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# SNAP Mixture and Alligation Questions PDF

Mixture and Alligation is an important topic in the Quant section of the SNAP Exam. You can also download this Free Mixture and Alligation Questions for SNAP PDF (with answers) by Cracku. These questions will help you to practice and solve the Mixture and Alligation questions in the SNAP exam. Utilize this PDF practice set, which is one of the best sources for practising.

Question 1:Â A vessel contains a mixture of milk and water in the respective ratio of 14 : 3. 25.5 litres of the mixture is taken out from the vessel and 2.5 litres of pure water and 5 litres of pure milk is added to the mixture. If the resultant mixture contains 20% water, what was the initial quantity of mixture in the vessel before the replacement? (in litres)

a)Â 51

b)Â 102

c)Â 68

d)Â 85

e)Â 34

Solution:

Let the total quantity of mixture in the vessel initially = $17x$ litres

=> Quantity of milk = $\frac{14}{17} \times 17x = 14x$ litres

Quantity of water = $17x – 14x = 3x$ litres

Acc. to ques,

=> $\frac{14x – (\frac{14}{17} \times 25.5) + 5}{3x – (\frac{3}{17} \times 25.5) + 2.5} = \frac{80}{20}$

=> $\frac{14x – 21 + 5}{3x – 4.5 + 2.5} = \frac{4}{1}$

=> $\frac{14x – 16}{3x – 2} = \frac{4}{1}$

=> $14x – 16 = 12x – 8$

=> $14x – 12x = 16 – 8$

=> $x = \frac{8}{2} = 4$

$\therefore$ Initial quantity of mixture in the vessel before the replacement = $17 \times 4 = 68$ litres

Question 2:Â A vessel contains 100 litres mixture of milk and water in the respective ratio of 22 : 3. 40 litres of the mixture is taken out from the vessel and 4.8 litres of pure milk and pure water each is added to the mixture. By what percent is the quantity of water in the final mixture less than the quantity of milk?

a)Â 78${1 \over 2}$

b)Â 79${1 \over 6}$

c)Â 72${5 \over 6}$

d)Â 76

e)Â 77${1 \over 2}$

Solution:

Quantity of milk in vessel = $\frac{22}{25} \times 100 = 88$ litres

=> Quantity of water = $100 – 88 = 12$ litres

40 litres of the mixture is taken out, i.e., $\frac{40}{100} = (\frac{2}{5})^{th}$

=> Milk left = $88 – \frac{2}{5} \times 88 = 52.8$ litres

Water left =Â $12 – \frac{2}{5} \times 12 = 7.2$ litres

Now, 4.8 lires of milk and water are added.

=> Quantity of milk in the vessel = 52.8 + 4.8 = 57.6 litres

Quantity of water in the vessel = 7.2 + 4.8 = 12 litres

$\therefore$ Required % = $\frac{57.6 – 12}{57.6} \times 100$

= $\frac{475}{6} = 79 \frac{1}{6} \%$

Question 3:Â Jar A has 60 litres of mixture of milk and water in the respective ratio of 2 : 1. Jar B which had 40 litres of mixture of milk and water was emptied into jar A, as a result in jar A, the respective ratio of milk and water became 13 : 7. What was the quantity of water in jar B?

a)Â 8 litres

b)Â 15 litres

c)Â 22 litres

d)Â 7 litres

e)Â 1 litre

Solution:

Jar A has 60 litres of mixture of milk and water in the respective ratio of 2 : 1

=> Quantity of milk in Jar A = $\frac{2}{3} \times 60 = 40$ litres

Quantity of water in Jar A = $60 – 40 = 20$ itres

Let quantity of water in Jar B = $x$ litres

=> Quantity of milk in Jar B = $(40 – x)$ litres

Acc. to ques, => $\frac{40 + (40 – x)}{20 + x} = \frac{13}{7}$

=> $560 – 7x = 260 + 13x$

=> $13x + 7x = 560 – 260$

=> $20x = 300$

=> $x = \frac{300}{20} = 15$ litres

Question 4:Â Jar A has 36 litres of mixture of milk and water in the respective ratio of 5 : 4. Jar B which had 20 litres of mixture of milk and water, was emptied into jar A, and as a result in jar A, the respective ratio of milk and water becomes 5: 3. What was the quantity of water in jar B?

a)Â 5 litres

b)Â 3 litres

c)Â 8 litres

d)Â 2 litres

e)Â 1 litre

Solution:

Jar A has 36 litres of mixture of milk and water in the respective ratio of 5 : 4

=> Quantity of milk in Jar A = $\frac{5}{9} \times 36 = 20$ litres

Quantity of water in Jar A = $36 – 20 = 16$ itres

Let quantity of water in Jar B = $x$ litres

=> Quantity of milk in Jar B = $(20 – x)$ litres

Acc. to ques, => $\frac{20 + (20 – x)}{16 + x} = \frac{5}{3}$

=> $120 – 3x = 80 + 5x$

=> $5x + 3x = 120 – 80$

=> $8x = 40$

=> $x = \frac{40}{8} = 5$ litres

Question 5:Â A vessel contains 100 litres mixture of milk and water in the respective ratio of 22 : 3. 40 litres of the mixture is taken out from the vessel and 4.8 litres of pure milk and pure water each is added to the mixture. By what percent is the quantity of water in the final mixture less than the quantity of milk?

a)Â 78${1 \over 2}$

b)Â 79${1 \over 6}$

c)Â 72${5 \over 6}$

d)Â 76

e)Â 77${1 \over 2}$

Solution:

Quantity of milk in vessel = $\frac{22}{25} \times 100 = 88$ litres

=> Quantity of water = $100 – 88 = 12$ litres

40 litres of the mixture is taken out, i.e., $\frac{40}{100} = (\frac{2}{5})^{th}$

=> Milk left = $88 – \frac{2}{5} \times 88 = 52.8$ litres

Water left =Â $12 – \frac{2}{5} \times 12 = 7.2$ litres

Now, 4.8 lires of milk and water are added.

=> Quantity of milk in the vessel = 52.8 + 4.8 = 57.6 litres

Quantity of water in the vessel = 7.2 + 4.8 = 12 litres

$\therefore$ Required % = $\frac{57.6 – 12}{57.6} \times 100$

= $\frac{475}{6} = 79 \frac{1}{6} \%$

Question 6:Â Jar A has 60 litres of mixture of milk and water in the respective ratio of 2 : 1. Jar B which had 40 litres of mixture of milk and water was emptied into jar A, as a result in jar A, the respective ratio of milk and water became 13 : 7. What was the quantity of water in jar B?

a)Â 8 litres

b)Â 15 litres

c)Â 22 litres

d)Â 7 litres

e)Â 1 litre

Solution:

Jar A has 60 litres of mixture of milk and water in the respective ratio of 2 : 1

=> Quantity of milk in Jar A = $\frac{2}{3} \times 60 = 40$ litres

Quantity of water in Jar A = $60 – 40 = 20$ itres

Let quantity of water in Jar B = $x$ litres

=> Quantity of milk in Jar B = $(40 – x)$ litres

Acc. to ques, => $\frac{40 + (40 – x)}{20 + x} = \frac{13}{7}$

=> $560 – 7x = 260 + 13x$

=> $13x + 7x = 560 – 260$

=> $20x = 300$

=> $x = \frac{300}{20} = 15$ litres

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Question 7:Â Jar A has 36 litres of mixture of milk and water in the respective ratio of 5 : 4. Jar B which had 20 litres of mixture of milk and water, was emptied into jar A, and as a result in jar A, the respective ratio of milk and water becomes 5: 3. What was the quantity of water in jar B?

a)Â 5 litres

b)Â 3 litres

c)Â 8 litres

d)Â 2 litres

e)Â 1 litre

Solution:

Jar A has 36 litres of mixture of milk and water in the respective ratio of 5 : 4

=> Quantity of milk in Jar A = $\frac{5}{9} \times 36 = 20$ litres

Quantity of water in Jar A = $36 – 20 = 16$ itres

Let quantity of water in Jar B = $x$ litres

=> Quantity of milk in Jar B = $(20 – x)$ litres

Acc. to ques, => $\frac{20 + (20 – x)}{16 + x} = \frac{5}{3}$

=> $120 – 3x = 80 + 5x$

=> $5x + 3x = 120 – 80$

=> $8x = 40$

=> $x = \frac{40}{8} = 5$ litres

Question 8:Â Jar A contains 78 litres of milk and water in the respective ratio of 6 : 7. 26 litres of the mixture was taken out from Jar A. What quantity of milk should be added to jarA, so that water constitutes 40% of the resultant mixture in jar A?

a)Â 8 litres

b)Â 36 litres

c)Â 12 litres

d)Â 14 litres

e)Â 18 litres

Solution:

Jar A has 78 litres of mixture of milk and water in the respective ratio of 6 : 7

=> Quantity of milk in Jar A = $\frac{6}{13} \times 78 = 36$ litres

Quantity of water in Jar A = $78 – 36 = 42$ litres

26 litres of the mixture was taken out from Jar A, i.e., $\frac{26}{78} = (\frac{1}{3})^{rd}$

=> Milk left = $36 – \frac{1}{3} \times 36 = 24$

Water leftÂ = $42 – \frac{1}{3} \times 42 = 28$

Let milk added to jar A = $x$ litres

Acc. to ques, => $\frac{24 + x}{28} = \frac{60}{40}$

=> $\frac{24 + x}{28} = \frac{3}{2}$

=> $48 + 2x = 84$

=> $2x = 84 – 48 = 36$

=> $x = \frac{36}{2} = 18$ litres

Question 9:Â A vessel contains 60 litres of milk. 6 litres of milk is taken out and 6 litres of water is added to the vessel. Again 6 litres of mixture from the vessel is withdrawn and 6 litres of water is added to the vessel. The ratio of milk and water in the resulting mixture in the vessel is

a)Â 81 : 19

b)Â 71 : 29

c)Â 61 : 39

d)Â 61 : 29

e)Â None of these

Solution:

If we are taking out 6 litres out of a 60 litre solution and replacing it with water,

=> We are replacing $\frac{1}{10}$th of the solution with water

=> Fraction of milk will become $\frac{9}{10}$th of original.

If the process is repeated ‘n’ times, fraction of milk will become $(\frac{9}{10})^n$ of the original.

Here, the process is done twice.

=> Final quantity of milk = $(\frac{9}{10})^2 \times 60$ = 48.6 litres

and Final quantity of water = 60 – 48.6 = 11.4 litres

$\therefore$ Required ratio = $\frac{48.6}{11.4}$ = 81Â : 19

Question 10:Â The milk and water in two vessels A and B are in the ratio 4 : 3 and 2 : 3 respectively. In what ratio the liquids in both the vessels be mixed to obtain a new mixture in vessel C consisting half milk and half water?

a)Â 8 : 3

b)Â 7 : 5

c)Â 4 : 3

d)Â 2 : 3

e)Â None of these

Solution:

Let mixture in vessel A = $x$ ml

and mixture in vessel B = $y$ ml

=> Milk in vesselÂ A = $\frac{4x}{7}$

Milk in vessel B = $\frac{2y}{5}$

Acc to ques,

=> $\frac{4x}{7} + \frac{2y}{5} = \frac{1}{2} (x + y)$

=> $\frac{4x}{7} – \frac{x}{2} = \frac{y}{2} – \frac{2y}{5}$

=> $\frac{x}{14} = \frac{y}{10}$

=> $\frac{x}{y} = \frac{14}{10} = \frac{7}{5}$

Question 11:Â Jar A contains ‘X’ litre of pure milk only. A 27 litre mixture of milk and water in the respective ratio of 4 : 5, is added to jar A. The new mixture thus formed in jar A contains 70% milk, what is the value of X ?

a)Â 23

b)Â 30

c)Â 27

d)Â 48

e)Â 28

Solution:

Quantity of milk in 27 litre mixture = $\frac{4}{4 + 5} \times 27 = 12$ litre

Quantity of water = $27 – 12 = 15$ litre

Ratio of milk and water in the new mixture = $70 : 30 = 7 : 3$

Acc to ques,

=> $\frac{X + 12}{15} = \frac{7}{3}$

=> $3X + 36 = 15 \times 7 = 105$

=> $3X = 105 – 36 = 69$

=> $X = \frac{69}{3} = 23$ litre

Question 12:Â In a vessel there is 40 litres mixture of milk and water. There is 15% water in the mixture. The milkman sells 10 litres of mixture to a customer and thereafter adds 12.5 litres of water to the remaining mixture. What is the respective ratio of milk and water in the new mixture ?

a)Â 2 : 3

b)Â 3 : 2

c)Â 3 : 4

d)Â 4 : 3

e)Â None of these

Solution:

Mixture remaining after selling 10 litres = 40 – 10 = 30 litres

Now, quantity of water in 30 litres of mixture = $\frac{15}{100} * 30$ = 4.5 litres

Milk = 30 – 4.5 = 25.5 litres

After adding 12.5 litres of water, total quantity of water = 12.5 + 4.5 = 17 litres

$\therefore$ Required ratio of milk and water = 25.5 : 17

= 1.5 : 1 = 3 : 2

Question 13:Â 18 litres of pure water was added to a vessel containing 80 litres of pure milk. 49 litres of the resultant mixture was then sold and some more quantity of pure milk and pure water was added to the vessel in the respective ratio of 2 : 1. If the resultant respective ratio of milk and water in the vessel was 4 : 1, what was the quantity of pure milk added in the vessel ? (in litres)

a)Â 4

b)Â 8

c)Â 10

d)Â 12

e)Â 2

Solution:

18 litres of pure water was added to a vessel containing 80 litres of pure milk.

Total mixture = 80 + 18 = 98 litres

Now, 49 litres i.e., $\frac{1}{2}$is removed, => Milk left = $\frac{80}{2}$ = 40 litres

WaterÂ left = $\frac{18}{2}$ =Â 9 litres

Let milk added be $2x$ litres and water added is $x$ litres

=> $\frac{40 + 2x}{9 + x} = \frac{4}{1}$

=> $40 + 2x = 36 + 4x$

=> $2x = 40 – 36 = 4$

=> $x = \frac{4}{2} = 2$

$\therefore$ Quantity of milk added = $2 \times 2 = 4$ litres

Question 14:Â A vessel contains a mixture of milk and water in the respective ratio of 10 : 3. Twenty-six litre of this mixture was taken out and replaced with 8 litre of water. If the resultant respective ratio of milk and water in the mixture was 5 : 2, what was the initial quantity of mixture in the vessel ? (in litre)

a)Â 143

b)Â 182

c)Â 169

d)Â 156

e)Â 130

Solution:

Let quantity of Milk and water be M and W respectively.
M : W =10 : 3
3M=10W
In 26 litre of mixture
M =26(10/13) = 20 litre and
W =26(3/13) = 6 litre
8 litre of water is added.
Resulting ratio of M and W is
M-20 : W-6+8 = 5 : 2
2(M-20) = 5(W+2)
2M – 40 = 5W + 10
Multiplyin all the terms by 2.
4M – 80 = 10 W + 20
Replacing 10W with 3M.
4M – 80 = 3M + 20
M = 100
Hence W would be 30.
Total quantity = 100+30 = 130.
Option E is the correect answer.

Question 15:Â In a 90 litres mixture of milk and water, percentage of water is only 30%. The milkman gave 18 litres of this mixture to a customer and then added 18 litres of water to the remaining mixture. What is the percentage of milk in the final mixture ?

a)Â 64

b)Â 48

c)Â 52

d)Â 68

e)Â 56

Solution:

In 90 liters of mixture,
Amount of water=$\frac{90\times30}{100}$.
=27 liters.
Amount of milk=$\frac{90\times70}{100}$.
=63 liters.
Similarly, in 18 liters of mixture,
Amount of water=$\frac{18\times30}{100}$.
=5.4 liters.
Amount of milk=$\frac{18\times70}{100}$.
=12.6 liters.
After removing 18 liters of solution,
Amount of water=27-5.4=21.6 liters.
Amount of milk=63-12.6=50.4 liters.
After adding 18 liters of water,
Amount of water in the solution=21.6+18=39.6 liters.
Hence, Percentage of milk in solution=$\frac{50.4}{50.4+39.6}\times 100$.
=56%.
Hence, Option E is correct.