### APTITUDE-ALLIGATION OR MIXTURES PROBLEMS

ALLIGATION OR MIXTURES

Important Facts and Formula:

1.Allegation:It is the rule that enables us to find the ratio in which two of more ingredients at the given price must be mixed to produce a mixture of a desired price.

2.Mean Price:The cost price of a unit quantity of the mixture is called the mean price.

3.Rule of Allegation:If two ingredients are mixed then

Quantity of Cheaper / Quantity of Dearer =
(C.P of Dearer â€“ Mean Price) /(Mean Priceâ€“C.P of Cheaper).

C.P of a unit quantity of cheaper(c) C.P of unit quantity of dearer(d)

Mean Price(m)

(d-m) (m-c)

Cheaper quantity:Dearer quantity = (d-m):(m-c)

4.Suppose a container contains x units of liquid from which y units
are taken out and replaced by water. After n operations the
quantity of pure liquid = x (1 â€“ y/x)n units.

SOLVED PROBLEMS

Simple problems:

1.In what ratio must rice at Rs 9.30 per Kg be mixed with rice
at Rs 10.80 per Kg so that the mixture be worth Rs 10 per Kg?

Solution:

C.P of 1 Kg rice of 1st kind 930 p C.P of 1 Kg rice of 2n d kind 1080p

Mean Price 1000p

80 70

Required ratio=80:70 = 8:7

2.How much water must be added to 60 liters of milk at 11/2 liters
for Rs 20 so as to have a mixture worth Rs 10 2/3 a liter?

Solution:C.P of 1 lit of milk = 20*2/3 = 40/3

C.P of 1 lit of water 0 C.P of 1 lit of milk 40/3

Mean Price 32/3

8/3 32/3

Ratio of water and milk =8/3 : 32/3 = 1:4
Quantity of water to be added to 60 lit of milk =1/4*60=15 liters.

3.In what ratio must water to be mixed with milk to gain 20% by selling the mixture at cost price?

Solution:Let the C.P of milk be Re 1 per liter
Then S.P of 1 liter of mixture = Re.1
Gain obtained =20%.
Therefore C.P of 1 liter mixture = Rs(100/120*1) =5/6

C.P of 1 liter of water 0 C.P of 1 liter of milk1

Mean Price 5/6

1/6 5/6

Ratio of water and milk =1/6 : 5/6 = 1:5.

4.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 per Kg?

Solution:

Cost of 1 Kg pulses of 1 kind 15 Cost of 1 Kg pulses of 2nd kind 20

Mean Price Rs 16.50

3.50 1.50

Required ratio =3.50 : 1.50 = 35:15 = 7:3.

5. 4Kg s of rice at Rs 5 per Kg is mixed with 8 Kg of rice at Rs 6 per Kg .Find the average price of the mixture?

Solution:

rice of 5 Rs per Kg rice of 6 Rs per Kg

Average price Aw

6-Aw Aw-5

(6-Aw)/(Aw-5) = 4/8 =1/2
12-2Aw = Aw-5
3Aw = 17
Aw = 5.66 per Kg.

6.5Kg of rice at Rs 6 per Kg is mixed with 4 Kg of rice to get a mixture costing Rs 7 per Kg. Find the price of the costlier rice?

Solution:Using the cross method:

rice at Rs 6 per Kg rice at Rs x per Kg

Mean price Rs 7 per Kg

5 4

x-7:1=5:4
4x-28 = 5
4x=33=>x=Rs 8.25.
Therefore price of costlier rice is Rs 8.25 per Kg

Medium Problems:

1.A butler stole wine from a butt of sherry which contained 40% of spirit and he replaced,what he had stolen by wine containing only 16% spirit. The butt was then of 24% strength only. How much of the butt did he steal?

Solution:
Wine containing 40%spirit Wine containing 16% spirit

Wine containing 24% spirit

8 16

They must be mixed in the ratio of =1:2.
Thus 1/3 of the butt of sherry was left and hence the
butler drew out 2/3 of the butt.

2.The average weekly salary per head of the entire staff of a factory consisting of supervisors and the laborers is Rs 60.The average salary per head of the supervisors is Rs 400 and that of the laborers is Rs 56.Given that the number of supervisors is 12.Find the number of
laborers in the factory.

Solution:

Average salary of laborer Rs 56 Average salary of supervisors Rs 400

Average salary of entire staff Rs 60

340 4

Number of laborer / Number of Supervisors = 340 / 4=85/1
Thus,if the number of supervisors is 1,number of laborers =85.
Therefore if the number of supervisors is 12 number of laborers
85*12=1020.

3.The cost of type 1 rice is Rs 15 per Kg and type 2 rice is Rs 20 per Kg. If both type1 and type 2 are mixed in the ratio of 2:3,then the price per Kg of the mixed variety of rice is?

Solution:Let the price of the mixed variety be Rs x per Kg.

Cost of 1 Kg of type 1 rice Rs 15 Cost of 1 Kg of type 2 rice Rs 20

Mean Price Rs x

20-x x-15

(20-x) /( x-15) = 2/3
=> 60-3x = 2x-30
5x = 90=>x=18.

4.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%?

Solution:S.P of 1 Kg of the mixture = Rs 68.20,gain =10%
S.P of 1 Kg of the mixture = Rs (100/110*68.20)=Rs 62.

Cost of 1 Kg tea of 1st kind 60 Cost of 1 Kg tea of 2nd kind 65

Mean Price Rs 62
3 2

Required ratio =3:2.

5.A dishonest milkman professes to sell his milk at cost price but he mixes t with water and there by gains 25% .The percentage of water in the mixture is?

Solution:Let C. P of 1 liter milk be Re 1.
Then S.P of 1 liter mixture=Re 1. Gain=25%
C.P of 1 liter mixture =Re(100/125*1) = Re 4/5.

C.P of 1 liter milk Re 1 C.P of 1 liter of water 0

Mean Price 4/5

4/5 1/5

Ratio of milk to water =4/5 : 1/5 = 4:1
Hence percentage of water n the mixture=1/5*100=20%.

12.A merchant has 1000Kg 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:

Profit on 1st part 8% Profit on 2nd part 18%

Mean Profit 14%

4 6

Ratio of 1st and 2nd parts =4:6 =2:3.
Quantity of 2nd ind =3/5*1000Kg =600 Kg.

6.A jar full of whiskey contains 40% alcohol. A part of this whiskey is replaced by another containing 19% alcohol and now the percentage of alcohol was found to be 26%. The quantity of whiskey replaced is?
Solution:

Strength of first jar 40% Strength of 2nd jar 19%

Mean Strength 26%

7 14

So,ratio of 1st and 2nd quantities =7:14 =1:2
Therefore required quantity replaced =2/3.

7.A container contains 40lit of milk. From this container 4 lit of milk was taken out and replaced by water. This process was repeated further two times. How much milk is now contained by the container?

Solution:Amount of milk left after 3 operations = 40(1-4/40)3lit
=(40*9/10*9/10*9/10)
= 29.16 lit