## Average -Mixture and Alligation For CAT 2018 students

I recently read a student blog on TG Town questioning whether it is correct by the instructors to tell students “not to take their mock scores seriously.” After all, if a student is not performing well in mocks, something must be wrong. The question really made me ask myself the reasons for my saying so. Is it even correct to say mock scores do not matter. I think there are several reasons instructors tell students not to take their mock scores seriously. Most important of them all is motivation. Students are never ready to recognize that scoring well in a test is more a matter of temperament and question selection than that of content. Infact an average student having the art of question-picking would perform twice better than a genius student who is out to solve questions in a near about serial order. Therefore, instructors want to keep their students motivated in order to keep their enthusiasm high. The second reason for not taking mocks seriously is that the level of preparedness of students at a particular point is different. Many of the institutes start their mocks by the month of May or June and most students are not ready by then. It is inevitable that they will perform badly. If the students take it seriously, they would spend rest of their time taking more and more tests to improve themselves instead of studying and then the real harm would be done. For me, when I entered the CAT preparation, my verbal didn’t need preparation, my quant took three or four months of tweaking, but my DI took more than a year to reach a decent level. The level of preparedness for every student is different. So the question is, what do you do with your mock scores? The better thing to do is to solve your mock paper, if you can solve it on your own, you don’t have a problem with content. If you cannot, get back to studies. So don’t look hard at your scores, look hard at the paper.

The concept of averages is hardly a new concept at all. If asked, all of you would give me the following formula for calculating

So far so good. But if I ask all of you to solve a simple problem many of you would reach for their pens.

The average score of three students A, B, and C is 50. When the score of another student D is added to the group, the average score become 47. What is the score of student D?

Answer: for most of you, the score of student D would be 4 × 47 – 3 × 50 = 38.

For me, the calculation would just be 47 – 9 = 38.

Some of you might have understood what I did. Let me start explaining through a simple example. Then we shall extend our explorations to more complex problems.

The two images above show two beam balances, one with equal arms and the other with the arms lengths in the ratio 1: 3. The dotted line in both cases shows the average value.

In the first beam balance, if you move any of the pan one unit towards the average value, the other pan would also move one unit in the direction of the average value to *keep the average constant*. For example, let the weights in the two pans be 50 kg and 60 kg. The average is 55 kg. If you increase the weight in 50 kg pan by one unit (i.e. 51) bringing it nearer to the average, you will have to decrease the weight in the other pan by one unit (i.e. 59), bringing it one unit nearer to the average, to keep the average constant.

In the second beam balance, if the arm of length 3 moves 1 unit towards the average, the arm of length 1 will have to move 3 units towards the average to keep the average constant. For example, let the weight in pan of 3 unit arm length be 40 kg and the weight in the pan of 1 unit arm length be 60 kg. Now if I increase the weight in the first pan by 1 unit, I shall have to decrease the weight in the second pan by 3 units to keep the average constant.

The same rule applies if I am moving a pan away from the average.

Now, understand this- the second beam balance can also be represented with a beam balance of equal arms but one arm having thrice the weight on the other arm. Savvy? Now let’s see the problem once again.

The average score of three students A, B, and C is 50. When the score of another student D is added to the group, the average score become 47. What is the score of student D?

Answer: Now there are three weights on one arm. The average of the four weights is 47. To move each of the three weight 3 units away from the average (50 – 47) I shall have to move the weight D 3 × 3 nine units away from the average. Therefore, weight D = 47 – 9 = 38.

Four friends have an average weight of 68. If Rahim is also included in the group, the average weight becomes 72. what is Rahim’s weight?

Answer: Same process: there are four people at a distance of 4 units from the average weight of 72. To balance them, we will have to place a person at 4 × 4 = 16 units from 72 on the other side. Therefore, Rahim’s weight = 72 + 16 = 88.

A batsman in his 20^{th} innings makes a score of 93 and thereby increases his average by 3. What is the average after 20 innings?

Answer: If you have understood what I have said so far, the new average is nothing but 93 – 57 = 36. Let the new average be A. Therefore, there are 19 scores at a distance of 3 units from A. To balance these scores we need one score (which is 93) 19 × 3 = 57 units from A on the other side. Therefore A = 93 – 57 = 36.

Now let’s see this funda of ‘balancing act’ applied in a small part of a DI set from CAT 2006. For more DI sets based on averages, you will have to visit our CAT CBT Club

What is Dipan’s score in paper II in English group?

Answer: Rather than any long and cumbersome method, we can do this question in a very short and sweet way- we just see the deviation of each group average from the overall average. Now the average of PCB group is 98 which is +2 from overall average. Mathematics group is -1, Social science group is -0.5, Vernacular group is -1. Therefore, total so far = 2 -1 -0.5 -1 = -0.5. Therefore to have deviation from overall average as 0, the English group average should have a deviation of +0.5, i.e. the average should be 96.5. Therefore, Dipan’s mark in paper II in English group are 97.

Let’s have a look at the unbalanced scale in the figure 2 again. The beam balance is shown below:

If you want to keep the average same while changing weights in both the pans, the weights in the pan would be inversely proportional to arm lengths, i.e. to increase the weight in the left pan by one unit, we shall have to increase the weight in the right pan by three units to keep the average constant. Here, the arm lengths signify the distance of the weights from the average. Let’s take a simple example:

For example, let’s mix two solutions one with 30% milk and the other with 75% milk. Let it be given that the mixture is of 50% milk. Now the distances (arm lengths) of both percentages from the average percentage are 50 – 30 = 20% and 75 – 50 = 25%. Since the arms lengths are in the ratio 20: 25 = 4: 5, the weights in the pan should be in the ratio 5: 4. Therefore, we are mixing the solutions in the ratio 5: 4.

Let’s see some examples now:

In what ratio must the rice at Rs 3.8 per kg be mixed with rice at Rs 4.5 per kg so that the price of the mixture is Rs 4.2 per kg?

Answer: By now, I believe you know what to do. The distances from the averages are 4.2 – 3.8 = 0.4 and 4.5 – 4.2 = 0.3. the distances are in the ratio 4: 3, therefore, the rice should be mixed in the ratio 3: 4.

A butler stole one-fourth of a wine bottle containing 60% alcohol and replaced it with water. Find the resultant concentration of the wine in the bottle.

Answer: Again an easy one with a little twist- now we are mixing three-fourth solution containing 60% alcohol with one-fourth solution containing 0%. The ratio of the quantities taken is 3/4 : 1/4 = 3: 1. Therefore, the ratio of the distances from the average would be 1: 3. Therefore, we Therefore, average value = 60% –.

In a wildlife sanctuary, the counting of Zebras and Ostriches is being done. It was found that there were 150 heads and 480 legs. How many Zebras were there in the sanctuary?

Answer: No mixture here? Yes but there is! Zebra being the milk and Ostriches being the water. How can you use Alligation here. Simple, you first need to determine of what value we can take the average. Also, if you see what is the basic difference between a Zebra and an Ostrich that has been considered in the problem here, you will realize it is the number of legs- Zebra has four and Ostrich ahs two. And what is the average number of legs per animal given? It is 480/150 = 3.2. The distances from the average are 3.2 – 2 = 1.2 and 4 – 3.2 = 0.8. These distances are in the ratio 3: 2 or the quantities are in the ratio 2: 3. Therefore, the number of Zebras = 150 × 3/5 = 90.

What percentage of the females polled said Yes?

Answer: Allegation over here? Yes sir, because we know the average percentage of males and females, i.e. 50% (number of males and females are equal). Now let’s just consider the percentage of males in those who said yes and those who said no. The corresponding percentages are 60% and 20% respectively. Let the ratio of the number of people who said yes and the number of people who said no be x : y. The ratio of distances of the percentages from the average percentage is (60 – 50): (50 – 20) = 1: 3 Þ x: y = 3 : 1. So let there be 200 people in all (100 males and 100 females). The ratio of people saying yes to people saying no is 3: 1. Therefore, number of people saying yes = 200 × 3/4 = 150, out of which females are 40% = 150 × 40/100 = 60 females said yes which is 60% of 100 females.

I am afraid I shall have to end here and leave the rest of it for my CBT Club students. I shall cover some DI sets based on averages and Alligation in the CBT Club .

## 0 Comments