Explanation / Important formulas:
Alligation: It is the rule that enables us to find the ratio in which two or more ingredients at the given price must be mixed to produce a mixture of desired price.
Mean Price: The cost of a unit quantity of the mixture is called the mean price.
Rule of Alligation: If two ingredients are mixed, then:
(Quantity of cheaper/Quantity of dearer) = (C.P of dearer – Mean price/Mean price – C.P of cheaper)
We present as under:
 C.P. of a unit quantity of cheaper (c)
 C.P of unit quantity (d)
 Mean price (m)
 (Cheaper quantity) : (Dearer quantity) = (d–m) : (m – c)
Suppose a container contains x of liquid from which y units are taken out and replaced by water. After n operations, the quantity of pure liquid = [x(1y/x)^{n}] units
More information: https://en.wikipedia.org/wiki/Alligation
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Test  Mixture and alligation
Mixture and Alligation  Question and Answers
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Question 1

A

16:5

B

14:5

C

16:7

D

19:5

Quantity of alcohol in vessel Q = 87.5/100 * 4 = 7/2 litres
Quantity of alcohol in the mixture formed = 5/4 + 7/2 = 19/4 = 4.75 litres
As 6 litres of mixture is formed, ratio of alcohol and water in the mixture formed = 4.75 : 1.25 = 19 : 5
Question 2

A vessel of capacity 90 litres is fully filled with pure milk. Nine litres of milk is removed from the vessel and replaced with water. Nine litres of the solution thus formed is removed and replaced with water. Find the quantity of pure milk in the final milk solution?
A

72

B

72.9

C

73.8

D

74.7

Let us say y litres of the mixture is taken out and replaced by water for n times, alternatively
Quantity of milk finally in the vessel is then given by [(T y) /T]n * T
For the given problem, T = 90, y = 9 and n = 2
Hence, quantity of milk finally in the vessel
= [(90  9) /90]2 (90) = 72.9 litres
Question 3

In a mixture of milk and water, the proportion of milk by weight was 80%. If, in a 180 gm mixture, 36 gms of pure milk is added, what would be the percentage of milk in the mixture formed?
A

83.33%

B

100%

C

84%

D

87.5%

Question 4

In a can, there is a mixture of milk and water in the ratio 4:5. If it is filled with an additional 8 litres of milk the can would be full and ratio of milk and water would become 6:5. Find the capacity of the can?
A

40

B

44

C

48

D

52

Quantity of milk in the mixture before adding milk = 4/9 (T  8)
After adding milk, quantity of milk in the mixture = 6/11 T
6T/11  8 = 4/9(T  8)
10T = 792  352 => T = 44
Question 5

A

5:6

B

3:4

C

7:8

D

8:9

By the rule of allegation, x/y = (87.5  7.50) / (7.50  6) = 5/6
Question 6

A

2

B

8

C

4

D

5

After adding water, milk would form 87 1/2% of the mixture
Hence, if quantity of mixture after adding x liters of water, (87 1/2) / 100 x = 63 => x = 72
Hence 72  70 = 2 litres of water must be added
Question 7

All the water in container A which was filled to its brim was poured into two containers B and C. The quantity of water in container B was 62.5% less than the capacity of container A. If 148 liters was now transferred from C to B, then both the containers would have equal quantities of water. What was the initial quantity of water in container A?
A

648

B

888

C

928

D

1184

Quantity of water in B = 8k  5k = 3k
Quantity of water in container C = 8k  3k = 5k
Container: A B C
Quantity of water: 8k 3k 5k
It is given that if 148 liters was transferred from container C to container B, then both the containers would have equal quantities of water
5k  148 = 3k + 148 => 2k = 296 => k = 148
The initial quantity of water in A = 8k = 8 * 148 = 1184 litres
Question 8

A

7 liters

B

15 liters

C

10 liters

D

9 liters

P liters of water added to the mixture to make water 25% of the new mixture
Total amount of water becomes (30 + P) and total volume of mixture is (150 + P)
(30 + P) = 25/100 * (150 + P)
120 + 4P = 150 + P => P = 10 liters
Question 9

A

9:1

B

4:7

C

7:1

D

2:5

If 10 liters of mixture are removed, amount of milk removed = 6 liters and amount of water removed = 4 liters
Remaining milk = 12  6 = 6 liters
Remaining water = 8  4 = 4 liters
10 liters of pure milk are added, therefore total milk = (6 + 10) = 16 liters
The ratio of milk and water in the new mixture = 16:4 = 4:1
If the process is repeated one more time and 10 liters of the mixture are removed, then amount of milk removed = 4/5 * 10 = 8 liters
Amount of water removed = 2 liters
Remaining milk = (16  8) = 8 liters
Remaining water = (4 2) = 2 liters
The required ratio of milk and water in the final mixture obtained = (8 + 10):2 = 18:2 = 9:1
Question 10

A

1:3

B

3:1

C

1:2

D

2:1

Mixture cost:Rs. 20/kg = (20  18)
2nd variety Rs. 24/kg
Mixture cost Rs. 20/kg = (24  20)
=> 4:2 = 2:1
Question 11

A

120 liters

B

180 liters

C

110 liters

D

160 liters

Mixture cost : Rs. 52/ltr = (52  40)
2nd Variety Rs.60/ltr
Mixture cost : Rs. 52/ltr = (60  52)
= 8:12 = 2:3
The two varities of oil should be mixted in the ratio 2:3. So, if 240 liters of the 2nd variety are taken, then the 1st variety should be taken as 160 liters
Question 12

A

Rs. 13.50

B

Rs. 14.50

C

Rs. 15.50

D

Rs. 16.50

Cost of 3x kg of A = 9(3x) = Rs. 27x
Cost of 7x kg of B = 15(7x) = Rs. 105x
Cost of 10x kg of the mixture = 27x + 105x = Rs. 132x
Cost of 5 kg of the mixture = 132x/10x (5) = Rs. 66
Profit made in selling 5 kg of the mixture = 25/100
(cost of 5 kg of the mixture) = 25/100 * 66 = Rs. 16.50
Question 13

A

1/3

B

1/4

C

1/5

D

1/7

Let x litres of this liquid be replaced with water.
Quantity of water in new mixture = (33x/8 +x) litres
Quantity of syrup in new mixture = (55x/8) litres
(3  3x/8 + x) = (55x/8)x
5x + 24 = 40  5x 10x = 16
x = 8/5
So, part of the mixture replaced = (8/5 x 1/8) = 1/5
Question 14

A

5:6

B

3:4

C

7:8

D

8:9

By the rule of allegation, x/y = (87.5  7.50) / (7.50  6) = 5/6
Question 15

A

Rs. 169.50

B

Rs. 170

C

Rs. 175.50

D

Rs. 180

So, their average price = Rs. (126 + 135)/2 = Rs. 130.50
So, the mixture is formed by mixing two varieties, one at Rs. 130.50 per kg and the other at say, Rs. x per kg in the ratio 2 : 2, i.e., 1:1
We have to find x
By the rule of allegation, we have:
Cost of 1 kg of 1st kind Cost of 1 kg tea of 2nd kind
Rs. 130.50 Mean Price
Rs. 153 Rs. x
(x  153) 22.50
x  153 = 1
22.50
x  153 = 22.50 x = 175.50
Question 16

A

10

B

20

C

21

D

25

Quantity of A in mixture left = (7x  7/12x 9)litres = (7x  21/4) litres
Quantity of B in mixture left = (5x  5/12x 9)litres = (5x  15/4) litres.
(7x  21/4) / ((5x  15/4) + 9) = 7/9
(28x  21) / (20x + 21) = 7/9
252x  189 = 140x + 147 112x = 336
x = 3. So, the can contained 21 litres of A.
Question 17

A

3 : 7

B

5 : 7

C

7 : 3

D

7 : 5

Cost of 1 kg pulses of 1st kind Cost of 1 kg pulses of 2nd kind
Rs. 15 Mean Price
Rs. 16.50 Rs. 20
3.50 1.50
Required rate = 3.50 : 1.50 = 7 : 3
Question 18

A

4%

B

6 1/4%

C

20%

D

25%

C.P. of 1 litre mixture = Re. ((100 x1) /125 = 4/5
By the rule of alligation, we have:
C.P. of 1 litre of milk C.P. of 1 litre of water
Re. 1 4/5 Mean Price Re. 4/5 0 1/5
Ratio of milk to water = 4/5 : 1/5 = 4 :1
5 5
Hence, percentage of water in the mixture = (1x 100) % = 20%
Question 19

A

26.34 litres

B

27.36 litres

C

28 litres

D

28 litres

= (40 x 9/10 x 9/10 x 9/10)
= 29.16 litres.
Question 20

A

1/3

B

2/3

C

2/5

D

3/5

Strength of first jar Strength of 2nd jar
40%  7
Mean Strength26%
19%  14
So, ratio of 1st and 2nd quantities = 7 : 14 = 1 : 2
Required quantity replaced = 2/3
Question 21

A

1:6

B

6:1

C

2:3

D

4:3

S.P. of 1 litre of mixture = Re.1, Gain = 50/3 %
C.P. of 1 litre of mixture = (100 x 3/350 x 1) = 6/7
By the rule of allegation, we have:
C.P. of 1 litre of water C.P. of 1 litre of milk
0
1
7
Mean Price
Re. 6/7
Re. 1
6/7
Ratio of water and milk = 1:6 = 1:6.
Question 22

A

18 litres

B

24 litres

C

32 litres

D

42 litres

Then, quantity of wine left in cask after 4 operations = [x(1  8/x) to the power 4] litres
(x(1 (8/x)to the power 4/x) = 16/81
(18/x)to the power 4 = (2/3) to the power 4
(x8)/x = 2/3
3x24 = 2x
x = 24
Question 23

A

1.0 litres

B

1.5 litres

C

2.0 litres

D

4.0 litres

That is 4.8 litres of milk and 7.2 litres of water
Now we are replacing the mixture with pure milk so that the amount of milk and water in the mixture is 50% and 50%
That is we will end up with 6 litres of milk and 6 litres of water
Water gets reduced by 1.2 litres
To remove 1.2 litres of water from the original mixture containing 60% water, we need to remove 1.20.61.20.6 litres of the mixture = 2 litres
Question 24

A

30

B

40

C

50

D

60

Then the number of pigeons = 80 − x
Each pigeon has 2 legs and each horse has 4 legs
Therefore, total number of legs = 4x + 2(80−x) = 260
⇒ 4x + 160 − 2x = 260
⇒ 2x = 100
⇒ x = 50
Question 25

A

26.34 litres

B

27.36 litres

C

28 litres

D

29.16 litres

[40(1)]
[40(1(4/40)^{3})] = (40 x 9/10 x 9/10 x 9/10) = 29.16 litres
Question 26

A

20%

B

16.66%

C

26.66%

D

8.33%

99 x (x+99)
Petrol kerosene Total volume
99 (x198) (x99)
now; [99/(x99)*100][99/(x = 99)*100] = 13.33
by simplifying [9900*198/(x^299^2)] = 40/3
x^299^2 = 147015 147015 can be written as (99^2*15)
Therefore x^2 = 99^2*16
x = 396
now; [ 99/(99+396)] *100 = 20%
Question 27

A

1 ml

B

6 ml

C

3 ml

D

9 ml

50/100*9 = 30/100*x
x = 15
since it is reduced by 9ml we have
15ml  9ml = 6ml
Question 28

A

81 kg

B

121 kg

C

98 kg

D

102 kg

Then type1 in 729 kg of mixture = (7/9 x 729 ) kg = 567 kg
And type2 in 729 kg of mixture = 729 – 567 = 162 kg
Let X be the quantity of type2 added to new mixture with the ratio 7:3
Quantity of type2 in the new mixture = (162 + X ) kg
Then 7 / 3 = 567 / (X + 162)
7(162 + X) = 3(567)
1134 + 7X = 1701
7X = 1701  1134
7X = 1134
X = 567/7 = 81 kg
Quantity of type2 added to new mixture = 81 kg
Question 29

A

2:1

B

5:1

C

99:61

D

227:133

Proportion of water in 3 samples is 1/3, 2/5, 3/8
Since equal quantities are taken,
Total proportion of milk is 2/3 + 3/5 + 5/8 = 227/120
Total proportion of water is 1/3 + 2/5 + 3/8 = 133/120
Proportion of milk and water in the solution is = 227:133
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