How to Prepare for Quantitative Aptitude for CAT (2014)

Block II: Averages and Mixtures

Chapter 4. ALLIGATIONS

INTRODUCTION

The chapter of alligation is nothing but a faster technique of solving problems based on the weighted average situation as applied to the case of two groups being mixed together. I have often seen students having a lot of difficulty in solving questions on alligation. Please remember that all problems on alligation can be solved through the weighted average method. Hence, the student is advised to revert to the weighted average formula in case of any confusion.

The use of the techniques of this chapter for solving weighted average problems will help you in saving valuable time wherever a direct question based on the mixing of two groups is asked. Besides, in the case of questions that use the concept of the weighted average as a part of the problem, you will gain a significant edge if you are able to use the techniques illustrated here.

THEORY

In the chapter on Averages, we had seen the use of the weighted average formula. To recollect, the weighted average is used when a number of smaller groups are mixed together to form one larger group.

If the average of the measured quantity was

We say that the weighted average, Aw is given by:

Aw = (n₁A₁ + n₂ A₂ + n₃ A₃ + ……. + n_kA_k)/ (n₁ + n₂ + n₃ … + n_k)

That is, the weighted average

=

In the case of the situation where just two groups are being mixed, we can write this as:

Aw = (n₁A₁ + n₂ A₂ )/(n₁ + n₂)

Rewriting this equation we get: (n₁ + n₂) Aw = n₁A₁ + n₂A₂

n₁(Aw – A₁) = n₂ (A₂ – Aw)

or n₁/n₂ = (A₂ – Aw)/(Aw – A₁) Æ 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 A₁ and the number of elements is n₁ and the average of the second group is A₂ and the number of elements is n₂, 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 A₁ < A₂. Then, by the principal of averages, we get A₁ < Aw < A₂.

Illustration 1

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.

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

fi 3Aw = 34 fi Aw = 11.33 `/kg.

Illustration 2

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.

Solution n₁/n₂ = (40 – 34)/(34 – 30) = 6/4 = 3/2

Graphical Representation of Alligation

The formula illustrated above can be represented by the following cross diagram:

[Note that the cross method yields nothing but the alligation equation. Hence, the cross method is nothing but a graphical representation of the alligation equation.]

As we have seen, there are five variables embedded inside the alligation equation. These being:

the three averagesÆ A₁, A₂ and Aw

and the two weightsÆ n₁ and n₂

Based on the problem situation, one of the following cases may occur with respect to the knowns and the unknown, in the problem.

Now, let us try to evaluate the effectiveness of the cross method for each of the three cases illustrated above:

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.

Let us illustrate through an example:

Illustration 3

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

(a)The ratio of students in the classes

(b)The number of students in the first class if the second class had 30 students.

Solution

(a)Hence, solution is 2 : 1.

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

Case 2: A₁, A₂, n₁ and n₂ are known, Aw is unknown.

Illustration 4

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.

Solution

= (6 – Aw) : (Aw – 5)

fi(6 – Aw)/(Aw – 5) = 4/8 Æ12 – 2 Aw = Aw – 5

3 Aw = 17

\ Aw = 5.66 `/kg. (Answer)

Task for student: Solve through the alligation formula approach and through the weighted average approach to get the solution. Notice, the amount of time required in doing the same.

Case 3: A₁, Aw, n₁ and n₂ are known; A₂ is unknown.

Illustration 5

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.

Solution Using the cross method:

= (x – 7) : 1

\ (x – 7)/1 = 5/4 Æ 4x – 28 = 5

\ x = ` 8.25.

Task for student: Solve through the alligation formula approach and through the weighted average approach to get the solution. Notice the amount of time required in doing the same.

The above problems can be dealt quite effectively by using the straight line approach, which is explained below.

The Straight Line Approach

As we have seen, the cross method becomes quite cumbersome in Case 2 and Case 3. We will now proceed to modify the cross method so that the question can be solved graphically in all the three cases.

Consider the following diagram, which results from closing the cross like a pair of scissors. Then the positions of A₁, A₂, Aw, n₁ and n₂ are as shown.

Visualise this as a fragment of the number line with points A₁, Aw and A₂ in that order from left to right.

Then,

(a)n₂ is responsible for the distance between A₁ and Aw or n₂ corresponds to Aw – A₁

(b)n₁ is responsible for the distance between Aw and A₂. or n₁ corresponds to A₂ – Aw

(c)(n₁ + n₂) is responsible for the distance between A₁ and A₂. or (n₁ + n₂) corresponds to A₂ – A₁.

The processes for the 3 cases illustrated above can then be illustrated below:

Illustration 6

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

(a)the ratio in which the classes were mixed.

(b)the number of students in the first class if the second class had 30 students.

Solution

Hence, ratio is 2 : 1, and the second class has 60 students.

Case 2 A₁, A₂, n₁ and n₂ are known; Aw is unknown.

Illustration 7

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.

Solution

is the same as

Then, by unitary method:

n₁ + n₂ corresponds to A₂ – A₁

Æ 1 + 2 corresponds to 6 – 5

That is, 3 corresponds to 1

\ n₂ will correspond to

In this case (1/3) × 2 = 0.66.

Hence, the required answer is 5.66.

Case 3: A₁, Aw, n₁ and n₂ are known; A₂ is unknown.

Illustration 8

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.

Using straight line method:

4 corresponds to 7 – 6 and 5 corresponds to x – 7.

The thought process should go like:

4 Æ1

\ 5 Æ1.25

Hence, x – 7 = 1.25

and x = 8.25

SOME TYPICAL SITUATIONS WHERE ALLIGATIONS CAN BE USED

Given below are typical alligation situations, which students should be able to recognize. This will help them improve upon the time required in solving questions. Although in this chapter we have illustrated problems based on alligation at level 1 only, alligation is used in more complex problems where the weighted average is an intermediate step in the solution process.

The following situations should help the student identify alligation problems better as well as spot the way A₁, A₂, n₁ and n₂ and Aw are mentioned in a problem.

In each of the following problems the following magnitudes represent these variables:

A₁ = 20, A₂ = 30, n₁ = 40, n₂ = 60

Each of these problems will yield an answer of 26 as the value of Aw.

1.A man buys 40 kg of rice at ` 20/kg and 60 kg of rice at ` 30/kg. Find his average price. (26/kg)

2.Pradeep mixes two mixtures of milk and water. He mixes 40 litres of the first containing 20% water and 60 litres of the second containing 30% water. Find the percentage of water in the final mixture. (26%)

3.Two classes are combined to form a larger class. The first class having 40 students scored an average of 20 marks on a test while the second having 60 students scored an average of 30 marks on the same test. What was the average score of the combined class on the test. (26 marks)

4.A trader earns a profit of 20% on 40% of his goods sold, while he earns a profit of 30% on 60% of his goods sold. Find his percentage profit on the whole. (26%)

5.A car travels at 20 km/h for 40 minutes and at 30 km/h for 60 minutes. Find the average speed of the car for the journey. (26 km/hr)

6.40% of the revenues of a school came from the junior classes while 60% of the revenues of the school came from the senior classes. If the school raises its fees by 20% for the junior classes and by 30% for the senior classes, find the percentage increase in the revenues of the school. (26%)

A Typical Problem

A typical problem related to the topic of alligation goes as follows:

4 litres of wine are drawn from a cask containing 40 litres of wine. It is replaced by water. The process is repeated 3 times

(a)What is the final quantity of wine left in the cask.

(b)What is the ratio of wine to water finally.

If we try to chart out the process, we get: Out of 40 litres of wine, 4 are drawn out.

This leaves 36 litres wine and 4 litres water. (Ratio of 9 : 1)

Now, when 4 litres are drawn out of this mixture, we will get 3.6 litres of wine and 0.4 litres of water (as the ratio is 9 : 1). Thus at the end of the second step we get: 32.4 litres of wine and 7.6 litres of water. Further, the process is repeated, drawing out 3.24 litres wine and 0.76 litres water leaving 29.16 litres of wine and 10.84 litres of water.

This gives the final values and the ratio required.

A closer look at the process will yield that we can get the amount of wine left by:

40 × 36/40 × 36/40 × 36/40 = 40 × (36/40)³

fi 40 × (1 – 4/40)³

This yields the formula:

Wine left : Capacity × (1 – fraction of wine withdrawn)ⁿ for n operations.

Thus, you could have multiplied:

40 × (0.9)³ to get the answer

That is, reduce 40 by 10% successively thrice to get the required answer.

Thus, the thought process could be:

40 – 10% Æ 36 – 10% Æ 32.4 – 10% Æ 29.16

LEVEL OF DIFFICULTY (I)

Level of Difficulty (I)

Solutions and Shortcuts

Level of Difficulty (I)

1.Solving the following alligation figure:

The answer would be 4.625/kg.

2.Mixing ` 4/kg and ` 5/kg to get ` 4.6 per kg we get that the ratio of mixing is 2:3. If the first oil is 40 kg, the second would be 60 kg.

3.Since by selling at ` 4.40 we want a profit of 10%, it means that the average cost required is ` 4 per kg. Mixing sugar worth ` 3.6/kg and `4.2/kg to get ` 4/kg means a mixture ratio of 1:2. Thus, to 8 kg of the second variety we need to add 4 kg of the first variety to get the required cost price.

4.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.

5.Since, Ravi earns ` 960 in 5 years, it means that he earns an interest of 960/5 = ` 192 per year. On an investment of 3600, an annual interest of 192 represents an average interest rate of 5.33%.

Then using the alligation figure below:

We get the ratio of investments as 1:2. Hence, he lent 1 × 3600/3 = 1200 at 4% per annum.

6.The ratio of boys and girls appearing for the exam can be seen to be 3:1 using the following alligation figure.

This means that out of 400 students, there must have been 100 girls who appeared in the exam.

7.The ratio of the cost of the cow and the calf would be 40:25 or 8:5 as can be seen from the following alligation figure:

Thus, the cost of the cow would be `800.

8.

From the figure 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.

9.The cost price of the mixture would have been ` 8 per liter for him to get a profit of 25% by selling at ` 10 per liter. The ratio of mixing would have been 1:4 (water is to milk) as can be seen in the figure:

Since we are putting in 5 liters of water, the amount of milk must be 20 liters. The amount of mixture then would become 25 liters.

10.Amount of water left = 50 × 9/10 × 9/10 = 40.5 liters. Hence, wine = 9.5 liters. Ratio of wine and water = 19:81. Option (c) is correct.

11.The amount of spirit left = 20 × 4/5 × 4/5 × 4/5 × 4/5 × 4/5 = 4096/625 = 6 (346/625).

12.Let the quantity of refined oil initially be Q. Then we have Q × ¾ × ¾ × ¾ × ¾ = 10 Æ Q = 2560/81 liters.

13.The ratio would be 1:2 as seen from the figure:

14.It can be seen from the ratio 36:13 that the proportion of liquid soap to water is 36/49 after two mixings. This means that 6/7^th of the liquid soap must have been allowed to remain in the container and hence 1/7^th of the conatiner’s original liquid soap would have been drawn out by the thief. Since he takes out 4 gallons every time, there must have been 28 gallons in the container. ( as 4 should be 1/7^th of 28).

15.In order to sell at a 25% profit by selling at 13.75 the cost price should be 13.75/1.25 = 11. Also since water is freely available, we can say that the ratio of water and soda must be 1:11.

16.The average value of a coin is 41 paise and there are only 20 paise and 50 paise coins in the sum. Hence, the ratio of the number of 20 paise coins to 50 paise coins would be 9:21 = 3:7. Since there are a total of 90 coins, the number of 20 paise coins would be 3 × 90/10 = 27 coins.

17.

The ratio of mixing required would be 1:4 which means that the percentage of adulterated fat would be 20%.

18.Solve using options. Initially there are 7 liters of water in 70 liters of the mixture. By mixing 2 liters of water we will have 9 liters of water in 72 liters of the mixture – which is exactly 12.5%.

19.Again solve this question using options. Initially there are 2 liters of water in 20 liters of the mixture. To take the water to 25% of the mixture we would need to add 4 liters of water as that would give us 6 liters of water in 24 liters of the mixture.

20.By selling at 300 if we need to get a profit of 25% it means that the cost price would be 300/1.25 = 240.

Thus, in 52 quintals of the second we need to mix 26 quintals of the first.

21.The requisite 11.11% profit can be got by mixing 0.111 liters of water in 1 liter of milk. In such a case the total milk quantity would be 1.111 liters and the price would be for 1 liter only. The profit would be 0.111/1 = 11.11%.

22.The percentage of honey in the new mixture would be:

(2 × 25 + 3 × 75)/5 = 275/5 =55%. The ratio of honey to water in the new mixture would be 55:45 = 11:9.

23.In order to mix two tin alloys containing 86.66% tin and 93.33% tin to get 90% tin, the ratio of mixing should be 1:1. Thus, each variety should be 25 kgs each.

24.Assume the capacity of the two containers is 198 liters each. When we mix 198 liters of the first and 198 liters of the second the amount of oil would be:

198 × 4/11 + 198 × 7/18 = 72 + 77 = 149 liters.

Consequently the amount of water would be 396 – 149 = 247 liters. Option (a) is correct.

25.The first vessel contains 25% spirit while the second vessel contains 37.5 % spirit. To get a 1:2 ratio we need 33.33% spirit in the mixture. The ratio of mixing can be seen using the following alligation figure:

26.Solve using options as that would be the best way to tackle this question. Option (b) fits the situation perfectly as if we take the price of the pen as ` 25, the cost of the pencil would be ` 10. The profit in selling the pen would be `5 while the loss in selling the pencil would be ` 1. The total profit would be ` 4 as stipulated by the problem.

27.The following alligation visualization would help us solve the problem:

Cost of cupboard = 5 × 18000/12 = 7500.

28.The following visualization would help:

From the figure we can see that the original mixture would be 40 liters and the petrol being mixed is 8 liters. Thus, the vessel capacity is 48 liters.

29.90% and 97% mixed to form 94% means that the mixing ratio is 3:4. The first solution would be 3 × 21/7 = 9 liters.

30.If all the animals were ducks we would have 180 heads and 360 legs. If we reduce the number of ducks by 1 to 179 and increase the number of deers by 1 to 1, we would get an incremental 2 legs.

Since, the number of legs we need to increment is 88 (448 – 360 = 88), we need to have 44 deers and 136 ducks.

31.Solve using options. Option (a) does not fit because 253 × 5000 + 47 × 2500 is not equal to 985000.

Similarly option (b) does not fit because 253 × 2500 + 47 × 5000 is not equal to 985000.

Option (c) fits as 94 × 5000 + 206 × 2500 = 985000.

32.In order to solve this we need to assume a value for the amounts in the vessels. If we assume 35 liters as the quantities in all the three vessels we will get:

28 liters + 25 liters + 30 liters = 83 liters of petrol and 22 liters of kerosene in 105 liters of the mixture.

The required ratio is 83:22.

33.The required ratio would be 7:3 as seen in the following figure.

34.We cannot determine the answer to this question as we do not know the price per kg of the other type of ghee. Hence, we cannot find the ratio of mixing which would be required in order to move further in this question.

35.To gain 20% by selling at cost price, milk should comprise 83.33% of the total mixture. The ratio of mixing that achieves this is 1:5.

36.20% spirit is mixed with 50% spirit to get 25% spirit. The ratio of mixing would be 5:1. This means he stole 5/6^th of the bottle or 83.33% of the bottle.

37.O + F + T = 95

O + 0.5F +0.25T =50.5

T = 1.333 O.

Solving we get: 24 coins of 25 paise each.

38.Annual interest income = 1904 /3.5 = 544. Interest of ` 544 on a lending of ` 6800 implies an 8% average rate of interest. This 8% is generated by mixing two loans @ 7.5% and 10% respectively. The ratio in which the two loans should be allocated would be 4:1. The amount lent at 10% would be 1 × 6800/5 = 1360.

39.Solve using options. If he travels 48 km on foot he would take 6 hours on foot. Also, in this case he would travel 32 km on bicycle @ 16kmph – which would take him 2 hours. Thus a total of 8 hours. Option (c) satisfies the conditions of the question.

40.Solving through options is the best way to tackle this question. Option (a) fits the conditions of the problem as if there are 11 liters in the first vessel, there would be 8 liters of spirit. Also it means that we would be taking 24 liters from the second vessel out of which there would be 20 liters of spirit. Thus, total spirit would be 28 out of 35 liters giving us 7 liters of water.

BLOCK REVIEW TEST

Review Test

Directions for Questions 10 to 12: There are 60 students in a class. These students are divided into three groups A, B and C of 15, 20 and 25 students each. The groups A and C are combined to form group D

Review Test