CAT Averages Mixtures Alligations Questions & Solutions

A sample of real CAT Averages Mixtures Alligations past-year questions, each with a worked solution and the correct answer highlighted. AthenaPrep has 53 Averages Mixtures Alligations questions in all — sign up free to practise them timed.

1. Q1.CAT 2025

   A container holds 200 litres of a solution of acid and water, having 30% acid by volume. Atul replaces 20% of this solution with water, then replaces 10% of the resulting solution with acid, and finally replaces 15% of the solution thus obtained, with water. The percentage of acid by volume in the final solution obtained after these three replacements, is nearest to

   • 23

   • 25

   • 29

   • 27

   Show solution

   200 L with (30%) acid $\Rightarrow$ acid = $0.30\times200=60 L$

   Replace 20% with water: remaining acid $=60\times(1-0.20)=60\times0.8=48 L$ .

   Replace 10% with pure acid: After removing 10% of the mixture, the acid becomes $48\times0.9=43.2 L$ , then adding back 20 L pure acid $\Rightarrow acid =43.2+20=63.2 L$ .

   Replace 15 with water: acid left $= 63.2\times(1-0.15)=63.2\times0.85=53.72 L$ .

   Final concentration $= \dfrac{53.72}{200}=0.2686\approx26.86$ .

   The answer is 27%

2. Q2.CAT 2025

   A mixture of coffee and cocoa, 16% of which is coffee, costs Rs 240 per kg. Another mixture of coffee and cocoa, of which 36% is coffee, costs Rs 320 per kg. If a new mixture of coffee and cocoa costs Rs 376 per kg, then the quantity, in kg, of coffee in 10 kg of this new mixture is

   • 5

   • 4

   • 2.5

   • 6

   Show solution

   Let coffee price = C Rs/kg and cocoa price = K Rs/kg.

   From the two given mixtures:

   $0.16C + 0.84K = 240$

   $0.36C + 0.64K = 320$

   Multiply both equations by 100 to remove decimals:

   $16C + 84K = 24000$  

   $36C + 64K = 32000$

   Subtract the first from the second:

   $(36C+64K)-(16C+84K)=32000-24000$

   $20C - 20K = 8000 \implies C - K = 400$
   

   Put $C=K+400$ into $16C+84K=24000$ :
   
   $16(K+400)+84K=24000$

   $16K+6400+84K=24000$

   $100K = 17600 \implies K = 176$
   
   So $C = 176 + 400 = 576$ (Rs/kg).

   For the new mixture priced at Rs 376/kg, let the coffee fraction be p. Then

   $p\cdot 576 + (1-p)\cdot 176 = 376$

   Upon solving  $p = \tfrac{1}{2}$
   

   Thus, coffee is (50%) of the new mixture. In 10 kg of this mixture, coffee = $10\times 0.5 = 5\text{kg}$

3. Q3.CAT 2024

   The average of three distinct real numbers is 28. If the smallest number is increased by 7 and the largest number is reduced by 10, the order of the numbers remains unchanged, and the new arithmetic mean becomes 2 more than the middle number, while the difference between the largest and the smallest numbers becomes 64.Then, the largest number in the original set of three numbers is

   Answer: 70

   Show solution

   We are given that average of three distinct integers is 28, that means the sum of these three integers is 28x3=84

   Let us write $x+y+z=84$
   x, y, z being the three distinct integers in ascending order. 

   If the smallest number is increased by 7 and the largest number is reduced by 10
   $(x+7)+(y)+(z-10)=81$

   New arithmetic mean will be  $\frac{81}{3}=27$
   And this is said to be 2 more than the middle number, meaning
   $27-2=y=25$
   

   $x+z=59$

   We are given that difference between the largest and the smallest numbers becomes 64, 
   $(z-10)-(x+7)=64$
   $z-x=81$

   Adding the two equations we get,  $2z=140$
   $z=70$
   

4. Q4.CAT 2023

   There are three persons A, B and C in a room. If a person D joins the room, the average weight of the persons in the room reduces by x kg. Instead of D, if person E joins the room, the average weight of the persons in the room increases by 2x kg. If the weight of E is 12 kg more than that of D, then the value of x is

   • 2

   • 0.5

   • 1

   • 1.5

   Show solution

   Let us assume that A, B, C, D, and E weights are a, b, c, d, and e.

   1st condition

   $\frac{\left(a+b+c\right)}{3}-\frac{\left(a+b+c+d\right)}{4}=x$

   2nd condition

   $\frac{\left(a+b+c+e\right)}{4}-\frac{\left(a+b+c\right)}{3}=2x$

   Adding both the equations, we get:

   $\frac{\left(e-d\right)}{4}=3x$

   => $\frac{\left(e-d\right)}{4}=3x$ => e - d = 12x

   Given that 12x = 12 => x = 1.

5. Q5.CAT 2022

   The average of three integers is 13. When a natural number n is included, the average of these four integers remains an odd integer. The minimum possible value of n is

   • 3

   • 4

   • 5

   • 1

   Show solution

   It is given that average of three numbers is 13.

   Sum = 3*13 = 39

   It is given,  $\ \frac{\ 39+n}{4}$ is a odd number.

   Minimum value  $\ \frac{\ 39+n}{4}$ can take such that n is a natural number is 11

   $\ \frac{\ 39+n}{4}=11$

   n = 5

   The answer is option C.

6. Q6.CAT 2022

   Five students, including Amit, appear for an examination in which possible marks are integers between 0 and 50, both inclusive. The average marks for all the students is 38 and exactly three students got more than 32. If no two students got the same marks and Amit got the least marks among the five students, then the difference between the highest and lowest possible marks of Amit is

   • 22

   • 21

   • 24

   • 20

   Show solution

   The average marks for all the students is 38.

   Sum = 5*38 = 190

   To find the minimum marks scored by Amit, we need to maximise the score of remaining students.

   Maximum scores sum of remaining students = 50 + 49 + 48 + 32 = 179

   Minimum possible score of Amit = 190 - 179 = 11

   It is given, Amit scored least. This implies maximum possible score of Amit is 31.

   Difference = 31 - 11 = 20

   The answer is option D.
   

7. Q7.CAT 2021

   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

   Answer: 200

   Show solution

   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% :  $\frac{A}{B}=\frac{21-17}{33-21}=\frac{4}{12}or\ \frac{1}{3}$

   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.  

8. Q8.CAT 2021

   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

   • 653

   • 688

   • 692

   • 675

   Show 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 = 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

9. Q9.CAT 2020

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

   • 90%

   • 94%

   • 92%

   • 89%

   Show solution

   Initially let's consider A and B as one component

   The volume of the mixture is doubled by 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 ,  $\frac{\left(72-60\right)}{x-72}=\frac{1}{1}$ => x= 84

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

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

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

   => x= 92%

10. Q10.CAT 2019

    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

    • 80

    • 70

    • 85

    • 75

    Show 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

11. Q11.CAT 2017

    A class consists of 20 boys and 30 girls. In the mid-semester examination, the average score of the girls was 5 higher than that of the boys. In the final exam, however, the average score of the girls dropped by 3 while the average score of the entire class increased by 2. The increase in the average score of the boys is

    • 9.5

    • 10

    • 4.5

    • 6

    Show solution

    Let, the average score of boys in the mid semester exam is A.
    Therefore, the average score of girls in the mid semester exam be A+5.
    Hence, the total marks scored by the class is 20\times (A) + 30\times (A+5) = 50\times A + 150

    The average score of the entire class is $\dfrac{(50\times A + 150)}{50} = A + 3$
    
    wkt, class average increased by 2, class average in final term $= (A+3) + 2 = A + 5$

    Given, that score of girls dropped by 3, i.e $(A+5)-3 = A+2$
    Total score of girls in final term  $= 30\times(A+2) = 30A + 60$
    

    

    
    Total class score in final term = (A + 5)\times50 = 50A + 250

    the total marks scored by the boys is $(50A + 250) - (30A - 60) = 20A + 190$
    Hence, the average of the boys in the final exam is $\dfrac{(20G + 190)}{20} = A + 9.5$
    
    

    

    Hence, the increase in the average marks of the boys is $(A+9.5) - A = 9.5$

12. Q12.CAT 2002

    Amol was asked to calculate the arithmetic mean of 10 positive integers, each of which had 2 digits. By mistake, he interchanged the 2 digits, say a and b, in one of these 10 integers. As a result, his answer for the arithmetic mean was 1.8 more than what it should have been. Then |b - a| equals

    • 1

    • 2

    • 3

    • None of these

    Show solution

    Let the actual average be n. So, the new average is n + 1.8
    Actual total = 10n
    New total = 10n + 18

    Let the number which was miswritten = ab(a is the tenth's digit and b is the units digit) = 10a+b

    and reversed number ba = 10b+a

    
    So, 10b + a - (10a + b) = 18
    => 9(b-a) = 18
    => b-a = 2

CAT Averages Mixtures Alligations previous year questions with solutions — AthenaPrep