ to $\frac{2}{3}$, so $1 - f = \frac{2}{3} \div \frac{5}{6} = \frac{4}{5}$ and $f = \frac{1}{5}$. That is $\frac{1}{5}$ of 90 = (d) 18 litres. Check: milk 75 L becomes 60 L and water becomes 30 L.

Question 15

Some of the milk in a can of pure milk is taken out and replaced with water. The same quantity of the mixture is then taken out and replaced with water again. The can now has 36% water. What percentage of the can's contents was replaced each time? (a) 18% (b) 20% (c) 16% (d) 36%

Answer. 36% water means 64% milk. Two equal replacements multiply the milk share by (1 − p) twice, so $(1 - p)^2 = 0.64$, 1 − p = 0.8 and p = (b) 20%. Check: 100 → 80 → 64.

Question 16

A tank holds 48 litres of pure milk. 8 litres is taken out and replaced with water. Then 12 litres of the mixture is taken out and replaced with water, and finally 16 litres of the mixture is taken out and replaced with water. How much milk is left? (a) 12 litres (b) 18 litres (c) 20 litres (d) 24 litres

Answer. The amounts differ, so I use each step's own fraction of the 48 litres: $\frac{8}{48} = \frac{1}{6}$, $\frac{12}{48} = \frac{1}{4}$ and $\frac{16}{48} = \frac{1}{3}$. The milk left = 48 × $\frac{5}{6}$ × $\frac{3}{4}$ × $\frac{2}{3}$ = (c) 20 litres.

Question 17

[Tier 2] A jar holds 40 litres of pure milk. 8 litres is taken out and 12 litres of water is poured in. Then 11 litres of the mixture is taken out and 8 litres of pure milk is poured in. What is the ratio of milk to water in the jar now? (a) 32 : 9 (b) 8 : 3 (c) 3 : 1 (d) 24 : 17

Answer. The amounts in and out are not equal, so I track litres. After step 1: milk 32 L and water 12 L, 44 L in all. The 11 litres taken out is $\frac{1}{4}$ of 44 L, so it carries 8 L milk and 3 L water, leaving 24 L milk and 9 L water. Adding 8 L of milk gives 32 L milk and 9 L water: (a) 32 : 9.

Question 18

A milkman has a can full of pure milk. Each day he sells $\frac{1}{5}$ of the contents of the can and fills it up again with water. What percentage of the can is milk at the end of the fourth day? (a) 20% (b) 59.04% (c) 40.96% (d) 51.2%

Answer. Each day is one removal and replacement in which $\frac{4}{5}$ of the milk stays. After four days the milk share is $\left(\frac{4}{5}\right)^4 = \frac{256}{625}$ = (c) 40.96%. 59.04% is the water, not the milk.

Question 19

A container is full of a 40% acid solution. 25% of the solution is taken out and replaced with water. Then 20% of the new solution is taken out and replaced with water. What is the strength of the final solution? (a) 22% (b) 24% (c) 18% (d) 26%

Answer. Each step keeps a fraction of the acid: 75%, then 80%. Strength = 40% × 0.75 × 0.8 = (b) 24%.

Question 20

[Tier 2] Jar A holds 36 litres of milk and jar B holds 36 litres of water. 9 litres of milk is poured from A into B and mixed well. Then 9 litres of the mixture is poured from B back into A. How much milk is now in jar B? (a) 9 litres (b) 1.8 litres (c) 6.75 litres (d) 7.2 litres

Answer. After the first pour, B has 9 L milk and 36 L water, 45 L in all, so its milk share is $\frac{1}{5}$. The 9 litres sent back carries 1.8 L milk and 7.2 L water. B keeps 9 − 1.8 = (d) 7.2 litres of milk. A gets exactly 7.2 L of water, which is the symmetry: the water in A equals the milk in B. Don't answer 1.8 litres: that is the milk that went back to A.

Question 21

A jar contains milk and water in the ratio 5 : 1. When 12 litres of the mixture is replaced with water, the ratio becomes 5 : 3. How many litres of milk did the jar contain at first? (a) 48 litres (b) 40 litres (c) 30 litres (d) 36 litres

Answer. The milk share falls from $\frac{5}{6}$ to $\frac{5}{8}$, so the fraction that stays is $\frac{5}{8} \div \frac{5}{6} = \frac{3}{4}$. That means $\frac{1}{4}$ was replaced, and 12 litres is $\frac{1}{4}$ of the jar. The jar holds 48 litres, so the milk at first was $\frac{5}{6}$ × 48 = (b) 40 litres. Check: 40 L milk becomes 30 L and water becomes 18 L, which is 5 : 3. 48 litres is the size of the jar, not the milk.

Question 22

A container holds a mixture of milk and water in the ratio 3 : 2. When 15 litres of the mixture is taken out and replaced with pure milk, the ratio becomes 2 : 1. What is the capacity of the container? (a) 90 litres (b) 75 litres (c) 60 litres (d) 45 litres

Answer. Milk is poured in, so the water is the part that shrinks. Its share falls from $\frac{2}{5}$ to $\frac{1}{3}$, so the fraction that stays is $\frac{1}{3} \div \frac{2}{5} = \frac{5}{6}$, and $\frac{1}{6}$ was replaced. 15 litres is $\frac{1}{6}$ of the container, which holds (a) 90 litres. Check: 54 L milk and 36 L water; removing 15 L leaves 45 L and 30 L; adding 15 L milk gives 60 : 30 = 2 : 1.