The document Caselets based questions LR Notes | EduRev is a part of the LR Course Logical Reasoning (LR) and Data Interpretation (DI).

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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 step-wise 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.

**1. Paragraph based on Reasoning**

Now letâ€™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 â€“ I and two matches in Stage â€“ 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 â€“ I and Stage â€“ II are as given below.

**Stage-I:**

- 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 Stage-I.

**Stage-II:**

- The leader of Stage-I lost the next two matches.
- Of the two teams at the bottom after Stage-I, one team won both matches, while the other lost both matches.
- One more team lost both matches in Stage-II.

1. The two teams that defeated the leader of Stage-I 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 Stage-II 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 step-wise 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 doesnâ€™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.

- Given leader in stage 1 lost 2 matches. Since A is the leader. A will be the one who will have lost all matches in next stage. Also, each team has a just single match against the other, therefore, A will lose against E and F.

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 |

- Now out of the two, losing team one won next two matches and one lost all. Since F won against A. Therefore, F will be the winning team and C, the losing team.
- Also given another team lost both matches and it canâ€™t be E, again as it won against A and it can neither be B as C to lose both matches require B to win against him. Therefore, D lost both too.

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 â‚¹100, while at the end of the fifth day it was priced at â‚¹110. At the end of each day, the MCS share price either went up by â‚¹10, or else, it came down by â‚¹10. 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 â‚¹110, then Michael sold 10 shares of MCS, while if it was below â‚¹90, 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) â‚¹ 90

(2) â‚¹100

(3) â‚¹110

(4) â‚¹120

(5) â‚¹130

2. If Chetan ended up with â‚¹1300 more cash than Michael at the end of day 5, what was the price of MCS share at the end of day 4?

(1) â‚¹90

(2) â‚¹100

(3) â‚¹110

(4) â‚¹120

(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)â‚¹90

(2) â‚¹100

(3) â‚¹110

(4) â‚¹120

(5) â‚¹130

4. If Michael ended up with â‚¹100 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 above-mentioned 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 â‚¹100 and end of the fifth day is â‚¹110.

5. Prices fluctuate every day either they went up by â‚¹10 or get down by â‚¹10. 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*10-100*10-90*10+100*10+110*10 = 1300 | ||

Michael | No share was bought or sold by him. |

Case 4 | |

Chetan | 110*10-100*10+110*10-100*10+110*10 = 1300 |

Michael | No share was bought or sold by him. |

Case 7 | |

Chetan | -90*10+100*10-90*10+100*10+110*10 = 1300 |

Michael | No share was bought or sold by him. |

Case 9 | |

Chetan | -90*10+100*10+110*10-100*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 â‚¹100.

**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+10-10-10-10 = x-10 |

Michael | x +10 = x+10 |

Thereâ€™s an only single possibility of Michael having 20 more shares that Chetan. Thus, the price at the end of day 3 isâ‚¹ 90.

**Caselets Question 4:**

We need to find out the cases where Michael has â‚¹100 less than Chetan. We will proceed as above.

Case 2 | Amount earned | No. of shares |

Chetan | 110*10+120*10-110*10-100*10+110*10 = 1300 | x â€“ 10-10+10+10-10 = x-10 |

Michael | 120*10 = 1200 | x â€“ 10 |

Case 10 | Amount earned | No. of shares |

Chetan | TRUE | x + 10-10-10-10+10 =x-10 |

Michael | 120* 10 = 1200 | x â€“ 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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