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Alligation and Mixtures for CUET Applied Maths

About 5 min to learn · 88 questions

Alligation and mixtures are part of Section B2 of CUET Maths, the Applied Mathematics section you answer if you choose Applied Mathematics instead of Section B1. One rule solves almost every question here, so a few minutes of practice turn it into sure marks. Learn each idea here, from the mean price of a mixture to the rule of alligation, mixing solutions and alloys, and repeated replacement, with worked examples, time-saving shortcuts and the traps CUET sets, then practise on free questions in CUET's own formats, each fully solved.

Syllabus area
Section B2: Applied Mathematics
On this page · 9 sections
  1. 01The mean price of a mixture
  2. 02The rule of alligation
  3. 03Mixing solutions and alloys
  4. 04Repeated replacement
  5. 05How CUET asks this chapter
  6. 06Exam shortcuts
  7. 07Common mistakes
  8. 08Quick revision
  9. 09Check your understanding

The mean price of a mixture

A tea stall buys two teas, one at ₹150 a kg and one at ₹240 a kg, and wants a blend it can count as costing ₹180 a kg. How much of each should go in? Before we can answer, we need to know what a mixture costs.

How do you find the mean price of a mixture?

Divide the total cost by the total quantity. The mean price is a weighted average: each price counts as many times as the quantity bought at it, so it always lies between the cheapest and the dearest price, nearer the one there is more of.

Example 1 · A weighted mean

3 kg of tea at ₹150 a kg is blended with 1 kg at ₹240 a kg. What does a kg of the blend cost?

  1. Total cost: 3×150+1×240=450+240=6903 \times 150 + 1 \times 240 = 450 + 240 = 690 rupees, for 4 kg.

  2. Mean price =6904=172.5= \frac{690}{4} = 172.5 rupees per kg.

Answer

₹172.50 per kg, nearer ₹150 because there is three times as much of that tea.

Watch out

The plain average 150+2402=195\frac{150 + 240}{2} = 195 is right only when equal quantities are mixed. And when equal sums of money are spent on each kind, the cheaper kind is bought in larger quantity, so the mean price falls below the plain average.

The rule of alligation

What is the rule of alligation?

When two kinds at prices cc (cheaper) and dd (dearer) are mixed to get a mean price mm, the quantities are in the ratio cheaper : dearer =(d−m):(m−c)= (d - m) : (m - c). Each kind's share is the other kind's distance from the mean.

Example 2 · The tea stall's blend

In what ratio should tea at ₹150 and tea at ₹240 be blended for a mean price of ₹180?

  1. Cheaper : dearer =(240−180):(180−150)=60:30= (240 - 180) : (180 - 150) = 60 : 30.

  2. That is 2:12 : 1. Check: 2×150+1×2403=5403=180\frac{2 \times 150 + 1 \times 240}{3} = \frac{540}{3} = 180.

Answer

2 kg of the ₹150 tea to every 1 kg of the ₹240 tea.

The rule works for any weighted average, not only prices: profit rates on two parts of a stock, the average marks of two groups, or speeds over two parts of a journey. With speeds, remember that an average speed is a mean over time, so the ratio found is a ratio of times.

Key idea

Water is a kind of milk that costs ₹0. Mixing milk at ₹48 a litre with water to cost ₹40 a litre gives water : milk =(48−40):(40−0)=1:5= (48 - 40) : (40 - 0) = 1 : 5.

Mixing solutions and alloys

How do you work out the strength of a mixed solution?

Add up the pure substance in each part and divide by the total volume. Strengths can also be alligated like prices, to find the ratio in which two solutions must be mixed.

Example 3 · Two acid solutions

20 litres of a 30% acid solution are mixed with 30 litres of a 50% solution. What is the strength of the mixture?

  1. Acid: 0.3×20+0.5×30=6+15=210.3 \times 20 + 0.5 \times 30 = 6 + 15 = 21 litres.

  2. Strength =2150=42%= \frac{21}{50} = 42\%.

Answer

42%

When only water, or only one part, is added, the other part stays the same: fix the amount that does not change, and find the new total it must be a share of.

Repeated replacement

What happens when part of a mixture is taken out and replaced again and again?

Each time a fraction xV\frac{x}{V} of a vessel of volume VV is drawn off and replaced, the original liquid is multiplied by 1−xV1 - \frac{x}{V}. After nn times, the original liquid left is V(1−xV)nV\left(1 - \frac{x}{V}\right)^{n}.

Example 4 · Milk replaced twice

A 50-litre can of milk has 5 litres drawn off and replaced with water, twice. How much milk is left?

  1. Each time 550=110\frac{5}{50} = \frac{1}{10} goes, so 910\frac{9}{10} of the milk stays.

  2. Milk left =50×(910)2=50×0.81=40.5= 50 \times \left(\frac{9}{10}\right)^{2} = 50 \times 0.81 = 40.5 litres.

Answer

40.5 litres of milk, and 9.5 litres of water.

Watch out

From the second draw on, what comes out is part water, so the milk lost is less each time. Subtracting 5 litres of milk twice, 40 litres, is wrong.

How CUET asks this chapter

  • Mean price: the price of a blend of two or three kinds, or with a free ingredient such as water.
  • Alligation: the ratio of mixing, a quantity from the ratio, or the same rule on profit rates, averages or speeds.
  • Solutions and alloys: the strength of a mixture, water to add, or the ratio for an alloy.
  • Replacement: what is left, the new ratio, or the amount replaced each time.

Exam shortcuts

  • Mean price = total cost ÷ total quantity.
  • Cheaper : dearer =(d−m):(m−c)= (d - m) : (m - c).
  • Water costs ₹0; a free ingredient alligates like any other.
  • After nn replacements: V(1−xV)nV\left(1 - \frac{x}{V}\right)^{n}.
  • Fix the part that does not change.

Common mistakes

  • Averaging prices as if equal quantities were mixed.
  • Turning the alligation ratio upside down.
  • Alligating with a selling price; the mean must be a cost price.
  • Subtracting the same amount of the original liquid at every replacement.

Quick revision

Mixing

  • Mean price: weighted average
  • Alligation: (d−m):(m−c)(d - m) : (m - c)
  • Works for strengths, rates, averages, speeds

Replacement

  • Original left: V(1−xV)nV\left(1 - \frac{x}{V}\right)^{n}
  • The rest of the vessel is the added liquid

Next in Section B2: boats and streams, pipes and cisterns.

Next · 08Quick revision

Check your understanding

88 questions, grouped by type, easy to hard. Pick an answer and check it. The worked solution opens when you ask for it.

The mean price of a mixtureQ1 of 88Q1

The mean price of a mixture

21 questions

Question 1

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5 kg of rice at ₹40 per kg is mixed with 3 kg of rice at ₹56 per kg. What is the price per kg of the mixture?

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AnswerB · ₹46

  1. Total cost: 5×40+3×56=200+168=3685 \times 40 + 3 \times 56 = 200 + 168 = 368 rupees, for 8 kg.

  2. Mean price =3688=46= \frac{368}{8} = 46 rupees per kg.

Worth noting

₹48 averages the two prices as if equal amounts were mixed; there is more of the cheaper rice.

Question 2

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4 kg of tea at ₹200 per kg is mixed with 6 kg at ₹250 per kg. What is the price per kg of the mixture?

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AnswerD · ₹230

  1. Total cost: 4×200+6×250=2,3004 \times 200 + 6 \times 250 = 2,300 rupees, for 10 kg.

  2. Mean price =2,30010=230= \frac{2,300}{10} = 230 rupees per kg.

Worth noting

₹225 averages the two prices as if equal amounts were mixed.

Question 3

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10 litre of mustard oil at ₹120 per litre is mixed with 15 litre at ₹100 per litre. What is the price per litre of the mixture?

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AnswerB · ₹108

  1. Total cost: 10×120+15×100=2,70010 \times 120 + 15 \times 100 = 2,700 rupees, for 25 litre.

  2. Mean price =2,70025=108= \frac{2,700}{25} = 108 rupees per litre.

Worth noting

₹110 averages the two prices as if equal amounts were mixed.

Question 4

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Two pulses costing ₹40 and ₹65 per kg are mixed in the ratio 3 : 2. What is the price per kg of the mixture?

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AnswerA · ₹50

  1. Take 5 kg: 3 kg cost 120 and 2 kg cost 130 rupees, 250 in all.

  2. 2505=50\frac{250}{5} = 50 rupees per kg.

Worth noting

₹55 puts the weight 3 on the dearer pulse: the ratio follows the order the prices are named.

Question 5

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A tea merchant has Assam tea at ₹360 a kg and Darjeeling tea at ₹480 a kg. He blends 30 kg of Assam with 10 kg of Darjeeling and sells the blend at ₹468 a kg.

What does a kg of the merchant's blend cost him?

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AnswerA · ₹390

  1. Cost: 30×360+10×480=10,800+4,800=15,60030 \times 360 + 10 \times 480 = 10{,}800 + 4{,}800 = 15{,}600 rupees for 40 kg.

  2. 15,60040=390\frac{15{,}600}{40} = 390 rupees a kg.

Worth noting

₹420 is the simple average of 360 and 480; there is three times as much Assam.

Question 6

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A milkman mixes 20 litres of milk costing ₹60 a litre with 5 litres of water, which costs him nothing. What does a litre of the mixture cost him?

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AnswerC · ₹48

  1. The cost is all in the milk: 20×60=1,20020 \times 60 = 1{,}200 rupees, now spread over 25 litres.

  2. 1,20025=48\frac{1{,}200}{25} = 48 rupees a litre.

Worth noting

₹30 averages ₹60 and ₹0 as though the two quantities were equal.

Question 7

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Three teas costing ₹120, ₹150 and ₹210 per kg are blended in the ratio 2 : 3 : 5. What is the price per kg of the blend?

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AnswerD · ₹174

  1. Take 10 kg of the blend: 2, 3 and 5 kg cost 240+450+1,050=1,740240 + 450 + 1{,}050 = 1{,}740 rupees.

  2. 1,74010=174\frac{1{,}740}{10} = 174 rupees per kg.

Worth noting

₹160 is the plain average of the three prices, ignoring the ratio.

Question 8

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A trader spends ₹600 on rice at ₹40 per kg and another ₹600 on rice at ₹60 per kg, and mixes the two. What is the price per kg of the mixture?

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AnswerA · ₹48

  1. Equal money buys unequal amounts: 60040=15\frac{600}{40} = 15 kg and 60060=10\frac{600}{60} = 10 kg.

  2. So ₹1,200 buys 25 kg: 1,20025=48\frac{1{,}200}{25} = 48 rupees per kg.

Worth noting

₹50 would be right for equal weights; equal money buys more of the cheaper rice.

Question 9

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Match each mixture in List I with its price per kg in List II.

List IList II
A. 3 kg at ₹70 with 1 kg at ₹90I. ₹65
B. 1 kg at ₹30 with 1 kg at ₹50II. ₹75
C. 2 kg at ₹50 with 3 kg at ₹75III. ₹25
D. 4 kg at ₹20 with 1 kg at ₹45IV. ₹40
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AnswerB · A-II, B-IV, C-I, D-III

  1. Mean price = total cost ÷ total weight: 210+904=75\frac{210 + 90}{4} = 75 and 30+502=40\frac{30 + 50}{2} = 40.

  2. 100+2255=65\frac{100 + 225}{5} = 65 and 80+455=25\frac{80 + 45}{5} = 25.

Question 10

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6 kg of rice at ₹50 per kg is mixed with 4 kg at ₹75 per kg. At what price per kg must the mixture be sold to make a 25% profit?

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AnswerB · ₹75

  1. Cost of the mixture: 300+30010=60\frac{300 + 300}{10} = 60 rupees per kg.

  2. For 25% profit: 60×1.25=7560 \times 1.25 = 75 rupees per kg.

Worth noting

₹60 is the cost price; the profit has not been added.

Question 11

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Which of the following statements are true?

I. The mean price of a mixture lies between the lowest and highest prices mixed.

II. When equal quantities are mixed, the mean price is the simple average of the prices.

III. When equal amounts of money are spent on each kind, the mean price is the simple average of the prices.

IV. Adding water to milk lowers the cost of a litre of the mixture.

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AnswerD · I, II and IV only

  1. I: a weighted mean never leaves the range of what is averaged. II: equal weights give the simple average. IV: water adds litres but no cost.

  2. III is false: equal money buys more of the cheaper kind, so the mean falls below the simple average.

Question 12

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3 kg of dal at ₹90 per kg is mixed with 2 kg at ₹120 per kg. What is the price per kg of the mixture?

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AnswerA · ₹102

  1. Total cost: 3×90+2×120=5103 \times 90 + 2 \times 120 = 510 rupees, for 5 kg.

  2. Mean price =5105=102= \frac{510}{5} = 102 rupees per kg.

Worth noting

₹105 averages the two prices as if equal amounts were mixed.

Question 13

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A juice seller mixes 30 litres of juice costing ₹80 a litre with 10 litres of water, which costs nothing. What does a litre of the mixture cost him?

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AnswerD · ₹60

  1. All the cost is in the juice: 30×80=2,40030 \times 80 = 2{,}400 rupees, spread over 40 litres.

  2. 2,40040=60\frac{2{,}400}{40} = 60 rupees a litre.

Worth noting

₹40 averages ₹80 and ₹0 as if the quantities were equal.

Question 14

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A spice blend uses 1 kg of a spice at ₹300 per kg, 2 kg at ₹240 per kg and 3 kg at ₹200 per kg. What is the price per kg of the blend?

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AnswerB · ₹230

  1. Cost: 300+480+600=1,380300 + 480 + 600 = 1{,}380 rupees for 6 kg.

  2. 1,3806=230\frac{1{,}380}{6} = 230 rupees per kg.

Worth noting

₹246.67 averages the three prices equally, ignoring that there is most of the cheapest spice.

Question 15

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8 kg of sugar at ₹40 per kg is mixed with 12 kg at ₹45 per kg. What is the price per kg of the mixture?

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AnswerC · ₹43

  1. Total cost: 8×40+12×45=8608 \times 40 + 12 \times 45 = 860 rupees, for 20 kg.

  2. Mean price =86020=43= \frac{860}{20} = 43 rupees per kg.

Worth noting

₹42.50 averages the two prices as if equal amounts were mixed.

Question 16

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A tea merchant has Assam tea at ₹360 a kg and Darjeeling tea at ₹480 a kg. He blends 30 kg of Assam with 10 kg of Darjeeling and sells the blend at ₹468 a kg.

What is the merchant's profit per cent on his blend?

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AnswerA · 20%

  1. Profit per kg =468−390=78= 468 - 390 = 78.

  2. 78390×100=20%\frac{78}{390} \times 100 = 20\%, reckoned on the cost.

Worth noting

16.67% divides the profit by the selling price; profit per cent is on the cost.

Question 17

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A tea merchant has Assam tea at ₹360 a kg and Darjeeling tea at ₹480 a kg. He blends 30 kg of Assam with 10 kg of Darjeeling and sells the blend at ₹468 a kg.

In what ratio should the merchant blend Assam and Darjeeling for a blend costing ₹420 a kg?

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AnswerD · 1 : 1

  1. Assam : Darjeeling =(480−420):(420−360)=60:60= (480 - 420) : (420 - 360) = 60 : 60.

  2. =1:1= 1 : 1: ₹420 is exactly halfway.

Worth noting

3:13 : 1 is the ratio he uses now, which gives ₹390.

Question 18

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A grocer mixes 6 kg of sugar at ₹38 per kg with 4 kg at ₹43 per kg and sells the mixture at ₹46 per kg. What is his profit per cent?

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AnswerC · 15%

  1. Cost price of the mixture: 6×38+4×4310=228+17210=40\frac{6 \times 38 + 4 \times 43}{10} = \frac{228 + 172}{10} = 40 rupees per kg.

  2. Profit is ₹6 on every ₹40, which is 640×100=15%\frac{6}{40} \times 100 = 15\%.

Worth noting

6% writes the profit in rupees as if it were a percentage; it has to be divided by the cost.

Question 19

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A shopkeeper spends ₹900 on rice at ₹30 per kg and another ₹900 on rice at ₹45 per kg, and mixes the two. What does a kg of the mixture cost him?

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AnswerA · ₹36

  1. ₹900 buys 30 kg of the first rice and 20 kg of the second: 50 kg for ₹1,800.

  2. 1,80050=36\frac{1{,}800}{50} = 36 rupees per kg.

Worth noting

₹37.50 averages the prices, which is right only for equal quantities; equal money buys more of the cheaper rice.

Question 20

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A grocer mixes 10 kg of tea at ₹60 per kg with 5 kg at ₹75 per kg and sells the mixture at ₹78 per kg. What is his profit per cent?

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AnswerC · 20%

  1. Cost of the mixture: 600+37515=65\frac{600 + 375}{15} = 65 rupees per kg.

  2. Profit =78−65=13= 78 - 65 = 13 on 65, that is 1365×100=20%\frac{13}{65} \times 100 = 20\%.

Worth noting

1313%13\frac{1}{3}\% works out the profit on the dearer tea's ₹75 alone; the cost is the mixture's mean price.

Question 21

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A tea merchant has Assam tea at ₹360 a kg and Darjeeling tea at ₹480 a kg. He blends 30 kg of Assam with 10 kg of Darjeeling and sells the blend at ₹468 a kg.

How much Darjeeling must the merchant add to his 40 kg blend to raise its cost to ₹420 a kg?

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AnswerC · 20 kg

  1. Alligate the blend at ₹390 with Darjeeling at ₹480: blend : Darjeeling =(480−420):(420−390)=60:30=2:1= (480 - 420) : (420 - 390) = 60 : 30 = 2 : 1.

  2. For 40 kg of blend, 20 kg of Darjeeling. Check: 15,600+9,60060=420\frac{15{,}600 + 9{,}600}{60} = 420.

Worth noting

40 kg comes from the 1:11 : 1 ratio of the two pure teas, but the blend already holds 30 kg of Assam.

The rule of alligation

27 questions

Question 22

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In what ratio must rice at ₹42 per kg be mixed with rice at ₹54 per kg to get a mixture worth ₹46 per kg?

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AnswerD · 2 : 1

  1. Alligation: cheaper : dearer = (dearer − mean) : (mean − cheaper).

  2. (54−46):(46−42)=8:4=2:1(54 - 46) : (46 - 42) = 8 : 4 = 2 : 1.

Worth noting

1 : 2 puts the differences the wrong way round; the mean is nearer the cheaper rice, so there is more of it.

Question 23

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Milk costs ₹50 a litre. In what ratio must water be mixed with milk so that the mixture costs ₹40 a litre?

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AnswerB · 1 : 4

  1. Water costs ₹0. Water : milk =(50−40):(40−0)=10:40= (50 - 40) : (40 - 0) = 10 : 40.

  2. That is 1:41 : 4: one litre of water to every four of milk.

Worth noting

4 : 1 is milk to water, the order reversed.

Question 24

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Milk costs ₹60 a litre. In what ratio must water be mixed with the milk so that the mixture costs ₹45 a litre?

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AnswerA · 1 : 3

  1. Water is the cheaper kind, at ₹0. Water : milk =(60−45):(45−0)=15:45= (60 - 45) : (45 - 0) = 15 : 45.

  2. In lowest terms, 1 : 3.

Worth noting

3 : 1 is milk : water, the ratio turned round.

Question 25

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A dairy buys 60 litres of milk at ₹54 a litre, adds some water, which costs nothing, and sells the mixture at ₹54 a litre.

If the dairy adds 15 litres of water, what does a litre of the mixture cost it?

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AnswerC · ₹43.20

  1. The cost is all in the milk: 60×54=3,24060 \times 54 = 3{,}240 rupees, now spread over 75 litres.

  2. 3,24075=43.20\frac{3{,}240}{75} = 43.20 rupees a litre.

Worth noting

₹39 takes ₹15 off the price, as if each litre of water lowered the cost by ₹1.

Question 26

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In what ratio must rice at ₹30 per kg be mixed with rice at ₹45 per kg to get a mixture worth ₹36 per kg?

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AnswerD · 3 : 2

  1. Cheaper : dearer =(d−m):(m−c)=(45−36):(36−30)=9:6= (d - m) : (m - c) = (45 - 36) : (36 - 30) = 9 : 6.

  2. In lowest terms, 3 : 2.

Worth noting

2 : 3 puts the ratio the wrong way round: the mean is nearer the cheaper price, so more of the cheaper kind is used.

Question 27

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In what ratio must coffee at ₹120 per kg be mixed with coffee at ₹180 per kg to get a mixture worth ₹160 per kg?

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AnswerC · 1 : 2

  1. Cheaper : dearer =(d−m):(m−c)=(180−160):(160−120)=20:40= (d - m) : (m - c) = (180 - 160) : (160 - 120) = 20 : 40.

  2. In lowest terms, 1 : 2.

Worth noting

2 : 1 puts the ratio the wrong way round: each kind's share is the other kind's distance from the mean.

Question 28

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How many kg of tea at ₹180 per kg must be mixed with 15 kg of tea at ₹240 per kg to make a blend worth ₹200 per kg?

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AnswerD · 30 kg

  1. Cheaper : dearer =(240−200):(200−180)=40:20=2:1= (240 - 200) : (200 - 180) = 40 : 20 = 2 : 1.

  2. So the cheaper tea is twice the dearer: 2×15=302 \times 15 = 30 kg.

Worth noting

7.5 kg uses the ratio upside down.

Question 29

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A merchant has 50 kg of sugar. He sells part of it at 8% profit and the rest at 18% profit, and gains 14% on the whole. How much does he sell at 18% profit?

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AnswerC · 30 kg

  1. Alligate the profit rates: 8% part : 18% part =(18−14):(14−8)=4:6=2:3= (18 - 14) : (14 - 8) = 4 : 6 = 2 : 3.

  2. The 18% part is 35×50=30\frac{3}{5} \times 50 = 30 kg.

Worth noting

20 kg is the part sold at 8%.

Question 30

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Read the two statements and choose the correct option.

I. The mean price of a mixture of two kinds always lies between the two prices.

II. The rule of alligation gives the ratio of the two prices.

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AnswerA · Only I

  1. I: a weighted mean of two prices can never fall outside them. True.

  2. II: alligation gives the ratio of the quantities mixed, not of the prices. False.

Question 31

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Assertion (A): To get rice worth ₹50 per kg from kinds at ₹40 and ₹70 per kg, they are mixed in the ratio 2 : 1.

Reason (R): The quantities mixed are in the inverse ratio of the differences of their prices from the mean price.

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AnswerA · Both A and R are true, and R is the correct explanation of A.

  1. The differences are 50−40=1050 - 40 = 10 and 70−50=2070 - 50 = 20; inverted, 20:10=2:120 : 10 = 2 : 1. A is true.

  2. R is the rule of alligation itself, and it is how A is found.

Question 32

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Juice costs ₹90 a litre. In what ratio must water be mixed with the juice so that the mixture costs ₹75 a litre?

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AnswerC · 1 : 5

  1. Water is the cheaper kind, at ₹0. Water : juice =(90−75):(75−0)=15:75= (90 - 75) : (75 - 0) = 15 : 75.

  2. In lowest terms, 1 : 5.

Worth noting

5 : 1 is juice : water, the ratio turned round.

Question 33

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How many kg of sugar at ₹50 per kg must be mixed with 12 kg of sugar at ₹80 per kg to get a mixture worth ₹60 per kg?

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AnswerC · 24 kg

  1. Cheaper : dearer =(80−60):(60−50)=20:10=2:1= (80 - 60) : (60 - 50) = 20 : 10 = 2 : 1.

  2. For 12 kg of the dearer sugar, 2×12=242 \times 12 = 24 kg of the cheaper.

Worth noting

6 kg uses the ratio the wrong way round, 1:21 : 2.

Question 34

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A trader has 60 kg of grain. He sells part of it at 10% profit and the rest at 25% profit, and gains 15% on the whole. How much did he sell at 10% profit?

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AnswerD · 40 kg

  1. Alligation on the profit rates: (25−15):(15−10)=10:5=2:1(25 - 15) : (15 - 10) = 10 : 5 = 2 : 1.

  2. So 23×60=40\frac{2}{3} \times 60 = 40 kg was sold at 10%.

Worth noting

20 kg is the part sold at 25%.

Question 35

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Read the two statements and choose the correct option.

I. For kinds at ₹cc and ₹dd (c<dc < d) and mean ₹mm, cheaper : dearer =(d−m):(m−c)= (d - m) : (m - c).

II. If the mean price is nearer the cheaper price, more of the dearer kind is used.

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AnswerA · Only I

  1. I is the rule of alligation. True.

  2. II is false: a mean near the cheaper price needs more of the cheaper kind to pull it there.

Question 36

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40 litres of milk cost ₹48 a litre. How much water must be added so that the mixture costs ₹40 a litre?

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AnswerB · 8 litres

  1. Water : milk =(48−40):(40−0)=8:40=1:5= (48 - 40) : (40 - 0) = 8 : 40 = 1 : 5.

  2. For 40 litres of milk, 405=8\frac{40}{5} = 8 litres of water. Check: 1,92048=40\frac{1{,}920}{48} = 40.

Worth noting

10 litres treats ₹8 off ₹48 as a fifth of the milk; the ratio is water : milk = 1 : 5.

Question 37

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A dairy buys 60 litres of milk at ₹54 a litre, adds some water, which costs nothing, and sells the mixture at ₹54 a litre.

With 15 litres of water added, what is the dairy's profit per cent?

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AnswerB · 25%

  1. It sells 75 litres at ₹54 for ₹4,050 against a cost of ₹3,240.

  2. Profit =810= 810, and 8103,240=25%\frac{810}{3{,}240} = 25\%.

Worth noting

20% is the water's share of the mixture, 1575\frac{15}{75}; profit is on the cost.

Question 38

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A dairy buys 60 litres of milk at ₹54 a litre, adds some water, which costs nothing, and sells the mixture at ₹54 a litre.

How much water must the dairy add to make a 20% profit?

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AnswerA · 12 litres

  1. For 20% profit the cost per litre must be 541.2=45\frac{54}{1.2} = 45.

  2. 3,24045=72\frac{3{,}240}{45} = 72 litres in all, so 72−60=1272 - 60 = 12 litres of water.

Worth noting

10 litres gives a cost of 3,24070≈46.29\frac{3{,}240}{70} \approx 46.29 and a profit of about 16.7%.

Question 39

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In what ratio must oil at ₹60 per litre be mixed with oil at ₹90 per litre to get a mixture worth ₹70 per litre?

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AnswerB · 2 : 1

  1. Cheaper : dearer =(d−m):(m−c)=(90−70):(70−60)=20:10= (d - m) : (m - c) = (90 - 70) : (70 - 60) = 20 : 10.

  2. In lowest terms, 2 : 1.

Worth noting

1 : 2 puts the ratio the wrong way round: the mean is nearer the cheaper price, so more of the cheaper kind is used.

Question 40

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In what ratio must salt at ₹8 per kg be mixed with salt at ₹13 per kg to get a mixture worth ₹11 per kg?

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AnswerD · 2 : 3

  1. Cheaper : dearer =(d−m):(m−c)=(13−11):(11−8)=2:3= (d - m) : (m - c) = (13 - 11) : (11 - 8) = 2 : 3.

  2. In lowest terms, 2 : 3.

Worth noting

3 : 2 puts the ratio the wrong way round: each kind's share is the other kind's distance from the mean.

Question 41

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The 40 students of a class average 62 marks. The boys average 60 and the girls 65. How many girls are in the class?

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AnswerA · 1616

  1. Alligate the averages: boys : girls =(65−62):(62−60)=3:2= (65 - 62) : (62 - 60) = 3 : 2.

  2. Girls =25×40=16= \frac{2}{5} \times 40 = 16.

Worth noting

24 is the number of boys.

Question 42

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A man covers 60 km in 5 hours, partly walking at 6 km/h and partly cycling at 15 km/h. How far does he walk?

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AnswerB · 10 km

  1. His average speed is 605=12\frac{60}{5} = 12 km/h. Alligate the speeds over time: walking time : cycling time =(15−12):(12−6)=1:2= (15 - 12) : (12 - 6) = 1 : 2.

  2. He walks for 13×5=53\frac{1}{3} \times 5 = \frac{5}{3} hours, covering 6×53=106 \times \frac{5}{3} = 10 km.

Worth noting

20 km splits the distance 1 : 2, but speeds average over time, so the ratio belongs to the times.

Question 43

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Rice at ₹50 per kg is mixed with rice at ₹65 per kg, and the mixture is sold at ₹66 per kg for a profit of 10%. In what ratio were the two mixed?

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AnswerD · 1 : 2

  1. First take out the profit: cost price of the mixture =66×100110=60= \frac{66 \times 100}{110} = 60 rupees per kg.

  2. Then alligate: cheaper : dearer =(65−60):(60−50)=5:10=1:2= (65 - 60) : (60 - 50) = 5 : 10 = 1 : 2.

Worth noting

2 : 1 puts the differences the wrong way round; ₹60 is nearer ₹65, so there is more of the dearer rice.

Question 44

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A shopkeeper sells part of his stock at a 10% loss and the rest at a 20% profit, and makes 5% profit on the whole. In what ratio were the two parts?

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AnswerA · 1 : 1

  1. Treat the loss as −10%-10\%: (20−5):(5−(−10))=15:15(20 - 5) : (5 - (-10)) = 15 : 15.

  2. So the parts are equal, 1:11 : 1.

Worth noting

1:31 : 3 or 3:13 : 1 comes from taking the loss as +10%+10\%; a loss sits below 0.

Question 45

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₹10,000 is lent, part at 8% and the rest at 12% simple interest a year. The total interest for one year is ₹1,040. How much is lent at 8%?

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AnswerB · ₹4,000

  1. The mean rate is 1,04010,000=10.4%\frac{1{,}040}{10{,}000} = 10.4\%. Alligation: (12−10.4):(10.4−8)=1.6:2.4=2:3(12 - 10.4) : (10.4 - 8) = 1.6 : 2.4 = 2 : 3.

  2. At 8%: 25×10,000=4,000\frac{2}{5} \times 10{,}000 = 4{,}000 rupees.

Worth noting

₹6,000 is the part at 12%.

Question 46

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30 workers earn ₹600 a day on average. Skilled workers earn ₹800 and unskilled workers ₹500. How many are skilled?

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AnswerA · 1010

  1. Unskilled : skilled =(800−600):(600−500)=200:100=2:1= (800 - 600) : (600 - 500) = 200 : 100 = 2 : 1.

  2. Skilled workers are 13×30=10\frac{1}{3} \times 30 = 10.

Worth noting

20 is the number of unskilled workers.

Question 47

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A traveller covers 90 km in 3 hours, partly by car at 40 km/h and partly by bicycle at 15 km/h. How far does he go by car?

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AnswerD · 72 km

  1. The mean speed is 903=30\frac{90}{3} = 30 km/h. Alligation on speeds gives the ratio of times: bicycle : car =(40−30):(30−15)=2:3= (40 - 30) : (30 - 15) = 2 : 3.

  2. Time by car =35×3=1.8= \frac{3}{5} \times 3 = 1.8 hours, so the distance by car is 1.8×40=721.8 \times 40 = 72 km.

Worth noting

Alligation on speeds splits the time, not the distance; 45 km splits the distance in half.

Question 48

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A dairy buys 60 litres of milk at ₹54 a litre, adds some water, which costs nothing, and sells the mixture at ₹54 a litre.

In what ratio must the dairy mix water and milk to make a 35% profit?

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AnswerD · 7 : 20

  1. The cost per litre must be 541.35=40\frac{54}{1.35} = 40.

  2. Water : milk =(54−40):(40−0)=14:40=7:20= (54 - 40) : (40 - 0) = 14 : 40 = 7 : 20.

Worth noting

7:137 : 13 puts water against the whole mixture instead of against the milk.

Mixing solutions and alloys

23 questions

Question 49

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30 litres of a 20% acid solution are mixed with 20 litres of a 45% acid solution. What is the strength of the mixture?

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AnswerC · 30%

  1. Acid in each: 0.2×30=60.2 \times 30 = 6 litres and 0.45×20=90.45 \times 20 = 9 litres, 15 litres in 50.

  2. 1550=30%\frac{15}{50} = 30\%.

Worth noting

32.5% averages the two strengths as if the volumes were equal.

Question 50

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5 litres of water are added to 20 litres of pure milk. What percentage of the mixture is milk?

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AnswerC · 80%

  1. The mixture has 25 litres, of which 20 are milk.

  2. 2025×100=80%\frac{20}{25} \times 100 = 80\%.

Worth noting

75% takes the water as 25% of the milk and subtracts it from 100%.

Question 51

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15 litres of a 40% sugar syrup are mixed with 25 litres of a 20% syrup. What is the strength of the mixture?

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AnswerC · 27.5%

  1. Sugar: 0.4×15+0.2×25=6+5=110.4 \times 15 + 0.2 \times 25 = 6 + 5 = 11 litres in 40 litres.

  2. 1140×100=27.5%\frac{11}{40} \times 100 = 27.5\%.

Worth noting

30% averages the strengths as if equal volumes were mixed.

Question 52

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A laboratory has 40 litres of a 25% acid solution in bottle A and a large supply of a 60% acid solution in bottle B.

How much acid is in bottle A?

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AnswerA · 10 litres

  1. 25% of 40 litres is acid.

  2. 0.25×40=100.25 \times 40 = 10 litres.

Worth noting

30 litres is the water in the bottle.

Question 53

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A laboratory has 40 litres of a 25% acid solution in bottle A and a large supply of a 60% acid solution in bottle B.

How much water must be added to bottle A to make it a 20% solution?

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AnswerD · 10 litres

  1. The 10 litres of acid must be 20% of the new total: 100.2=50\frac{10}{0.2} = 50 litres.

  2. So add 50−40=1050 - 40 = 10 litres of water.

Worth noting

50 litres is the new total, not the water added.

Question 54

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How much water must be added to 12 litres of a 60% alcohol solution to make it 40% alcohol?

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AnswerB · 6 litres

  1. The alcohol stays 0.6×12=7.20.6 \times 12 = 7.2 litres; it must be 40% of the new volume VV: 0.4V=7.20.4V = 7.2, so V=18V = 18.

  2. Water to add =18−12=6= 18 - 12 = 6 litres.

Worth noting

18 litres is the new volume, not the water added.

Question 55

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In what ratio must a 10% salt solution be mixed with a 25% salt solution to get a 16% solution?

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AnswerD · 3 : 2

  1. Alligation works on strengths just as on prices: 10% : 25% =(25−16):(16−10)=9:6= (25 - 16) : (16 - 10) = 9 : 6.

  2. 9:6=3:29 : 6 = 3 : 2.

Worth noting

2 : 3 reverses the ratio; 16% is nearer 10%, so more of the weak solution is needed.

Question 56

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A 40-litre mixture of milk and water is 25% water. How much milk must be added so that water is 20% of the mixture?

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AnswerC · 10 litres

  1. Water: 0.25×40=100.25 \times 40 = 10 litres, and adding milk does not change it.

  2. 10 litres must be 20% of the new total: total =50= 50 litres, so add 50−40=1050 - 40 = 10 litres of milk.

Worth noting

2 litres treats 25% − 20% = 5% of 40 litres as the milk needed.

Question 57

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50 litres of a mixture have milk and water in the ratio 7 : 3. How much water must be added to make the ratio 7 : 5?

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AnswerB · 10 litres

  1. Milk is 710×50=35\frac{7}{10} \times 50 = 35 litres and water 15 litres. The milk stays 35.

  2. For 7 : 5, water must be 57×35=25\frac{5}{7} \times 35 = 25 litres, so add 25−15=1025 - 15 = 10 litres.

Worth noting

20 litres turns 3 parts into 5 as if each part were 10 litres, but one part is 357=5\frac{35}{7} = 5 litres.

Question 58

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50 litres of brine are 10% salt. How much water must evaporate to make it 25% salt?

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AnswerD · 30 litres

  1. Salt stays at 0.1×50=50.1 \times 50 = 5 litres; it must be 25% of what remains.

  2. Remaining =50.25=20= \frac{5}{0.25} = 20 litres, so 50−20=3050 - 20 = 30 litres evaporate.

Worth noting

20 litres is what is left after evaporation.

Question 59

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2 kg of an alloy that is 30% silver is melted with 3 kg of an alloy that is 60% silver. What percentage of the new alloy is silver?

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AnswerC · 48%

  1. Silver: 0.3×2+0.6×3=0.6+1.8=2.40.3 \times 2 + 0.6 \times 3 = 0.6 + 1.8 = 2.4 kg in 5 kg.

  2. 2.45×100=48%\frac{2.4}{5} \times 100 = 48\%.

Worth noting

45% is the simple average of 30% and 60%; there is more of the richer alloy.

Question 60

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How much water must be added to 30 litres of an 80% alcohol solution to make it 60% alcohol?

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AnswerB · 10 litres

  1. Alcohol stays at 0.8×30=240.8 \times 30 = 24 litres; it must be 60% of the new total.

  2. New total =240.6=40= \frac{24}{0.6} = 40 litres, so add 40−30=1040 - 30 = 10 litres.

Worth noting

40 litres is the new total, not the water added.

Question 61

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In what ratio must a 15% sugar solution be mixed with a 35% sugar solution to get a 20% solution?

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AnswerD · 3 : 1

  1. Alligation on strengths: weaker : stronger =(35−20):(20−15)=15:5= (35 - 20) : (20 - 15) = 15 : 5.

  2. =3:1= 3 : 1.

Worth noting

1:31 : 3 is turned round: 20% is close to 15%, so most of the mixture is the weaker solution.

Question 62

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60 litres of a mixture have milk and water in the ratio 3 : 1. How much water must be added to make the ratio 3 : 2?

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AnswerC · 15 litres

  1. Milk =34×60=45= \frac{3}{4} \times 60 = 45 litres and water 15 litres.

  2. For 3 : 2, water must be 23×45=30\frac{2}{3} \times 45 = 30 litres: add 30−15=1530 - 15 = 15 litres.

Worth noting

30 litres is the water needed in all, not the amount to add.

Question 63

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40 litres of a mixture have milk and water in the ratio 5 : 3. How much milk must be added to make the ratio 2 : 1?

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AnswerA · 5 litres

  1. Milk 25 litres, water 15 litres.

  2. Water stays at 15, so for 2 : 1 milk must be 30 litres: add 5 litres.

Worth noting

30 litres is the milk needed in all; 25 are already there.

Question 64

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A laboratory has 40 litres of a 25% acid solution in bottle A and a large supply of a 60% acid solution in bottle B.

All of bottle A is mixed with 20 litres from bottle B. What is the strength of the mixture?

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AnswerC · 1103\frac{110}{3}%

  1. Acid: 10+0.6×20=2210 + 0.6 \times 20 = 22 litres in 60 litres.

  2. 2260×100≈36.67%\frac{22}{60} \times 100 \approx 36.67\%.

Worth noting

42.5% averages 25% and 60% as if equal volumes were mixed.

Question 65

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A laboratory has 40 litres of a 25% acid solution in bottle A and a large supply of a 60% acid solution in bottle B.

How much of bottle B must be added to all of bottle A to make a 40% solution?

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AnswerB · 30 litres

  1. A : B =(60−40):(40−25)=20:15=4:3= (60 - 40) : (40 - 25) = 20 : 15 = 4 : 3.

  2. For 40 litres of A, 40×34=3040 \times \frac{3}{4} = 30 litres of B. Check: 10+1870=40%\frac{10 + 18}{70} = 40\%.

Worth noting

53.3 litres uses the ratio the wrong way round, 3:43 : 4.

Question 66

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Alloy A has gold and copper in the ratio 3 : 2; alloy B has them in the ratio 1 : 4. In what ratio must A and B be melted together to get an alloy with equal amounts of gold and copper?

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AnswerA · 3 : 1

  1. The share of gold is 35\frac{3}{5} in A, 15\frac{1}{5} in B and 12\frac{1}{2} in the mixture.

  2. A : B =(12−15):(35−12)=310:110=3:1= \left(\frac{1}{2} - \frac{1}{5}\right) : \left(\frac{3}{5} - \frac{1}{2}\right) = \frac{3}{10} : \frac{1}{10} = 3 : 1.

Worth noting

1 : 3 swaps the two differences.

Question 67

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One jar has milk and water in the ratio 3 : 1, another in the ratio 5 : 3. Equal amounts from the two jars are poured into a bowl. What is the ratio of milk to water in the bowl?

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AnswerD · 11 : 5

  1. Milk is 34\frac{3}{4} of the first jar and 58\frac{5}{8} of the second. With equal amounts, the bowl's share of milk is their average: 12(68+58)=1116\frac{1}{2}\left(\frac{6}{8} + \frac{5}{8}\right) = \frac{11}{16}.

  2. Water is 516\frac{5}{16}, so milk : water =11:5= 11 : 5.

Worth noting

8 : 4 adds the two ratios term by term, which treats a quarter jar and an eighth of a jar as equal parts.

Question 68

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How much pure acid must be added to 20 litres of a 25% acid solution to make it 40% acid?

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AnswerB · 5 litres

  1. Acid now: 0.25×20=50.25 \times 20 = 5 litres. Adding aa litres of pure acid: 5+a=0.4(20+a)5 + a = 0.4(20 + a).

  2. 5+a=8+0.4a5 + a = 8 + 0.4a, so 0.6a=30.6a = 3 and a=5a = 5 litres.

Worth noting

3 litres raises the acid to 8 litres but forgets that the added acid also increases the total.

Question 69

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Alloy X has copper and zinc in the ratio 3 : 1, and alloy Y in the ratio 1 : 1. They are melted together in the ratio 2 : 3 by weight. What is the ratio of copper to zinc in the new alloy?

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AnswerD · 3 : 2

  1. Take 2 kg of X and 3 kg of Y. Copper: 2×34+3×12=32 \times \frac{3}{4} + 3 \times \frac{1}{2} = 3 kg; zinc: 2×14+3×12=22 \times \frac{1}{4} + 3 \times \frac{1}{2} = 2 kg.

  2. Copper : zinc =3:2= 3 : 2.

Worth noting

2:12 : 1 adds the ratio terms, (3+1):(1+1)(3 + 1) : (1 + 1), ignoring how much of each alloy is used.

Question 70

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A milkman buys milk at ₹40 a litre, adds water, and sells the mixture at ₹40 a litre, making a 25% profit. In what ratio are water and milk in the mixture?

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AnswerA · 1 : 4

  1. A 25% profit means the cost of a litre of the mixture is 401.25=32\frac{40}{1.25} = 32 rupees.

  2. Water : milk =(40−32):(32−0)=8:32=1:4= (40 - 32) : (32 - 0) = 8 : 32 = 1 : 4.

Worth noting

1:51 : 5 treats 25% profit as water being a fifth of the mixture; profit is reckoned on the cost, so water is a quarter of the milk.

Question 71

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Jar A has milk and water in the ratio 2 : 1, jar B in the ratio 3 : 2. Equal amounts from the two jars are mixed. What is the ratio of milk to water in the mixture?

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AnswerB · 19 : 11

  1. Take 1 litre from each. Milk: 23+35=1915\frac{2}{3} + \frac{3}{5} = \frac{19}{15}; water: 13+25=1115\frac{1}{3} + \frac{2}{5} = \frac{11}{15}.

  2. So milk : water =19:11= 19 : 11.

Worth noting

5:35 : 3 adds the ratio terms, (2+3):(1+2)(2 + 3) : (1 + 2), which is wrong because the jars' parts are of different sizes.

Repeated replacement

17 questions

Question 72

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Read the two statements and choose the correct option.

I. Each replacement takes away the same fraction of the original liquid that is still in the vessel.

II. After two replacements of 10% each, 20% of the original liquid is gone.

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AnswerA · Only I

  1. I: each draw takes the same share of whatever the vessel holds, so the original liquid is multiplied by the same factor each time. True.

  2. II: (0.9)2=0.81(0.9)^{2} = 0.81, so 19% is gone, not 20%: the second 10% is taken from less. False.

Question 73

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A 50-litre drum is full of pure mustard oil. Every morning 5 litres are drawn off and replaced with a cheaper oil.

How much pure mustard oil is in the drum after two mornings?

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AnswerC · 40.5 litres

  1. Each morning keeps 4550=0.9\frac{45}{50} = 0.9 of the pure oil.

  2. 50×0.92=40.550 \times 0.9^{2} = 40.5 litres.

Worth noting

40 litres takes out 5 litres of pure oil both times; the second draw is partly the cheaper oil.

Question 74

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A can holds 40 litres of milk. 4 litres are taken out and replaced with water, and this is done three times in all. How much milk is left?

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AnswerC · 29.16 litres

  1. Each time, 440=110\frac{4}{40} = \frac{1}{10} of what is in the can goes, so 910\frac{9}{10} of the milk stays.

  2. After three times: 40×(910)3=40×0.729=29.1640 \times \left(\frac{9}{10}\right)^{3} = 40 \times 0.729 = 29.16 litres.

Worth noting

28 litres takes out 4 litres of pure milk every time, but from the second time the 4 litres are part water.

Question 75

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A vessel holds 64 litres of pure oil. A certain amount is taken out and replaced with a cheaper oil, and this is done twice. 49 litres of the pure oil are left. How much was taken out each time?

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AnswerC · 8 litres

  1. If xx litres go each time, 64(1−x64)2=4964\left(1 - \frac{x}{64}\right)^{2} = 49, so (1−x64)2=4964\left(1 - \frac{x}{64}\right)^{2} = \frac{49}{64}.

  2. 1−x64=781 - \frac{x}{64} = \frac{7}{8}, so x=8x = 8 litres.

Worth noting

7.5 litres halves the 15 litres lost, as if pure oil left both times.

Question 76

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A jug holds 20 litres of pure juice. 5 litres are poured out and replaced with water, and then 5 litres of the mixture are poured out and replaced with water. What percentage of the jug is now juice?

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AnswerB · 56.25%

  1. Each time 520=14\frac{5}{20} = \frac{1}{4} of the contents go, so 34\frac{3}{4} of the juice stays.

  2. After two times, (34)2=916=56.25%\left(\frac{3}{4}\right)^{2} = \frac{9}{16} = 56.25\% is juice.

Worth noting

50% takes away a quarter of the full jug twice, 10 litres of juice in all.

Question 77

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A can holds 20 litres of oil. Each time, 2 litres are drawn off and replaced with a cheaper oil. After how many such replacements are 16.2 litres of the original oil left?

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AnswerB · 22

  1. 20(910)n=16.220\left(\frac{9}{10}\right)^{n} = 16.2, so (910)n=0.81\left(\frac{9}{10}\right)^{n} = 0.81.

  2. 0.81=0.920.81 = 0.9^2, so n=2n = 2.

Worth noting

Removing 2 litres twice would leave 16; the second draw takes out only 1.8 litres of the original oil.

Question 78

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A 40-litre mixture is 80% milk. 10 litres of it are taken out and replaced with water. What percentage of the mixture is milk now?

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AnswerB · 60%

  1. Milk now: 32 litres. The 10 litres taken out are a quarter of the mixture, so they take a quarter of the milk, 8 litres.

  2. 32−840=2440=60%\frac{32 - 8}{40} = \frac{24}{40} = 60\%.

Worth noting

55% takes all 10 litres out of the milk, as if the mixture were pure milk.

Question 79

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A barrel holds 60 litres of wine. 6 litres are taken out and replaced with water, and this is done twice in all. How much wine is left?

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AnswerC · 48.6 litres

  1. Each replacement keeps 1−660=9101 - \frac{6}{60} = \frac{9}{10} of the wine.

  2. After 2 times: 60(910)2=243560\left(\frac{9}{10}\right)^{2} = \frac{243}{5} litres.

Worth noting

48 litres takes out 6 litres of pure wine every time; after the first, each draw is part water.

Question 80

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A jug holds 40 litres of pure juice. 8 litres are poured out and replaced with water, twice. What percentage of the jug is now juice?

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AnswerD · 64%

  1. Each replacement keeps 3240=45\frac{32}{40} = \frac{4}{5} of the juice.

  2. After two: (45)2=1625=64%\left(\frac{4}{5}\right)^2 = \frac{16}{25} = 64\%.

Worth noting

60% subtracts 20% twice as if each draw took pure juice.

Question 81

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A tank holds 100 litres of milk. 10 litres are taken out and replaced with water, and this is done three times in all. How much milk is left?

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AnswerB · 72.9 litres

  1. Each replacement keeps 1−10100=9101 - \frac{10}{100} = \frac{9}{10} of the milk.

  2. After 3 times: 100(910)3=72910100\left(\frac{9}{10}\right)^{3} = \frac{729}{10} litres.

Worth noting

70 litres takes out 10 litres of pure milk every time; after the first, each draw is part water.

Question 82

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A 50-litre drum is full of pure mustard oil. Every morning 5 litres are drawn off and replaced with a cheaper oil.

After two mornings, what is the ratio of pure mustard oil to the cheaper oil in the drum?

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AnswerD · 81 : 19

  1. Pure oil is 40.5 litres, so the cheaper oil is 50−40.5=9.550 - 40.5 = 9.5 litres.

  2. 40.5:9.5=81:1940.5 : 9.5 = 81 : 19.

Worth noting

4:14 : 1 is 40 : 10, from taking pure oil out both times.

Question 83

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A 50-litre drum is full of pure mustard oil. Every morning 5 litres are drawn off and replaced with a cheaper oil.

How much pure mustard oil is in the drum after three mornings?

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AnswerB · 36.45 litres

  1. 50×0.93=50×0.72950 \times 0.9^{3} = 50 \times 0.729.

  2. =36.45= 36.45 litres.

Worth noting

35.5 litres subtracts a full 5 litres of pure oil on the third morning.

Question 84

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A cask holds 80 litres of juice. 8 litres are drawn off and replaced with water, and then 8 litres of the mixture are drawn off and replaced with water again. What is now the ratio of juice to water?

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AnswerA · 81 : 19

  1. Juice left =80×(7280)2=80×0.81=64.8= 80 \times \left(\frac{72}{80}\right)^{2} = 80 \times 0.81 = 64.8 litres; water =80−64.8=15.2= 80 - 64.8 = 15.2 litres.

  2. 64.8:15.2=81:1964.8 : 15.2 = 81 : 19.

Worth noting

4 : 1 takes out 8 litres of juice both times, 64 litres to 16.

Question 85

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From a can full of milk, 4 litres are taken out and replaced with water, twice. Milk and water are then in the ratio 16 : 9. How much does the can hold?

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AnswerD · 20 litres

  1. Milk is now 1625\frac{16}{25} of the can, so (1−4V)2=1625\left(1 - \frac{4}{V}\right)^2 = \frac{16}{25}.

  2. 1−4V=451 - \frac{4}{V} = \frac{4}{5}, so 4V=15\frac{4}{V} = \frac{1}{5} and V=20V = 20 litres.

Worth noting

25 litres reads the ratio's total, 16+916 + 9, as the capacity.

Question 86

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A vessel holds 81 litres of milk. A certain amount is taken out and replaced with water, and this is done twice. 64 litres of milk are left. How much was taken out each time?

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AnswerC · 9 litres

  1. 81(1−x81)2=6481\left(1 - \frac{x}{81}\right)^2 = 64, so 1−x81=891 - \frac{x}{81} = \frac{8}{9}.

  2. x81=19\frac{x}{81} = \frac{1}{9}, so x=9x = 9 litres.

Worth noting

8.5 litres splits the 17 litres lost evenly between the two draws; the second draw takes less milk.

Question 87

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A container holds 50 litres of milk. 10 litres are taken out and replaced with water, and then this is done once more. What is the ratio of milk to water now?

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AnswerA · 16 : 9

  1. Milk left: 50(45)2=3250\left(\frac{4}{5}\right)^2 = 32 litres, so water is 18 litres.

  2. 32:18=16:932 : 18 = 16 : 9.

Worth noting

3:23 : 2 takes 20 litres of pure milk out, leaving 30 : 20.

Question 88

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A 50-litre drum is full of pure mustard oil. Every morning 5 litres are drawn off and replaced with a cheaper oil.

After how many mornings does the pure mustard oil first fall below 40 litres?

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AnswerB · 33

  1. After 2 mornings: 40.5 litres, still not below 40.

  2. After 3 mornings: 36.45 litres, below 40.

Worth noting

2 mornings would be the answer if pure oil were removed each time; the cheaper oil drawn out slows the fall.

NTA's Mathematics / Applied Mathematics syllabus has three sections: Section A1, which every candidate answers, then Section B1 (Mathematics) or Section B2 (Applied Mathematics), of which a candidate answers one. Each section lists units and topics but no chapters, so the chapters here are ours. Where both B sections list a topic that Section A also lists, its questions sit once, in Section A.