Mixtures & Alligations CAT Previous Year Questions with Answer PDF

2024

Q1: Rajesh and Vimal own 20 hectares and 30 hectares of agricultural land, respectively, which are entirely covered by wheat and mustard crops. The cultivation area of wheat and mustard in the land owned by Vimal are in the ratio of 5 : 3. If the total cultivation area of wheat and mustard are in the ratio 11 : 9, then the ratio of cultivation area of wheat and mustard in the land owned by Rajesh is 
(a) 1 : 1
(b) 4 : 3
(c) 7 : 9
(d) 3 : 7

Ans: c
Sol: According to the question,
Rajesh -> 20 hectares
Vimal -> 30 hectares
The entire land is cultivated by 2 crops i.e., wheat and mustard
We have, cultivation area of wheat and mustard on Vimal’s land = 5: 3
Total cultivation area of wheat and mustard = 11: 9
So, for finding the cultivation area of wheat and mustard on Rajesh’s land here, we can use the concept of alligation, (Assume the ratio to be x: y)

Hence, the ratio of cultivation area of wheat and mustard in the land owned by Rajesh = 7: 9

2023

Q1: A mixture P is formed by removing a certain amount of coffee from a coffee jar and replacing the same amount with cocoa powder. The same amount is again removed from mixture P and replaced with same amount of cocoa powder to form a new mixture Q. If the ratio of coffee and cocoa in the mixture Q is 16 : 9, then the ratio of cocoa in mixture P to that in mixture Q is  [2023]
(a) 4 : 9
(b) 1 : 3
(c) 5 : 9
(d) 1 : 2

Ans: (c)
Sol: Every time 20% of the mixture is replace with a pure adulterant (here cocoa), the concentration of the mixture becomes 80% of the initial concentration.
Similarly when a proportion p of a mixture is replaced with pure adulterant, the concentration of the mixture becomes (1 - p) times the previous concentration. (0 ≤ p ≤ 1)
Initially there was pure coffee, so the strength of the mixture is 100% or 1 (in terms of proportion)
Let a proportion p of the mixture be replaced each time… Since after repeating the process twice, we have the concentration of coffee as   at the end of the two replacements.

Or, 20% of the mixture is replaced each time.
The ratio of cocoa powder in the mixture P and Q will then be

Therefore, the ratio is 5 : 9

Q2: A container has 40 liters of milk. Then, 4 liters are removed from the container and replaced with 4 liters of water. This process of replacing 4 liters of the liquid in the container with an equal volume of water is continued repeatedly. The smallest number of times of doing this process, after which the volume of milk in the container becomes less than that of water, is 

Ans: 7
Sol: Every time 10% of the mixture is replace with a pure adulterant (here water), the concentration of the mixture becomes 90% of the initial concentration.
Similarly when a proportion p of a mixture is replaced with pure adulterant, the concentration of the mixture becomes (1 - p) times the previous concentration. (0 ≤ p ≤ 1)
Here p = 4/40 = 0.1 
Initially there was pure milk, so the strength of the mixture is 100% or 1 (in terms of proportion)
After repeating the process n times the concentration of Milk in the mixture will be 1×0.9ⁿ
For the volume of milk to be less than the volume of water, p should be less than 0.5
So we are looking for the smallest n, such that 1 × 0.9ⁿ < 0.5
This happens when n = 7

2022

Q1: A mixture contains lemon juice and sugar syrup in equal proportion. If a new mixture is created by adding this mixture and sugar syrup in the ratio 1 : 3, then the ratio of lemon juice and sugar syrup in the new mixture is  [2022]
(a) 1 : 4
(b) 1 : 5
(c) 1 : 6
(d) 1 : 7

Ans: (d)
Sol: Lemon juice : sugar syrup in the mixture is 1:1, i.e. 50% Lemon juice and 50% sugar syrup.
 In sugar syrup, 100% is sugar syrup.
 These two are mixed in the ratio 1:3.

 Required ratio = 50 : 350 = 1 : 7

Q2: There are two containers of the same volume, first container half-filled with sugar syrup and the second container half-filled with milk. Half the content of the first container is transferred to the second container, and then the half of this mixture is transferred back to the first container. Next, half the content of the first container is transferred back to the second container. Then the ratio of sugar syrup and milk in the second container is [2022]
(a) 5 : 6
(b) 5 : 4
(c) 6 : 5
(d) 4 : 5

Ans: (a)
Sol: We know that the process of shifting half volumes is happening three times overall.

 So it would be wise to assume the initial volume to be some multiple of 8 so that we don’t have to deal with fractions later on.

The ratio of Sugar Syrup to Water in the second container is 50/60 which is equal to 5:6

2021

Q1: 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 [2021]

Ans: 200
Sol: Let Bottle A have an indigo solution of strength 33% while 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% : A/B = 21-17/33-21 = 4/12 or 1/3
Hence, three parts of the solution from Bottle 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.  

Q2: From a container filled with milk, 9 litres of milk are drawn and replaced with water. Next, from the same container, 9 litres are drawn and again replaced with water. If the volumes of milk and water in the container are now in the ratio of 16 : 9, then the capacity of the container, in litres, is 

Ans: 45
Sol: Let initial volume be V, final be F for milk.
The formula is given by : is the number of times the milk is drawn and replaced.

 V =5, but this is not possible because 9 liters is drawn every time.

Q3: A box has 450 balls, each either white or black, there being as many metallic white balls as metallic black balls. If 40% of the white balls and 50% of the black balls are metallic, then the number of non-metallic balls in the box is 

Ans: 250
Sol: Let the number of white balls be x and black balls be y
So we get x+y =450 (1)
Now metallic black balls = 0.5y
Metallic white balls = 0.4x
From condition 0.4x=0.5y
we get 4x-5y=0 (2)
Solving (1) and (2) we get
x=250 and y =200
Now number of Non Metallic balls = 0.6x+0.5y = 150+100 = 250

Q4: A person buys tea of three different qualities at ₹ 800, ₹ 500, and ₹ 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 ₹ 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) 653
(b) 688
(c) 692
(d) 675

Ans: b
Sol: 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 = 12kg, B = 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

Q5: 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) 3.5
(b) 2.5
(c) 3
(d) 4

Ans: c
Sol: Let the alloy contain x Kg silver and y kg copper
Now when mixed with 3Kg Pure silver

we get 10x+30 =9x+9y+27

9y-x=3 (1)
Now as per condition 2 silver in 2nd alloy = 2(0.9) =1.8

we get 21y-4x =3 (2)

solving (1) and (2) we get y= 0.6 and x =2.4
so x+y = 3

2020

Q1: 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  [2020]
(a) 70
(b) 96
(c) 48
(d) 84

Ans: (d)
Sol: Required weight of C =
= 84 kg

Q2: A solution, of volume 40 litres, has dye and water in the proportion 2 : 3. Water is added to the solution to change this proportion to 2 : 5. If one-fourths of this diluted solution is taken out, how many litres of dye must be added to the remaining solution to bring the proportion back to 2 : 3?  [2020]

Ans: 8
Sol: Original quantity of dye and water in the original solution i.e., 16 litres  and 24 litres (i.e. = 40 - 16)
Quantity of water added = 16 litres (As 1 part = 8 litres). Quantity of dye and water 
removed =  i.e., 4 litres and  i.e., 10 litres. Final quantity of dye and 
water is 12 litres and 30 litres.
∴ Quantity of dye to be added to make the ratio of dye and water again 2: 3 i.e., 8 litres. 

2019

Q1: 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  [2019]
(a) 70
(b) 85
(c) 80
(d) 75

Ans: (c)
Sol: In Liquid L1 - 1L = 1000 grams
So, 1000 mL = 1000 grams
1 mL = 1 gram
x mL = x grams
In Liquid L2 - 1L = 800 grams
1000 mL = 800 grams
1mL = 0.8 grams
(500-x) mL = (500-x) x 0.8 grams
Total mass = x + 400 - 0.8x = 480 grams
0.2x = 80 grams
x = 400 grams
Therefore, Liquid 1 has 400 mL and Liquid 2 has 500 - 400 = 100 Ml
Therefore, Percentage of Liquid 1 = (400/500) x 100 = 80%

Q2: The strength of a salt solution is p% if 100 ml of the solution contains p grams of salt. Each of three vessels A, B, C contains 500 ml of salt solution of strengths 10%, 22%, and 32%, respectively. Now, 100 ml of the solution in vessel A is transferred to vessel B. Then, 100 ml of the solution in vessel B is transferred to vessel C. Finally, 100 ml of the solution in vessel C is transferred to vessel A. The strength, in percentage, of the resulting solution in vessel A is  [2019]
(a) 15
(b) 12
(c) 13
(d) 14

Ans: (d)
Sol: Vessels A, B and C contains salt solution of strengths 10%, 22% and 32% respectively
It is also given that the amount of salt solution = 500 ml
So, Vessels A, B and C contains salt of 50 grams, 110 grams and 160 grams respectively
100 ml of Solution is transferred from A to B:
A would have 400 ml, B would have 600 ml of solution
Amount salt from A which is transferred to B = (Initial salt amount/5) = 10 grams
So, Total salt in B = 110 + 10 = 120 grams (After first transfer)
Total salt in A = 40 grams (After first transfer)
Now, 100 ml from Vessel B is transferred to Vessel C
So similarly, (1/6)th of salt would transfer from B to C
Total Salt in B = 120 - 20 = 100 grams (After second transfer)
Total Salt in C = 160 + 20 = 180 grams (After second transfer)
Now, 100 ml from Vessel C is transferred to Vessel A
So similarly, (1/6)thof salt would transfer from C to A
Total Salt in C = 160 - 30 = 130 grams (After third transfer)
Total Salt in A = 40 + 30 = 70 grams (After third transfer)
So, Vessel A contains 70 grams Salt in 500 ml solution
Strength of Salt Solution in Vessel A = 14%

2018

Q1: 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  [2018]
(a) 20
(b) 16
(c) 22
(d) 26

Ans: (a)
Sol: Given, The amount of B in mixture ≤ Amount of A in mixture
 Selling Price = Rs. 264 and Profit = 10 %
 264 = 1.1 × CP
 CP = (264/1.1) = Rs. 240 for 10 litres
 CP per litre = Rs. 24
 We also know that, CP per litre of A = CP per litre of B + 8 Let CP per litre of B = x So, CP per litre of A = x + 8

 Given Amount of B ≤ Amount of A,
 Maximum possible cost of B occurs when B = A
 We need to choose values for A and B in such a way that the quantities remain same.But A should be 8 more than B
 Therefore, A should have a CP per litre of Rs. 28 and B should have a CP per litre of Rs 20

 B cannot be assigned any more than this as the amount of B would become more than A
 Example: Let us assign B to be Rs. 21 and A to be Rs. 29 which would result in a ratio of 3:5 where B amounts more than A, which doesn’t satisfy the condition
 Max possible cost of Paint B = Rs. 20 per litre

Q2: 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?   [2018]
(a) 84
(b) 86
(c) 96
(d) 98

Ans: (c)
Sol: Let the wholesaler bought walnuts and peanuts at Rs 3x and Rs x respectively
He sold 8kgs of Peanuts to the shopkeeper at 10% Profit
Cost price of the Peanuts bought by the shopkeeper = Rs. 1.1x per kg
Similarly, he sold 16 kgs of Walnuts to the shopkeeper at 20% Profit
Cost price of the Walnuts bought by the shopkeeper = 1.2 × 3x = Rs. 3.6x per Kg
He lost 5 kgs of Walnuts and 3 Kgs of Peanuts in transit
Remaining = 16 - 5 = 11 kgs of Walnuts & 8-3 = 5 kgs of Peanuts
He mixes them together and sells them at Rs. 166 per kg, making an overall Profit of 25%
Selling Price = 5/4 × Cost Price
Overall Cost Price = Rs.(3.6 × 16x + 1.1 × 8x)
Overall Selling Price = Rs. 16 × 166
Overall Selling Price = (5/4) x Overall Cost Price
(3.6 × 16x + 1.1 × 8x) × (5/4) = 16 × 166
((57.6 × 8.8) x ) × 5/4 = 16 × 166
x = Rs. 32
Cost price per Kg of Walnuts bought by the Wholesaler = Rs. 3x = 3 × 32 = >Rs. 96

Q3: 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 [2018]
(a) 251 : 163
(b) 239 : 161
(c) 220 : 149
(d) 229 : 141

Ans: b

Sol: 

Given that two drums each containing a mixture of paints A and B. In drum 1, A and B are in the ratio 18 : 7. Drums 1 and 2 are mixed in the ratio 3 : 4 and in the final mixture A and B are in the ratio 13 : 7. For these kinds of questions do not consider separately as A and B. Deal with either A or B as a share of overall

 Here A is (18/25)  of overall

 In D2, let us assume the proportion of A with respect to the overall mixture is x. 3 parts of D1 is mixed with parts of D2 to give proportion of A of which is (13/20)th of the overall mixture. From here on, it is weighted averages.

In drum 2, A is (239/400) so B should be the remaining. Therefore, A : B = 239 : 161.

Q4: 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  [2018]
(a) 50%
(b) 55%
(c) 48%
(d) 52%

Ans: (a)
Sol: Given that 20% ethanol solution is mixed with another ethanol solution, say, S of unknown concentration in the proportion 1:3 by volume

This M 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 has to be found
 
Equal quantities of mixture and 20% ethanol solution are mixed in equal ratio to get 31.25%
The mixture and 20% ethanol solution are mixed in equal ratio to get 31.25%. This 31.25% is 11.25% more than this 20%. This mixture must be 11.25% more than 31.25% so mixture is equal to 42.5%
 
 This mixture 42.5% is mixed in the ratio 1:3
 
 Using allegations we can find this 22.5% which should be in the ratio 1:3 so the other one is 7.5%
 S = 42.5 + 7.5
 S = 50%
 Here to find the solution, do the second thing first and then go to the first thing
 The second one is mixed in the ratio 1 : 1 so that the final thing should be bang in the middle

Q5: 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  [2018]
(a) 25.4
(b) 20.5
(c) 30.3
(d) 35.2

Ans: (d)
Sol: 

Given that a jar contains a mixture of 175 ml water and 700 ml alcohol.
It is given that 10% of the mixture is removed and it is substituted by water of the same amount and the process is repeated once again
Now we have to find the percentage of water in the mixture.
Since the mixture is removed and substituted with water, we can deal with alcohol and the second step we can find how much amount of alcohol is retained and not about how much amount of alcohol is removed
As 10% of alcohol is removed, 90% of alcohol is retained
So alcohol remaining = 700 × 90% × 90%
⇒  700 × 0.9 × 0.9 = 567
We totally have 875 ml overall mixture and of this 567 ml is alcohol.
Remaining 875 – 567 = 308 is the amount of water.
We have to find the percentage of water in the mixture i.e. 308/875
Approximately 308 is 30% of 1000 so by this we know that 308 is more than 30%
Hence 35.2% is the percentage of water in the given mixture.

Q6: 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  [2018]
(a) 3 : 10
(b) 1 : 3
(c) 2 : 5
(d) 1 : 4

Ans: (b)
Sol: Given that the strength of the salt solution is p% if 100 ml of the solution contains p grams of salt
It is also given that three salt solutions A , B , C are mixed in the proportion 1 : 2 : 3, then the resulting solution has strength 20%.
So  = 0.2 ⇒ A + 2B + 3C = 1.2 ...(1)
If instead the proportion is 3 : 2 : 1, then the resulting solution has strength 30%
So,  = 0.3 ⇒ 3A +2B + C = 1.8 ...(2)
It is given that 4^th solution D is produced by mixing B and C in the ratio 2 : 7
So D =
We have to find the ratio of the strength of D : A
Subtracting equations 1 and 2, we get
2A – 2C = 0.6 or A – C = 0.3
Since we could not find anything from the above methods, we can eliminate the number part and get the ratio going
A + 2B + 3C = 1.2
3A + 2B + C = 1.8
So let us multiply eqn 1 and 2 with 3 and 2,
3A + 6B + 9C = 6A + 4B + 2C
2B + 7C = 3A
It is given that D =

Hence D = A/3
Therefore the ratio D : A = 1 : 3.

2017

Q1: 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?  [2017]
(a) 27 : 14
(b) 27 : 13
(c) 27 : 16
(d) 27 : 18

Ans: (b)
Sol: 

Bottle 1 and bottle 2 contains mixture of milk and water in the ratio 7 : 2 and 9 : 4 respectively.
We have to find by what ratio bottle 1 and Bottle 2 should be combined to obtain a mixture of milk and water in the ratio 3 : 1
 For this we can consider milk to the whole mixture or else water to the whole mixture.

⇒

Taking LCM we get

LCM is 52 so

Bottle 1 and 2 are mixed in the ratio 27 : 13.

Q2: Consider three mixtures - 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  [2017]
(a) The same amount of water and liquid B
(b) The same amount of liquids B and C
(c) More water than liquid B
(d) More water than liquid A

Ans: (c)
Sol: There are three mixtures ,the first having water and liquid A in the ratio 1 : 2,
 so we can consider only the liquid part which is 2/3.
 The second having the water and liquid in the ratio 1 : 3 = 3/4.
 The third having the liquid and water in the ratio 1 : 4 = 4/5


 Hence by this we can infer that more water than liquid B is true