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## ALLIGATIONS

If A1,A2,A3,...,An are the groups of people with total members n1,n2,..,nn

Then the weighted average is given by n1∗A1+n2∗A2+...+nn∗Ann1+n2+..+nn

Assume two groups are there we can rewrite average as Aw = n1A1∗n2∗A2n1∗n2

We rewrite the equation as : (n1∗n2)Aw = n1A1+n2A2

n1(Aw−A1)=n2(A2−Aw)

or n1n2=A2−AwAw−A1. This is the Alligation equation.

** The Alligation Situation**

Two groups of elements are mixed together to form a third group containing the elements of both the groups

If the average of the first group is A

_{1}and the number of elements is n_{1}and the average of the second group is A_{2}and the number of elements is n_{2}, then to find the average of the new group formed, we can use either the weighted average equation or the alligation equation.

As a convenient convention, we take A

_{1}< A_{2}. Then, by the principal of averages, we get A_{1}< Aw < A_{2}.

** Some Keys to spot A _{1}, A_{2} and A_{w} and differentiate these from n_{1} and n_{2}**

Normally, there are 3 averages mentioned in the problem, while there are only 2 quantities. This isn’t foolproof though, since at times the question might confuse the student by giving 3 values for quantities representing n

_{1}, n_{2}and n_{1}+ n_{2}respectively

A

_{1}, A_{2}and Aw are always rate units, while n_{1}and n_{2}are quantity units.

The denominator of the average unit corresponds to the quantity unit (i.e. unit for n

_{1}and n_{2}).

All percentage values represent the average values.

Q.

Two varieties of rice at ` 10 per kg and ` 12 per kg are mixed together in the ratio 1 : 2. Find the average price of the resulting mixture

11

12

11.33

11.45

Ans .11.33

Explanation :

1/2 = (12 – Aw)/(Aw – 10) Æ Aw – 10 = 24 – 2Aw

fi 3Aw = 34 fi Aw = 11.33 `/kg

Q.

On combining two groups of students having 30 and 40 marks respectively in an exam, the resultant group has an average score of 34. Find the ratio of the number of students in the first group to the number of students in the second group.

3/2

4/5

1/2

4/9

Ans .3/2

Explanation :

n1

/n2 = (40 – 34)/(34 – 30) = 6/4 = 3/2

Case 1: A1, A2, Aw are known; may be one of n1 or n2 is known. To find : n1 : n2 and n2 if n1 is known OR n1 if n2 is known

Q.

On mixing two classes of students having average marks 25 and 40 respectively, the overall average obtained is 30 marks. Find The ratio of students in the classes

6 : 1

2 : 3

2 : 1

3 : 1

Ans .2 : 1

Explanation :

Hence, solution is 2 : 1

Q.

On mixing two classes of students having average marks 25 and 40 respectively, the overall average obtained is 30 marks. Find The number of students in the first class if the second class had 30 students.

50

80

60

90

Ans .60

Explanation :

If the ratio is 2 : 1 and the second class has 30 students, then the first class has 60 students

Q.

4 kg of rice at ` 5 per kg is mixed with 8 kg of rice at ` 6 per kg. Find the average price of the mixture.

6.66

5.66

3.66

1.66

Ans .5.66

Explanation :

= (6 – Aw) : (Aw – 5)

fi(6 – Aw)/(Aw – 5) = 4/8 So 12 – 2 Aw = Aw – 5

Illustration 5

Thus 3 Aw = 17

We get Aw = 5.66

Q.

5 kg of rice at ` 6 per kg is mixed with 4 kg of rice to get a mixture costing ` 7 per kg. Find the price of the costlier rice

8.75

5.25

8.25

6.25

Ans .8.25

Explanation :

= (x – 7) : 1

(x – 7)/1 = 5/4 Æ 4x – 28 = 5

x = 8.25

Q.

On mixing two classes of students having average marks 25 and 40 respectively, the overall average obtained is 30 marks. Find the ratio in which the classes were mixed

5:1

3:1

2:1

4:1

Ans .2:1

Explanation :

Hence, ratio is n1 = 40-30 and n2 = 30-25; ratio is 2 : 1, and the second class has 60 students

Q.

On mixing two classes of students having average marks 25 and 40 respectively, the overall average obtained is 30 marks. Find the the number of students in the first class if the second class had 30 students.

40

20

30

60

Ans .60

Explanation :

Hence, ratio is n1 = 40-30 and n2 = 30-25; ratio is 2 : 1, and the second class has 60 students

Q.

4 kg of rice at ` 5 per kg is mixed with 8 kg of rice at ` 6 per kg. Find the average price of the mixture.

6.66

5.66

3.66

4.66

Ans .5.66

Explanation :

n1=4, n2=8, A1 = 5, A2 = 6. So by solving we get the required answer is 5.66

Q.

A mixture of 125 gallons of wine and water contains 20% water. How much water must be added to the mixture in order to increase the percentage of water to 25% of the new mixture?

10

8.5

8

8.33

Ans .

8.33

Explanation :

In 125 gallons we have 25 gallons water and 100 gallons wine. To increase the percentage of water to 25%, we need to reduce the percentage of wine to 75%. This means that 100 gallons of

wine = 75% of the new mixture. Thus the total mixture = 133.33 gallons. Thus, we need to mx

133.33 – 125 = 8.33 gallons of water in order to make the water equivalent to 25% of the mixture

Q.

400 students took a mock exam in Delhi. 60% of the boys and 80% of the girls cleared the cut off in the examination. If the total percentage of students qualifying is 65%, how many girls appeared in the examination?

300

200

100

50

Ans .100

Explanation :

The ratio of boys and girls appearing for the exam can be seen to be 3:1 as A1 = 60, A2 = 80 and Aw=65

Q.

The average salary per head of all employees of a company is ` 600. The average salary of 120 officers is ` 4000. If the average salary per head of the rest of the employees is ` 560, find the total number of workers in the company.

10200

10320

10500

10680

Ans .10320

Explanation :

A1=560, A2=4000 and Aw=600 so solving we get it is clear that the ratio of the number of officers to the number of other employees

would be 40:3400. Since there are 120 officers, there would be 3400 × 3 = 10200 workers in the

company. Thus the total number of employees would be 10200 + 120 = 10320

Q.

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

8 : 7

8 : 4

8 : 2

8 : 6

Ans .8:7

Explanation :

Required ratio = 80 : 70 = 8 : 7. we get as A1 = 9.3 and Aw=10 and A2=10.8

Q.

A mixture of 125 gallons of wine and water contains 20% water. How much water must be added to the mixture in order to increase the percentage of water to 25% of the new mixture?

10

8.5

8

8.33

Ans .8.33

Explanation :

In 125 gallons we have 25 gallons water and 100 gallons wine. To increase the percentage of water to 25%, we need to reduce the percentage of wine to 75%. This means that 100 gallons of

wine = 75% of the new mixture. Thus the total mixture = 133.33 gallons. Thus, we need to mx

133.33 – 125 = 8.33 gallons of water in order to make the water equivalent to 25% of the mixture

Q.

400 students took a mock exam in Delhi. 60% of the boys and 80% of the girls cleared the cut off in the examination. If the total percentage of students qualifying is 65%, how many girls appeared in the examination?

300

200

100

50

Ans .100

Explanation :

The ratio of boys and girls appearing for the exam can be seen to be 3:1 as A1 = 60, A2 = 80 and Aw=65

Q.

The average salary per head of all employees of a company is ` 600. The average salary of 120 officers is ` 4000. If the average salary per head of the rest of the employees is ` 560, find the total number of workers in the company.

10200

10320

10500

10680

Ans .10320

Explanation :

A1=560, A2=4000 and Aw=600 so solving we get it is clear that the ratio of the number of officers to the number of other employees

would be 40:3400. Since there are 120 officers, there would be 3400 × 3 = 10200 workers in the

company. Thus the total number of employees would be 10200 + 120 = 10320

Q.

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

8 : 7

8 : 4

8 : 2

8 : 6

Ans .8:7

Explanation :

Required ratio = 80 : 70 = 8 : 7. we get as A1 = 9.3 and Aw=10 and A2=10.8