Quantitative Aptitude Notes: Allegations and Mixtures
1. Allegation: Allegation
is the rule which 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 specified price.
2. Mean Price: Mean price
is the cost price of a unit quantity of the mixture
3. Suppose a container contains
x units of a liquid from which y units are taken out and replaced by water. After
n operations, quantity of pure liquid is
x (1 – y/x)^{n}
4. Rule of Allegation: If two
ingredients are mixed, then
(Quantity
of cheaper)/(Quantity of dearer) = (CP of dearer – Mean Price)/(Mean Price – CP
of Cheaper)
The above
formula can be represented with the help of the following diagram which is
easier to understand.
=>
(Cheaper quantity) : (Dearer quantity) = (d  m) : (m  c)
5. Replacement of Part of
Solution: Suppose a container contains a solution from which some quantity of solution
is taken out and replaced with one of the ingredients. This process is repeated
n times then,
Final
Amount of ingredient that is not replaced = Initial Amount × [Vol. after
removal/Vol. after replacing]^{n}
Above
formula is not only true for absolute amounts but for ratios as well. So
following formula is also valid:
Final ratio
of ingredient not replaced to total = Initial Ratio × [Vol. after removal/Vol.
after replacing]^{n}
6. Mixture of More than
two elements: These questions may seem a little tricky at first,
but it is similar concept applied repeatedly.
In order to
calculate final ratio of ingredients when mixture contains more than two
ingredients,
a) Take two
ingredients such that 1st ingredient is LOWER than the mean value and the other
one is HIGHER than the mean value.
b)
Calculate the ratio of ingredients
c) Repeat
for all possible pairs
d) Final
ratio is the ratio obtained from step 2 (if an ingredient is common in the
ratios, add values for this particular ingredient)
MEMORY
BASED SOLVED EXAMPLES BASED ON VARIOUS TYPES
Type 1 – Based on Quantity
1. A goldsmith has two qualities of gold, one of 10 carats and
another of 15 carats purity. In what proportion should mix both to make an
ornament of 12 carats purity?
Solution:
Given,
One quality
of gold = 10 carats
Another
quality of gold = 15 carats
By
allegation rule,
Required
ratio = 15 – 12: 12 – 10 = 3: 2
2. How many kilograms of dearer sugar costing Rs. 5.75 per kilogram
are mixed with 75 kilogram of cheaper sugar costing Rs. 4.50 per kilogram so
that the mixture is worth Rs. 5.50 per kilogram?
Solution:
Cost price
of 1 kilogram of cheaper sugar = Rs. 4.50
Cost price
of 1 kilogram dearer sugar = Rs. 5.75
Mean price
= Rs. 5.50
Quantity of
cheaper sugar = 75 kilogram
Here, Rs.
5.75 – Rs. 5.50 = Rs. 0.25
Rs. 5.50 –
Rs. 4.50 = Rs. 1.00
Therefore,
required ratio is
(Quantity
of cheaper sugar)/(Quantity of dearer sugar) = 0.35/1 = ¼
=> 75/(Quantity
of dearer sugar) = ¼ => Quantity of dearer sugar = 75 × 4 = 300
⇒ Quantity
of dearer sugar = 75 x 4 = 300 kilograms
Therefore,
Quantity of dearer sugar = 300 kilograms
3. Two varieties of wheat  A and B costing Rs. 9 per kg and Rs. 15
per kg were mixed in the ratio 3: 7. If 5 kg of the mixture is sold at 25%
profit, find the profit made?
Solution:
Let the
quantities of A and B mixed be 3x kg and 7x kg.
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
4. The ratio of expenditure and savings is 3: 2. If the income
increases by 15% and the savings increases by 6%, then by how much percent
should his expenditure increases?
Solution:
6

X


15


6 = 2k

3k = 9

Therefore x
= 21%
5. A merchant has 1000 kg of sugar, part of which he sells at 8%
profit and the rest at 18% profit. He gains 14% on the whole. The quantity sold
at 18% profit is:
Solution:
By the rule
of allegation, we have:
Profit on 1^{st} part
8%

Profit on 2^{nd} Part
18%


Mean Profit
14%


4

6

Ratio of
1st and 2nd parts = 4 : 6 = 2 : 3
Quantity of
2nd kind = 3/5 x 1000kg = 600 kg.
Type 2 – Based on mixer of two or more varieties:
1. Teas worth Rs. 126 per kg and Rs. 135 per kg are mixed with a
third variety in the ratio 1: 1: 2. If the mixture is worth Rs 153 per Kg, the
price of the third variety per Kg will be?
Solution:
Since first
and second varieties are mixed in equal proportions.
So the
average price = (126 + 135)/2 = Rs.130.5
So, the
mixture is formed by mixing two varieties, one at Rs. 130.50 per kg and the
other at say, Rs.xper kg in the ratio 2 : 2, i.e., 1 : 1.
By the rule
of allegation, we have:
Cost of 1Kg of 1^{st} kind
Rs.130.50

Cost of 1kg of 2^{nd} Kind
Rs.x


Mean Price
Rs.153


x153

22.50

=> (x –
153)/22.50 = 1 => x – 153 = 22.50 => x = Rs. 175.50
2. A milk vendor has 2 cans of milk .The first contains 25% water
and the rest milk. The second contains 50% water. How much milk should he mix
from each of the container so as to get 12 litres of milk such that the ratio
of water to milk is 3:5?
Solution:
Let the
cost of 1 litre milk be Rs. 1
Milk in 1
litre mix. in 1st can = ¾ litre CP of 1litre mixture in 1^{st} can Rs. ¾
Milk in 1
litre mix. in 2nd can = ½ litre CP of 1litre mixture in 1^{st} can Rs. ½
Milk in
1litre of final mix. = 5/8 litre, Mean price = Rs.5/8
By the rule
of allegation, we have:
So,
quantity of mixture taken from each can =1/2 x 12= 6 litres.
3. How many kgs of Basmati rice costing Rs.42/kg should a
shopkeeper mix with 25 kgs of ordinary rice costing Rs.24 per kg so that he
makes a profit of 25% on selling the mixture at Rs.40/kg?
Solution:
Let the
amount of Basmati rice being mixed be x kgs.
As the
trader makes 25% profit by selling the mixture at Rs.40/kg, his cost per kg of
the mixture = Rs.32/kg.
i.e.
(x×42)+(25×24)=32(x+25)
⇒ 42x+600=32x+800
⇒ 10x=200
⇒ x=20kgs.
4. How many litres of oil at Rs.40 per litre should be mixed with
240 litres of a second variety of oil at Rs.60 per litre so as to get a mixture
whose cost is Rs.52 per litre?
Solution:
The two verities
of oil should be mixed in the ratio 2 : 3. So, if 240 litres of the 2^{nd}
variety are taken, then the 1^{st} variety should be taken as 160
litres
5. Two vessels P and Q contain 62.5% and 87.5% of alcohol
respectively. If 2 litres from vessel P is mixed with 4 litres from vessel Q,
the ratio of alcohol and water in the resulting mixture is?
Solution:
Quantity of
alcohol in vessel P = 62.5/100 * 2 = 5/4 litres
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.
Type 3 – Based on addition or removal in mixture:
1. A vessel is filled with liquid, 3 parts of which are water and 5
parts of syrup. How much of the mixture must be drawn off and replaced with
water so that the mixture may be half water and half syrup?
Solution:
Let the
quantity of the liquid in the vessel = 8 litre.
Then,
quantity of water in the liquid = 3 litre,
and
quantity of syrup in the liquid = 5 litre.
Suppose x
litre of the mixture is drawn off and replaced with water. Then,
Quantity of
water in the new mixture = 3− 3x/ 8 + x
Quantity of
syrup in the new mixture =5−5x/ 8
Given that
in the new mixture, quantity of water = quantity of syrup
⇒3−3x/8+x=5−5x/8
⇒10x/8=2⇒5x/4=2⇒x=8/5
i.e., if
the quantity of the liquid is 8 litre, 8/5 litre of the mixture needs to be
drawn off and replaced with water so that the mixture may be half water and
half syrup.
It means
1/5 or 20% of the mixture needs to be drawn off and replaced with water so that
the mixture may be half water and half syrup.
2. In a mixture of milk and water, there is only 26% water. After
replacing the mixture with 7 litres of pure milk, the percentage of milk in the
mixture become 76%. The quantity of mixture is:
Solution:
Initially
water is 26% and finally water is 24%
This means,
2% water is removed due to the removal of 7 litre of the mixture.
In other
words, 7 litre of the mixture contains 2% of water
Therefore,
to contain 26% of water, mixture quantity =7×26/2=91 litre
3. The ratio of petrol and kerosene in the container is 3:2 when 10
litres of the mixture is taken out and is replaced by the kerosene, the ratio
become 2:3. Then total quantity of the mixture in the container is:
Solution:
petrol :
kerosene
3 :
2(initially)
2 : 3(after
replacement)
Therefore
the total quantity of the mixture in the container is 30 liters.
4. From a container, 6 litres milk was drawn out and was replaced
by water. Again 6 litres of mixture was drawn out and was replaced by the
water. Thus the quantity of milk and water in the container after these two
operations is 9:16. The quantity of mixture is:
Solution:
Let
quantity of mixture be x liters.
Suppose a
container contains x units of liquid from which y units are taken out and
replaced by water. After operations, the quantity of pure liquid = x (1 – y/x)^{n}
units, Where n = no of operations.
So,
Quantity of Milk = x (1 – 6/x)^{2}
Given that,
Milk : Water = 9 : 16
=> Milk
: (Milk + Water) = 9 : (9+16)
=> Milk
: Mixture = 9 : 25
Therefore, [x
(1 – 6/x)^{2}]/x = 9/25 => x = 15 litres
5. A can contains a mixture of two liquids A and B in the ratio 7:5
when 9 litres of mixture are drawn 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?
Solution:
Suppose the
can initially contains 7x and 5x of mixtures A and B
respectively.
Quantity of
A in mixture left = [7x – (7/12) × 9] litres = [7x – 21/4] litres
Quantity of
B in mixture left = [5x – (5/12) × 9] litres = [5x – 15/4] litres
Therefore,
[7x – 21/4]/ [5x – 15/4] = 7/9 => (28x – 21)/(20x + 21) = 7/9
=> 252x –
189 = 140x + 147 => 112x = 336 => x = 3
So, the can
contained 21 litres of A.
6. A painter mixes blue paint with white paint so that the mixture
contains 10% blue paint. In a mixture of 40 litres paint how many litres blue
paint should be added so that the mixture contains 20% of blue paint.
Solution:
10%

100%


20%


80

10

Ratio = 80
: 10 or 8 : 1
In 40
litres of mixture, 5 litres of blue paint must be mixed, Ratio becomes 40 : 5
or 8 : 1
7. 10 gallons are drawn from a container full of alcohol and filled
with water again. 10 gallons of mixture are again drawn and the container is
filled with water again. If the ratio of alcohol and water left in the
container is 49: 32, then find how much quantity does the container hold?
Solution:
=> First
assume that the initial quantity of alcohol is ‘A’.
=> We
are given that, the ratio of alcohol and water is 49: 32
=> Assume
initial quantity of alcohol in the container = 49 + 32 = 81  (This is
because we have assumed that initial quantity of alcohol = final quantity of
water and alcohol)
=> Subtract
the quantity of alcohol replaced by water from the initial quantity of alcohol
(A – B).
As this
operation is repeated n times, therefore (A – B)n
Therefore,
(Quantity
of alcohol left after nth operaio)/(Initial quantity of alcohol) = [(A – B)^{n}/A]
=> 49/81
= [(A – 10)^{2}/A]
Solving, we
can find the value of A (initial quantity of alcohol) A = 45 gallons
8. In a mixture of milk and water, the proportion of milk by weight
was 80%. If, in a 180 gm mixture, 36 grams of pure milk is added, what would be
the percentage of milk in the mixture formed?
Solution:
Percentage
of milk in the mixture formed = [80/100 (180) + 36] / (180 + 36) * 100%
= (144 +
36)/216 * 100% = 5/6 * 100% = 83.33%.
9. A mixture of 70 liters of milk and water contains 10% water. How
many liters of water should be added to the mixture so that the mixture
contains 12 1/2% water?
Solution:
Quantity of
milk in the mixture = 90/100 (70) = 63 litres.
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
10. A vessel contains 20 liters of a mixture of milk and water in
the ratio 3:2. 10 liters of the mixture are removed and replaced with an equal
quantity of pure milk. If the process is repeated once more, find the ratio of
milk and water in the final mixture obtained?
Solution:
Milk = 3/5
* 20 = 12 liters, water = 8 liters
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.
Type 4 – Ratio of ratios
1. The milk and water in two vessels S and T are in the ratio 4:
3and 2: 3 respectively. In what ratio, the liquids in both the vessels be mixed
to obtain a new mixture in vessel U containing 1/2 milk and ½ water?
Solution:
Given,
Ratio of milk and water in vessel S = 4: 3
Ratio of
milk and water in vessel T = 2: 3
Let cost
price of milk be Rs. 1 litre
Milk in 1
litre mixture of S = 4/7 litre
Milk in 1
litre mixture of T = 2/5 litre
Milk in 1
litre mixture of U = 1/2 litre
Therefore, Cost
price of 1 litre mixture in S = Rs. 4/7
Cost price
of 1 litre mixture in T = Rs. 2/5
Mean price
= Rs. 1/2
By the rule
of allegation,
Required
ratio = 2/5 – 1/2: 1/2 – 4/7 ⇒ 1/10:1/14⇒ 7: 5
2. Two containers P and Q contain milk and water in the ratio of 5:
2 and 7: 6 respectively. Find the ratio in which these two mixtures can be
mixed so that a new mixture formed in the container R is in the ratio of 8: 5.
Solution:
Let the
cost price of milk be Rs. 1 per litre.
Therefore,
cost of milk in 1 litre of mixture in
Container A
(Milk : Water = 5 : 2) = (5/7) x Rs.1 = Rs.5/7
Container B
(Milk : Water = 7 : 6) = (7/13) x Rs.1 = Rs.7/13
Container C
(Milk : Water = 8 : 5) = (8/13) x Rs.1 = Rs.8/13
Now use the
rule of allegation, to find the required ratio
The
required ratio of milk and water: 1/13 : 9/91 = 7 : 9
3. Three vessels whose capacities are in the ratio of 6:3:2 are
completely filled with milk and water. The ratio of milk and water in the
mixture 2:3, 4:2 and 5:2. Taking 1/4 of first, 1/2 of second and 1/2 of third,
new mixture kept in a new vessel. What is the percentage of water in the new
mixture?
Solution:
4. The ratio of milk and water in four solutions are 1:2, 2: 3, 3:2
and 7: 8respectively. If their equal quantities are mixed together, then the
ratio of milk and water in the new solution will be:
Solution:
Milk in the
resultant mixture = 1/3 + 2/5 + 3/5 + 7/15 = 27/15
Water in
the resultant mixture = 4 – 27/15 = 33/15
Therefore,
Ratio of milk and water = 27/15: 33/15 = 27: 33 = 9: 11
5. Two vessels are full of milk with milkwater ratio 1: 3 and 3: 5
respectively. If both are mixed in the ratio 3: 2, what is the ratio of milk
and water in the new mixture?
Solution:
Quantity of
milk in two mixtures = 1/4 and 3/8
Quantity of
mixture taken from two vessels 3/5 and 2/5 respectively.
Quantity of
milk in resultant mixture = 3/5 * 1/4 + 2/5*3/8 = 3/20 + 3/20 = 3/10
Quantity of
water in resultant mixture = 13/10 = 7/10
Ratio of
milk and water = 3/10: 7/10 = 3: 7