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Here you can download CMAT 2022 – important Allegation and Mixture Questions PDF by Cracku. Very Important Allegation and Mixture Questions for CMAT 2022 based on asked questions in previous exam papers. These questions will help your CMAT preparation. So kindly download the PDF for reference and do more practice.

Question 1:Â Two varieties of dal – Rs 30 per kg and Rs 48 per kg – are mixed in a particular ratio, such that when the shopkeeper sells 10 kg of the new variety of dal for Rs 410 in total, he makes a profit of Rs 50. What is the ratio of the two varieties of dal in the mixture?

a)Â 5 : 3

b)Â 3:2

c)Â 2:1

d)Â None of the above

Instructions

The following pie â€“ charts provide information about the percentage composition of two alloys â€“ A and B.

Question 2:Â If the weight of aluminium and tin in certain mixture of sample of alloy A and alloy B is the same, then what is the ratio of weight of the same samples of alloy A and alloy B?

a)Â 2:1

b)Â 1:2

c)Â 2:3

d)Â 3:1

Question 3:Â In what ratio must a person mix three varieties of rice which have a price of Rs 5, 8 and 12 such that the price of the resultant mixture is Rs 10 per kg?

a)Â 2:1:6

b)Â 1:2:3

c)Â 2:2:7

d)Â More than one of the above

Question 4: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 liquid 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 27:10

d)Â 27:19

Question 5:Â A 5 kg alloy made of zinc and copper in the ratio 1:3 is mixed with 3 kg of another alloy of zinc and copper such that in the resultant alloyÂ the ratio of zinc and copper becomes 25:39. Find the ratio of zinc and copper in the second alloy.

a)Â 3:1

b)Â 4:3

c)Â 5:4

d)Â 5:3

Question 6:Â A shopkeeper mixes two types of sugar and sells it at the cost price of costlier sugar. If he mixes both sugars in ratio 3:4, he gets a profit of 28%. If mixes it in a ratio 5:2, then he gets a profit of x%. Find the value of x.

Question 7:Â A mixture of 150 L of wine and water contains 20% water. How many litres of water must be added so that water may be 25% of the mixture?

a)Â 12 L

b)Â 8 L

c)Â 10 L

d)Â 6 L

e)Â None of these

Question 8:Â Two types of rice (type 1 and type 2) were mixed in the respective ratio of 1 : 3. The mixture was then sold @ 75.60 per kg to gain a profit of 20%. If the price of type 1 rice is Rs. 75 per kg, what is the price of type 2 price per kg?

a)Â Rs. 55

b)Â Rs. 53

c)Â Rs. 59

d)Â Rs. 57

e)Â Rs. 62

Question 9:Â A vessel contains a mixture of Grape, Pineapple and Banana juices in the respective ratio of 4 : 6 : 5. 15 litres of this mixture is taken out and 8 litres of grape juice and 2 litres of pineapple juice is added to the vessel. If the resultant quantity of grape juice is 10 litres less than the resultant quantity of pineapple juice. what was the initial quantity of mixture in the vessel ? (in litres)

a)Â 120

b)Â 150

c)Â 105

d)Â 135

e)Â 90

Question 10:Â A vessel contains 64 litres of mixture of milk and water in the ratio 7 : 3 respectively. 8 litres of mixture is replaced by 8 litres of milk. What is the ratio of milk and water in the resulting mixture ?

a)Â 59 : 21

b)Â 35 : 22

c)Â 64 : 23

d)Â 65 : 21

e)Â None of these

The shopkeeper makes a profit of Rs 50. So, the cost price of 10 kg of rice = 410 – 50 = Rs 360

So, CP of 1 kg of dal = Rs 36

The required ratio of the two varieties of dal = (48-36) : (36-30) = 12 : 6 = 2 : 1

Let the weight of the alloy A and alloy B that must be mixed be a and b respectively.
As per the information given in the question: 0.2a + 0.25b = 0.3a + 0.2b => 0.05b = 0.1a
Required ratio = 1:2.

Assume a:b:c is the ratio.
Possible pairs of ingredients which can give mixture of price 10.
(5,12), using alligation, the ratio = 2:5 = 2x:5x
(8,12) using alligation, the ratio = 1:1= y:y
Now, a:b:c = 2x:y:5x+y
If x=1 and y=1, we get the option a.
and if x=1 and y=1, we get the option c.

let the amount of milk and water in first solution be 7x,2x and in bottle 2 be 9y, 4y

So, (7x+9y)/(2x+4y)=3/1
7x+9y=6x+12y
x=3y
.’. Ratio= 9x/13y
=27:13

In 5 kg of alloy, the amount of zinc =Â $\ \frac{\ 5\times\ 1}{1+3}=\frac{\ 5}{4}$
In 5 kg of alloy, the amount of copper =Â $\ \frac{\ 5\times3}{1+3}=\frac{\ 15}{4}$
Total amount of the final mixture = 5+3 = 8 kg
In the final mixture, the amount of zinc = 8*25/(25+39)$\ \frac{\ 8\times25}{25+39}=\frac{\ 25}{8}$
The amount of copper = 8 –Â $\frac{\ 25}{8}$ =Â $\frac{\ 39}{8}$
Now the amount of zinc in the second alloy = $\frac{25}{8}-\frac{5}{4}$ =Â $\frac{15}{8}$
The amount of copper in the second alloy = $\frac{39}{8}-\frac{15}{4}$Â =Â $\frac{9}{8}$
The required ratio =Â $\ \dfrac{\ \ \frac{\ 15}{8}}{\frac{9}{8}}=\ \frac{\ 5}{3}$

Assume the cost price of costlier sugar is a/kg and the cost price of cheaper sugar is b/kg. Also, consider the total amount as 7kg.

So in the first mixture, the cost price = 3a+4b

Then his profit = 7a-3a-4b = 4a-4b

The profit percentage =Â $\ \frac{\ 4a-4b}{7a}$ =Â $\ \ \frac{\ 28}{100}$

In the second mixture, the cost price = 5a+2b

Then his profit = 7a-5a-2b = 2a-2b

The profit percentage = $\ \frac{\ 2a-2b}{7a}$ = $\ \ \frac{\ 14}{100}$

The mixture contains 20% water and 80% of wine.
By the rule of allegation, Ratio of wine : water = 75 : 5 = 15 : 1
For 15 L wine, 1 L water is to be added.
âˆ´ For 150 L wine,$\frac{(150*1)}{15}=10$ L to be added.

Profit = 0.2 CP
Profit = SP – CP
1.2CP = SP
Hence, CP = (SP/1.2) = 75.6/1.2 = 63
Now, let the cost of type 1 rice is T1 and cost of type 2 be T2.
63 = $\frac{T1 + 3T2}{4}$
= (75 + 3*x)/4
x = 59
Therefore, cost of type 2 price is 59.

let the amount of grape juice ,pineapple juice and banana juice in vessel be 4y ,6y,5y respectively

Now when we removed 15 ltr from vessel the juice will be removed in their given ratio i.e 4 ltr of grape juice will be removed and 6 ltrs of pineapple will be removed and 5 ltrs of banana juice will be removed and hence new quantities are

Grape juice = 4y-4

Pineapple juuce = 6y- 6

Banana juice = 5y- 5

Niw 8 ltrs of grape juuce is added and 2 ltrs of pineapple juice is added so new quantities of Juices in vessel are

Grape juice = 4y+4

Pineapple juuce = 6y-4

It is given that grape juice amount is 10 ltrs less than pineapple juice quantity .

So

6y-4 – 4y-4 = 10

2y= 18

y= 9

Initial quantity in vessel = 15 y = 15Ã—9=135 ltrs

Solution of milk and water in vessel = 64 litres

ration of Milk:Water = 7:3

using

$\frac{water concentration final}{total} = \frac{initial water conc.}{total}(1- \frac{removed volume}{total})^n$

$\frac{water concentration final}{total} = \frac{3}{10}(1- \frac{8}{64})^1$

$\frac{water concentration final}{total} = \frac{3}{10}(1- \frac{1}{8})^n$

$\frac{water concentration final}{total} = \frac{21}{80}$

water : milk in new solution after replacement = 21: 59