Average and Alligation  
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. In fact an average
student having the art of questionpicking 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 average: . 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 onefourth 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 threefourth solution containing 60% alcohol with onefourth 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: Alligation 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 this week.

Re: Average and Alligation  
As usual, another piece of gem from TG...you are awesome!!! 
Re: Average and Alligation  
Dear Sir Awesome as usual Ronak 
Re: Average and Alligation  
hi sir, thanx 4 the article, i was waiting 4 dis coz i m just completing my basics nice day everybody........ 
Re: Average and Alligation  
Thanks TG, Your way of seeing things in different ways is excellent. 
thanx  
keep going sirji..thanx a lot.. atlast there is something to cheer even after india's defeat to pak 
Re: Average and Alligation  
Awesome post Sir.. 
Re: Average and Alligation  
I'd attempted the '06 DI ques in routine way. bt this way wz pure TG magic. Marriage poll ques. is subtle. Tnx TG . 
Re: Average and Alligation  
i guess that iS the differnce b\w mortals n the great TG..LOLS.. 
Re: Average and Alligation  
good one 
Re: Average and Alligation  
Hi Kirubakaran, As the computerbased CAT is going to have a onequestionperpage format the question selection skill needs to be sharpened more than ever. The biggest challenge a student is going to face is whether he/she should attempt a particular question or move on to the next one. I see you are in CBT Club already. We will talk about the strategy there only shortly. Total Gadha 
Re: Average and Alligation  
TG sir , why " there only ". We are also deprived and old student of yours. Need your benovalence.. 
Re: Average and Alligation  
hello.... didnt got last Q ... how 1 :3 also ratio of yes :no person....???? shoulnt be 1:3 ratio of only MEN yes:No........howcome womens too included...? 
Re: Average and Alligation  
Hi Mansi, When you submit the test you would see the options with white background , they are the correct answers.

Re: Average and Alligation  
Totally awesome sir.... 
Re: Average and Alligation  
I liked the simple and elegant example for allegation(distance from avg. ) 
Re: Average and Alligation  
Thank you so much sir.. 
Re: Average and Alligation  
fantastic post sir! thank you so much 
Re: Average and Alligation  
thank you very much..... 
Re: Average and Alligation  
Harish, Example please? Total Gadha 
Re: Average and Alligation  
Kaushik its 1) 48.63 
Re: Average and Alligation  
Damn....that was too easy...I was making it difficult... Thanx Rohit 
Re: Average and Alligation  
@Saurabh My dear friend in the question it is mentioned that no. of males is equal to no. of females, but in your approach they are different. mayank 
Re: Average and Alligation  
hi, answer will be 98 as old average=36*60 and 36*60=34*x+y+Z 58*34=34*x therefore x=58 z=90 therefore y=98 
Re: Average and Alligation  
@ saurabh From ur solution wat u are inferring is that there were 200 people and 100 said yes and 100 said no which is a wrong assumption 
Re: Average and Alligation  
hii abhishek.. thanx for the ans...ya ans is ryt...its 27:6:1... 
Re: Average and Alligation  
hii abhishek... typin mistake..i meant its wrong...nyways thanx... 
Re: Average and Alligation  
Hi Deepti Can you plz explain the answer? 
Re: Average and Alligation  
Thanks dude.Gr8 equation...!!! 
Re: Average and Alligation  
Gr8 article. Hats off!!!!! 
Re: Average and Alligation  
Hi RITABAN SENGUPTA See that there are 35 students in all and their expediture per head is getting reduced by 1. That means the new 7 students has 35 more rupees (taken 1 from each of 35 students) other than the increased expenditure of 42 rupees for their own wellbeing. So in all these 7 students have 35 + 42 = 77 rupees which they can share 77/7 = 11 each. I hope this is clear now. Kamal Lohia 
Re: Average and Alligation  
Thank u so much.. G for genius 
Re: Average and Alligation  
nice article.. 
Re: Average and Alligation  
@Navneet: sol> > (10*2 + 20*3 + 30 N)/(2 + 3 + N) = 23 > solve this equation > N = 5

Re: Average and Alligation  
even simpler way is... If 100 liters contian 60 percnt i.e 60 l alcohol then in 75 percent i.e 75 l. Vil have how much..... Hence,60*75/100 is 45... 
Re: Average and Alligation  
even simpler way is... If 100 liters contian 60 percnt i.e 60 l alcohol then in 75 percent i.e 75 l. Vil have how much..... Hence,60*75/100 is 45... 
Re: Average and Alligation  
@ Siddharth.. Is it 40 gallons? 
Re: Average and Alligation  
yes.. it is 40 gallons, eqn used3/5=5x10/3x6+16,this gives x=5 so original=3x+5x=40 hence the answer 
Re: Average and Alligation  
yus it is 40 .. Thnx Mr. N Ms. A... 
Re: Average and Alligation  
how do i know when to take lcm in ratio proportion questions.. 
Re: Average and Alligation  
Hi Tarun Post some questions where you have doubt. Then it'll be easier to explain. Kamal Lohia 
Re: Average and Alligation  
i did not get it ,, can u plzzz explain step by step?? 
Re: Average and Alligation  
Hi Sir , Thanks for your reply.. But i think, i did not understand the concept properly.. Can you shed some light on it.. Will be great help. Neha 
Re: Average and Alligation  
Some important Questions for CAT please 
Re: Average and Alligation  
to select 3 from 6 c(6,3)=20; now its obvious that for all conc=50 so rule out (50,70,80)so ans= 205=15 
Re: Average and Alligation  
Hi Shivashankar, is it 100L?? 