# Important CAT Mixture and Alligation Questions (with Notes) PDF

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Mixtures and Alligation is an important topic in the CAT Quants section. If we look at the previous year’s papers, CAT Mixture and Alligation made a recurrent appearance in the CAT Quants. It is an important topic and hence must not be avoided by the aspirants. In this article, we will look into some important CAT Mixture and Alligation Questions (with Notes) PDF. These are a good source for practice; If you want to practice these questions, you can download these CAT Mixture and Alligation Questions for CAT PDF, which is completely Free.

• CAT Mixture and Alligation Questions – Tip 1: The topic of Mixtures and Alligation is basically an application of the Averages concept in CAT. The theory in this topic is less and hence students should Not avoid this topic.
• CAT Mixture and Alligation Questions – Tip 2: Based on our analysis of the previous year’s CAT questions,Â  this was the CAT Mixture and Alligation weightage: 1-2 questions were asked on this topic (in CAT 2021).
• If you’re starting the prep, firstly understand the CAT Arithmetic Syllabus; For more practice take these Free mock tests for CAT. Practice these CAT mixture and alligation questions PDF. Learn all the major formulae from these concepts. You can check out the Important CAT mixture and alligation questions & Formulas PDF here.

Question 1:Â A milkman mixes 20 litres of water with 80 litres of milk. After selling one-fourth of this mixture, he adds water to replenish the quantity that he had sold. What is the current proportion of water to milk?
[CAT 2004]

a)Â 2 : 3

b)Â 1 : 2

c)Â 1 : 3

d)Â 3 : 4

Solution:

After selling 1/4th of the mixture, the remaining quantity of water is 15 liters and milk is 60 liters. So the milkman would add 25 liters of water to the mixture. The total amount of water now is 40 liters and milk is 60 liters. Therefore, the required ratio is 2:3.

Question 2:Â Two liquids A and B are in the ratio 5 : 1 in container 1 and 1 : 3 in container 2. In what ratio should the contents of the two containers be mixed so as to obtain a mixture of A and B in the ratio 1 : 1?

a)Â 2 : 3

b)Â 4 : 3

c)Â 3 : 2

d)Â 3 : 4

Solution:

Fraction of A in contained 1 = $\frac{5}{6}$

Fraction of A in contained 2Â = $\frac{1}{4}$

Let the ratio of liquid required from containers 1 and 2 be x:1-x

x($\frac{5}{6}$) + (1-x)($\frac{1}{4}$) =Â $\frac{1}{2}$

$\frac{7x}{12}$ = $\frac{1}{4}$

=> x = $\frac{3}{7}$

=> Ratio = 3:4

Question 3:Â There are two containers: the first contains 500 ml of alcohol, while the second contains 500 ml of water. Three cups of alcohol from the first container is taken out and is mixed well in the second container. Then three cups of this mixture is taken out and is mixed in the first container. Let A denote the proportion of water in the first container and B denote the proportion of alcohol in the second container. Then,

a)Â A > B

b)Â A < B

c)Â A = B

d)Â Cannot be determined

Solution:

Let the volume of the cup be V.
Hence, after removing three cups of alcohol from the first container,

Volume of alcohol in the first container is 500-3V
Volume of water in the second container is 500 and volume of alcohol in the second container is 3V.
So, in each cup, the amount of water contained is $\frac{500}{500+3V}*V$

Hence, after adding back 3 cups of the mixture, amount of water in the first container is $0+\frac{1500V}{500+3V}$
Amount of alcohol contained in the second container is $3V – \frac{9V^2}{500+3V} = \frac{1500V}{500+3V}$

So, the required proportion of water in the first container and alcohol in the second container are equal.

Instructions

DIRECTIONS for the following two questions: The following table presents the sweetness of different items relative to sucrose, whose sweetness is taken to be 1.00.

Question 4:Â What is the maximum amount of sucrose (to the nearest gram) that can be added to one-gram of saccharin such that the final mixture obtained is atleast 100 times as sweet as glucose?

a)Â 7

b)Â 8

c)Â 9

d)Â 100

Solution:

For the mixture to be 100 times as sweet as glucose, its sweetness relative to the mixture should be at least 74.

1 gm of saccharin = 675

Let the number of grams of sucrose to be added be N. Thus, the total weight of the mixture = N + 1.

So, (675 + N) / (N+1) = 74

=> 675 + N = 74N + 74

=> 601 = 73N => N = 8.23

When N=9, sweetness will be S = (675+9)/10 = 684/10 = 68.4

When N=8, sweetness will be S = (675+8)/9 = 683/9 = 75.8

So, option b) is the correct answer.

Question 5:Â Approximately how many times sweeter than sucrose is a mixture consisting of glucose, sucrose and fructose in the ratio of 1: 2: 3?

a)Â 1.3

b)Â 1

c)Â 0.6

d)Â 2.3

Solution:

The relative sweetness of the mixture is (1*0.74 + 2*1 + 3*1.7) / (1+2+3) = 7.84/6 = 1.30

Option a) is the correct answer.

Question 6:Â Bottle 1 contains a mixture of milk and water in 7: 2 ratio and Bottle 2 contains a mixture of milk and water in 9: 4 ratio. In what ratio of volumes should the liquids in Bottle 1 and Bottle 2 be combined to obtain a mixture of milk and water in 3:1 ratio?

a)Â 27:14

b)Â 27:13

c)Â 27:16

d)Â 27:18

Solution:

The ratio of milk and water in Bottle 1 is 7:2 and the ratio of milk and water in Bottle 2 is 9:4
Therefore, the proportion of milk in Bottle 1 is $\frac{7}{9}$ and the proportion of milk in Bottle 2 is $\frac{9}{13}$

Let the ratio in which they should be mixed be equal to X:1.

Hence, the total volume of milk is $\frac{7X}{9}+\frac{9}{13}$
The total volume of water is $\frac{2X}{9}+\frac{4}{13}$
They are in the ratio 3:1

Hence,Â $\frac{7X}{9}+\frac{9}{13} = 3*(\frac{2X}{9}+\frac{4}{13})$
Therefore, $91X+81=78X+108$

Therefore $X = \frac{27}{13}$

Question 7:Â Consider three mixtures â€” the first having water and liquid A in the ratio 1:2, the second having water and liquid B in the ratio 1:3, and the third having water and liquid C in the ratio 1:4. These three mixtures of A, B, and C, respectively, are further mixed in the proportion 4: 3: 2. Then the resulting mixture has

a)Â The same amount of water and liquid B

b)Â The same amount of liquids B and C

c)Â More water than liquid B

d)Â More water than liquid A

Solution:

The proportion of water in the first mixture is $\frac{1}{3}$
The proportion of Liquid A in the first mixture is $\frac{2}{3}$

The proportion of water in the second mixture is $\frac{1}{4}$
The proportion of Liquid B in the second mixture is $\frac{3}{4}$

The proportion of water in the third mixture is $\frac{1}{5}$
The proportion of Liquid C in the third mixture is $\frac{4}{5}$

As they are mixed in the ratio 4:3:2, the final amount of water is $4 \times \frac{1}{3} + 3 \times \frac{1}{4} + 2 \times \frac{1}{5} = \frac{149}{60}$
The final amount of Liquid A in the mixture is $4\times\frac{2}{3} = \frac{8}{3}$
The final amount of Liquid B in the mixture is $3\times\frac{3}{4} = \frac{9}{4}$
The final amount of Liquid C in the mixture is $2\times\frac{4}{5} = \frac{8}{5}$

Hence, the ratio of Water : A : B : C in the final mixture is $\frac{149}{60}:\frac{8}{3}:\frac{9}{4}:\frac{8}{5} = 149:160:135:96$

From the given choices, only option C Â is correct.

Question 8:Â Two types of tea, A and B, are mixed and then sold at Rs. 40 per kg. The profit is 10% if A and B are mixed in the ratio 3 : 2, and 5% if this ratio is 2 : 3. The cost prices, per kg, of A and B are in the ratio

a)Â 17 : 25

b)Â 18 : 25

c)Â 19 : 24

d)Â 21 : 25

Solution:

The selling price of the mixture is Rs.40/kg.
Let a be the price of 1 kg of tea A in the mixture and b be the price per kg of tea B.
It has been given that the profit is 10% if the 2 varieties are mixed in the ratio 3:2
Let the cost price of the mixture be x.
It has been given that 1.1x = 40
x = 40/1.1
Price per kg of the mixture in ratio 3:2 =Â $\frac{3a+2b}{5}Â$
$\frac{3a+2b}{5} = \frac{40}{1.1}$
$3.3a+2.2b=200$ ——–(1)

The profit is 5% if the 2 varieties are mixed in the ratio 2:3.
Price per kg of the mixture in ratio 2:3 =Â $\frac{2a+3b}{5}$
$\frac{2a+3b}{5} = \frac{40}{1.05}$
$2.1a+3.15b=200$Â ——(2)

Equating (1) and (2), we get,
$3.3a+2.2b = 2.1a+3.15b$
$1.2a=0.95b$
$\frac{a}{b} = \frac{0.95}{1.2}$
$\frac{a}{b} = \frac{19}{24}$

Therefore, option C is the right answer.

Question 9:Â A wholesaler bought walnuts and peanuts, the price of walnut per kg being thrice that of peanut per kg. He then sold 8 kg of peanuts at a profit of 10% and 16 kg of walnuts at a profit of 20% to a shopkeeper. However, the shopkeeper lost 5 kg of walnuts and 3 kg of peanuts in transit. He then mixed the remaining nuts and sold the mixture at Rs. 166 per kg, thus making an overall profit of 25%. At what price, in Rs. per kg, did the wholesaler buy the walnuts?

a)Â 96

b)Â 98

c)Â 86

d)Â 84

Solution:

Let the price of peanuts be Rs. 100x per kg
Then, the price of walnuts = Rs. 300x per kg
Cost price of peanuts for the shopkeeper = Rs. 110x per kg
Cost price of walnuts for the shopkeeper = Rs. 360x per kg
Total cost incurred to the shopkeeper while buying = Rs.(8 * 110x + 16 * 360x) = Rs. 6640x
Since, 5kg walnut and 3kg peanuts are lost in transit, the shopkeeper will be remained with (16-5)+(8-3)=16kgs of nuts
Total selling price that the shopkeeper got = Rs. (166 * 16) = Rs. 2656
Profit = 25%
So, cost price = Rs. 2124.80
Therefore, 6640x = 2124.80
On solving, we get x = 0.32
Therefore, price of walnuts = Rs. (300 * 0.32) = Rs. 96 per kg.
Hence, option A is the correct answer

Question 10:Â A trader sells 10 litres of a mixture of paints A and B, where the amount of B in the mixture does not exceed that of A. The cost of paint A per litre is Rs. 8 more than that of paint B. If the trader sells the entire mixture for Rs. 264 and makes a profit of 10%, then the highest possible cost of paint B, in Rs. per litre, is

a)Â 16

b)Â 26

c)Â 20

d)Â 22

Solution:

Let the price of paint B be x.
Price of paint A = x+8

We know that the amount of paint B in the mixture does not exceed the amount of paint A. Therefore, paint B can at the maximum compose 50% of the mixture.

The seller sells 10 litres of paint at Rs.264 earning a profit of 10%.
=> The cost price of 10 litres of the paint mixture = Rs. 240

Therefore, the cost of 1 litre of the mixture = Rs.24

We have to find the highest possible cost of paint B.
When we increase the cost of paint B, the cost of paint A will increase too. If the cost price of the mixture is closer to the cost of paint B, then the amount of paint B present in the mixture should be greater than the amount of paint A present in the mixture.
The highest possible cost of paint B will be obtained when the volumes of paint A and paint B in the mixture are equal.

=> (x+x+8)/2 = 24
2x = 40
x = Rs. 20

Therefore, optionÂ C is the right answer.

Question 11:Â A 20% ethanol solution is mixed with another ethanol solution, say, S of unknown concentration in the proportion 1:3 by volume. This mixture is then mixed with an equal volume of 20% ethanol solution. If the resultant mixture is a 31.25% ethanol solution, then the unknown concentration of S is

a)Â 30%

b)Â 40%

c)Â 50%

d)Â 60%

Solution:

Let the volume of the first and the second solution be 100 and 300.
When they are mixed, quantity of ethanol in the mixture
= (20 + 300S)
Let this solution be mixed with equal volume i.e. 400 of third solution in which the strength of ethanol is 20%.
So, the quantity of ethanol in the final solution
= (20 + 300S + 80) = (300S + 100)
It is given that, 31.25% of 800 = (300S + 100)
or, 300S + 100 = 250
or S = $\frac{1}{2}$ = 50%
Hence, 50 is the correct answer.

Question 12:Â A jar contains a mixture of 175 ml water and 700 ml alcohol. Gopal takes out 10% of the mixture and substitutes it by water of the same amount. The process is repeated once again. The percentage of water in the mixture is now

a)Â 30.3

b)Â 35.2

c)Â 25.4

d)Â 20.5

Solution:

Final quantity ofÂ alcohol in the mixture = $\dfrac{700}{700+175}*(\dfrac{90}{100})^2*[700+175]$ = 567 ml

Therefore, final quantity of water in the mixture = 875 – 567 = 308 ml

Hence, we can say that the percentage of water in the mixture = $\dfrac{308}{875}\times 100$ = 35.2 %

Question 13:Â There are two drums, each containing a mixture of paints A and B. In drum 1, A and B are in the ratio 18 : 7. The mixtures from drums 1 and 2 are mixed in the ratio 3 : 4 and in this final mixture, A and B are in the ratio 13 : 7. In drum 2, then A and B were in the ratio

a)Â 251 : 163

b)Â 239 : 161

c)Â 220 : 149

d)Â 229 : 141

Solution:

It is given that in drum 1, A and B are in the ratio 18 : 7.

Let us assume that in drum 2, A and B are in the ratio x : 1.

It is given thatÂ drums 1 and 2 are mixed in the ratio 3 : 4 and in this final mixture, A and B are in the ratio 13 : 7.

By equating concentration of A

$\Rightarrow$ $\dfrac{3*\dfrac{18}{18+7}+4*\dfrac{x}{x+1}}{3+4} = \dfrac{13}{13+7}$

$\Rightarrow$ $\dfrac{54}{25}+\dfrac{4x}{x+1} = \dfrac{91}{20}$

$\Rightarrow$ $\dfrac{4x}{x+1} = \dfrac{239}{100}$

$\Rightarrow$ $x = \dfrac{239}{161}$

Therefore, we can say that in drum 2,Â A and B are in the ratio $\dfrac{239}{161}$ : 1 or 239 : 161.

Question 14:Â The strength of a salt solution is p% if 100 ml of the solution contains p grams of salt. If three salt solutions A, B, C are mixed in the proportion 1 : 2 : 3, then the resulting solution has strength 20%. If instead the proportion is 3 : 2 : 1, then the resulting solution has strength 30%. A fourth solution, D, is produced by mixing B and C in the ratio 2 : 7. The ratio of the strength of D to that of A is

a)Â 3 : 10

b)Â 1 : 3

c)Â 1 : 4

d)Â 2 : 5

Solution:

Let ‘a’, ‘b’ and ‘c’ be the concentration of salt in solutions A, B and C respectively.

It is given thatÂ three salt solutions A, B, C are mixed in the proportion 1 : 2 : 3, then the resulting solution has strength 20%.

$\Rightarrow$ $\dfrac{a+2b+3c}{1+2+3} = 20$

$\Rightarrow$ $a+2b+3c = 120$ … (1)

If instead the proportion is 3 : 2 : 1, then the resulting solution has strength 30%.

$\Rightarrow$ $\dfrac{3a+2b+c}{1+2+3} = 30$

$\Rightarrow$Â $3a+2b+c = 180$ … (2)

From equation (1) and (2), we can say that

$\Rightarrow$ $b+2c = 45$

$\Rightarrow$ $b = 45 – 2c$

Also, on subtracting (1) from (2), we get

$a – c = 30$

$\Rightarrow$ $a = 30 + c$

In solution D, B and C are mixed in the ratio 2 : 7

So, the concentration of salt in D = $\dfrac{2b + 7c}{9}$ =Â $\dfrac{90 – 4c + 7c}{9}$ =Â $\dfrac{90 + 3c}{9}$

Required ratio = $\dfrac{90 + 3c}{9a}$ =Â $\dfrac{90 + 3c}{9 (30 + c)}$ = $1 : 3$

Hence, option B is the correct answer.

Question 15:Â A chemist mixes two liquids 1 and 2. One litre of liquid 1 weighs 1 kg and one litre of liquid 2 weighs 800 gm. If half litre of the mixture weighs 480 gm, then the percentage of liquid 1 in the mixture, in terms of volume, is

a)Â 80

b)Â 70

c)Â 85

d)Â 75

Solution:

The weight/volume(g/L) for liquid 1 = 1000

The weight/volume(g/L) for liquid 2 = 800

The weight/volume(g/L) of the mixture = 480/(1/2) = 960

Using alligation the ratio of liquid 1 and liquid 2 in the mixture = (960-800)/(1000-960) = 160/40 = 4:1

Hence the percentage of liquid 1 in the mixture = 4*100/(4+1)=80

Question 16:Â An alloy is prepared by mixing three metals A, B and C in the proportion 3 : 4 : 7 by volume. Weights of the same volume of the metals A. B and C are in the ratio 5 : 2 : 6. In 130 kg of the alloy, the weight, in kg. of the metal C is

a)Â 48

b)Â 84

c)Â 70

d)Â 96

Solution:

Let the volume of Metals A,B,C we 3x, 4x, 7x

Ratio weights of given volume be 5y,2y,6y

.’. 15xy+8xy+42xy=130 => 65xy=130 => xy=2.

.’.`The weight, in kg. of the metal C is 42xy=84.

Question 17:Â Two alcohol solutions, A and B, are mixed in the proportion 1:3 by volume. The volumeÂ of the mixture is then doubled by adding solution A such that the resulting mixtureÂ has 72% alcohol. If solution A has 60% alcohol, then the percentage of alcohol inÂ solution B is

a)Â 90%

b)Â 94%

c)Â 92%

d)Â 89%

Solution:

Initially let’s consider A and B as one component

The volume of the mixture is doubled byÂ adding A(60% alcohol) i.e they are mixed in 1:1 ratio and the resultant mixture has 72% alcohol.

Let the percentage of alcohol in component 1 be ‘x’.

Using allegations ,Â $\frac{\left(72-60\right)}{x-72}=\frac{1}{1}$ => x= 84

Percentage of alcohol inÂ A = 60% => Let’s percentage of alcohol in B = x%

The resultant mixture has 84% alcohol. ratio = 1:3

Using allegations ,Â $\frac{\left(x-84\right)}{84-60}=\frac{1}{3}$

=> x= 92%

Question 18:Â The strength of an indigo solution in percentage is equal to the amount of indigo in grams per 100 cc of water. Two 800 cc bottles are filled with indigo solutions of strengths 33% and 17%, respectively. A part of the solution from the first bottle is thrown away and replaced by an equal volume of the solution from the second bottle. If the strength of the indigo solution in the first bottle has now changed to 21% then the volume, in cc, of the solution left in the second bottle is

Solution:

Let Bottle A haveÂ an indigo solution of strength 33%Â whileÂ Bottle B have an indigo solution of strength 17%.

The ratio in which we mix these two solutions to obtain a resultant solution of strength 21% :Â $\frac{A}{B}=\frac{21-17}{33-21}=\frac{4}{12}or\ \frac{1}{3}$

Hence, three parts of the solution from BottleÂ B is mixed with one part of the solution from Bottle A. For this process to happen, we need to displace 600 cc of solution from Bottle A and replace it with 600 cc of solution from Bottle B {since both bottles have 800 cc, three parts of this volume = 600cc}.As a result, 200 cc of the solution remains in Bottle B.

Hence, the correct answer is 200 cc.

Question 19:Â A person buys tea of three different qualities at â‚¹ 800, â‚¹ 500, and â‚¹ 300 per kg, respectively, and the amounts bought are in the proportion 2 : 3 : 5. She mixes all the tea and sells one-sixth of the mixture at â‚¹ 700 per kg. The price, in INR per kg, at which she should sell the remaining tea, to make an overall profit of 50%, is

a)Â 653

b)Â 688

c)Â 692

d)Â 675

Solution:

Considering the three kinds of tea are A, B, and C.

The price of kind A = Rs 800 per kg.

The price of kind B = Rs 500 per kg.

The price of kind C = Rs 300 per kg.

They were mixed in the ratio of 2 : 3: 5.

1/6 of the total mixture is sold for Rs 700 per kg.

Assuming the ratio of mixture to A =Â 12kg, B =Â 18kg, C =30 kg.

The total cost price is 800*12+500*18+300*30 = Rs 27600.

Selling 1/6 which is 10kg for Rs 700/kg the revenue earned is Rs 7000.

In order to have an overall profit of 50 percent on Rs 27600.

Thes selling price of the 60 kg is Rs 27600*1.5 = Rs 41400.

Hence he must sell the remaining 50 kg mixture for Rs 41400 – Rs 7000 = 34400.

Hence the price per kg is Rs 34400/50 = Rs 688

Question 20:Â If a certain weight of an alloy of silver and copper is mixed with 3 kg of pure silver, the resulting alloy will have 90% silver by weight. If the same weight of the initial alloy is mixed with 2 kg of another alloy which has 90% silver by weight, the resulting alloy will have 84% silver by weight. Then, the weight of the initial alloy, in kg, is

a)Â 3.5

b)Â 2.5

c)Â 3

d)Â 4

Solution:

Let the alloy contain x Kg silver and y kg copper
Now when mixed with 3Kg Pure silver
we getÂ $\frac{\left(x+3\right)}{x+y+3}=\frac{9}{10}$
we get 10x+30 =9x+9y+27
9y-x=3Â Â Â  (1)
Now as per condition 2
silver in 2nd alloy = 2(0.9) =1.8
so we get$\frac{\left(x+1.8\right)}{x+y+2}=\frac{21}{25}$
we get 21y-4x =3 Â  (2)
solving (1) and (2) we get y= 0.6 and x =2.4
so x+y = 3

So, these are some of the most important CAT mixture and alligation questions. For more practice take these Free mock tests for CAT. Practice these CAT mixture and alligation questions PDF. Learn all the major formulae from the CAT mixture and alligation. You can also check out the Important CAT mixture and alligation questions & Formulas PDF here.