Mixtures — SSC CGL Quant Notes

Concepts and rules

If a mixture has two components in the ratio a : b, their fractions of the whole are a/(a + b) and b/(a + b). In 60 L with milk : water = 7 : 3, milk = 7/10 × 60 = 42 L and water = 18 L.

Ratios cannot be added directly. To combine two mixtures, find the actual quantity of each component in each mixture, add component by component, and then form the new ratio.

Adding a pure component (or removing it by evaporation) changes only that component. The other component stays fixed, so use it as the anchor.

Concentration % = quantity of the component × 100/total quantity. When water evaporates, the solute is unchanged and its percentage rises.

Mixing two mixtures in the volume ratio m : n: fraction of a component = (m × f1 + n × f2)/(m + n), where f1 and f2 are its fractions in the two mixtures. For equal volumes this is (f1 + f2)/2.

Taking some mixture out (without replacement) leaves the ratio unchanged; it only reduces each component in the same proportion.

Method and shortcuts

Step 1: write the actual litres or kilograms of each component. Step 2: apply each operation (add, remove, evaporate, mix) to these quantities in order. Step 3: write the final ratio and simplify.

Anchor shortcut: if only water is added to milk : water = a : b to make it c : d, milk is fixed, so new water = milk × d/c and water added = new water − old water.

LCM shortcut for equal volumes: make the ratio sums equal. 3 : 1 (sum 4) and 5 : 3 (sum 8) become 6 : 2 and 5 : 3; adding gives 11 : 5.

When it fails: the LCM trick needs EQUAL quantities of the two mixtures. For unequal quantities scale each ratio to its actual quantity first.

Example 1: Adding water to change the ratio

Question: A 60 L mixture has milk and water in the ratio 7 : 3. How much water must be added to make the ratio 3 : 2?

Solution: Milk = 42 L, water = 18 L. Milk stays 42, so for 3 : 2 the water must be 42 × 2/3 = 28 L. Water to add = 28 − 18 = 10 L.

Answer: 10 L

Example 2: Equal volumes of two mixtures

Question: Two vessels contain milk and water in the ratios 3 : 1 and 5 : 3. Equal quantities are poured into a third vessel. Find the ratio of milk to water in it.

Solution: Take 8 L from each. First: 6 L milk, 2 L water. Second: 5 L milk, 3 L water. Total: 11 L milk and 5 L water, so the ratio is 11 : 5.

Answer: 11 : 5

Example 3: Evaporation

Question: 40 L of a salt solution contains 20% salt. How much water must evaporate to make it a 25% salt solution?

Solution: Salt = 20% of 40 = 8 L and it does not change. For 25%, total = 8 × 100/25 = 32 L. Water evaporated = 40 − 32 = 8 L.

Answer: 8 L

Example 4: Mixing in a given volume ratio

Question: Vessel A has milk and water in the ratio 5 : 2 and vessel B has them in the ratio 3 : 4. If A and B are mixed in the volume ratio 2 : 3, what is the ratio of milk to water in the mixture? (a) 17 : 18 (b) 16 : 19 (c) 19 : 16 (d) 8 : 7

Solution: Take 14 L of A and 21 L of B (ratio 2 : 3, multiples of 7). A: 10 L milk, 4 L water. B: 9 L milk, 12 L water. Total: 19 L milk and 16 L water.

Answer: (c) 19 : 16

Example 5: Remove, then add water

Question: A 90 L mixture has milk and water in the ratio 7 : 2. 18 L of the mixture is taken out and then 12 L of water is added. What is the new ratio of milk to water? (a) 7 : 4 (b) 2 : 1 (c) 5 : 3 (d) 9 : 5

Solution: Removing mixture keeps the ratio 7 : 2, so 90 − 18 = 72 L remains with milk = 7/9 × 72 = 56 L and water = 16 L.

Adding 12 L of water gives water = 28 L. New ratio = 56 : 28 = 2 : 1.

Answer: (b) 2 : 1

Example 6: Three containers combined

Question: Three containers hold 15 L, 20 L and 21 L of alcohol–water mixtures in the ratios 2 : 3, 1 : 3 and 3 : 4. All are poured into one drum. How much pure alcohol must be added to make the ratio 1 : 1? (a) 12 L (b) 14 L (c) 18 L (d) 16 L

Solution: Alcohol = 2/5 × 15 + 1/4 × 20 + 3/7 × 21 = 6 + 5 + 9 = 20 L. Water = 9 + 15 + 12 = 36 L. The drum has alcohol : water = 20 : 36 = 5 : 9.

Water is the anchor at 36 L, so alcohol must reach 36 L. Alcohol to add = 36 − 20 = 16 L.

Answer: (d) 16 L

Common mistakes

Adding ratios: 3 : 1 and 5 : 3 do not combine to 8 : 4. Convert to quantities first.

Changing the anchor. When only water is added, the milk quantity must stay exactly the same in the new ratio.

Thinking that removing some mixture changes the ratio. Removal alone never changes the ratio; adding something afterwards does.

In evaporation questions keeping the water fixed. It is the salt (solute) that stays fixed.

Using the LCM trick for unequal quantities. It works only when equal volumes are mixed.

Practice questions

Q1. A 45 L mixture has milk and water in the ratio 3 : 2. How much milk must be added to make the ratio 9 : 4? (a) 13.5 L (b) 12 L (c) 15 L (d) 10.5 L

Q2. Two alloys contain gold and copper in the ratios 7 : 2 and 7 : 11. Equal weights of the two are melted together. What is the ratio of gold to copper in the new alloy? (a) 5 : 7 (b) 7 : 5 (c) 9 : 5 (d) 7 : 4

Q3. 30 kg of a salt solution contains 10% salt. How much salt must be added to make it 25% salt? (a) 4.5 kg (b) 5 kg (c) 6 kg (d) 7.5 kg

Answers

Q1. (a) Milk 27 L, water 18 L. Water is the anchor: new milk = 18 × 9/4 = 40.5 L, so add 40.5 − 27 = 13.5 L.

Q2. (b) Take 18 kg of each: gold 14 + 7 = 21, copper 4 + 11 = 15, ratio 21 : 15 = 7 : 5.

Q3. (c) Water = 27 kg stays fixed and must be 75% of the new total, so total = 36 kg and salt added = 36 − 30 = 6 kg.

Quick revision

Ratios become quantities first; add like components; then simplify.

The component that is not added or removed is the anchor.

Evaporation: solute fixed; new total = solute × 100/new %.

Equal volumes: equalise the ratio sums (LCM) and add.

Concepts and rules

Replacement means taking out some quantity of a mixture and putting back the same quantity of another liquid (usually water), so the total volume V stays constant.

Each time x litres are drawn, the fraction x/V of EVERY component is removed. So the liquid that is not put back is multiplied by (1 − x/V) at each operation.

After n equal operations: Original liquid left = Initial quantity × (1 − x/V)ⁿ. If the vessel started pure, initial quantity = V.

If a fraction f of the contents is replaced each time, the factor is (1 − f). Replacing 1/4 of it 3 times leaves (3/4)³ = 27/64 of the original liquid.

If different amounts x1, x2, … are drawn, multiply the separate factors (1 − x1/V)(1 − x2/V)…, provided the vessel is refilled to V each time.

The replacing liquid = V − original liquid left (for two components). The ratio original : replacing = (1 − x/V)ⁿ : 1 − (1 − x/V)ⁿ when the vessel started pure.

The formula needs the replacing liquid to contain none of the original liquid. If a mixture is poured back instead, work step by step.

Method and shortcuts

Step 1: identify V, x and n, and which liquid is NOT being put back. Step 2: compute the factor (1 − x/V) as a fraction. Step 3: multiply the starting quantity of that liquid by the factor n times.

Reverse questions: final/initial = (1 − x/V)ⁿ. Take the n-th root of the ratio (it is usually a perfect square or cube) to get 1 − x/V, then find x or V.

Starting with a mixture: apply the factor only to the component that is drawn and not replaced; get the other by subtraction from V.

Check: for n = 1 the formula gives V − x, which is plain subtraction. This is a quick way to test that the factor is right.

When it fails: if the volume drawn is not replaced, or the replacement is a different quantity, the total changes and the formula does not hold.

Example 1: Two replacements

Question: A can has 40 L of pure milk. 4 L is drawn and replaced with water, and this is done once more. How much milk is left?

Solution: Factor = 1 − 4/40 = 9/10. Milk left = 40 × 9/10 × 9/10 = 32.4 L. Step check: after the first round 36 L milk; the second draw removes 10% of it, 3.6 L, leaving 32.4 L.

Answer: 32.4 L

Example 2: Replacing a fraction

Question: A cask holds 80 L of pure juice. One-fourth of its contents is replaced with water, and this is repeated until it has been done 3 times. How much juice remains?

Solution: Factor = 3/4 each time. Juice left = 80 × (3/4)³ = 80 × 27/64 = 33.75 L.

Answer: 33.75 L

Example 3: Finding the quantity drawn

Question: From 60 L of pure milk, x litres are drawn and replaced with water. After this is done twice, 38.4 L of milk remains. Find x.

Solution: (1 − x/60)² = 38.4/60 = 0.64, so 1 − x/60 = 0.8 and x/60 = 0.2, giving x = 12 L.

Answer: 12 L

Example 4: Starting from a mixture

Question: A 75 L mixture has milk and water in the ratio 4 : 1. 15 L of it is drawn and replaced with water. What is the new ratio of milk to water? (a) 3 : 2 (b) 9 : 16 (c) 16 : 9 (d) 12 : 13

Solution: Milk = 60 L initially. Milk left = 60 × (1 − 15/75) = 60 × 4/5 = 48 L. Water = 75 − 48 = 27 L. Ratio = 48 : 27 = 16 : 9.

Answer: (c) 16 : 9

Example 5: Three replacements of a fraction

Question: A vessel is full of pure milk. One-fifth of it is replaced with water, and this is done 3 times in all. What is the final ratio of milk to water? (a) 64 : 61 (b) 61 : 64 (c) 16 : 9 (d) 64 : 125

Solution: Milk left = (4/5)³ = 64/125 of the vessel, so water = 1 − 64/125 = 61/125 of the vessel.

Milk : water = 64 : 61. Option (d) 64 : 125 is milk : whole vessel, a common trap.

Answer: (a) 64 : 61

Example 6: Finding the capacity

Question: From a vessel full of pure milk, 9 L is drawn and replaced with water. This is done once more, and the milk to water ratio becomes 16 : 9. What is the capacity of the vessel? (a) 36 L (b) 45 L (c) 54 L (d) 40 L

Solution: Milk is 16/(16 + 9) = 16/25 of the vessel, so (1 − 9/V)² = 16/25 and 1 − 9/V = 4/5.

Hence 9/V = 1/5 and V = 45 L. Check: 45 → 36 L milk → 36 × 4/5 = 28.8 L milk, water 16.2 L, and 28.8 : 16.2 = 16 : 9.

Answer: (b) 45 L

Common mistakes

Subtracting x each time. The second draw removes less milk because the vessel is no longer pure; use the factor, not repeated subtraction.

Applying the factor to the liquid that is being added. The factor applies only to the liquid that is drawn and not put back.

Giving milk : whole vessel when milk : water is asked, as in option (d) of Example 5.

Using the total V as the starting milk when the vessel started as a mixture. Start from the actual milk quantity.

Miscounting n. 'Done once more' after the first operation means n = 2 in total.

Practice questions

Q1. A tank holds 50 L of pure acid. 5 L is drawn and replaced with water, and this is done 3 times in all. How much acid remains? (a) 36.45 L (b) 35 L (c) 40.5 L (d) 37.5 L

Q2. From a 64 L vessel of pure milk, x litres are drawn and replaced with water twice, leaving 49 L of milk. What is x? (a) 6 (b) 7 (c) 8 (d) 9

Q3. A 120 L mixture has milk and water in the ratio 3 : 1. 24 L is drawn and replaced with water, and this is done once more. What is the final ratio of milk to water? (a) 13 : 12 (b) 12 : 13 (c) 3 : 2 (d) 6 : 5

Answers

Q1. (a) 50 × (9/10)³ = 50 × 0.729 = 36.45 L.

Q2. (c) (1 − x/64)² = 49/64, so 1 − x/64 = 7/8 and x = 8.

Q3. (b) Milk = 90 × (4/5)² = 57.6 L, water = 62.4 L, ratio 576 : 624 = 12 : 13.

Quick revision

Left = Initial × (1 − x/V)ⁿ for the liquid that is not put back.

Fraction f replaced each time ⇒ factor (1 − f).

Reverse: take the n-th root of final/initial to get 1 − x/V.

Other component = V − original liquid left.

Concepts and rules

Alligation finds the ratio in which two ingredients of values c (cheaper) and d (dearer) must be mixed to give a mixture of mean value m. Rule: Cheaper : Dearer = (d − m) : (m − c).

Why it works: the gain of the cheaper part (m − c per unit) must balance the loss of the dearer part (d − m per unit). So quantity × distance is equal on both sides, and each quantity is proportional to the OTHER distance.

The mean m must lie strictly between c and d; otherwise no mixture of the two can give it.

The rule works for any weighted average: price per kg, % concentration, average marks, rate of interest, average speed. The ratio it gives is in the unit the values are 'per': per kg gives kg, per litre gives litres, km/h gives hours.

Water or any free ingredient has value 0. Mixing water with milk of price d to get price m needs water : milk = (d − m) : m.

Alligation works with cost prices. If a selling price and a profit per cent are given, first find the mean cost price = SP × 100/(100 + profit%).

Replacing part of a solution by another: if a fraction of a solution of strength c is replaced by strength d to get m, then (part kept) : (part replaced) = (d − m) : (m − c).

Method and shortcuts

Step 1: write the cheaper value c and the dearer value d on top and the mean m in the middle. Step 2: write d − m under c and m − c under d (cross differences). Step 3: these give the quantity ratio; scale to the given quantity.

Reverse alligation: if the mixing ratio is known and the mean is asked, use the weighted average (n1c + n2d)/(n1 + n2).

Speed questions: alligation of speeds gives the ratio of TIMES, because speed is distance per hour. Convert to distances afterwards.

More than two ingredients: pair each ingredient below the mean with one above it, apply the rule to each pair, and add the parts.

When it fails: alligation needs the mean of the SAME kind of value. Never put a selling price as the mean against cost prices.

Example 1: Two grades of rice

Question: In what ratio must rice at ₹40 per kg be mixed with rice at ₹55 per kg so that the mixture costs ₹45 per kg?

Solution: Cheaper : Dearer = (55 − 45) : (45 − 40) = 10 : 5 = 2 : 1. Check: (2 × 40 + 1 × 55)/3 = 135/3 = 45.

Answer: 2 : 1

Example 2: Mixing water with milk

Question: In what ratio must water be mixed with milk costing ₹60 per litre to get a mixture worth ₹50 per litre?

Solution: Water costs 0. Water : Milk = (60 − 50) : (50 − 0) = 10 : 50 = 1 : 5.

Answer: 1 : 5

Example 3: Finding a quantity

Question: How many kg of tea at ₹180 per kg must be mixed with 30 kg of tea at ₹240 per kg so that the mixture costs ₹220 per kg?

Solution: Cheaper : Dearer = (240 − 220) : (220 − 180) = 20 : 40 = 1 : 2. Dearer = 30 kg is 2 parts, so cheaper = 15 kg.

Answer: 15 kg

Example 4: Mean from selling price and profit

Question: A grocer mixes sugar costing ₹36 per kg with sugar costing ₹48 per kg and sells the mixture at ₹46.20 per kg, making a profit of 10%. In what ratio are the two mixed? (a) 2 : 1 (b) 1 : 2 (c) 3 : 2 (d) 1 : 1

Solution: Mean cost price = 46.20 × 100/110 = ₹42. Cheaper : Dearer = (48 − 42) : (42 − 36) = 6 : 6 = 1 : 1. Using ₹46.20 as the mean would wrongly give 1.8 : 10.2.

Answer: (d) 1 : 1

Example 5: Part of a solution replaced

Question: Some part of a 30% alcohol solution is replaced by an equal amount of a 55% alcohol solution, and the strength becomes 40%. What fraction of the original solution was replaced? (a) 1/3 (b) 2/5 (c) 3/5 (d) 1/2

Solution: The final solution is a mixture of the 30% part that stayed and the 55% part that came in. Kept : Replaced = (55 − 40) : (40 − 30) = 15 : 10 = 3 : 2.

So the replaced part is 2/(3 + 2) = 2/5 of the whole. Check: 3/5 × 30 + 2/5 × 55 = 18 + 22 = 40.

Answer: (b) 2/5

Example 6: Alligation of speeds

Question: Kiran covers 61 km in 9 hours, partly on foot at 4 km/h and partly by cycle at 9 km/h. How far did she walk? (a) 20 km (b) 12 km (c) 16 km (d) 18 km

Solution: Average speed = 61/9 km/h. Speed is per hour, so alligation gives the ratio of TIMES: foot : cycle = (9 − 61/9) : (61/9 − 4) = 20/9 : 25/9 = 4 : 5.

Time on foot = 4/9 × 9 = 4 hours, so distance walked = 4 × 4 = 16 km. Check: cycle 5 hours × 9 = 45 km, and 16 + 45 = 61.

Answer: (c) 16 km

Common mistakes

Writing the differences on the same side. Each quantity takes the OTHER ingredient's distance from the mean.

Using the selling price as the mean. Convert to the mean cost price first, as in Example 4.

Treating the speed alligation ratio as a ratio of distances. It is a ratio of times.

Forgetting that water has value 0 and leaving it out of the alligation.

In replacement-by-solution questions reporting the ratio 3 : 2 instead of the fraction replaced, 2/5.

Practice questions

Q1. Two varieties of pulses costing ₹62 and ₹72 per kg are mixed in the ratio 3 : 2. What is the cost of the mixture per kg? (a) ₹66 (b) ₹67 (c) ₹65 (d) ₹68

Q2. A milkman buys milk at ₹48 per litre, adds water and sells the mixture at ₹48 per litre, making a profit of 20%. In what ratio did he mix water and milk? (a) 1 : 4 (b) 1 : 6 (c) 1 : 5 (d) 1 : 3

Q3. How many litres of a 30% acid solution must be added to 25 L of a 12% acid solution to get an 18% acid solution? (a) 10 L (b) 15 L (c) 12 L (d) 12.5 L

Answers

Q1. (a) Weighted average = (3 × 62 + 2 × 72)/5 = (186 + 144)/5 = 330/5 = ₹66.

Q2. (c) Mean cost = 48 × 100/120 = ₹40; water : milk = (48 − 40) : (40 − 0) = 8 : 40 = 1 : 5.

Q3. (d) 12% : 30% = (30 − 18) : (18 − 12) = 12 : 6 = 2 : 1, so 30% solution = 25/2 = 12.5 L.

Quick revision

Cheaper : Dearer = (d − m) : (m − c); m must lie between c and d.

Water has value 0; use mean cost price, not selling price.

Speeds give a ratio of times; prices per kg give a ratio of kg.

Part replaced: kept : replaced = (d − m) : (m − c).