When two or more things are mixed together, they are said to be "mixed." Finding the average of something (cost, quantity, percentage, etc.) in a mixture is the goal of mixing questions. Some liquids are mixed together to make a mixture with a certain percentage. Items with different prices are sometimes mixed together. When two or more things are mixed in the same amount, you can use the idea of easy means to solve problems. Weighted averages are used to solve problems when two or more things have different weights.
Mixture and Allegation is a very important subject for any MBA Entrance Test like CAT, XAT, MHCET, IIFT etc. When it comes to questions about this topic, ratios and proportions, and sometimes the basics of percentages, come up. also
Most mixtures fall into two groups.
- When you mix two different things together. It is called a simple mixture, like when you mix water and milk together.
- Compound mixtures are made when two or more simple mixtures are put together to make a new mixture.
Concepts
In general, if the average of group 1 be A1 and the number of the elements be n1 and the average of the group 2 be A2 and the number of element be n2 then the weighted average
\(A_w=\frac{n_1A_1+n_2A_2}{n_1+n_2}\)
- n1Aw + n2Aw = n1A1 + n2A2
- n1 (Aw − A1) = n2 (A2 − Aw)
\(\frac{n_1}{n_2}=\frac{A_2-A_w}{A_w-A_1}\) (Alligation Equation)
Represented in a graphical form:
As we've already talked about, you can use the weighted average formula, the Alligation equation, or the graphical representation method to solve questions related to the topic.
While calculating concentration of liquid after replacing. The following formula can be used:
Final Concentration = Initial Concentration \((1-\frac{Amount \ being \ replaced \ in \ each \ operation}{Total \ Amount})^n\)
where n is the number of times that the same thing is done. The ''amount being changed'' could be pure or a split, depending on the situation. In the same way, ''total amount'' could be either pure or a mix. Here, the amount being refilled is the amount that needs to be taken out each time.
Previous year Mixtures and Alligation CAT questions
Ques 1: 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 (CAT 2022 slot 1)
- 1:4
- 1:6
- 1:5
- 1:7
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Ans: (D)
Let 10 units of blend and 30 units of sugar syrup be mixed together.
For every 10 units of blend, there are 5 units of sugar syrup and 5 units of lemon juice. The end mix has a total of 5 units of lemon juice.
5 units of sugar syrup plus 30 units of sugar syrup equals 35 units in the end mixture.
The finished mixture has 5:35 = 1:7 parts lemon juice to sugar syrup.
Ques 2: 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 (CAT 2022 slot 2)
- 5:6
- 4:5
- 5:4
- 6:5
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Ans:
Let there be 100 litres of sugar in container A
100 litres of milk in container B
After 1st transfer:-
Sugar left in A is 50 litres
Sugar added to B is 50 litres from A.
Ratio of Sugar to Milk in B = 1:2
After 2nd transfer:-
When 75 litres is transferred to A then, it is transferred in the same ratio as that of B in 1:2.
i.e., 50 litres of milk and 25 litres of sugar .
Resultant amount of sugar in A = 50 +25 =75
Resultant amount of milk in A = 0 +50 = 50
So, Sugar : Milk = 75:50= 3 : 2.
After 3rd transfer:-
When half of amount in A is transferred i.e., 62.5 litres then,
Amount of sugar transferred= (3/5) x 62.5= 37.5 litres
Amount of milk transferred = (2/5) x 62.5= 25 litres
Resultant amount of sugar in B = 25+37.5 = 62.5 litres
Resultant amount of milk in B = 50 + 25 = 75 litres
Ques 3: A glass contains 500 cc of milk and a cup contains 500 cc of water. From the glass, 150 cc of milk is transferred to the cup and mixed thoroughly. Next, 150 cc of this mixture is transferred from the cup to the glass. Now, the amount of water in the glass and the amount of milk in the cup are in the ratio (CAT 2022 Slot 3)
- 10:13
- 10:3
- 3:10
- 1:1
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Ans: (D)
When two containers have the same amount of milk and water, and then equal amounts are moved from the first to the second, and then from the second to the first,
The quantity of milk in the first becomes equal to the amount of water in the second, and the amount of water in the first becomes equal to the amount of milk in the second.
Ques 4: 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 (CAT 2021 slot 1)
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Ans:
Let x amount is removed from first and replaced with solution from second bottle.
Where 33% and 17% are the strengths of the indigo solutions and
average value after mixing = 21%
Using alligation rule,
\(\frac{800-x}{x}= \frac{33-21}{21-17}\)
x=600
Therefore the solution left in the second bottle = 800 – 600 = 200 cc
Ques 5: From a container filled with milk, 9 lites of milk are drawn and replaced with water. Next, from the same container, 9 lites are drawn and again replaced with water. 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 (CAT 2021 slot 2)
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Ans:
Assuming the container to be full of milk and the capacity is V
Ratio of quantity when 9 litres of milk is drawn = 9/V
Using the concentration formula:-
Fc= Ic(1- \(\frac{amount\ replaced\ in \ each\ operation}{Total\ volume}\))n
\(\frac{16}{16+25}\)=V (1 - \(\frac{9}{V}\))2 [n=2, n= the number of replacements with water]
After solving the equation, V = 45 litres.
Ques 6: 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 (CAT 2021 slot 3)
- 2.5
- 3.5
- 3
- 4
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Ans: (C)
Assuming the alloy of silver and copper be x kg and p% silver
After the mixing of 3kg of silver
\(\frac{a}{3}= \frac{100-90}{90-p}= \frac{10}{90-p}\) …………...(i)
Initial alloy is mixed with 2kg , 90% silver concentration
\(\frac{a}{x2}= \frac{90-84}{84-p}= \frac{6}{84-p}\)…………...(ii)
Solving equation (i) and (ii) we get,
\(\frac{30}{90-p}= \frac{12}{84-p}\)
P= 80%
a= \(\frac{30}{90-p}\) = 3 kg
Ques 7: A solution, of volume 40 lites, 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 lites of dye must be added to the remaining solution to bring the proportion back to 2 : 3? (CAT 2020 slot 1)
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Ans:
Dye and Water are in the ratio 2:3
Dye= (2/5)x 40 = 16 litres
Water= (3/5) x 40 = 24 litres
After adding x litres of water into the mixture.
\(\frac{16}{24+x}= \frac{2}{5}\)
x= 16 litres.
Resultant amount of water in the solution = 24 + 16 = 40 litres.
When (1/4)th of the solution is removed
Amount of dye left = (3/4) x 16 = 12 litres
Amount of water left = (3/4) x 40 = 30 litres
Addition of y amount of dye will result into the dye:water = 2:3
\(\frac{12+y}{30}= \frac{2}{3}\)
y= 8 litres
Ques 8: 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 (CAT 2020 slot 1)
- 70
- 84
- 96
- 48
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Ans: (B)
Let the weights of same volume of A, B and C are 5kg, 2kg and 6 kg respectively.
After mixing of A, B and C in proportion of 3:4:7
Let 3, 4 and 7 litres of A, B and C are mixed respectively.
Resultant weight of 14 litres of solution= 3 x 5 + 4 x 2 + 7 x 6 = 65 kgs
42 kgs of C [7 x 6] is present in 65 kgs of solution.
So, in 130 kgs, amount C = 42 x 2 = 84 kgs.
Ques 9: 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 (CAT 2020 slot 3)
- 92%
- 90%
- 89%
- 94%
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Ans: (A)
Assuming A is 10 litres and B is 30 litres, are mixed
Total volume = 10 + 30 = 40 litres
Resultant A after adding of 40 litres of the same solution = 10 + 40 = 50
Concentration of alcohol in the mixture is 72%.
Concentration of alcohol in A = 60 %
Assuming the concentration of alcohol in B is X%
80 x 72 = 50 x 60 + 30 x X%
X= 92%
Ques 10: 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 (CAT 2019 slot 2)
- 15
- 12
- 13
- 14
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Ans: (14)
Strength of A in the solution = 10% x 500 = 50
Strength of B in the solution = 22% x 500 = 110
Strength of C in the solution = 32% x 500 = 160
After transfer of 100 ml from A to B.
Resultant A = 400ml and B = 600ml
Amount of salt transferred to B from A = (Initial salt amount) / 5 = 10 grams
Total salt in B = 110 +10 = 120 gms
Total salt in A= 50 -10 = 40 gms
Transfer 2 : - 100 ml from B is transferred to C
Total salt in C = 160 + (1/6)x120 =180 gms
Total salt in B = 120 – (1/6)x120 = 100 gms
Transfer 3:- 100ml from C to A
Total salt in A = 40 + (1/6) x 180 = 70 gms
Total salt in C = 160 – (1/6) x 180 = 130 gms
Salt concentration in. Vessel A = 70/500 = 14%
Ques 11: 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 (CAT 2018 slot 1)
- 17:25
- 18:25
- 21:25
- 19:24
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Ans: (D)
Let cost of tea A and B be a and b respectively.
When 3 kg of tea A is mixed with 2 kg of tea B,
(3a + 2b) x 1.1 = 40x 5 = 200 (i)
When 2 kg of tea A is mixed with 3 kg of tea B,
(2a + 3b) x 1.05 = 40x 5 = 200 (ii)
Equating Eq. (i) and (ii)
(3a + 2b) × 1.1 = (2a + 3b) x 1.05
3a + 2.2b = 2.1a + 3.15b
a/b = 19/24
a:b = 19 : 24
Ques 12: 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 lite 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 (CAT 2018 slot 1)
- 20
- 26
- 16
- 22
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Ans: (A)
Let the x be the no of litres of A in the mixture of A and B.
Cost of paint B is y
Cost of paint A is (y+8)
Based on the question;
x \( \geq\)(10 – x)
x \( \geq\) 5
Cost Price of the mixture at a profit of 10 % = \(\frac{264}{1+0.1}\)= Rs. 240
Therefore,
muItiplying Quantity with price,
((y+8) . x) + ((10-x).y) = 240
10y+8x = 240
y= 24- 0.8x
At minimum x, y will be maximum.
So,
Y= 24 – 0.8(0.5) = Rs 20
Ques 13: 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 (CAT 2018 slot 2)
- 25.4
- 20.5
- 35.2
- 30.3
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Ans: (C)
Ratio of water to alcohol = 175/700 = 1:4
So the initial concentration of alcohol is 80% and that of water is 20%
After removal of 10% from mixture and replacing with water
Final concentration of alcohol = 80 x (1- \(\frac{1}{10}\) )= 64.8%
Concentration of water in the mixture = 100- 64.8 = 35.2%
Ques 14: 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? (CAT 2018 slot 2)
- 1:3
- 1:4
- 2:5
- 3:10
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Ans: (A)
When the salt concentration is 20%, ratio between A, B and C is 1:2:3
Resultant equation will be;
\(\frac{a+2b+3c}{1+2+3}= 20\)
a+2b+3c = 120 ……………. (i)
When the salt concentration is 30%, ratio between A, B and C is 3:2:1
Resultant equation will be:
\(\frac{3a+2b+c}{3+2+1}= 30\)
3a+2b+c = 120 ……………. (ii)
Multiplying Eq (ii) with 3 and (ii) with 2 and equating both
3a + 6b + 9c = 6a + 4b + 2c
= 26 + 70 = 3a
Concentration of D = \(\frac{2b+7c}{9}= \frac{3a}{9}= \frac{a}{3}\)
Concentration of D to A = \(\frac{a/3}{a}= \frac{1}{3}\)
Ques 15: 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 (CAT 2018 slot 2)
- 220 : 149
- 251: 163
- 229: 141
- 239 : 161
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Ans: (D)
Let the mixture in drum 1 is 300litres and drum 1 is 400 litres
Total volume = 300 + 400 = 700 litres
Volume of A in final mixture = (13/20)x700 = 455 litres
Volume of B in final mixture= (7/20)x700 = 245 litres
Paint A in Drum 1= (18/25)x300 = 216 litres
Paint B in Drum 1 = 300-216 = 84 litres
Paint A in drum 2 = 455 – 216 = 239 litres
Paint B in drum 2 = 245 – 84 = 161 litres
Ratio of A:B = 239:161
How to Prepare Mixtures & Alligation Questions for CAT?
- Start with the weighted average and alligation formula for easy calculation.
- After being familiar with the two types of approaches start attempting questions.
- There will be few questions which will be solved faster by alligation or graphical methods faster.
After solving enough questions, try to club the concept of mixtures and alligations with profit, loss and discount for preparing for practical based que
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