4 Mixtures and Alligations

Introduction ⇒ x(Aw – A) = y(B – Aw)

The questions on mixtures and alligation

x B Aw
are repeatedly tested in CAT, XAT and other ⇒
y Aw A
management exams. One can expect one or
two questions from this topic. Generally, the [Known as Alligation equation]
questions are not asked directly from these
concepts, but the application of mixtures
and alligation can be utilised to solve dif- Key Points
ficult problems in a simple way. One must
consider the concepts of mixtures and alli- The structure of alligation can be
gation as a tool to ease the problem solving. used to find the ratio of the weights
associated with two groups.
Weighted Average and its relation with the
structure of alligation:
Note: Weighted average (Aw) of the two
When two groups of items are combined
groups lies between A and B, i.e.,
together, then we can talk about the aver-
A ≤ Aw ≤ B.
age of the entire group. This average of the
entire group is called weighted average. The Diagrammatically this can be represented as
weighted average is simply the more gener- below:
alized form of average.
Let us understand weightage average and its
linkage to alligation:
Let the average of group 1 be ‘A’ and the
number of people in group 1 be x and the
average of group 2 be ‘B’ and the number of
people in group 2 be y. Note: Here A and B are the averages and x
Here, we make an assumption that the av- and y are the weights associated with
erage of group 2 is more than the average of A and B respectively.
group 1, i.e., B > A.
The weighted average is given by Let us see how can we ease the
calculation while dealing with
Ax By
⇒ Aw the structure of alligation.
x y
If x = mp and y = np
By cross multiplication we get, If there is any common factor among ‘x’
⇒ x × Aw + y × Aw = Ax + By and ‘y’, then let’s see the final outcome as-
⇒ x × Aw – Ax = By – y × Aw sociated with this.

Mixtures and Alligations 119

 A × x B× y A × mp B × np Ax By
Aw By using weighted average: Aw =
x y mp np x y

p(A × m B × n) Where,
⇒
p(m n) Aw = Weighted Average = 15
Am Bn A = Average Weight of girls = 10
⇒
m n B = Average Weight of boys = 20
As one can see that ultimately the common x = Number of girls = a
factor among x and y gets cancelled so we y = Number of boys = 20
can utilize this concept to make the calcu- 10 × a 20 × 20
⇒ 15
lation faster. a 20
Let us understand a basic example to ease ⇒ 15a + 300 = 10a + 400
our calculation based on the above learning. ⇒ 5a = 100
a = 20
For Example:
Number of girls = 20
If the average weight of class A of 24 students
is 25 kg and that of class B of 36 students is
30 kg. Find the average of both the classes
when combined.
A = 25 kg, B = 30 kg, x = 24 students and
y = 36 students.
x : y = 24 : 36 = 2 : 3
A × x B× y
Aw Number of girls 1
x y
Number of boys 1
25 × 2 30 × 3
⇒
2 3 a
1 ⇒ a 20
140 20
⇒ 28
5 Example 2:
Example 1: If 20 kg of rice costing Rs. 16/kg is mixed
The average weight of girls in a class is 10 kg with 16 kg of the second variety of rice cost-
and the average weight of boys of the same ing Rs. 25/kg, then what is the average cost
class is 20 kg. The average weight of the per kg of the resulting mixture?
whole class is 15 kg. If the number of boys
are 20, then the number of girls are: Solution: 20/kg
(A) 5 (B) 10 By using weighted average.
(C) 15 (D) 20 Ax By
Aw =
Solution: (D) x y

Mixtures and Alligations 120

Where, 720
Aw =
A = Cost of first variety rice = 16/kg 36

B = Cost of second variety rice = 25/kg Aw = 20/kg

x = Quantity of first variety rice = 20 kg

Alligation
y = Quantity of second variety rice = 16 kg
16 × 20 25 × 16 Alligation is a method or a rule for the solu-
Aw =
20 16 tion of problems concerning the compound-
320 400 ing or mixing of ingredients differing in price
Aw ⇒
36 or quality. Recently we have learned the as-
720 sociation of weightage average with respect
Aw ⇒
36 to alligation.
Aw = Rs. 20/kg This rule connects quantities and prices in
a mixture.
x B Aw
Alternate Solution: [structure of Alligation]
y Aw A
x B Aw
By using the Alligation equation The representation of the structure of
y Aw A
20 25 Aw Alligation as follows:
⇒
16 Aw 16 Quantity of cheaper
Quantity of Dearer
20 Aw – 320 = 400 – 16 Aw
Dearer price Average price
36 Aw = 720
Average price Cheaper price

Pictorially this can be represented as below,

Mixtures and Alligations 121

Lets understand how the alligation can Example 4:
be applied using some examples: There is a 100-litre solution of milk and wa-
ter in which milk forms 60 %. How much wa-
Example 3:
ter must be added to this solution to make
How many kilograms of coffee costing it a solution in which milk forms 40 %?
Rs. 20/kg must be mixed with 20 kg of cof-
Solution: 50 litres
fee costing Rs. 15/kg, so that the resulting
We will apply the alligation rule here, taking
mixture cost Rs. 18/kg?
60 % concentration solution (of milk) mixed
Solution: 30 kg
with a 0 % concentration solution of pure
By using the rule of Alligation, water (100 % water) to give 40 % concentra-
Quantity of cheaper
tion solution.
Quantity of Dearer
Dearer price Average price
Average price Cheaper price
20 20 18
x 18 15
∴ x = 30 kg
Let the quantity of water added be x units
Alternate Solution: The ratio of the quantity of initial solution to
“By using the rule of alligation”. the quantity of water = 2 : 1
100 2
= ⇒ x = 50 litres
x 1

Application of Alligation
Alligation and its application are extensively
The ratio of the quantity of Dearer to the applied in solving the problem of ratio-
quantity of Cheaper = 3 : 2 proportion, S.I - C.I., Profit-Loss, Time-
Speed-Distance and some miscellaneous
Rack Your Brain topics as well.

Two solution of acid (Acid + Water) 1. Alligation in Ratio

with concentration of 70% and 80%
Example 5:
respectively are mixed in a certain ratio
The cost of type 1 rice is Rs. 22 per kg and
to get a 64% acid solution. This solution
type 2 rice is Rs. 18 per kg. If both type 1 and
is mixed with 10L of water to get back
type 2 are mixed in the ratio of 3 : 1, then the
50% acid solution. How much of the
price (in Rs.) per kg of the mixed variety of
80% solution has been used in the
rice is?
entire process?
Solution: 21

Mixtures and Alligations 122

Let the average price per kg be ` x.
Previous Years’ Question

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 drum 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
Ratio of quantities of Type 1 and Type 2 = 3 : 1 ‘B’ were in the ratio?

x 18 3 (A) 239 : 161 (B) 251 : 166

22 x 1 (C) 273 : 167 (D) 239 : 157
⇒ x – 18 = 66 – 3x
⇒ 4x = 84
2. Alligation in Compound Interest and
⇒ x = 21
Simple Interest
Average price = Rs. 21 per kg.
Example 7:
Example 6:
Ashwani lent a certain sum of money to
The ratio of expenditure and savings is 9 : 11. Raghav at 12 % p.a. S.I. and lent the remain-
If the income increases by 20 % and the sav- ing amount to Neeti at 7 % p.a. S.I. overall
ings increase by 11 %, then by how much per- he gets the interest of 8.5 % p.a. Find the
cent does the expenditure increase? amount of money Ashwani lent to Raghav, if
Solution: 31% Ashwani has a total of Rs. 50,000.
Let the % increase in expenditure is x % Solution: Rs. 15,000
By using the rule of alligation. By using the rule of Alligation

The ratio of expenditure and saving = 9 : 11.

9% 9 The ratio of the amount lent to Raghav and
So,
(x 20)% 11 Neeti = (3 : 7).
∴ x = 31 %

Mixtures and Alligations 123

Let the amount lent to Raghav and Neeti be
3K and 7K respectively, then Previous Years’ Question

3K + 7K = 50000 Two alcohol solutions ‘A’ and ‘B’ are mixed

K = 5,000 in proportions 1 : 3 by volume. The volume
of the mixture is then doubled by adding
Amount lent to Raghav = 3K solution ‘A’ such that the resulting mixture
⇒ 3 × 5000 has 72% alcohol. If solution ‘A’ has 60%
⇒ Rs. 15,000 alcohol, then the percentage of alcohol in
solution ‘B’ is?
3. Alligation in Profit and Loss (A) 92% (B) 83%
(C) 62% (D) 74%
Example 8:
Anil sold 40 % of the book at a profit of 40 %
and 60 % of the book at 20 % profit. Find the Example 9:
average profit percent, if Anil sells only these Simba purchased a table for `13,000 and
two kinds of books. sold it at 30 % profit. He again purchased
Solution: 28 % a chair at `4,000 and sold it at 5 % profit,
The ratio of the number of books sold at then find the overall profit percent earned
40 % profit and 20 % profit is: by him.
⇒ 40 % : 60 % Solution: 24.12 %
⇒ 2:3 Let overall profit percent earned = P %
By using the rule of alligation we get,
Let Aw be the average profit percentage.
By using the rule of Alligation.

Aw 20% 2
13000 P 5
40% Aw 3 ⇒
4000 30 P
3Aw – 60 % = 80 % – 2Aw
⇒ 17P = 410
5Aw = 140 %
⇒ P = 24.12 %
Aw = 28 %

Mixtures and Alligations 124

4. Alligation in Time , Speed and Distance = 10 hours + 40 min
1 28
=9+ hours
Example 10: 3 3

Keerti usually takes 10 hours to ride from By using the rule of alligation we get,
Agra to New Delhi. One day her car had a
technical issue at Greater Noida, so she has 10 8
to stop for 20 minutes to resolve the issue.
After that, she increases her speed by 25 %
and reaches New Delhi 20 minutes before 28
her scheduled time. What is the ratio of the 3
distance between Agra to Greater Noida to
Greater Noida to New Delhi?
Solution: 2 : 1
28 28
1 8 10
25 % means = (increase of 1 unit over 3 3
4
4 unit).
Let the usual speed = 4x Distance from agra to Greater Noida
Distance from Greater Noida to New delhi
Increased speed = 5x
28
Total distance = speed × time = 4x × 10 = 40x 8
3
Time is taken by Keerti if she travels the 28
10
whole distance at increased speed 3
4
distance 40x
Time = 8 hours 3 4
speed 5x 2 2
Now, the average time is taken by her to 3

cover the whole distance Hence, the required ratio will be

4 : 2 or 2 : 1

Rack Your Brain 5. Alligation in Compound Mixtures

A solution of volume 40 litre, has dye

Example 11:
and water in the proportion 2 : 3. Water
In a vessel, the ratio of acid and water is
is added to the solution to change this
2 : 5. In another vessel, the ratio of water and
proportion to 2 : 5. If one-fourth of this
acid is 6 : 1. In what ratio the contents from
diluted solution is taken out, how many
both vessels should be mixed, so that the
litres of dye must be added to the
ratio of acid and water becomes 3 : 11?
remaining solution to bring the proportion
back to 2 : 3? (A) 3 : 2 (B) 1 : 1

(A) 8 litres (B) 6 litres (C) 4 : 3 (D) 5 : 2

(C) 7.5 litres (D) 9 litres Solution: (B)

Mixtures and Alligations 125

We can solve such a by applying the method of alligation either on acid or on water

TIP:
2 1 3
As the fraction involved are , ,
7 7 14
To get rid of fraction, take the L.C.M of (7, 7, 14) = 14

Mixture 2. Compound Mixture: When two or more

than two simple mixtures are mixed to-
When two or more than two pure substanc-
gether, a compound mixture is formed.
es/mixtures are mixed, then a mixture is
formed.
Mixing without Replacement
When two items of different concentrations
In this type of mixing, two or more than two
are mixed, the concentration of the result-
substances are mixed without any part of
ant mixture lies in between the concentra-
any mixture being replaced.
tions of the other two.
Let’s understand this concept with the help
Type of Mixture: of some examples.
There are two types of mixtures:
1. Simple Mixture: When two or more than Example 12:
two different things are mixed together, In a mixture of 600 litres, the ratio of milk
a simple mixture is formed. and water is 7 : 8 respectively. Now, 120 litres

Mixtures and Alligations 126

of water is added to the mixture. What is the Alternate Solution: (Rule of Constant)
ratio of milk and water in the final mixture? When the quantity of one element in the
mixture does not change while adding
Solution: 7 : 11 another element to the first mixture.
Let the volume of milk and water in the 15 litres = 30 % of first solution = 10 % of the
solution be 7K and 8K. new solution
15 K = 600 10 % of the new solution = 15 litres
K = 40 The total quantity of new solution
Volume of milk = 7 × 40 = 280 litres
15
Volume of water = 8 × 40 = 320 litres = × 100% 150 litres
10%
When 120 L of water is added.
Quantity of fresh water added
Volume of water = 320 + 120 = 440 litres
= 150 – 50 = 100 litres
Hence, the ratio of milk and water
= 280 : 440 = 7 : 11
Mixing with Replacement
Example 13: It is a type of mixing when two substances
How many litres of water should be mixed are mixed by replacing some part of a mix-
with 50 litres of 30 % wine solution so ture. In this type of mixing total volume may
that the resultant solution is a 10 % wine or may not be the same.
solution? In this type of mixing, three types of ques-
Solution: x = 100 litres tions can be asked.
By using the rule of Alligation with respect
Type 1:
to wine,
In this case, the quantity withdrawn and
quantity replaced are of the same volume
and this process can be repeated ‘n’ times.
Suppose from ‘V’ litres of pure milk (or any
pure thing), ‘A’ litres of milk is replaced by A
litres of water, then the volume of the milk
in the final solution after ‘n’ such operation
n
A
is given by, V′ = V 1
Volume of solution of 30% wine 1 V
=
Volume of solution of water 2 Example 14:
50 1 A person has 90 litres of pure alcohol solu-
=
x 2 tion and 9 litres of alcohol is replaced with
x = 100 litres 9 litres of water. This operation is performed
two more times. Find the quantity of alcohol
100 litres of water added to make the result-
left.
ant solution, 10 % wine solution.

Mixtures and Alligations 127

Solution: 65.61 litres Solution: 72 litres
Here, initial volume V = 90 litres Here, V = 90 litres; A = 9 litres; B = 10
Volume of alcohol replaced by water litres
A = 9 litres A B
V′ = V 1 1
V V
n = 3 times
9 8
9
3 V′ = 90 × ×
V′ = 90 × 1 10 9
90
V′ = 72 litres
729
V′ = 90 × V′ ⇒ 65.61 litres
1000

Previous Years’ Question

Rack Your Brain
Each of 3 vessels A, B, C contains 500
Two glass P and Q were partially filled ml of salt solution of strengths 10%, 22%
with water. 70% of the water in P was and 32% respectively. Now, 100 ml of the
transferred to Q. After that, 50% of the solution in vessel ‘A’ is transferred to vessel
water in Q was transferred to P. At this ‘B’. Then 100 ml of the solution in vessel ‘B’
stage, the volume of water in glass P and is transferred to vessel ‘C’. Finally, 100 ml
Q are in the ratio of 5 : 3. Find the ratio of the solution in vessel ‘C’ is transferred
of the initial volume of water in P and Q. to vessel ‘A’. The strength, in percentage of
the resulting solution in vessel ‘A’ is.
(A) 14% (B) 12%
Type 2:
(C) 86% (D) 72%
In this case, the quantity withdrawn and the
quantity replaced are of the same volume,
but the withdrawn quantity and replace
quantity may vary subsequently. Rack Your Brain
Suppose from ‘V’ litres of pure milk (or any
A sample of ‘a’ litres from a tank having
liquid), ‘A’ litres of pure milk is replaced by
60 litres mixture of alcohol and water
A litres of water, again ‘B’ litres of mixture is
containing alcohol and water in the ratio
replaced by ‘B’ litres of water and so on.
of 2 : 3 is replaced with pure alcohol,
The final volume of milk is given by,
so that the tank will have alcohol and
A B C
V′ = V 1 1 1 ... water in equal proportion. What is the
V V V
value of a?
Example 15:
A milkman has 90 litres of milk and 9 litres Type 3:
of milk is taken out and replaced by 9 litres
In this case, the quantity withdrawn and
of water. Again, 10 litres of mixture is taken
the quantity replaced are not of the same
out and replaced by 10 litres of water. Find
volume.
the volume of milk in the final solution.

Mixtures and Alligations 128

Let’s understand this concept with examples. After 2 : 5) × 5Multiply
first ratio by (2) and second by (5) we get,
Some Miscellaneous Examples Wine : Water

Example 16: 10 : 14

8 litres of alcohol solution contains 60 % al- 10 : 25

cohol. 3 litres of water is added to this solu- Amount of water poured = 44 litres
tion. What is the percentage of alcohol in 25K – 14K = 44
the new solution? K=4
Quantity of wine
Solution: 43.63 %
⇒ 10K ⇒ 10 × 4 = 40 litres
The volume of alcohol in a solution
60
= 8× 4.8 litres
100 Rack Your Brain
The volume of water in a solution = 8 – 4.8
There is 40 litres of milk in the container
⇒ 3.2 litres
four litres of milk is taken out and
3 litres of water added to this solution
replaced with 5L of water. After this 6L of
Volume of water in a new solution = 3.2 + 3 mixture is replaced with 7 litres of water
⇒ 6.2 litres and finally 8 litres of mixture is replaced
Total volume of a new solution with 9 litres of water. How much of the

= 8 + 3 = 11 litres milk is there in the vessel now?

% of alcohol in a solution
Example 18:
4.8
= × 100 43.63%
11 Jar A contains 20 % acid and Jar B contains
50 % acid. In what ratio should Jar A be mixed
Example 17:
with Jar B to obtain a mixture of 38 % acid?
The wine and water in a mixture are in the
Solution: 2 : 3
ratio of 5 : 7 respectively. If 44 litres of water
is added to it, then the ratio of wine and wa- By using rule of alligation with respect to
ter in the new mixture becomes 2 : 5. What acid.
is the total quantity (in litres) of the wine in
the new mixture?
Solution: 40 litres
Here, the quantity of wine in the mixture re-
mains constant. So, the value of the quantity
of wine in both the ratio must be same.

Wine : Water
Before 5 : 7) × 2

Mixtures and Alligations 129

Hence, Jar A and Jar B must be mixed in the Alternate Solution:
ratio of 2 : 3.
Formula Based:

Example 19: The initial volume of Alcohol V = 50 litres

Sharad Chand has 50 litres of alcohol in a Replaced volume A = 5 litres

cask. He drinks 5 litres of alcohol and re- Number of times operation performed
placed it with 5 litres of water. He per- (n) = 3
formed this operation 2 more times. Find The volume of alcohol left = V′
the amount of alcohol left after third n
A
operation. V′ = V × 1
V
3
Solution: 36.45 litres 5
V′ = 50 × 1
50
Traditional Method:
⇒ 36.45 litres
Initially, there are 50 litres of alcohol, and
5 litres of alcohol is replaced with 5 litres of
water. Previous Years’ Question
Now, there will be 45 litres of alcohol and 5
litres of water. The strength of salt solution is P% if 100
ml of the solution contains ‘P’ grams
The quantity of Alcohol and water being
of salt. If 3 salt solutions A, B, C are
withdrawn here will be in the ratio of
mixed in the proportion 1 : 2 : 3, then the
(45 : 5) = 9 : 1
resulting solution has strength 20%. If
So, the quantity of alcohol withdrawn instead the proportion is 3 : 2 : 1., then
9 the resulting solution has strength 30%.
= ×5 4.5 litres
10 A fourth solution, D, is produced by
So, the volume of alcohol mixing B and C in the ratio 2 : 7. The ratio
= 45 – 4.5 = 40.5 litres of the strength of ‘D’ to that of ‘A’ is:
And the volume of water = 5 + 4.5 = 9.5 litres (A) 1 : 2 (B) 1:3
Now, again 5 litres of mixture is replaced (C) 1 : 4 (D) 1:5
with 5 litres of water.
The quantity of Alcohol and water being
Alternate Solution:
withdrawn here will be in the ratio of
Percentage Change Method:
(40.5 : 9.5) = 81 : 19.
The quantity of alcohol withdrawn % of alcohol withdrawn
81 5
= × 5 ⇒ 4.05 litres = × 100 ⇒ 10% (each time)
100 50
The volume of alcohol 10 % decrement means decrement of 1 over
= 40.5 4.05 36.45 litres 10.

Mixtures and Alligations 130

 Alcohol before operation 1000 = 3 × 24 + 12 = 84 litres
Alcohol after operation 729 So, now when 22 litres solution is removed,
the solution will be removed in the ratio of
50 1000
x 729 4 : 7.

⇒ x = 36.45 litres 7
Water removed = × 22 14 litre
11
Example 20: So, new volume of water = 84 – 14 = 70 litre
When 12 litres of water is added to the
vessel having milk and water in the ratio Alternate Solution:
of 2 : 3, the ratio changes to 4 : 7. Find the
new volume of the water in the vessel if
22 litres of solution is removed from the
same vessel.
(A) 63 (B) 56
(C) 70 (D) 84
Solution: (C)
Let the initial milk and water in the vessel 1 part = 12 litres
be 2x and 3x
7 part = 84 litres (after adding water)
So, when 12 litre water is added
2x 4 22 litres mixture is removed
; Solving it we get x = 24 7
(3x 12) 7 Water quantity removed = × 22 14 litres
11
So, new volume of the water in the vessel
New volume of water = 84 – 14 = 70 litres
now is (3x + 12)

Mixtures and Alligations 131

 Practice Exercise – 1

Level of Difficulty – 1 to obtain a new mixture in ves-

sels C consisting of half Milk and half
1. A mixture of 100 litres of Mango juice and Water?
water contains 15 % water. How many
(A) 9 : 1
litres of water should be added to the
(B) 7 : 3
mixture, so that the mixture contains
(C) 15 : 8
25 % water?
(D) 8 : 5
(A) 15 litres
(B) 13.33 litres 5. If a woman buys 1 litres of vinegar for
(C) 14 litres Rs. 12, mixes it with 300ml water and
(D) 12 litres sells it for Rs. 15, then what is her profit
percentage?
2. A Juice seller claim to sell Juice at
(A) 26 %
a cost price, still he is making a prof-
(B) 27 %
it of 37.5%, since he has mixed some
(C) 25 %
amount of water in the Juice. What is the
(D) 22 %
percentage of Juice in the mixture?
(A) 62.5 % Level of Difficulty – 2
(B) 66.66 %
(C) 72.72% 6. Raghav bought 3 kg grapes. The ratio of
(D) 87.5 % water to pulp was 3 : 2. When his moth-
er crushed these grapes, then some
3. Two solutions of hydrochloric acid are amount of water gets wasted. Now, the
mixed in the ratio of 1 : 9. The concen- ratio of water is to pulp is 1 : 4. What is
tration of hydrochloric acid in the first the quantity of crushed grapes?
and second solutions are 10 % and 20 % (A) 1.5 kg
respectively. What is the concentra- (B) 2.5 kg
tion of hydrochloric acid in the resulting
(C) 2.75 kg
solution?
(D) 1.75 kg
(A) 19 %
(B) 17 % 7. Two chocolate powders costing
(C) 18 % Rs. 300/kg and Rs. 180/kg were blended
(D) 20 % in different ratios to get chocolate blends
P and Q. P and Q were mixed in the ratio
4. The Milk and Water in two vessels 1 : 1 to form blend R. R was sold at a price
A and B are in the ratio of 6 : 7 and of Rs. 294/kg at 20 % profit. If P costs
11 : 2 respectively. In what ratio the Rs. 210/kg. Find the ratio in which the
liquid in both the vessels be mixed two powders were mixed to form Q.

Mixtures and Alligations 132

 (A) 5 : 2 Level of Difficulty – 3
(B) 5 : 1
11. Dye and Urea are two fertilizers. Dye is
(C) 5 : 4
consists of N, P and K and Urea consists
(D) 5 : 7
of only N and P. A mixture of Dye and
8. Two alloys are made up of gold and sil- Urea is prepared in which the ratio of N,
ver. The ratio of gold and silver in the first P and K is 27 %, 65 % and 8 % respective-
alloy is 1 : 4 and in the second alloy, it is ly. The ratio of N, P and K in Dye is 30 %,
2 : 3. In what ratio the two alloys should 55 % and 15 % respectively. What is the
be mixed to obtain a new alloy in which ratio of N and P in the Urea?
the ratio of silver and gold be 9 : 5? (A) 33 : 107
(A) 4 : 11 (B) 33 : 105
(B) 3 : 9 (C) 34 : 107
(C) 3 : 11 (D) 33 : 109
(D) 3 : 15
12. There is 120 litres of Gastrol and 180 litres
of Mastrol oil. The price of Gastrol is Rs.
9. In two alloys, the ratio of silver and tin is
90/litre and price of Mastrol is Rs. 70/li-
5 : 7 and 9 : 4. How many kilograms of the
tre. Equal amount of Gastrol and Mastrol
first alloy and the second alloys, respec-
is taken out and poured in Mastrol and
tively, should be melted together to ob-
Gastrol respectively. Now, the price of
tained 1333 kg of a new alloy with equal
both the mixture becomes the same.
silver and tin?
What is the amount of oil taken out form
(A) 900 kg and 433 kg
each vessel?
(B) 910 kg and 423 kg
(A) 90 litres
(C) 920 kg and 413 kg
(B) 84 litres
(D) 930 kg and 403 kg
(C) 80 litres
10. Two varieties of rice are mixed in the ra- (D) 72 litres
tio 2 : 3. The cost price of each kg of the
13. Milk and water are mixed in the ratio 7 : 4
first variety of rice is Rs. 5 more than the
to form a solution. From the solution, 22
cost price of each kg of the second va-
litres is taken out and 4 of water is added
riety of rice. The mixture is sold at 20 %
in the solution, which result in ratio of
profit at Rs. 30/kg. Find the CP of the
milk and water in the final mixture as 12
first variety in Rs./kg
: 7. Find the quantity of milk (in litres) in
(A) Rs. 28.5/kg
the initial solution.
(B) Rs. 28.75/kg
(A) 320
(C) Rs. 28/kg
(B) 334
(D) Rs. 20/kg
(C) 350
(D) 360

Mixtures and Alligations 133

14. There are three vessels filled to their ca- a sip of 4ml of the resulting mixture and
pacities with a mixture of water and al- finds it bitter in taste. He then adds 8ml
cohol in the ratio of 3 : 5, 5 : 9 and 7 : 11 of water more to the same mixture and
respectively. These are all poured into a still finds it bitter after taking a sip of
big vessel. What proportion of the mix- 28ml. He adds 20 ml of water more and
ture in the big vessels should be replaced finally gets a perfect peg. Find the ratio
by water, so that the alcohol forms 50 % of alcohol and water in the resulting mix-
of the resulting mixture? ture peg.
(A) 193 : 947 (A) 1 : 1
(B) 191 : 947 (B) 1 : 2
(C) 189 : 947 (C) 2 : 1
(D) 185 : 947 (D) 3 : 4

15. Shubham sits for a drink and adds 20ml

water to 60 of alcohol in a glass. He takes

Solutions
1. (B)
Method 1:

Traditional Method = (Rule of constant)

Quantity of fruit juice in the mixture
15
= 100 100 × = 85 Litres
100
After adding water, Juice would form
75 % of the mixture.
75 % of mixture = 85
85 Quantity of mixture : Water added = 15 : 2
1 % of mixture =
75 100 15
100 % of mixture Water added 2
85 Water added = 13.33 litres
⇒ × 100
75
2. (C)
⇒ 113.33 litres
Let the Juice he bought is 1000 ml.
Quantity of water added = 113.33 – 100
Let C.P. of 1000 ml is Rs. 100
= 13.33 litres
Juice seller sells the mixture of Juice
Method 2: and water at CP.
So, we can conclude that he sells the
By rule of alligation
water, which is freely available at the
On applying the rule of alligation with re-
price of milk.
spect to water.

Mixtures and Alligations 134

 So, the overall profit the Juice seller gains
because of the amount of water he mixed.
WATER
3
37.5% profit in fraction =
8 JUICE
Juice : Water
8 : 3
8
Percentage of Juice = × 100 72.72%
11
3. (A)
By using the rule of alligation with re-
spect to hydrochloric acid. The required ratio of the quantity of ves-
Let the average percentage of hydrochlo- sel A and vessel B.
ric acid in the solution is Aw. 9 1
Vessel A : Vessel B = : 9:1
26 26
5. (C)
Here, it is not directly given but we can
conclude from the statement that she
sells whole quantity of vinegar for Rs. 15.
CP of vinegar = Rs. 12
SP of vinegar = Rs. 15
Profit = SP – CP = Rs. 3
Profit 3
Profit % = × 100 = × 100 25%
CP 12
Given that, 6. (A)
Quantity of solution 1 : Quantity of solu-
Water Pulp
tion 2 = 1 : 9.
Initial 3x 2x
20 Aw 1
Final y 4y
Aw 10 9
But 2x = 4y since the quantity of pulp
180 – 9Aw = Aw – 10
remains the same.
10 Aw = 190
Water Pulp
Aw = 19 %
Initial 6y 4y
4. (A) Final y 4y
6 Given that initially, Raghav bought 3 kg
Concentration of Milk in mixture A =
13 grapes.
11
Concentration of Milk in mixture B = 6y + 4y = 3 kg
13 3
y= kg
Concentration of Milk in resultant mix- 10
1
ture = Final amount of crushed grapes = y + 4y
2
3
By using rule of allegation with respect = 5y = 5 × kg = 1.5 kg
10
to Milk.

Mixtures and Alligations 135

7. (B) q – 245 = 35
(Traditional Method) q = 280/kg
Let the two blends be mixed in the ratio CP of ‘Q’ = 280 Rs/kg.
of x : y. Now, again using alligation among two
Let Q’s cost be q Rs/kg. types of powder with respect to Q.
Given R was sold at 20 % profit at 294/kg
294
CP of R =
1.2
CP of R = 245/kg
P and Q mixed in 1 : 1
210 × 1 1 × q
Weighted Average ⇒ 245
2
210 + q = 490
q = 280/kg.
Now, again using weighted average The two blends were mixed in 5 : 1 to
among two types of powder with respect produced Q.
to Q we get,
300 × x 180 × y 8. (C)
WeightedAverage = 280 1
x y Concentration of gold in alloy 1 =
5
300x + 180y = 280x + 280y 2
Concentration of gold in alloy 2 =
20x = 100y 5
x 5 5
y 1 Concentration of gold in final alloy =
14
Two blends were mixed in the ratio of By using alligation with respect to gold.
5 : 1 to produce ‘Q’.
Alternate Method:
CP of
294
R= Rs. 245 / kg ( 20% profit)
1.2
By using Alligation in P, Q, R.

The ratio of the quantity of Alloy 1 : Alloy

3 11
2= : ⇒ 3 : 11
Quantity of P 1 70 70
Quantity of Q 1 Hence, the two alloys must be mixed in
q 245 1 the ratio of 3 : 11
35 1

Mixtures and Alligations 136

9. (D) 2 x× (y+ 5) + 3 x× y
CP of the 1 kg mixture =
Let x and y kg of the first alloy and the 2 x+ 3 x
second alloy are taken. 2xy 10 x 3 xy
= = (y + 2).
5 5x
The concentration of silver is Alloy 1 =
12 CP of 1 kg of mixture
The concentration of silver is Alloy 30
9 = 25 / kg ( 20% profit)
2= 1.2
13
y + 2 = 25
The concentration of silver in final Alloy
1 y = 23
= CP of first variety of rice
2
By using the rule of alligation. = y + 5 = 23 + 5 = Rs. 28/kg

11. (A)

x 5 / 26 This 8 % of K is obtained only from Dye.

y 1 / 12 To make the concentration of k the same
in both Dye and Mixture, multiply the
x : y = 30 : 13
quantity of Dye by 8 and mixture by 15
Amount of Alloy 1 in mixture
we get,
30
= × 1333 930kg
43
Amount of Alloy 2 in mixture
13
= × 1333 403kg
43
(930 kg, 403 kg)

10. (C)
Let the quantities of two varieties of rice
be 2x and 3x kg respectively.
Let the CP of the second variety of rice ∴ Urea + Dye = Mixture
be Rs. y/kg. NUrea + NDye = NMixture
So, CP of the first variety of rice
x + 240 = 405
= (y + 5)/kg
x = 165

Mixtures and Alligations 137

 Similarly, PUrea + PDye = PMixture This implies, after removal of 22 litres
y + 440 = 975 solution,
y = 535 Ratio of milk and water = 7 : 4
Ratio of N : P in urea = 165 : 535 water added
Now, m : w m: w
33 : 107
7:4 12 : 7
12. (D) We can solve this by making same value
Traditional Method: of milk as it quantity is not changing.
Let the amount of oil withdrawal be x. +4l water
Therefore, m : w m: w
Gastrol = 120 litre at Rs. 90/litres
Mastrol = 180 litre at Rs. 70/litres 84 : 48 84 : 49
+1 part of water
After withdrawal and pouring.
Now, 1 part of water = 4.
Gastrol Mastrol ∴ total solution (84 + 48 parts)
= 132 parts = 132 × 4 = 528.
(120 – x) x
total solution before removal (in starting)
Mastrol Gastrol = (528 + 22) litres = 550 litres
Now, quantity of milk in the solution at
(180 – x) x 7
this stage = × 550 350 litres
11
Price of Gastrol = Price of Mastrol. 14. (B)
(120 x) × 90 x × 70 (180 x) × 70 x × 90 In the new vessel water and alcohol are
120 180 in the ratio
3 5 7 5 9 11
10, 800 90x 70x 12,600 70x 90x :
8 14 18 8 14 18
2 3
189 180 196 315 324 308
:
⇒ (10, 800 20x) × 3 2 × (12,600 20x) 504 504
565 : 947
⇒ 32, 400 60x 25, 200 40x
Let’s assume the quantity of water,
⇒ 100x = 7,200 alcohal be 565 k and 947 k respectively.
⇒ x = 72 Thus, total solution
Hence, 72 of oil was taken out from the = 565 k + 947 k = 1512 k
vessel. Let’s assume the fraction of mixture
withdrawn and replaced with water = x
13. (C)
Alcohal = 947 × (1 – x) = 50% of total
Since the mixture of milk and water is
1
a homogeneous mixture, therefore ratio = × 1512 756
2
will remain same if we withdraw some 756 191
⇒ x 1
solution from it. 947 947

Mixtures and Alligations 138

15. (A)
Shubham adds 20 ml water to 60ml of alcohol when he sits for drink

Alcohol : Water

60ml : 20ml

Ratio 3 : 1

3 & 1
Qty left. ×76 ×76 (Takes 4ml sip)
4 4

57ml & 19ml

57 & (19+8) (Adds 8ml water)

57ml & 27ml

Ratio 19 : 9

19 9 (Takes 28ml sip)

Qty. Removed ×28 & ×28
28 28

Qty. left (57–19) & (27–9)

38ml & 18ml

38ml & (18+20)ml (Addition of 20ml water)

38ml & 38ml

(Alcohol) : (Water)

Therefore, Required ratio = 38 : 38 = 1 : 1

Hence, Option (A) is the correct.

Mixtures and Alligations 139

 Practice Exercise – 2

Level of Difficulty – 1 (C) 1000

(D) 900
1. In a cooking competition, Team A and
Team B (First two winners) made mock- 4. From a vessel containing 750 ml of pure
tails. Mocktail of Team A contains a mix- milk, 160 ml pure milk was drawn out
ture of orange juice and blueberry juice and distributed equally between two
in the ratio 11 : 3 and Team B’s mocktail beakers P and Q. The milk in both the
contains a mixture of orange juice and beakers was diluted by adding water
blueberry juice in the ratio 7 : 4. What in different proportions. After that, the
should be the ratio of volumes of mock- contents of P and Q were added back
tails of Team A and Team B be taken so to the vessel. The concentration of milk
as to obtain a mocktail of Orange juice in the vessel now is 75%. Had the con-
and blueberry juice in the ratio 5 : 2? tents of beakers P and Q been mixed
(A) 6 : 11 with each other instead of adding it
(B) 1 : 14 into the vessel, what would be the ap-
(C) 3 : 7 proximate concentration of milk in the

(D) 12 : 11 mixture?
(A) 42%
2. The ratio of copper and zinc by weight in
(B) 39%
two alloys x and y is 2 : 7 and 5 : 4. How
(C) 35%
many kilograms of the alloy x and y are
(D) 36%
required to make 42 kg of new alloy z in
which the ratio of copper and zinc is the 5. Shreyas took pure milk and started to
same? dilute it with water. He first replaced 14
(A) 6 kg and 36 kg litres from the beaker full of pure milk
(B) 10 kg and 32 kg with 14 litres of water. He performed this
(C) 7 kg and 35 kg process three more times. Finally, the
(D) None of these ratio of milk left in the beaker to that
of water became 81 : 544. Find out how
3. A, B, C are 3 containers, which contain much pure milk (in litres) was there in
Milk, Sugar, and Water in varying propor- the beaker initially.
tions. Container ‘A’ contains x% milk and
(A) 21
y% sugar. Container ‘B’ contains y% milk
(B) 28
and y% water. Container ‘C’ contains x%
(C) 35
sugar and y% water. When A, B, C are
(D) 42
mixed in the ratio 2 : 3 : 5, then the fi-
nal mixture has 18% milk and 23% sugar. 6. In what ratio should water which costs
Find the product of x and y. Rs. 10 per litre be mixed with a concen-
(A) 1500 trated acid costing Rs 25/litre to make
(B) 1200 a profit of 30% by selling the resultant

Mixtures and Alligations 140

 liquid at Rs 18/litre? Given that any acid 10. A vessel is filled with milk. 12 litres of
should have at least 50% acid content or milk is withdrawn and replaced with wa-
else it will be rejected. ter. Next, from the same vessel, 12 litres
(A) 1.2 of the mixture is withdrawn and replaced
(B) 1.3 with water. If the volumes of milk and
(C) 2.9 water in the vessel are now in the ratio
(D) Not possible of 81 : 19, then the capacity of the vessel
(in litres) is:
7. Two alloys ‘P’ and ‘Q’ of Tin and lead are (A) 40
prepared by mixing the respective metals (B) 80
in the proportion of 5: 2 and 5 : 11 respec-
(C) 90
tively. If the two alloys are melted and
(D) 120
mixed to form a 3rd alloy ‘R’, having 1 : 1
proportion of Tin and Lead, what is the Level of Difficulty – 2
ratio of alloys ‘P’ and ‘Q’ in the mixture?
(A) 7 : 8 11. In what ratio should two varieties of tea
(B) 8 : 7 costing Rs. 600 per kg and Rs. 900 per kg
(C) 5 : 7 respectively be mixed so that the result-
ing mixture when mixed with another va-
(D) 7 : 5
riety of tea costing Rs. 1000 per kg in the
8. In 320 litres of the mixture, the ratio of ratio 5 : 4, would yield a mixture costing
milk and water is 5 : 3. If 80 litres of the Rs. 800/kg?
mixture is taken out and some quantity (A) 2 : 13
of water is added to the mixture, then the (B) 13 : 2
ratio between milk and water becomes 5 (C) 5 : 13
: 4 respectively. Find the amount of wa- (D) 9 : 17
ter that is added to the mixture.
(A) 30 litres 12. Two types of spirits solutions ‘P’ and ‘Q’
(B) 50 litres are mixed in the proportion 3: 5 by vol-
(C) 80 litres ume. The volume of the mixture is then
doubled by adding solution ‘P’ such that
(D) 20 litres
the resulting mixture has 60% spirit. If
9. If 40 litres of milk solution having 36 % of solution ‘P’ has 70% spirit, then find the
milk and remaining water is given, then % of spirit in solution ‘Q’ ?
how many litres of water (in litres) must (A) 36%
be added to this solution to make it a (B) 38%
30% solution of milk? (C) 40%
(A) 4 (D) 42%
(B) 6
(C) 8 13. When 20 litres of milk is added to 60
litres of a milk and water solution, the
(D) 10

Mixtures and Alligations 141

 new concentration of milk in the solution 16. A milk distribution tanker contains 320
formed is the same as the new concen- litres mixture of milk and water with 70%
tration of water, when 30 litres of water of milk. In order to increase the purity,
is added to the same 60 litres of initial the distributor decided to add more milk
milk and water solution. What is the ratio to the mixture. Find the quantity of milk
of milk and water in the original milk-wa- (in litres) that should be added to the
ter solution? mixture so that the percentage of milk is
(A) 3:2 increased to 80%.
(B) 8:9 (A) 120
(C) 9:8 (B) 130
(D) 5:3 (C) 140
(D) 160
14. A jar contains 40 litres of pure milk, 20%
of this is taken out and replaced with 17. A 30% ethanol solution is mixed with an-
water in the first cycle. In the 2nd cycle, other ethanol solution, say, K of unknown
30% of the resultant mixture is taken concentration in the proportion 3 : 5 by
out and replaced with water and finally volume. This mixture is then mixed with
in the third cycle 20% of the mixture is an equal volume of 12.5 % ethanol solu-
taken out and replaced with water. Find tion. If the resultant mixture is an 18.75%
the ratio of milk and water in the final ethanol solution, then the unknown con-
mixture. centration of K is:
(A) 56 : 69 (A) 20%
(B) 71 : 79 (B) 22%
(C) 38 : 43 (C) 24%
(D) 45 : 56 (D) 18%

15. Mixture ‘x’ has been formed by two 18. Three vessels of equal capacities contain
types of raw elements A and B by mix- the three liquids L, M & N in the ratio 1
ing them in the ratio of 3 : 5. Similarly, : 2 : 3, 3 : 4 : 5 & 5 : 6 : 7 respectively.
mixture ‘y’ has been formed by mixing The mixtures from these vessels are tak-
two raw elements ‘A’ and ‘C’ in the ra- en in the ratio 1 : 2 : 3 and poured into
tio of 3: 2. In what ratio should both a fourth vessel with a capacity equal to
the mixtures be mixed to obtain a solu- the capacity of the given three vessels.
tion which has a concentration of ‘B’ as The ratio of the liquids L, M and N in the
40%? resulting mixture is:
(A) 9 : 1 (A) 16 : 21 : 26
(B) 13 : 2 (B) 3 : 4 : 5
(C) 16 : 9 (C) 1 : 2 : 3
(D) 5 : 9 (D) None of these

Mixtures and Alligations 142

19. A 20 kg alloy made up of copper and zinc 22. A can contains a solution of milk and wa-
in the ratio 4 : 1 is mixed with 12 kg of an- ter. After adding 50 litres of water into
other alloy of copper and zinc, such that the can, the concentration of the milk
in the resultant alloy, the ratio of copper in the can becomes 40%. Now if further
and zinc becomes 5 : 3. Find the ratio of 60 litres of more water is added to the
copper and zinc in the 2nd alloy. can, the concentration of the milk in the
(A) 1 : 2 can reduces by 10 percentage points.
(B) 2 : 1 How much more water (in litres) must be
(C) 3 : 2 added to the can so that the concentra-
(D) 2 : 3 tion of milk in the can becomes 15%?
(A) 80
20. Two vessels whose volumes are in (B) 120
the ratio of 3: 4 contain milk and wa-
(C) 240
ter mixtures. If 40% of the first mix-
(D) 360
ture, which is of 60% milk concen-
tration, is mixed with 60% of the 2nd 23. A mixture ‘P’ is produced by mixing the
mixture, which is of 50% milk concen- chemical ‘A’ and ‘B’ in the ratio of 3 : 5.
tration, then find the milk concentra- Chemical ‘A’ is prepared by mixing two raw
tion of the final mixture or the resultant materials ‘m’ and ‘n’ in the ratio of 2 : 3.
mixture. Chemical ‘B’ is prepared by mixing raw ma-
(A) 53.33% terials ‘n’ and ‘K’ in the ratio of 3 : 2. Then
(B) 42.28% the final mixture is prepared by mixing 128
(C) 19.75% litres of ‘P’ with some litres of water. If the
(D) 80% concentration of the raw material ‘n’ in the
final mixture is 40%, how much water (in
Level of Difficulty – 3 litres) has been added to the mixture ‘P’?

21. If a certain weight of an alloy of copper 24. A vessel contains 240 litres of milk and
and gold is mixed with 6 kg of pure gold, water solution in the ratio 1 : 5 respec-
the resulting alloy will have 90% gold by tively. 60 litres of the contents of the
weight. If the same weight of the initial vessel are taken out and ‘x’ litres of pure
alloy is mixed with 4 kg of another alloy milk is added to the vessel. Due to this,
which has 90% gold by weight, the re- the ratio of milk and water in it gets re-
sulting alloy will have 80% gold by weight. versed. Then, ‘y’ litre of water is added to
Then, the weight of the initial alloy, in kg the vessel due to which the ratio of milk
is. and water once again gets reversed. The
(A) 0.82 kg value of (y + x) is equal to (in litres):

(B) 5/13 kg (A) 4200

(C) 8/13 kg (B) 4230

(D) 2 kg (C) 4320

(D) None of these

Mixtures and Alligations 143

25. A vessel has a 900-litre mixture of spirit of alcohol in both the solutions is equal,
and water which has 40% spirit in it. How then find the amount of solution (in li-
many litres of another mixture with 60% tres) that was taken out and mixed with
spirit content may be added so that spir- the other solution.
it content in the resulting mixture will be (A) 11.42
more than 45% but less than 50% ? (B) 12.33
(A) More than 300 litres but less than (C) 11.60
900 litres (D) 12.2
(B) More than 800 litres but less than
1200 litres 28. A mixture, of volume 120 litres, has milk
and water in the proportion 2 : 3. When
(C) More than 500 litres but less than
water is added to this mixture, the ratio
900 litres
of milk and water changes to 2 : 5. If 1/5th
(D) More than 200 litres but less than
part of this mixture is taken out, then
600 litres
how many litres of milk must be add-
26. Ganpat was an alcoholic. One day, ed to the remaining mixture so that the
he mixed alcohol and water in equal ratio of milk and water again becomes
proportions to get 300ml of solution. He 2: 3?
tasted 30ml of the solution and felt that (A) 28.7 litres
he should mix some water. He mixed 30ml (B) 24.5 litres
of water and tasted the solution again. He (C) 25.6 litres
kept on doing it until the concentration (D) 26.7 litres
of alcohol was at most 30%. How much
alcohol (in millilitres) did he consume in 29. The profit is 20% when two types of puls-
the process ? es, ‘x’ and ‘y’ are mixed in the ratio 3 : 2
Mark the answer to the closest integral and then sold at Rs. 60/kg. The profit is
value. 5%, when the same two types of pulses
(A) 61 ‘x’ and ‘y’ are mixed in the ratio 2 : 3 and

(B) 62 then sold at Rs. 60/kg. The cost prices,

per kg, of x and y, are in the ratio:
(C) 88
(A) 41 : 43
(D) 89
(B) 5 : 3
27. Two solutions P and Q made up of alcohol (C) 1 : 2
and soda water have different percent- (D) 2 : 3
ages of alcohol in it. The total volume of
P and Q are 40 and 16 litres respectively. 30. A milkman has a 100L tank of milk. He
Equal volume of solutions of P and Q are removes 5 L from the container and re-
taken. Solution taken from P was mixed places it with water. He keeps on repeat-
with Q and the solution taken from Q ing it till the content of milk is a maxi-
was mixed with P. If the final percentage mum of 80%.

Mixtures and Alligations 144

 Statement 1: The milkman has the de- (A) Both statements are correct
sired proportion when he has added 20 (B) Only statement 1 is correct
L of water. (C) Only statement 2 is correct
Statement 2: The milkman has the de- (D) Both statements are false
sired proportion when he has added 25
L of water.

Solutions

1. (D)
Fraction of Orange juice in mocktail of
Team A = 11/4
Fraction of Orange juice in mocktail of
Team B = 7/11
Fraction of orange juice in resultant mix-
ture =5/7
Using alligation rule:
∴ amount of X = (1/6) × 42 = 7 kg
and Amount of Y = (5/6) × 42 = 35 Kg
Hence, option (C) is the correct answer.

3. (B)
Let’s assume the quantity of A, B and C
are 200 litres, 300 litres and 500 litres
respectively.
Container ‘A’ contains x% milk and y%
sugar.
Container ‘B’ contains y% milk and y%
Required ratio = (6/77 : 1/14) = 12 : 11 water.
Hence, option (D) is the correct answer. Container ‘C’ contains x% sugar and y%
water.
2. (C)
It is given that the concentration of milk
in the final mixture is 18%.
2x + 3y = 18% of 1000 = 180 …(1)
Also the concentration of sugar in the fi-
nal mixture is 23%.
2y + 5x = 230 …(2)
On solving (1) and (2) we get
Copper = 2/9 = 4/18 Copper = 5/9 = 10/18
x = 30 and y = 40
Now use alligation:

Mixtures and Alligations 145

 x × y = 30 × 40 = 1200 Cost of water: 10y
Hence, option (B) is the correct answer. Total Cost: 25x + 10y
Total SP = Cost + Profit
4. (B)
= (25x + 10y) + 30% of (25x + 10y)
According to the question, when diluted
= 1.3 (25x + 10y) = 32.5x + 13y = 18 (x + y)
contents (milk + water) of both the beak-
14.5x = 5y
ers P as well as Q are added back to the y 14.5 29
vessel, the milk volume will be equal to x 5 10
750 ml and this 750 ml represents 75%
But since the content of water in the final
milk concentration
mixture cannot be above 50%, the ratio
So, 750 ml of pure milk corresponds to
of water to acid should be less than 1.
75% concentration.
But 2.9 is not less than 1.
100% = 1000 ml
Thus, our answer will be not possible
Volume of water in the vessel = 250 ml
Hence, option (D) is the correct answer.
If the contents of beaker P and Q are
mixed then there would be 160 ml of pure 7. (A)
milk and 250 ml water will be present. Let’s assume x kg of alloy ‘P’ and y kg of
% of milk = (161/410) x 100 = 39% alloy ‘Q’ are mixed to form 3rd alloy ‘R’.
approximately Now, given that the ratio of tin and lead
Hence, option (C) is the correct answer. in the mixture ‘R’ is 1 : 1
Solving which, we will get x/y = 7 : 8
5. (C)
Hence, option (A) is the correct answer.
If a vessel contains ‘q’ amount of solution
and ‘p’ amount of solution is replaced for 8. (A)
‘n’ times by another solution, then When 80 litres of the mixture is removed,
n
final amount left p then the remaining mixture
1
Initial amount q = (320 – 80) = 240 litres
4 When the mixture is removed from the
81 14
So, 1 given mixture then the ratio will not
625 q
change.
14 3
⇒ 1− = Therefore, the quantity of milk, when 80
q 5
litres of the mixture is removed
14 2
⇒ = = (5/8) x 240 = 150 litres
q 5
Also, Quantity of water
⇒ q = 35 = 240 – 150 = 90 litres
Hence, option (C) is the correct answer. Let ‘x’ litres of water is added to the mix-
6. (D) ture after that ratio becomes 5 : 4
150/(90 + x) = 5/4
Let the quantity of acid be x L and the
x = 30 litres
quantity of water be y L.
Hence, option (A) is the correct answer.
Cost of acid: 25x

Mixtures and Alligations 146

9. (C) 2
81 12
1
100 IQ

9 12
1
10 IQ

12 9
1
IQ 10

12 1
IQ 10

IQ = 120 litres
Hence, option (D) is the correct answer.

11. (B)
Let the price of the resulting mixture
Let ‘x’ litres of water is added in the be Rs. x/kg and this resulting mixture is
solution. mixed with another variety of tea costing
14.4 30 Rs. 1,000/kg in the ratio 5 : 4. The price of
Then, = the mixture is:
25.6 x 70
x × 5 + 1000 × 45 + 4 = 800
7 × 1.44 = 3x 25.6 + 3x
5x + 4000 = 7200
100.8 = 76.8 + 3x
5x = 3200
3x = 24 x = Rs. 640/kg
x = 8 litres
Now, apply the allegation method in the
Thus, 8 litres of water is added first given statement:
Hence, option (C) is the correct answer.

10. (D)
Since we know the formula:
n
U
FQ = IQ 1
IQ

Where, FQ = Final quantity of the milk.

IQ = Initial quantity of the milk.
U = Amount of substance taken out.
n = Number of times the process is
repeated.
n
FQ U
1
IQ IQ Hence the two varieties of tea cost Rs.
600/kg and Rs. 900/kg are mixed in the

Mixtures and Alligations 147

 ratio of 13 : 2, so that desired mixture can M = 540/17 so W = 60 – 540/17 = 480/17
be obtained. Thus, the initial Ratio of M and W
Hence, Option (B) is the correct answer. = 540/17 : 480/17 = 9 : 8
Hence, Option (C) is the correct answer.
12. (B)
Let solution ‘P’ has 3x volume and solu- 14. (A)
tion ‘Q’ has 5x volume. The volume of milk in the final mixture
Since solution ‘P’ is only added not solu- = 40 (1 – 0.2) (1 – 0.3) (1 – 0.4) = 17.92
tion ‘Q’ therefore, the volume of solution litres
‘Q’ will remain the same from start to Therefore, volume of water
end. = 40 litres – 17.92 litres = 22.08 litres
Original quantity of mixture = 8x Now, find the ratio of milk and water in
Quantity of mixture after volume final mixture.
doubled = 16x Milk : Water = 17.92 : 22.08 = 56 : 69
Amount of P in this mixture = 3x + 8x Hence, option (A) is the final answer
= 11x and amount of Q would be 5x
15. (C)
Let solution ‘Q’ have a% spirit in it.
Let the amount of mixture ‘x’ be ‘P’ units
∴ P Q
and the amount of mixture ‘y’ be ‘Q’
11x : 5x ⇒ 16x
units.
70% a% 60%
Amount of ‘B’ in mixture ‘x’ = 5/8 x P units
Therefore, 11x × 70% + 5x × a% = 16x ×
Also, amount of ‘B’ in mixture ‘y’ = 0 units.
60%
Concentration of
11x × 70 + 5a = 16 × 60
‘B’ = (5P/8)/(P + Q) = 40%
770 + 5a = 960
5P/8 = 0.4(P + Q)
5a = 190
0.625 P = 0.4P + 0.4Q
a = 38%
0.225 P = 0.4 Q
Hence, solution ‘Q’ has 38% spirit in it.
P/Q = 16/9
Hence, option (B) is the correct answer.
Hence, option (C) is the correct answer.
13. (C)
16. (D)
Let’s initial quantity of Milk and water in
In 320 litres, 70% is milk. But in pure
the 60 litres of milk-water solution be M
milk, there will be 100% milk only.
and W respectively.
Using alligation:
According to the question
M 20 W 30
80 90
⇒ 9M + 180 = 8W + 240
⇒ 9M − 8W = 60 …(1)
Also, M + W = 60
So, 8M + 8W = 480 …(2)
Adding (1) and (2), 17M = 540

Mixtures and Alligations 148

 Ratio = 20 : 10 = 2 : 1 Similarly, the amount of liquid L, M, N in
Milk that should be added to 320 litres of the second vessel is 9, 12 and 15 litres
mixture = 320/2 = 160 litres respectively.
Alternate Method: And the amount of liquid L, M, N in
In 320 litres 30% is water. the third vessel is 10, 12 and 14 litres
That is 30/100 x 320 = 96 litres. respectively.
So, if we add some amount of pure milk Now, the mixtures are taken out in the
(say, x), that won’t change the amount of ratio 1 : 2 : 3; 6 litres, 12 litres and 18
water. It will still be 96 litres but these litres mixture are drawn from the three
96 litres will now be 20% of the new total casks respectively and poured in the
quantity. fourth vessel.
20% of (320 + x) = 96 litres Therefore,
20/100 (320 + x) = 96 Amount of Liquid L in fourth vessel
(320 + x) = 480 =1+3+5=9
Amount of liquid M in fourth vessel
Therefore, x = 160 litres
= 2 + 4 + 6 = 12
Hence, Option (D) is the correct answer.
Amount of liquid N in fourth vessel
17. (B) = 3 + 5 + 7 = 15
Hence, the ratio of the liquids in the
resulting mixture
= 9 : 12 : 15 = 3 : 4 : 5
Hence, Option (B) is the correct answer.

Equate ethanol on both sides 19. (A)

(90x + 5xK) + 100x = 18.75% of 1600x
90 + 5K + 100 = (3/16) x 1600 = 300
5K = 110
K = 22
Hence, Option (B) is the correct answer.

18. (B) 3: 4: 5
The total amount of final mixture when
Let the capacities of the four casks be 36
20 kg of 1st alloy is mixed with
litres each.
12 kg, of 2nd alloy = 20 + 12 = 32 kg
Then,
∴ Amount of copper (Cu) in resultant
Amount of Liquid L in first vessel
alloy
= (1/6) × 36 = 6 litres 5
= × 32 = 20 kg
Amount of Liquid M in first vessel 8
= (1/6) × 36 = 12 litres Also, the amount of Zinc (Zn) in the
Amount of Liquid N in first vessel resultant alloy
3
= (1/6) × 36 = 18 litres = × 32 = 12 kg
8

Mixtures and Alligations 149

 Now, the amount of copper in the 2nd al- 4b − a = 2 …(2)
loy = 20 kg – 16kg = 4 kg. If we solve (1) and (2) equation we well
Also, amount of Zinc in 2nd alloy = 12 – 4 get b = 4/5
= 8 kg Also, 4b – a = 2
⇒ The ratio of copper (Cu) and Zinc in (16/5) − a = 2
2 nd
alloy is a = (16/5) − 2
Cu : Zinc a = 6/5
4 : 8 Therefore, the weight of the initial alloy =
1 : 2 (a + b) kg = (4/5) + (6/5) = 2
Hence, option (A) is the correct answer. Hence, option (D) is the correct answer.

20. (A) 22. (C)

Since, concentration Let the initial quantity of milk and water
Total amount of milk be P and Q litres respectively
× 100
Total quantity of mixture After adding 50 litres of water, milk will
3 × 40% × 60% 4 × 60% × 50% be P litres but water will be Q + 50 litres
3 × 40% 4 × 60% and according to the question milk con-
3 × 0.4 × 0.6 4 × 0.6 × 0.50
centration becomes 40 %, which means
3 × 0.40 4 × 0.60
1.92 1.92 milk to water ratio would be 40 : 60 or
× 10 × 100 = 53.33%
1.2 2.4 3.6 2:3
Therefore, concentration of the final mix- So, P/(Q + 50) = 2/3 …(1)
ture is 53.33% After adding 60 litres of more water,
Hence, Option (A) is the correct answer. milk will be P litres but water will be Q +
110 litres and according to the question
21. (D) milk concentration becomes 30%, which
Let the alloy initially contain ‘a’ kg gold means milk to water ratio would be 30 :
and ‘b’ kg copper. 70 or 3 : 7
When it is mixed with ‘b’ kg pure gold. So, P/(Q + 110) = 3/7 …(2)
(a 6) 90
Then, Solving I and II we will get Q = 58 and P
(a b 6) 100
= 72
10a + 60 = 9a + 9b + 54
Now we have 72 litres of milk, 168 (58 +
a − 9b = −6
110) litres of water and total solutions of
9b − a = 6 …(1)
240 (168 + 72) litres
Now, again according to the 2nd condition
Now according to the question, suppose
given in the question-
y litres of more water is added so that
Gold in 4kg alloy = 90% × 4 kg = 3.6 kg
(a 3.6) 80 the milk concentration becomes 15% or
Thus, milk to water ratio becomes 15 : 85 or 3
(a b 4) 100
a 3.6 4 : 17
a b 4 5 So, 72/(168 + y) = 3/17 solving which we
5a + 18 = 4a + 4b + 16 will get y = 240 litres
a − 4b = −2

Mixtures and Alligations 150

 Hence, 240 litres of water must be added After 60 litres of solution is taken out,
to make the milk concentration 15% the remaining would have 30 litres of
Hence, Option (C) is the correct answer. milk and 150 litres of water.
Now, x litres of milk is added and the ra-
23. 64
tio becomes M : W = 5 : 1
Let the amount of chemical ‘A’ in mixture
‘P’ = 3x units (30 + x) : 150 = 5 : 1
Also, the amount of chemical ‘B’ in mix- i.e. 30 + x = 750
ture ‘P’ = 5x units i.e. x = 720
Since chemical ‘A’ is prepared by mixing So, milk = 750 and water = 150
two raw materials ‘m’ and ‘n’ in the ratio Now, y litres water is added and ratio
of 2 : 3. again gets reversed
Quantity of ‘n’ raw material in chemical 750 : (150 + y) = 1 : 5
‘A’ = 3/5 x (3x) = 1.8x units i.e. y = 3600
Similarly, chemical ‘B’ is prepared by Therefore, y + x = 3600 + 720 = 4320
mixing raw materials ‘n’ and ‘K’ in the ra- Hence, option (C) is the correct answer.
tio of 3 : 2.
25. (A)
Quantity of ‘n’ in chemical ‘B’ =3/5 x (5x)
= 5x units Amount of spirit content in the first mix-
ture = 40% of 900 = 360 litres
The quantity of ‘n’ in mixture ‘P’ = (1.8x +
Let the amount of another mixture be ‘a’
3x) units = 4.8 x units
litres will be added with the first mixture.
Now, according to the question, 128 litres
Therefore, The proportion of spirit con-
of mixture ‘P’ is mixed with water to pre-
tent in the mixture now
pare the final mixture.
Total quantity of ‘P’ = (3x + 5x) units 360 60% of a 360 0.60a
8x = 128 litres 900 a 900 a

x = 16 litres Since the spirit contained in the result-

Also, it is given in the question that, if ing mixture is 45%.
the concentration of raw material ‘n’ in 360 0.6a 45
∴
the final mixture is 40%. 900 a 100
Quantity of n = Quantity of final mixture 360 0.6a 9
⇒
× 40% = 4.8x 900 a 20
Quantity of final mixture × (2/5) = 4.8 × 7200 + 12a = 8100 + 9a
16 litres 3a = 900
Quantity of final mixture = 192 litres a = 300 litres
Therefore, the quantity of water added =
Again, when spirit content will be 50% in
192 – 128 = 64 litres.
the resulting mixture, then the amount
24. (C) of 2nd mixture will be:
Ratio of milk and water in the vessel = 360 0.6a 1
1: 5. 900 a 2

Mixtures and Alligations 151

 720 + 1.2 a = 900 + a 2x (40 – k) + 2yk = 5y (16 – k) + 5xk
0.2 a = 180 80x – 2xk + 2yk = 80y – 5yk + 5xk
a = 900 litres 80 (x – y) = 7k (x – y)
Hence, the quantity of the 2nd mixture 80 = 7k ……….. as (x is not equal to y)
to be added should be more than 300 k = 80/7 = 11.42
litres but less than 900 litres. Hence, Option (A) is the correct answer.
Hence, Option (A) is the correct answer.
28. (C)
26. (A)
As the alcohol and water are in the same
proportion, the amount of alcohol would
be 150 ml.

Let n be the number of replacements.

n
30
150 1 ≤ 30% of 300
100
n M: W ⇒ 5R → 120 litres
1 90
⇒ 1 ≤
10 150 2 : 3 1R → 24 litres
+ 2R
1
n
90 2 : 5 2R → 24 × 2 = 48 li-
⇒ 150 1 ≤ tres
10 150
⇒ (0.9)n ≤ 0.6
As we know that
⇒ (0.9)2 = 0.81
⇒ (0.9)4 = (0.81)2 = 0.6561
Four replacements have already been done. Therefore, water added in the mixture is
Now, if we multiply 0.6561 with 0.9, the 48 litres.
product would be 0.59049 (which is less Now, the total mixture becomes
than 0.6) = 120 litre + 48 litres = 168 litres
The remaining quantity of alcohol = 150 × Since 1/5th part of this mixture is taken
(0.9)5 = 88.57 ml (approx.) out, then the remaining mixture
So, Ganpat must have consumed 150 − = 168 x (4/5) = 134.4 litres
88.57 = 61.42 ml ~ 61 ml of alcohol. Again, some quantity of milk is added to
Hence, option (A) is the correct answer. the remaining mixture.
27. (A) Then,
Let the % of alcohol in solutions P and Q
are x% and y% respectively. Let the vol-
ume taken from both the solutions be
‘k’ litres. Equating the % of alcohol from
both P and Q
(x% of (40 – k) + y% of k)/40 = (y% of (16
–k) + x% of k)/16

Mixtures and Alligations 152

 Now, solving equation 1 and 2.
3a + 2b = 250
2.1a + 3.15b = 300
6.3a + 4.2b = 525
⇒ 21R → 134.4 litre ⇒ 6.3a + 9.45b = 900
134.4
1R → litre = 6.4 litre –5.25b = –375
21
b = 500/7
∴ 4R → 4 × 6.4 = 25.6 litre Put the value of b = 500/7 in equation (1),
Hence, 25.6 litres of milk is added to the we will get a = 250/7
mixture, so that ratio of milk and water Therefore, the cost prices, per kg of x
becomes 2: 3. and y are in the ratio
Hence, option (C) is the correct answer. x : y = 250/7 : 500/7 = 1 : 2
29. (C) Hence, option (C) is the correct answer.

Let ‘a’ be the price of 1 kg of pulse ‘x’ is Statement 1: The milkman has the de-

the mixture and ‘b’ be the price per kg of sired proportion when he has added 20

pulse ‘y’. L of water.

Case-1 Statement 2: The milkman has the de-

Since the selling price of the mixture is sired proportion when he has added 25

Rs. 60/kg and a profit of 20% is given. L of water.

Let the cost price of the mixture is CP.

30. (C)
CPmix1 × 120% 60
For a content, less than 80 % of
60 100 litres, the quantity of milk should be
CPmix1 Rs. 50 / kg
120 below 80 litres
Price per kg of the mixture when pulses After 1st time: Tank has 95 litres milk and
are mixed in the ratio 3 : 2. 5 litres water

3×a 2×b For, 2nd time, when 5 litres mixture is

⇒ 50 removed and 5 litres water is added
5
Milk: 95% of previous: 95% of 95 litres milk
3a + 2b = 250 …(1)
= 90.25L milk
Case-2
Remaining 100 – 90.25 litres = 9.75 litres
The profit is 5% if the two vertices of
is water
pulses are mixed in 2 : 3.
For, 3rd time, when 5 litres mixture is
CPmix2 × 105% 60
removed and 5 litres water is added
60 Milk: 95% of previous: 95% of 90.25 litres
CPmix2
105 milk = 85.7375 litres milk
∴ Price per kg of the mixture when Remaining 100 – 85.7375.25 litres
pulses are mixed in the ratio 2 : 3. = 14.2625 litres is water
2a 3b 60
For, 4th time, when 5 litres mixture is
5 105
removed and 5 litres water is added
2.1a + 3.15b = 300 …(2)

Mixtures and Alligations 153

 Milk: 95% of previous: 95% of 85.7375
litres milk = 81.45 litres milk
Remaining 100 – 81.45 litres = 18.55 litres is water
For, 5th time, when 5 litres mixture is removed and 5 litres water is added
Milk: 95% of previous: 95% of 81.45L milk = 77.37 litres milk
Remaining 100 – 77.37 litres = 22.63 litres is water
Thus, it needs at least 5 steps to reach the desired concentration.
Any further step will lower milk content which also lies in the range i.e. less than 80%.
Thus, he needs to add water at least 5 times i.e. 5 × 5 = 25 litres
Hence, statement 1 is false and statement 2 is true.
Hence, option (C) is the correct answer.

MIND MAP

Mixtures and Alligations 154

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