Dear students in the series of Study Notes this week our topic
is Mixture and allegation which is extensively used as a short
cut while solving problems of other chapters. So we can say
this is very important chapter specially in bank exams.
So lets start
Alligation
I. Alligation : It enables us to find the ratio in which two
or more ingredients at the given price must be mixed to
produce a mixture of a desired price.
II. Mean price : The cost price of a unit quantity of the
mixture is called the mean price.
III. Basic Formula : If two ingredients are A and B are
mixed of price x and y respectively, then the price of
resultant mixture is M mean price. The ratio in which
ingredients are mixed is given by rule of alligation
The above formula can be respented as
Thus the required ratio is R
Ex. In what ratio must rice costing Rs.8.50 per kg be mixed
with rice costing Rs.13 per kg so that the mixture be
worth Rs.10 per kg?
Sol.
Ex. In what ratio must a grocer mix two varities of sugr
costing Rs.60 per kg and Rs.65 per kg, so that on selling
the mixture at Rs.68.20 per kg he may gain 10%?
Sol. Cost price of 1 kg of mixture
Required Ratio = 3 : 2
Ex. 729 Litres of mixtures containing milk and water in the
ratio 7 : 2. How much more water should be added so
that the new mixture contains milk and water in the ratio
7 : 3 ?.
Sol. 729 litre quantity of water =
Pure water = 1
Quantity of water in mixture =
APPLYING ALLIGATION:
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Required Ratio = 9 : 1
9 = 729 litre
1 = ×1= 81 Litres
IV. Mixture of more than two elements:
These question may seems to be a little tricky at first,
but similar concept is applied repeatedly in order to
calculate final ratio of ingredients when mixture
contains more than two ingredients.
1. Take two ingredients in such away that first ingredient
is lower than the mean value and the other one is higher
than mean value.
2. Calculate the ratio of ingredients.
3. Repeat for all possible pairs.
4. Final ratio is the ratio obtained from step 2 (if an
ingredients is common in the ratios, add value for this
particular ingredient)
Ex. Three types of Rice of Rs. 1.27, Rs. 1.29 and Rs. 1.32
per kg are mixed together to be sold at Rs. 1.30 per kg.
in what ratio should this rice be mixed.
Sol.
Hence final ratio is 2 : 1 : 3 + 2 = 2 : 1 : 5
V. Mixture containing 4 ingredients:
Ex. How much a shop owner mixture 4 types of rice worth
Rs. 95, Rs. 60, Rs. 90 & Rs. 50 perkg so that he can
make the mixture of these rice worth Rs. 80 perkg.
Sol.
I : II : III : IV = 4 : 4 : 5 : 1
VI. Concept of replacement:
Suppose a container contains a – units of liquid from
which b -units are taken out and replaced by water- after
k-operations, the quantity of pure liquid
Ex. 8 litres are drawn from a cask full of wine and is then
filled with water. This operation is performed three
more times. The ratio of the quantity of wine now left
in cask to that of the water is 16 :65. How much wine
did the cask hold originally ?
Sol. Let initial quantity of wine in a cask = x lit
So, After 4 process,
Q.1. In 330 litres of mixtures of milk and water, water is only
24%. The milkman sold 80 litres of this mixture and
then he added 60 litres of pure milk and 26 litres of pure
water in the remaining mixture. What is the percentage
of water in the final mixture?
A39.70%
B25. 59%
C34. 68%
D37. 34%
E16. 89%
Sol: Since, milkman sold 80 litre of mixture
So, remaining mixture
= 330 – 80 = 250 litre
Quantity of water
= 250 × 24/100 = 60 litre
Quantity of milk
= 250 – 60 = 190 litre.
Now, milkman made new mixture in
which
water = 60 + 26 = 86 litre
milk = 190 + 60 = 250 litre
Percentage of water in the new mixture
= 86/(86 + 250) × 100
Q.2. A paint maker has three different qualities of
paint. 806 barrels of 1st quality, 930 barrels of 2nd
quality and 992 barrels of 3rd quality. Find the
least possible number of buckets of equal size in
which different paint of different qualities can be
filled without mixing?
(A) 34
(B) 46
(C) 26
(D) 44
(E) None of these
Sol : Oops! Wrong answer!
It is given that paint of, Ist quality = 806 barrels
IInd quality = 930 barrels
IIIrd quality = 992 barrels
Since,
Least number of buckets of equal size will be
possible, when we have bottle having highest of
largest size.
Largest size having highest or largest size.
Largest size bottle can be found by finding HCF
of 806, 930 & 992.
HCF = 2×31 = 62
Therefore, Total numbers of barrels required
=13+15+16 = 44
Q.3. Two vessels A and B contain milk and water mixed
in the ratio 4 : 3 and 2 : 3. The ratio in which these
mixtures be mixed to form a new mixture
containing half milk and half water is
(A) 7 : 5
(B) 6 : 5
(C) 5 : 6
(D) 4 : 3
Sol.
Q.4. A milkman makes 20% profit by selling milk
mixed with water at Rs. 9 per litre. If the cost price
of 1 litre pure milk is Rs. 10, then the ratio of milk
and water in the said mixture is
(A) 3 : 1
(B) 4 : 1
(C) 3 : 2
(D) 4 : 3
So: Short Trick:
SP of Mixture = 9 Rs
120% CP = 9 Rs
CP of Mixture = (9/120)×100 = 7.5 Rs
CP of Milk = 10 Rs
CP of Water = 0 Rs
Using (A) lligation on CP
Basic Method:
Milk : Water = K : 1
S.P. = (K + 1) × 9
C.P. = 10 K
Gain = 9 – K
Gain
Ratio = 3 : 1
Q.5. In two types of stainless steel, the ratio of
chromium and steel are 2 : 11 and 5 : 21
respectively. In what proportion should the two
types be mixed so that the ratio of chromium to
steel in the mixed type becomes 7 : 32?
(A) 2 : 3
(B) 3 : 4
(C) 1 : 2
(D) 1 : 3
Sol: Stainless Steel I II III
Let’s take part of Chromium
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1/78 : 1/39
Required ratio =
Q.6. A vessel has 60 litres of solution of acid and
water having 80% acid. How much water be
added to make it a solution in which acid forms
60%?
(A) 48 litres
(B) 20 litres
(C) 36 litres
(D) None of these
Sol. In 60 litres of solution,
Water =
On adding litres of water,
litres
Q.7. A mixture contains alcohol and water in the ratio
4 : 3. If 5 litres of water is added to the mixture,
the ratio becomes 4 : 5. The quantity of alcohol in
the given mixture is
(A) 3 litres
(B) 4 litres
(C) 15 litres
(D) 10 litres
Sol. In original mixture,
Alcohol = 4x litres
Water = 3x litres
On adding 5 litres of water,
Q.8. A Can contains a mixture of two liquids A and B
in the ratio 7:5. When 9 litres of mixture are
drained off and the Can is filled with B, the ratio
of A and B becomes 7 : 9. How many litres of
liquid A was contained by the Can initially?
(A) 10
(B) 20
(C) 21
(D) 25
Sol. Suppose the can initially contains 7x and 5x of A
and B respectively
Quantity of A in mixture left=7x-(7/12 ×9)
=7x-21/4
Quantity of B in mixture left = 5x-(5/12 ×9)
=5x-15/4
Thus (7x-21/4) ÷ (5x-15/4)+9=7/9
x=3
The can contained 21 litres of A
Q.9. The ratio of spirit and water in two mixtures of 20
litre and 36 litre is 3 : 7 and 7 : 5 respectively.
Both the mixtures are mixed together. Now the
ratio of the spirit and water in the new mixture is
(A) 25 : 29
(B) 9 : 10
(C) 27 : 29
(D) 27 : 31
Sol. In 20 litres of mixture,
Spirit = litre
Water = 14 litre
In 36 litres of mixture
Spirit = litre
Water = 15 litre
Required ratio = (21 + 6) : (14 + 15)
= 27 : 29
Q.10. A litre of pure alcohol is added to 6 litres of
30% alcohol solution. The percentage of
water in the solution is
(A) 50%
(B) 65%
(C) 60%
(D) 40%
Sol. In 30% alcohol solution,
Alcohol = litre
Water = 4.2 litre
On mixing 1 litre of pure alcohol,
Percentage of water =
Q.11. In what ratio must a mixture of 30% alcohol
strength be mixed with that of 50% alcohol
strength so as a mixture of 45% alcohol
strength?
(A) 3:1
(B) 1:3
(C) 2:3
(D) 4:1
Sol.
Hence, the ratio of 30% alcohol: 50% alcohol
= 1:3
Q.12. A shopkeeper mixed two varieties of rice at
Rs. 20/kg and Rs. 30/kg in the ratio 2 : 3 and
sell the mixture at 10% profit. Find the price
per kg at which he sold the mixture?
(A) Rs. 26
(B) Rs. 28.8
(C) Rs. 28
(D) Rs. 28.6
Sol. Cost Price of Mixture = (20×2+30×3)/(2+3) =
130/5 = 26 Rs/Kg
Selling Price of Mixture = (26×110)/100 =
28.60 Rs/Kg
Q.14. In a container A, ratio of milk and water is 5:3 &
in container B, ratio of milk and water is 11:5. If
they are mixed in a ratio of 2:3 in another
container, find the ratio of milk and water in the
new mixture?
(A) 53:27
(B) 27:53
(C) 53:80
(D) 27:80
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Sol.
Hence, the amount of milk in new mixture
(X) = (5/8×2 + 11/16×3)/(2+3) = (5/4
+33/16)/ 5
= 53/80
The ratio of milk and water in the new
mixture = 53/(80-53) = 53:27
Q.15. A vessel full of whisky contains 40% alcohol. A
part of this whisky is replaced by another
containing 19% alcohol and now the percentage
of alcohol was found to be 26%. The quantity of
whisky replaced is :
fa
A1/3
B2/3
C2/5
D3/5
Sol.
Ratio of 1st and 2
nd quantities= 7:14=1:2
So, Required Quantity replaced = 2/3
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