Alligation And Mixture

Mixture: Mixing of two or more than two type of

quantities gives us a mixture.

Example:

Quantities of these elements can be expressed as

percentage or ratio.(20% of sugar in water)

Fraction ( A solution of sugar and water such that sugar

: water = 1:4)
Alligation : Alligation is a rule which is used to
solve the problems related to mixture and its
ingredient.

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.
Alligation Rule
When two elements are mixed to make a mixture and
one of the elements is cheaper and other one is
costlier then,
Diagram Representation
Practice Questions
Q1. In what ratio must a grocer mix two varieties of
pulses costing Rs. 15 and Rs. 20 per kg respectively
so as to get a mixture worth Rs. 16.50 kg?

A.7:3
B.4:5
C.6:4
D.None
 Option A
• 15 20
16.5
3.5 1.5

35/15=7:3 Ans
Verification
7 * 15 + 3 * 20 = 165 = 10 * 16.5
Q2. How much water be added to 14 litres of milk
worth Rs. 5.40 a litre so that the value of the mixture
may be Rs. 4.20 a litre ?

A. 7 litres
B. 6 litres
C. 5 litres
D. 4 litres
 Option D
W M
0 5.40
4.20
1.2 4.20
So Ratio of Cheaper (Water) / Costlier (Milk) = 1.2/4.2=2/7
W:M
2 : 7 if milk is 14 litres water must be 4 litresAns
Verification
Let total mixture be x litres
Cp per kg of mixture =4.2
x*4.2=14*5.4
X=18
Q3. How many kg of tea worth Rs. 20/kg must be
blended with 30 kg of tea worth Rs. 30/kg so that by
selling the blended variety at Rs. 30/kg there should
be a gain of 25%?

A. 45 kg
B. 40 kg
C. 50 kg
D. None
 Option A
% increase 25%=1/4
% decrease = 1/5 so CP of Mixture / Kg = 30-1/5*30=24
Or
SP=125/100 *CP i.e 30=125/100 *CP
so CP of mixture =24
20 30
24
6 4 i.e Quantity of Cheaper/ Quantity of Costlier = 6/4=3/2
So Ans should be 45Kg because 3/2=x/30
Verification
Total mixture 45 Kg + 30 Kg = 75 Kg Total Cp of mixture = 75*24=1800
Total SP of mixture = 75 *30=2250
clearly profit % is 25% as Given
Q4. In what ratio must water be added to spirit to gain
10% by selling it at the cost price?

A. 1 : 11
B. 1:5
C. 1 : 10
D. 1:9
 Option A
% increase=10%=1/10
% decrease = 1/11 Ans
Verification
%decrease=100/11=9.09%
If Cp of 100 l Spirit was Rs100
now he is selling 100-9.09=90.91l for rs 100
So cp of 90.9 l=Rs 90.9
P= 100-90.9=9.1 p%=9.1/90.9 * 100 = 10% as
given
Q. In what ratio must a grocer mix two varieties of
tea worth Rs. 60 a kg and Rs. 65 a kg so that by
selling the mixture at Rs. 68.20 a kg he may gain
10%?

A.3:2
B.3:5
C.3:6
D.None
 Option A
• SP=110/100 * cp
• 68.2=110/100 * cp cp (mixture)=62/kg

• 60 65
62
3 2
Verification
Total CP=3Kg @ 60/kg + 2 kg @65/kg=310 or 5Kg@62=310
Total SP= 5Kg@68.2/kg = 341 clearly 10 % profit
Q. Sea water contains 5% salt by weight. How many
kilograms of fresh water must be added to 40kg of sea
water for the salt content of the solution to be 2%?

A. 50
B. 60
C. 65
D. 70
 Option B
• 40 Kg of sea water salt = 5% of 40 = 2 Kg
• Now for Salt to be 2% i.e 2 with respect to 100

• Add 60 Kg Ans
Q. In a mixture of 60 L, the ratio of milk and water is
2:1, If the rate of milk and water is to be 1:2, then the
amount of water to be further added must be ?

A. 12 L
B. 20 L
C. 40 L
D. 60 L
 Option D
• 60 L contain M = 40 L and W= 20 L
• now W should be double than M i.e W=80 L
• So ans is 60L
Q. An alcohol water mixture of 729 L is in the ratio 7:2,
how much more water is to be added to get a new
alcohol and water ratio of 7:3?

A. 81 L
B. 72 L
C. 80 L
D. 78 L
Q. A mixture of certain quantity of juice with 32 liters
of water is worth 1.50 per liter. If pure juice is to be
worth 4.50 per liter, how much juice is there in the
mixture?

A. 18 L
B. 14 L
C. 16 L
D. 20 L
• 16 * 4.5=72
• (16+32)*1.5 is also 72
Q. A mixture of 45 L of spirit and water contains 20%
of water in it. How much water must be added to it to
make the water 25% in the new mixture?

A. 3L
B. 4L
C. 5L
D. 6L
 Option A
• 45 L contains 9 L water 36 L spirit
• So 9 +3 =12 which is 25% of 45 + 3 =48
Q. One mixture contains 25% milk and other contains
30% milk and the rest water. A jar is filled with 6 parts
of first mixture and 4 parts of second mixture. The
percentage of milk in the mixture is

A. 23%
B. 67%
C. 27%
D. 16%
 Option C
• Mix 1 In 100 parts 25 part milk
in 6 parts 25/100 * 6 =1.5 part milk

Mix 2 In 100 parts 30 part milk

in 4 parts 30/100 * 4 = 1.2 part milk

So in 10 parts 2.7 part milk

In 100 parts 27 part milk
Q. A dishonest milkman professes to sell his milk at
cost price but he mixes it with water and thereby
gains 25%. The percentage of water in the mixture is:

A.20%
B.10%
C.11%
D.None
• % increase = 1/n
• % decrease = 1/(n+1)
Q. In what ratio must a person mix three kinds of tea
costing Rs.60/kg, Rs.75/kg and Rs.100 /kg so that the
resultant mixture when sold at Rs.96/kg yields a profit
of 20%?

A. 1:2:4
B. 3:7:6
C. 1:4:2
D. None of these
• Total CP=60*1+ 75 * 4 + 100 * 2= 560
• or
• Total CP=7*80(CP of mixture) = 560
• Total SP= 7 * 96=672 profit is clearly 20 %
Removal and Replacement
If a vessel contains “x” liters of liquid A and if “y” liters
be withdrawn and replaced by liquid B, then if “y” liters
of the mixture be withdrawn and replaced by liquid B,
and the operation is repeated „n‟ times in all, then :

F.C = I.C(1-y/x)n
FC= Final concentration
IC= Initial concentration
y = no. of liters replaced
x = Total concentration
n = total number of iterations
Q. A vessel contains 125 litres of wine. 25 litres of wine
was taken out of the vessel and replaced by water. Then,
25 litres of mixture was withdrawn and again replaced
by water. The operation was repeated for third time.
How much wine is now left in the vessel?

A. 49 L
B. 64 L
C. 72 L
D. 56 L
• 125 * ( 1 – 25 /125 ) 3
Q A container contains 40 liters of milk. From this
container 4 liters of milk was taken out and replaced
with water. This process was repeated further three
times. How much milk is now contained by the
container?

A. 27.36 L
B. 26.24 L
C. 29.36 L
D. 24.36 L
• 40 * ( 1 – 4 / 40 ) 4
Q. From a cylindrical can containing milk of 30 liters, 5
liters are drawn and replaced with water. If the same
process is repeated for 3 more times, what is the amount
of milk left in the cylindrical can?

A. 14.46 L
B. 15.56 L
C. 17 L
D. 20 L
• 30(1 - 5/30)4