Sunday, September 30th, 2018
Data Interpretation section of CAT, as the name suggests is all about analyzing substantial data and interpret meaningful or useful results from that huge pile of unessential data. With exploding population, interlinked economies and the recent result of globalization companies are not just dealing with the population of their city or countries but of hundreds of cities and many countries and as a result we have this enormous data that doesn’t make much of sense. The data needs to be categorized, simplified and summarized into numbers, numbers that make sense and form the basis of rational decision making. And this is the reason why as future business managers or entrepreneurs we need to understand and learn how to translate the huge data into quantifying relevant facts and figures. And thus, DI section holds a lot of significance and have become a part of CAT curriculum. In exam data will be given in organized and compressed forms using pie charts, Bar graphs, case lets, scatter plots etc. and following which few questions will be asked on that organized data. Today we will talk about one such important topic of Data interpretation that consistently comes in exam i.e. Caselets. It is one of the most difficult concepts in Data Interpretation. Since, in Caselet information is given in paragraph form, unlike graphs, charts, and tables that make data easy to read and identify useful information from that. Caselet is comparatively much more arduous and challenging task than pie charts or histograms. Later in this blog, we will discuss how to simplify this burdensome task and interpret Caselet in a timesaving manner.
1. Paragraph based on Reasoning.
2. Paragraph based on numerical Data.
Now to understand how to interpret the data we will use examples and try to find a stepwise solution that may help to solve Caselet questions in the exam. Before hopping on to examples keep in mind the following points that will assist you while devising a solution to the problem of Caselet.
· Read the paragraph carefully and recognize the variables around which the whole paragraph revolves and questions are asked. Note down all the important points.
· Try to formulate relationships between the variables pictographically using tables, symbols or Venn diagrams. Tables help to define multivariate relationships more clearly so try using them more often.
· Data interpretation usually requires numerical and arithmetic calculations such as averages, ratios, percentages etc. Be thorough with their concepts and use shortcuts and tricks for faster calculations, it will save you a lot of time.
· Do not assume information that is not given and use logic and reasoning to find out the hidden information that is given in paragraph.
· Do not indulge into troublesome lengthy calculations when approximations or relative values are asked. Calculate only what is asked.
Now let’s move on to an example first of paragraph based on reasoning. Consider this question that came in CAT 2008.
In a sports event, six teams (A, B, C, D, E, and F) are competing against each other. Matches are scheduled in two stages. Each team plays three matches in stage – I and two matches in Stage – II. No team plays against the same team more than once in the event. No ties are permitted in any of the matches. The observations after the completion of Stage – I and Stage – II are as given below.
StageI:
• One team won all the three matches.
• Two teams lost all the matches.
• D lost to A but won against C and F.
• E lost to B but won against C and F.
• B lost at least one match.
• F did not play against the top team of StageI.
StageII:
• The leader of StageI lost the next two matches.
• Of the two teams at the bottom after StageI, one team won both matches, while the other lost both
matches.
• One more team lost both matches in StageII.
1. The two teams that defeated the leader of StageI are:
(1) F & D (2) E & F (3) B & D (4) E & D (5) F & D
2. The only team(s) that won both matches in StageII is (are)
(1) B (2) E & F (3) A, E & F (4) B, E & F (5) B & F
3. The teams that won exactly two matches in the event are:
(1) A, D & F (2) D & E (3) E & F (4) D, E & F (5) D & F
4. The team(s) with the most wins in the event is (are):
(1) A (2) A & C (3) F (4) E (5) B & E
Now let us devise a stepwise solution to above question. First, we will note down all key points given in the question.
· There are 6 teams: A, B, C, D, E and F.
· There are 3 matches in stage 1 and 2 matches in stage 2.
· Each team plays against other once only.
· There are no ties in the game.
Keeping these points in mind and using the information given about stage 1 we will construct a table for it.
A  B  C  D  E  F  
A  
B  
C  
D  
E  
F 
One by one we will interpret all the points given in stage 1 and use x to denote no match between two teams and won & loss for signifying winning and losing teams. The first statement is
· One team won all 3 matches. But at this moment we have no other information about which team has lost or won so we will get back to this point later.
· Two teams lost all matches. Though it is a useful piece of information as out of 6 teams 2 lost all but we have no further info about which team hence we will move on.
· Next is D lost to A. Thus, we will write lost in row 5 and column 1. Also, we will rule out D as the team who won all matches. Also, it won against C and F.
A  B  C  D  E  F  
A  x  
B  x  
C  x  Lost  
D  Lost  x  Won  X  x  Won 
E  x  
F  Lost  X 
· Since no team can play against each other. Therefore, we have put x there. Also as all teams play only 3 matches. There will be no match between D & B and D & E.
· Again, as given E lost to B but won against C & F. Therefore, E is also ruled out of the one who won all matches or lost all matches. Thus, there would be no match of E & A and E & D.
A  B  C  D  E  F  
A  x  
B  x  X  Won  
C  x  Lost  Lost  
D  Lost  x  Won  X  x  Won 
E  x  Lost  Won  X  x  Won 
F  Lost  Lost  x 
· Given B has lost at least one match. Therefore, B is not all winning team. And B will not be the losing team too. Since, all B, C, D, E, and F has lost one match at least thus, A is the only team left and hence became the all winning team.
· F doesn’t play against the winning team i.e. A.
· Thus, C and F becomes all losing team. And this will be the table formed.
A  B  C  D  E  F  
A  x  Won  Won  Won  x  x 
B  Lost  x  x  X  Won  Won 
C  Lost  x  x  Lost  Lost  x 
D  Lost  x  Won  X  x  Won 
E  X  Lost  Won  X  x  Won 
F  X  Lost  x  Lost  Lost  x 
Now we will move on to stage 2 and move on to form a table.
A  B  C  D  E  F  
A  x  x  x  X  Lost  Lost 
B  x  x  x  x  
C  x  x  X  x  
D  x  x  X  x  
E  Won  x  x  x  x  
F  Won  x  Won  X  x  x 
A  B  C  D  E  F  
A  x  x  x  X  Lost  Lost 
B  x  x  Won  Won  x  x 
C  x  Lost  x  X  x  Lost 
D  x  Lost  x  X  Lost  x 
E  Won  x  x  Won  x  x 
F  Won  x  Won  X  x  x 
Now we are in position to answer any question regarding this problem. Hence, we can simply look at these tables and answer the above question easily. Using similar approach, we can solve many such Caselet reasoning questions.
To explain these type of Caselet once more we will make use of an example. Consider this problem of CAT 2006:
Two traders, Chetan and Michael, were involved in the buying and selling of MCS shares over five trading
days. At the beginning of the first day, the MCS share was priced at ₹100, while at the end of the fifth day
it was priced at ₹110. At the end of each day, the MCS share price either went up by ₹10, or else, it
came down by ₹10. Both Chetan and Michael took buying and selling decisions at the end of each
trading day. The beginning price of MCS share on a given day was the same as the ending price of the
previous day. Chetan and Michael started with the same number of shares and amount of cash, and had
enough of both. Below are some additional facts about how Chetan and Michael traded over the five trading
days.
· Each day if the price went up, Chetan sold 10 shares of MCS at the closing price. On the other
hand, each day if the price went down, he bought 10 shares at the closing price.
· If on any day, the closing price was above ₹110, then Michael sold 10 shares of MCS, while if it
was below ₹90, he bought 10 shares, all at the closing price.
1. If Chetan sold 10 shares of MCS on three consecutive days, while Michael sold 10 shares only once
during the five days, what was the price of MCS at the end of day 3?
(1) ₹ 90 (2) ₹100 (3) ₹110 (4) ₹120 (5) ₹130
2. If Chetan ended up with ₹1300 more cash than Michael at the end of day 5, what was the price of
MCS share at the end of day 4?
(1) ₹90 (2) ₹100 (3) ₹110 (4) ₹120 (5) Not uniquely determinable
3. If Michael ended up with 20 more shares than Chetan at the end of day 5, what was the price of the
share at the end of day 3?
(1)₹90 (2) ₹100 (3) ₹110 (4) ₹120 (5) ₹130
4. If Michael ended up with ₹100 less cash than Chetan at the end of day 5, what was the difference
in the number of shares possessed by Michael and Chetan (at the end of day 5)?
(1)Michael had 10 less shares than Chetan.
(2) Michael had 10 more shares than Chetan.
(3) Chetan had 10 more shares than Michael,
(4) Chetan had 20 more shares than Michael.
(5) Both had the same number of shares.
To solve the above caselet and questions on them we will keep all the abovementioned points and proceed similarly as in the previous example. Again, this time we will construct a table using significant key points
3. In this case, there are two people Michael and Chetan.
4. The price at the beginning of the first day is ₹100 and end of the fifth day is ₹110.
5. Prices fluctuate every day either they went up by ₹10 or get down by ₹10. And the ending price of that day becomes the beginning price of next day.
6. Using the above points there could be drawn 10 different cases and a table can be constructed like this:
At the end of  Day 1  Day 2  Day 3  Day 4  Day 5 
Case 1  110  100  90  100  110 
Case 2  110  120  110  100  110 
Case 3  110  120  130  120  110 
Case 4  110  100  110  100  110 
Case 5  110  100  110  120  110 
Case 6  110  120  110  120  110 
Case 7  90  100  90  100  110 
Case 8  90  80  90  100  110 
Case 9  90  100  110  100  110 
Case 10  90  100  110  120  110 
Now with help of two points given in problem about Michael and Chetan and their reaction to decrease and increase in prices. We will directly solve all the following questions of the problem.
It is being told that Chetan sold 10 shares on 3 consecutive days and Chetan only sells shares if prices went up. Thus, coinciding cases in relevance to Chetan is Case 3, Case 8, Case 10. Also, Michael sold 10 shares only once during all 5 days where Chetan sold thrice. And Michael sells only if the closing price is above 110. Now by comparing all the 3 cases and adding the Michael factor to it, we can easily conclude to solitary case 10. Thus, our solution Case is Case 10. Therefore, the answer to the question is 110.
If Chetan has 1300 more cash than Michael at the end of the fifth day. The possibility of this happening could be
Case 1  
Chetan  110*10100*1090*10+100*10+110*10 = 1300

Michael  No share was bought or sold by him. 
Case 4  
Chetan  110*10100*10+110*10100*10+110*10 = 1300

Michael  No share was bought or sold by him. 
Case 7  
Chetan  90*10+100*1090*10+100*10+110*10 = 1300

Michael  No share was bought or sold by him. 
Case 9  
Chetan  90*10+100*10+110*10100*10+110*10 = 1300

Michael  No share was bought or sold by him. 
Now in all these cases price of the shares at the end of the 4^{th} day is ₹100.
Let us assume both Chetan and Michael started with x no. of shares. Now at the end of the 5th day, Michael had 20 more shares than Chetan. We will do similar reasoning as in previous questions but now instead of the amount earned, we will calculate no. of shares.
Case 8  No. of shares at the end of 5^{th} day 
Chetan  x + 10+10101010 = x10

Michael  x +10 = x+10 
There’s an only single possibility of Michael having 20 more shares that Chetan. Thus, the price at the end of day 3 is₹ 90.
We need to find out the cases where Michael has ₹100 less than Chetan. We will proceed as above.
Case 2  Amount earned  No. of shares 
Chetan  110*10+120*10110*10100*10+110*10 = 1300

x – 1010+10+1010 = x10 
Michael  120*10 = 1200  x – 10 
Case 10  Amount earned  No. of shares 
Chetan  90*10+100*10+110*10+120*10110*10 = 1300

x + 10101010+10 =x10 
Michael  120* 10 = 1200  x – 10 
Now as we can see in both case, Michael and Chetan ended up with equal no. of shares. Thus, option (5) is correct.
The above example has now given a vivid picture of how to decode the information given in the problem and then work out on the solution. There can be thousands distinctive formation of these question but you can solve them using the method explained above. It is not mandatory to use tabular representation you can other forms too that makes easy to decipher the problem. You can also try interpreting Business Magazine and newspaper. They are real life scenario problems. If you can understand them solving such questions in the exam would be simpler. You need to practice a lot to have a tight grip Caselet as “Knowledge is of no value unless you put it into practice”. So, keep practicing!
This article was contributed by Sejal Khurana. If you want to write for us, please email us on [email protected]
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