Caselet can be of two forms:
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.
1. Paragraph based on Reasoning
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:
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.
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
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 
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 
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.
2. Paragraph based on Numerical Data
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.
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.
Caselets Question 1:
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.
Caselets Question 2:
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.
Caselets Question 3:
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 5th 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.
Caselets Question 4:
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  TRUE  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.
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