Reading a Mixture: Ratio, Fraction and Percentage

A mixture question can describe the same thing in three ways. "Milk and water in the ratio 5 : 3", "water is 38\frac{3}{8} of the mixture" and "the mixture has 37.5% water" all describe one and the same mixture. The first skill is to move between these forms without stopping to think.

From a ratio to fractions. If milk : water = a : b, the mixture has a + b equal parts. Milk is aa+b\frac{a}{a+b} of it and water is ba+b\frac{b}{a+b} of it. For 5 : 3 there are 8 parts, so milk is 58\frac{5}{8} = 62.5% and water is 38\frac{3}{8} = 37.5%.

Quantity of each part. Divide the total by the number of parts to get one part. A 96-litre mixture with milk : water = 5 : 3 has 8 parts, so one part is 12 litres. Milk = 5 × 12 = 60 litres and water = 3 × 12 = 36 litres.

From a percentage to a ratio. Write both percentages and cancel. A solution with 45% acid has 55% water, so acid : water = 45 : 55 = 9 : 11. A mixture with 20% water is 80 : 20 = 4 : 1.

Concentration or strength. For a solution, the percentage of the active part is called its strength or concentration. 80 litres of a 15% acid solution holds 15% of 80 = 12 litres of acid and 68 litres of water. A question about "strength" is a percentage question in disguise.

Part of the other part, or part of the whole? Read the base of the percentage carefully.

  • "Water is 25% of the mixture" means water : milk = 1 : 3.
  • "Water is 25% of the milk" means water : milk = 1 : 4, so water is only 15\frac{1}{5} = 20% of the mixture.

Lines like "1 litre of water for every 4 litres of milk", "milk is 4 times the water" and "water is one-fourth of the milk" all describe the second mixture. SSC puts both answers in the options, so when an option simply repeats the percentage in the question, check the base.

Three questions settle almost every case.

  1. Which part changes and which part stays fixed? (Pouring in water leaves the milk fixed.)
  2. Is some mixture being taken out? (Then both parts leave in the mixture's own ratio.)
  3. Am I adding real quantities and not ratio terms? (Ratios never add directly.)

The parts below take these cases one at a time.

Adding Water or Milk: Only One Part Changes

When you pour in water, only the water changes. The milk in the vessel is exactly what it was. So hold the fixed part steady and let the other part move.

Adding water for a new ratio. "A 63-litre mixture has milk and water in the ratio 5 : 2. How much water must be added to make the ratio 3 : 2?"

  • One part = 63 ÷ 7 = 9 litres, so milk = 45 L and water = 18 L.
  • Milk stays 45 L. For 3 : 2, water must be 45 × 23\frac{2}{3} = 30 L.
  • Water to add = 30 − 18 = 12 litres.

The unit way. Make the milk terms the same in both ratios: 5 : 2 = 15 : 6 and 3 : 2 = 15 : 10. Milk stays at 15 units and water goes from 6 to 10 units, so 4 units are added. The old mixture is 21 units = 63 L, so 1 unit = 3 L and 4 units = 12 L. This is the quicker way when the question gives the change and asks for the total.

Adding milk instead. Now the water is fixed. To turn the same mixture into 4 : 1, the water (18 L) stays, so milk must be 4 × 18 = 72 L. Milk to add = 72 − 45 = 27 litres.

Finding the original mixture. "A mixture has acid and water in the ratio 5 : 3. When 10 litres of water is added, the ratio becomes 1 : 1. How much acid is there?" Acid is fixed, so write 1 : 1 as 5 : 5. Water goes from 3 units to 5 units, so 2 units = 10 L and 1 unit = 5 L. Acid = 25 litres, water was 15 L, and the mixture was 40 L at the start.

Both parts added. "To 32 litres of milk and water in the ratio 3 : 1, 4 litres of milk and some water are added, and the ratio becomes 7 : 3. How much water was added?" Milk = 24 + 4 = 28 L. For 7 : 3, water must be 28 × 37\frac{3}{7} = 12 L. It was 8 L, so 4 litres of water went in. When both parts change, work with real litres, never with ratio terms.

Don't compare 5 : 2 with 3 : 2 as they stand and say "the milk went down by 2 parts"; the milk did not change at all. The units of the two ratios are different sizes until the fixed part's terms are equal.

Reaching a Target Percentage by Adding One Part

Percentages work the same way as ratios here. The part you do not touch keeps its quantity; only its share of the mixture changes. Divide the fixed quantity by its new share and you get the new total.

Adding water. "A 75-litre mixture of milk and water has 16% water. How much water must be added so that water becomes 30% of the mixture?"

  • Milk = 84% of 75 = 63 L, and it stays 63 L.
  • In the new mixture milk is 70%, so the new total is 63 ÷ 0.7 = 90 L.
  • Water to add = 90 − 75 = 15 litres.

Adding milk. Same 75 litres, but now we want water to fall to 12%. The water (12 L) is fixed. If 12 L is 12% of the new mixture, the new total is 100 L, so 25 litres of milk is added.

Adding the pure substance to a solution. "48 litres of a solution is 25% acid. How much pure acid must be added to make it 40% acid?" This time the water is fixed: 36 L. In the new solution water is 60%, so the total is 36 ÷ 0.6 = 60 L, and 12 litres of acid is added.

Two formulas for checking. For a mixture of V litres, when the water share rises from p1%p_1\% to p2%p_2\% by adding water,

Water to add=V×p2−p1100−p2\text{Water to add} = V \times \frac{p_2 - p_1}{100 - p_2}

and when the water share falls from p1%p_1\% to p2%p_2\% by adding milk,

Milk to add=V×p1−p2p2\text{Milk to add} = V \times \frac{p_1 - p_2}{p_2}

Check: 75 × 1470\frac{14}{70} = 15 and 75 × 412\frac{4}{12} = 25. For acid poured into a solution, read "acid" in place of "water" in the first formula: 48 × 1560\frac{15}{60} = 12.

Ratio given, percentage asked. "A 72-litre mixture has milk and water in the ratio 7 : 2. How much water must be added so that water is 30% of the mixture?" Milk = 56 L and it must be 70% of the new total, so the total becomes 80 L and 8 litres of water is added.

Forward questions. CGL also asks it the other way round. "A 90-litre mixture has 40% water. 30 litres of milk is added. What percentage of the new mixture is water?" Water = 36 L, the new total is 120 L, and 36 ÷ 120 = 30%. Only the total changed; the water did not.

[SSC Tip] Do not take the change in percentage of the old volume. Going from 16% to 30% water is not "add 14% of 75 litres" (10.5 L). The new water has to be 30% of the new, bigger total, which is why the answer is 15 L.

Taking Some Mixture Out, Then Adding One Part

When you pour out some of the mixture, milk and water leave together in the mixture's own ratio. So the ratio does not change; only the amount does. A 4 : 3 mixture is still 4 : 3 after you take out a jugful.

Remove, then add water. "84 litres of a mixture has milk and water in the ratio 4 : 3. 21 litres of the mixture is taken out and then 9 litres of water is added. Find the new ratio."

  • After the removal 63 L is left, still 4 : 3: milk 36 L and water 27 L.
  • Add 9 L of water: milk 36 L and water 36 L, so the ratio is 1 : 1.

The slip is to take the 21 litres out of the milk alone, or out of the water alone. Both parts leave together, so after any removal write down what is left of each part before you do anything else.

Finding how much was taken out. "From 70 litres of a mixture with milk : water = 5 : 2, some mixture is taken out. Then 9 litres of water is added, and the ratio becomes 5 : 3. How much mixture was taken out?" After the removal the milk does not change again, so compare the ratios with the same milk term: 5 : 2 and 5 : 3. Water grew by 1 unit, and that was the 9 litres. The mixture left after the removal was 7 units = 63 L, so 7 litres was taken out.

Remove, then reach a percentage. "From 50 litres of a mixture with 20% water, 10 litres is taken out. How much water must now be added to make the water 50%?" The 40 litres left still has 20% water, so milk = 32 L. Milk must be 50% of the new total, which is then 64 L, so 24 litres of water is added.

Evaporation in ratio form. Boiling or drying removes only water; the dissolved part stays. "45 kg of sugar syrup has sugar and water in the ratio 2 : 7. How much water must be boiled off to make the ratio 2 : 5?" Sugar = 10 kg and water = 35 kg. Sugar stays 10 kg, so for 2 : 5 the water must be 25 kg. 10 kg of water must evaporate.

Taking out and putting back the same amount. If the 21 litres in the first case had been replaced by 21 litres of water, that is a removal-and-replacement question. Those, done once or many times, are in Removal and Replacement.

Mixing Two or Three Mixtures

The rule: find each part's real quantity in each vessel, add milk to milk and water to water, then write the ratio. Adding ratio terms is the classic wrong answer; it works only when the vessels are the same size and the two ratios already have the same total.

Given amounts. "20 litres of a mixture with milk : water = 7 : 3 is mixed with 48 litres of a mixture with milk : water = 1 : 3. Find the new ratio."

  • First vessel: 14 L milk and 6 L water.
  • Second vessel: 12 L milk and 36 L water.
  • Together: 26 L milk and 42 L water, so 13 : 21.

Adding the ratios (7 + 1 : 3 + 3 = 8 : 6 = 4 : 3) is wrong.

Equal amounts and the LCM trick. Two vessels of the same size hold mixtures in the ratios 3 : 2 and 5 : 4. You cannot add 3 + 5 and 2 + 4, because one part of a 5-part ratio is not the same size as one part of a 9-part ratio. Make the totals equal first. The LCM of 5 and 9 is 45:

  • 3 : 2 = 27 : 18 and 5 : 4 = 25 : 20, both out of 45.
  • Now add: 52 : 38 = 26 : 19.

Mixed in a given ratio. If the same two mixtures are mixed in the ratio 2 : 3 by volume, equalise the totals as above and then multiply each by its share: 2 × (27 : 18) = 54 : 36 and 3 × (25 : 20) = 75 : 60. Together 129 : 96 = 43 : 32. Vessels of unequal size work the same way: sizes 2 : 3 are simply a mixing ratio of 2 : 3.

The fraction way gives the same: milk is 35\frac{3}{5} of the first and 59\frac{5}{9} of the second, so milk = 2 × 35\frac{3}{5} + 3 × 59\frac{5}{9} = 4315\frac{43}{15} out of 5 units, and water is 3215\frac{32}{15}.

Three vessels. Three equal glasses hold milk and water in the ratios 1 : 1, 2 : 1 and 3 : 1. The totals are 2, 3 and 4, with LCM 12, so write them as 6 : 6, 8 : 4 and 9 : 3. All together: 23 : 13.

Part-filled glasses. "Two identical bottles are 25\frac{2}{5} and 13\frac{1}{3} full of juice. Both are filled up with water and emptied into a jug." Juice = 25+13=1115\frac{2}{5} + \frac{1}{3} = \frac{11}{15} of a bottle, and the jug holds 2 bottles, so water = 2−1115=19152 - \frac{11}{15} = \frac{19}{15}. Juice : water = 11 : 19.

Pouring from Two Containers into a Third

Litres taken out of a container carry that container's ratio with them. So split the litres taken from each container in that container's own ratio, then add part by part.

The standard question. "Container A holds 66 litres of milk and water in the ratio 5 : 1. Container B holds 45 litres in the ratio 4 : 5. 30 litres from A and 18 litres from B are poured into an empty container C. Find the ratio of milk to water in C."

  • From A: 30 L in the ratio 5 : 1, so 25 L milk and 5 L water.
  • From B: 18 L in the ratio 4 : 5, so 8 L milk and 10 L water.
  • C holds 33 L milk and 15 L water, so the ratio is 11 : 5.

The sizes of A and B (66 L and 45 L) play no part. Only the ratios and the litres taken matter. SSC adds the container sizes to slow you down.

Then adding to C. "How much water must be added to C to make the ratio 3 : 2?" The milk in C is 33 L and stays, so water must be 33 × 23\frac{2}{3} = 22 L. Add 7 litres of water.

Pouring one container into another. "Vessel P has 24 litres of acid and water in the ratio 5 : 3. Vessel Q has 30 litres in the ratio 1 : 2. 8 litres from P is poured into Q." The 8 litres carries 5 L acid and 3 L water. Q had 10 L acid and 20 L water, so now it has 15 L acid and 23 L water: 15 : 23. P is left with 16 litres, still 5 : 3.

Strengths instead of ratios. "12 litres of a 40% acid solution and 18 litres of a 15% acid solution are poured into an empty jar." Acid = 4.8 + 2.7 = 7.5 L in 30 L, which is 25%. The jar's strength is a weighted average of 40% and 15%, so it must lie between them, closer to the one used more.

Checks before you answer.

  • The litres in C must equal the litres poured in (30 + 18 = 48 here).
  • The milk share in C must lie between the milk shares of A and B (56\frac{5}{6} and 49\frac{4}{9} here; C has 1116\frac{11}{16}).

When the question turns this round ("how many litres from each to make 20 litres of a 2 : 1 mixture?"), use alligation instead.

[SSC Tip] Adding the ratio terms (5 + 4 : 1 + 5 = 9 : 6 = 3 : 2) gives the right answer only when the litres taken are in the same proportion as the ratio totals, 6 : 9. Here they are 30 : 18, so 3 : 2 is a wrong option. Split the litres properly every time; it takes ten seconds.

Alloys: Metals in Ratio and Percentage

An alloy is a mixture of metals. Brass is copper and zinc, bronze is copper and tin, and jewellery gold is gold mixed with copper or silver. The rules are the same as for milk and water, with kilograms in place of litres.

Quantity from the ratio. A 45 kg alloy has copper : zinc = 7 : 2. One part = 5 kg, so copper = 35 kg and zinc = 10 kg.

One metal as a percentage of another. "In an alloy, zinc is 60% of the copper by weight. How much copper is there in 40 kg of the alloy?" Zinc = 35\frac{3}{5} of copper, so copper : zinc = 5 : 3. Copper = 58\frac{5}{8} × 40 = 25 kg. Taking 60% of 40 kg as the zinc is the trap; the 60% is of the copper, not of the alloy.

Mixing two alloys by weight. 20 kg of an alloy with copper : zinc = 3 : 1 is melted with 30 kg of an alloy with copper : zinc = 2 : 3. The first gives 15 kg copper and 5 kg zinc; the second gives 12 kg copper and 18 kg zinc. The new alloy has 27 kg copper and 23 kg zinc: 27 : 23.

Equal weights. One alloy has gold : silver = 5 : 1 and another has 3 : 5. Equal weights are melted together. The totals are 6 and 8, with LCM 24, so write them as 20 : 4 and 9 : 15. Together: 29 : 19.

Alloys with different metals. "10 kg of an alloy with copper : zinc = 2 : 3 is melted with 14 kg of an alloy with copper : tin = 3 : 4. Find copper : zinc : tin in the new alloy." The first gives 4 kg copper and 6 kg zinc. The second gives 6 kg copper and 8 kg tin. Copper adds up; zinc and tin come from one alloy each. New alloy: 10 : 6 : 8 = 5 : 3 : 4.

Adding a pure metal. This is the "only one part changes" idea again. "32 kg of an alloy has copper : tin = 5 : 3. How much tin must be added to make the ratio 1 : 1?" Copper = 20 kg and stays. Tin must also become 20 kg; it is 12 kg now, so add 8 kg of tin.

Gold purity. Pure gold is called 24 carat, so 22 carat gold is 2224\frac{22}{24} gold, which is about 91.7%. A 10 g ornament of 18 carat gold has 1824\frac{18}{24} × 10 = 7.5 g of gold. SSC rarely asks this, but when it does it is a straight ratio question.

In a three-metal answer, the terms must add up to the total weight in some unit. Here 5 + 3 + 4 = 12 units for 24 kg, so 1 unit = 2 kg. An option whose terms do not divide the total weight evenly is usually wrong.

Average Price of a Mix

When varieties at different prices are mixed, the price of the mix is the total cost divided by the total quantity. This is a weighted average, with the quantities as weights.

Price of the mix=q1p1+q2p2+⋯q1+q2+⋯\text{Price of the mix} = \frac{q_1 p_1 + q_2 p_2 + \cdots}{q_1 + q_2 + \cdots}

Given quantities. 25 kg of rice at ₹44 per kg and 15 kg at ₹60 per kg cost 1,100 + 900 = ₹2,000 for 40 kg, so the mix costs ₹50 per kg. The plain average, ₹52, is wrong because there is more of the cheaper rice.

Given a ratio. Sugar at ₹42 and ₹50 per kg is mixed in the ratio 3 : 5. Treat the ratio terms as kilograms: (3 × 42 + 5 × 50) ÷ 8 = (126 + 250) ÷ 8 = 376 ÷ 8 = ₹47 per kg.

The base-price shortcut. Measure each price from the cheapest. Here the gaps are 0 and 8, and the weighted gap is 5 × 8 ÷ 8 = 5, so the price is 42 + 5 = ₹47. With big prices this saves most of the multiplying.

Three or more varieties. Teas at ₹140, ₹160 and ₹220 per kg are mixed in the ratio 2 : 3 : 5. Total cost = 280 + 480 + 1,100 = ₹1,860 for 10 kg, so the mix costs ₹186 per kg. With the base ₹140: the gaps are 0, 20 and 80, the weighted gap is (3 × 20 + 5 × 80) ÷ 10 = 46, and 140 + 46 = 186.

Equal quantities or equal money? Read the line twice.

  • Equal quantity of each kind: the price of the mix is the plain average.
  • Equal money on each kind: you get more of the cheaper kind, so the price leans towards it. For two prices x and y,

Price of the mix=2xyx+y\text{Price of the mix} = \frac{2xy}{x + y}

Oil at ₹60 and ₹90 per litre, bought for equal sums: 2×60×90150\frac{2 \times 60 \times 90}{150} = ₹72 per litre, not ₹75.

  • Three prices with equal money: choose a spend that every price divides. For pulses at ₹30, ₹40 and ₹60 per kg, spend ₹120 on each. That buys 4 + 3 + 2 = 9 kg for ₹360, so the mix costs ₹40 per kg.

Going the other way, finding the mixing ratio for a target price, is alligation; see The Alligation Rule and Mixing for Profit.

The price of a mix must lie between the cheapest and the dearest price, and nearer the variety used more. Any option outside that range, or on the wrong side of the plain average, can be crossed out at once.

Mixes with Free Water: Cost per Litre

Water costs nothing, but it adds volume. So the cost of a milk and water mix is just the cost of the milk, spread over more litres.

Cost per litre of the mix=milk price×milkmilk+water\text{Cost per litre of the mix} = \text{milk price} \times \frac{\text{milk}}{\text{milk} + \text{water}}

Milk and water in a ratio. Milk costs ₹56 per litre and is mixed with water in the ratio 7 : 1. Eight litres of the mix contain 7 litres of milk costing 7 × 56 = ₹392, so one litre of the mix costs 392 ÷ 8 = ₹49. By the formula: 56 × 78\frac{7}{8} = 49.

Water given as a share of the milk. "Water equal to 25% of the milk is added to milk costing ₹60 per litre." Milk : water = 4 : 1, so the mix costs 60 × 45\frac{4}{5} = ₹48 per litre. A common slip is 60 × 0.75 = ₹45, which treats the water as 25% of the mix.

Two kinds of milk and some water. A dairy mixes 12 litres of milk at ₹45, 18 litres at ₹55 and 6 litres of water. The cost is 540 + 990 + 0 = ₹1,530 for 36 litres, so the mix costs ₹42.50 per litre. The water counts in the litres but adds nothing to the cost. Dividing ₹1,530 by 30 (leaving the water out) gives ₹51, which is the cost of the milk alone.

How much water for a target cost. "Milk costs ₹60 per litre. How much water must be mixed with 40 litres of it so that the mix costs ₹50 per litre?" Milk must be 5060=56\frac{50}{60} = \frac{5}{6} of the mix, so milk : water = 5 : 1. With 40 litres of milk, add 8 litres of water. Check: 40 × 60 = ₹2,400 for 48 litres, which is ₹50 a litre.

Several liquids with prices. A drink seller mixes 10 litres of juice at ₹80, 6 litres of syrup at ₹40 and 4 litres of water. The cost is 800 + 240 = ₹1,040 for 20 litres, which is ₹52 per litre. Water is just one more variety, priced at ₹0.

Selling such mixes at a profit (for example, a milkman who sells at the cost price of pure milk and gains because of the water) is covered in The Alligation Rule and Mixing for Profit.

[SSC Tip] When water is in the mix, the cost per litre must be below the cheapest milk price used. If an option sits above it, the water was left out of the litres.

Solved Examples

Question 1

A 135-litre mixture contains milk and water in the ratio 11 : 4. How many litres of water does it contain? (a) 36 litres (b) 99 litres (c) 45 litres (d) 33.75 litres

Answer. There are 11 + 4 = 15 parts, so one part is 135 ÷ 15 = 9 litres. Water is 4 parts = 36 litres (and milk is 99 litres). So water is 415\frac{4}{15} of the mixture, which is 2623%26\frac{2}{3}\%. The answer is (a) 36 litres.

Question 2

A 77-litre mixture has milk and water in the ratio 8 : 3. How much water must be added to make the ratio 4 : 3? (a) 14 litres (b) 28 litres (c) 21 litres (d) 35 litres

Answer. Milk = 56 L and water = 21 L. The milk does not change, so for 4 : 3 the water must be 56 × 34\frac{3}{4} = 42 L. I add 42 − 21 = (c) 21 litres. Unit way: 4 : 3 = 8 : 6, so water goes from 3 units to 6 units; 1 unit is 77 ÷ 11 = 7 L, and 3 units = 21 L.

Question 3

An 80-litre mixture has milk and water in the ratio 3 : 2. How much milk must be added to make the ratio 2 : 1? (a) 24 litres (b) 16 litres (c) 8 litres (d) 32 litres

Answer. Milk = 48 L and water = 32 L. Now the water is fixed, and for 2 : 1 the milk must be twice the water, 64 L. So I add 64 − 48 = (b) 16 litres of milk. Check: 64 : 32 = 2 : 1.

Question 4

A mixture contains acid and water in the ratio 3 : 2. When 10 litres of water is added, the ratio becomes 2 : 3. What was the quantity of the mixture at first? (a) 25 litres (b) 30 litres (c) 12 litres (d) 20 litres

Answer. Acid does not change, so I make its terms equal: 3 : 2 = 6 : 4 and 2 : 3 = 6 : 9. Water goes from 4 units to 9 units, so 5 units = 10 L and 1 unit = 2 L. At first there were 6 + 4 = 10 units = (d) 20 litres (12 L acid and 8 L water). Check: 12 : 18 = 2 : 3.

Question 5

A 64-litre solution contains 15% alcohol. How much water must be added so that alcohol becomes 12% of the solution? (a) 16 litres (b) 12.8 litres (c) 20 litres (d) 8 litres

Answer. Alcohol = 15% of 64 = 9.6 L, and adding water does not change it. If 9.6 L is 12% of the new solution, the new total is 9.6 ÷ 0.12 = 80 L. So I add 80 − 64 = (a) 16 litres. Formula check: 64 × 15−1212\frac{15 - 12}{12} = 16. Dividing by 15 instead of 12 gives the trap 12.8.

Question 6

A 45-litre mixture of milk and water contains 20% water. How much milk must be added so that water becomes 15% of the mixture? (a) 12 litres (b) 2.25 litres (c) 15 litres (d) 9 litres

Answer. Water = 20% of 45 = 9 L, and it stays 9 L when milk is poured in. If 9 L is 15% of the new mixture, the new total is 9 ÷ 0.15 = 60 L. So I add 60 − 45 = (c) 15 litres of milk.

Question 7

A 75-litre mixture of milk and water contains 36% water. If 15 litres of milk is added to it, what is the percentage of water in the new mixture? (a) 21% (b) 30% (c) 32% (d) 45%

Answer. Water = 36% of 75 = 27 L, and adding milk does not change it. The new total is 75 + 15 = 90 L, so water is 27 ÷ 90 = (b) 30%.

Question 8

From 120 litres of a mixture with milk : water = 7 : 5, 24 litres of the mixture is taken out, and then 12 litres of milk is added. What is the new ratio of milk to water? (a) 41 : 25 (b) 29 : 25 (c) 7 : 5 (d) 17 : 10

Answer. Taking out mixture keeps the ratio 7 : 5, so the 96 L left has 56 L milk and 40 L water. Adding 12 L milk gives 68 : 40 = (d) 17 : 10. If you skip the removal, you get 41 : 25.

Question 9

From 90 litres of a mixture with milk : water = 7 : 2, some mixture is taken out. Then 12 litres of water is added, and the ratio becomes 7 : 4. How much mixture was taken out? (a) 36 litres (b) 54 litres (c) 30 litres (d) 24 litres

Answer. After the removal the mixture is still 7 : 2, and after that only water changes. With the milk term fixed at 7, water goes from 2 units to 4 units, so 2 units = 12 L and 1 unit = 6 L. The mixture left after the removal was 9 units = 54 L. So (a) 36 litres was taken out. Check: 54 L has 42 L milk and 12 L water; with 12 L more water it is 42 : 24 = 7 : 4.

Question 10

75 kg of a salt solution contains salt and water in the ratio 2 : 13. How much water must evaporate so that the ratio of salt to water becomes 1 : 4? (a) 15 kg (b) 25 kg (c) 35 kg (d) 20 kg

Answer. Salt = 75 × 215\frac{2}{15} = 10 kg and water = 65 kg. Evaporation takes only water, so the salt stays 10 kg. For 1 : 4 the water must be 40 kg, so (b) 25 kg of water must evaporate.

Question 11

28 litres of a mixture with milk : water = 5 : 2 is mixed with 33 litres of a mixture with milk : water = 7 : 4. What is the ratio of milk to water in the new mixture? (a) 2 : 1 (b) 20 : 41 (c) 41 : 20 (d) 35 : 26

Answer. First mixture: 20 L milk and 8 L water. Second: 21 L milk and 12 L water. Together: 41 L milk and 20 L water, so (c) 41 : 20.

Question 12

Two vessels of equal capacity are full of milk and water mixtures in the ratios 3 : 1 and 4 : 5. Both are emptied into a large container. What is the ratio of milk to water in it? (a) 7 : 6 (b) 29 : 43 (c) 3 : 2 (d) 43 : 29

Answer. The totals 4 and 9 are different, so I make them both 36: 3 : 1 = 27 : 9 and 4 : 5 = 16 : 20. Now I can add: 43 : 29. The answer is (d) 43 : 29. Adding the ratios straight (3 + 4 : 1 + 5 = 7 : 6) is wrong because a part of the first ratio is not the same size as a part of the second.

Question 13

Two mixtures of milk and water, with milk : water = 3 : 1 and 3 : 5, are mixed in the ratio 2 : 3 by volume. What is the ratio of milk to water in the result? (a) 21 : 19 (b) 1 : 1 (c) 15 : 17 (d) 19 : 21

Answer. First I make the totals equal: 3 : 1 = 6 : 2, and 3 : 5 is already out of 8. Then I take 2 shares of the first and 3 shares of the second: 2 × (6 : 2) = 12 : 4 and 3 × (3 : 5) = 9 : 15. Together: 21 : 19. The answer is (a) 21 : 19. 15 : 17 is what you get if you multiply the ratios without making their totals equal first.

Question 14

[Tier 2] Three vessels whose capacities are in the ratio 2 : 3 : 4 are full of milk and water mixtures in the ratios 3 : 1, 1 : 2 and 5 : 3 respectively. All three are emptied into one tank. What is the ratio of milk to water in the tank? (a) 3 : 2 (b) 41 : 31 (c) 5 : 4 (d) 4 : 5

Answer. I take the capacities as 2, 3 and 4 litres. Milk = 2 × 34\frac{3}{4} + 3 × 13\frac{1}{3} + 4 × 58\frac{5}{8} = 1.5 + 1 + 2.5 = 5 litres out of 9. Water = 4 litres. So the tank has (c) 5 : 4.

Question 15

Two identical jugs are 34\frac{3}{4} and 23\frac{2}{3} full of milk. Both are filled up with water, and the contents are poured into a large pot. What is the ratio of milk to water in the pot? (a) 17 : 7 (b) 7 : 17 (c) 5 : 2 (d) 3 : 1

Answer. Milk = 34+23=1712\frac{3}{4} + \frac{2}{3} = \frac{17}{12} of a jug. The pot holds 2 full jugs, so water = 2−1712=7122 - \frac{17}{12} = \frac{7}{12} of a jug. Milk : water = (a) 17 : 7.

Question 16

Can A holds 40 litres of milk and water in the ratio 5 : 3, and can B holds 54 litres in the ratio 4 : 5. 16 litres from A and 27 litres from B are poured into an empty can C. What is the ratio of milk to water in C? (a) 9 : 8 (b) 22 : 21 (c) 21 : 22 (d) 25 : 24

Answer. The 16 litres from A split as 5 : 3, which is 10 L milk and 6 L water. The 27 litres from B split as 4 : 5, which is 12 L milk and 15 L water. C has 22 L milk and 21 L water, so (b) 22 : 21. The sizes 40 L and 54 L do not matter.

Question 17

Alloy P has copper and zinc in the ratio 4 : 1, and alloy Q has them in the ratio 3 : 5. 25 kg of P is melted with 16 kg of Q. What is the ratio of copper to zinc in the new alloy? (a) 7 : 6 (b) 15 : 26 (c) 26 : 15 (d) 4 : 1

Answer. P gives 20 kg copper and 5 kg zinc. Q gives 6 kg copper and 10 kg zinc. The new alloy has 26 kg copper and 15 kg zinc, so (c) 26 : 15.

Question 18

An alloy of copper and zinc has them in the ratio 3 : 2, and an alloy of copper and tin has them in the ratio 4 : 1. 20 kg of the first is melted with 30 kg of the second. What is the ratio of copper, zinc and tin in the new alloy? (a) 7 : 2 : 1 (b) 18 : 4 : 3 (c) 9 : 2 : 3 (d) 18 : 3 : 4

Answer. The first alloy gives 12 kg copper and 8 kg zinc. The second gives 24 kg copper and 6 kg tin. Copper adds up to 36 kg, zinc is 8 kg and tin is 6 kg. So copper : zinc : tin = 36 : 8 : 6 = (b) 18 : 4 : 3. 7 : 2 : 1 is what equal weights would give; here the weights are 20 kg and 30 kg.

Question 19

In an alloy of copper and nickel, the weight of nickel is 35% of the weight of copper. How much copper is there in 54 kg of the alloy? (a) 35.1 kg (b) 18.9 kg (c) 40 kg (d) 14 kg

Answer. Nickel = 0.35 × copper, so the alloy weighs 1.35 times the copper. Copper = 54 ÷ 1.35 = (c) 40 kg, and nickel is 14 kg. Check: 35% of 40 is 14. Taking 35% of 54 as the nickel treats the 35% as a share of the alloy, which it is not.

Question 20

Rice costing ₹38 per kg and rice costing ₹53 per kg are mixed in the ratio 3 : 2. What is the cost of the mixture per kg? (a) ₹45.50 (b) ₹44 (c) ₹47 (d) ₹42

Answer. I treat the ratio as 3 kg and 2 kg. Cost = 3 × 38 + 2 × 53 = 114 + 106 = ₹220 for 5 kg, so the mix costs (b) ₹44 per kg. Shortcut: the dearer rice is ₹15 more and makes up 25\frac{2}{5} of the mix, so 38 + 6 = 44.

Question 21

A shopkeeper spends ₹2,160 on tea at ₹240 per kg and another ₹2,160 on tea at ₹360 per kg, and mixes the two. What is the cost of the mixture per kg? (a) ₹300 (b) ₹280 (c) ₹312 (d) ₹288

Answer. He gets 2160 ÷ 240 = 9 kg of the first tea and 2160 ÷ 360 = 6 kg of the second. That is 15 kg for ₹4,320, so the mix costs (d) ₹288 per kg. With equal money, 2×240×360240+360=288\frac{2 \times 240 \times 360}{240 + 360} = 288 gives it straight away.

Question 22

A dairy mixes 15 litres of milk costing ₹54 per litre and 25 litres costing ₹62 per litre, and then adds 10 litres of water. What is the cost per litre of the final mixture? (a) ₹59 (b) ₹46.40 (c) ₹47.20 (d) ₹58

Answer. Water is free, so the cost is only the milk: 15 × 54 + 25 × 62 = 810 + 1,550 = ₹2,360. That is spread over 15 + 25 + 10 = 50 litres, so one litre costs (c) ₹47.20. Dividing by 40 instead of 50 leaves the water out and gives ₹59.