Caselet can be of two forms:
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.
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.
One team won all the three matches.
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)
(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):
(2) A & C
(5) B & E
Now let us devise a step-wise solution to the 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 the 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.
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:
Now we will move on to stage 2 and move on to form a table.
Now we are in a position to answer any question regarding this problem. Hence, we can simply look at these tables and answer the above question easily. Using a similar approach, we can solve many such Caselet reasoning questions.
From this caselet, it is important to extract the necessary data first. If the total were 75, 80% of that would be- 80/ 100 x 75 => 60. Now, calculating the total number of students who scored less than 60 but more than 15 will give the total number of students who scored between 15-60 marks. Here, the data “70% score more than 50%” can be ignored. Therefore, Students who scored > 60 = 50% of 60 = 30 —–(i) Students who scored <= 60 = 30 ——(ii) Students who secured < 15 = 6 ——(iii) Now, students who score between 15 to 60 can be easily obtained by subtracting equation (ii) and (iii). Therefore, Students scoring between 15 to 60 marks are (30 – 6) = 24.
From this caselet, it is important to extract the necessary data first. If the total were 75, 80% of that would be-
80/ 100 x 75
Now, calculating the total number of students who scored less than 60 but more than 15 will give the total number of students who scored between 15-60 marks. Here, the data “70% score more than 50%” can be ignored.
Students who scored > 60 = 50% of 60 = 30 —–(i)
Students who scored <= 60 = 30 ——(ii)
Students who secured < 15 = 6 ——(iii)
Now, students who score between 15 to 60 can be easily obtained by subtracting equation (ii) and (iii).
Therefore, Students scoring between 15 to 60 marks are (30 – 6) = 24.
Example 1: Directions for Q. 1 to 5: Refer to the following information and answer the following questions.
To explain these types 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 Rs. 100, while at the end of the fifth day it was priced at Rs. 110. At the end of each day, the MCS share price either went up by Rs.10, or else, it came down by Rs.10. Both Chetan and Michael took buying and selling decisions at the end of each trading day. The beginning price of the 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) Rs. 90
(2) Rs. 100
(3) Rs. 110
(4) Rs. 120
(5) Rs. 130
2. If Chetan ended up with Rs.1300 more cash than Michael at the end of day 5, what was the price of MCS share at the end of day 4?
(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?
4. If Michael ended up with Rs.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 above-mentioned points and proceed similarly as in the previous example. Again, this time we will construct a table using significant key points.
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:
Now in all these cases price of the shares at the end of the 4th day is Rs.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.
There’s an only single possibility of Michael having 20 more shares that Chetan. Thus, the price at the end of day 3 is Rs.90.
We need to find out the cases where Michael has Rs.100 less than Chetan. We will proceed as above.
Now as we can see in both case, Michael and Chetan ended up with equal no. of shares. Thus, option (5) is correct.
People Power Corporation presently employs three Managers (A, B, and C) and five recruitment agents (D, E, F, G, and H). The company is planning to open a new office in San Jose to manage the placement of software professionals in the US. It is planning to relocate two of the three managers and three of the five recruitment agents to the office in San Jose. As it is an organization that is highly people-oriented the management wants to ensure that the individuals who do not function well together should not be made as a part of the team going to the US.
The following information was available to the HR department of People Power Corporation:
Q.1. If D goes to the new office which of the following is (are) true?
I. C cannot go
II. A cannot go
III. H must also go
(a) I only
(b) II and III only
(c) I and III only
(d) I, II and III
Q.2. If A is to be moved as one of the Managers, which of the following cannot be a possible working unit?
Q.3. If C and F are moved to the new office, how many combinations are possible?
Q.4. Given the group dynamics of the Managers and the recruitment agents, which of the following is sure to find a berth in the San Jose office?
Q.5. If C is sent to the San Jose office which member of the staff cannot go with C?
Answers: 1. (c) 2. (d) 3. (b) 4. (a) 5. (b)
Example 2: Ghosh Babu took voluntary retirement in Dec. 1991 and received a certain amount of money as retirement benefits. On Jan 1, 1992, he invested the entire amount in shares. At the end of the month, he sold all his shares and realised 25% profit. On Feb 1, he reinvested the entire amount in shares which he sold at the end of the month at a loss of 20%. Again, he invested the entire amount on Mar 1 in a new company. At the end of the month, he sold the new company to a friend and realised a profit of 20% in the process. He invested the entire amount in shares on Apr 1, which he sold at the end of the month for Rs. 1,08,000 incurring a loss of 10%.
Q.1. What is the amount of retirement benefits received by Ghosh Babu?
(a) Rs. 1,08,000
(b) Rs. 1,25,000
(c) Rs. 1,20,000
(d) Rs. 1,00,000
Correct Answer is option (d).
Let the amount received by Ghosh Babu in Dec. 1991 be Rs. x, as retirement benefits:
- Therefore, investment in the month of Jan 1992 = 100
- Profit of 25% at the end of Jan 1992.
- Hence, investment in the month of Feb 1992 = 125
- Loss of 20% at the end of Feb 1992
- Hence, investment in the month of March 1992 = 100
- Profit of 20% at the end of March 1992
- Hence, investment in the month of April 1992 = 120
- Loss of 10% at the end of April 1992
- Therefore the amount left at the end of April 1992 = 108
- Amount at the end of April 1002 = Rs. 1,08,000
- Therefore, simply equating figures, he would have started with Rs 1,00,000
Q.2. The percentage profit received by Ghosh Babu between Jan 1 and Apr 30 is:
(c) - 10.00%
(d) None of these
Correct Answer is option (a).
% Profit between Jan 1 and Apr 30 = (1.08x - x/x) X 100
Q.3. The amount of loss incurred by Ghosh Babu based on his operation in Apr 1992 is:
(a) Rs. 25,000
(b) Rs. 12,000
(c) Rs. 20,000
(d) Rs. 8,000
Correct Answer is option (b).
Investment in the month of April = Rs. 1,20,000
Amount received at end of April = Rs. 1,08,000
Therefore, Loss = Rs. 12,000
Q.4. The maximum amount invested by Ghosh Babu in any one month was in:
Correct Answer is option (b).
Maximum amount invested by Ghosh Babu is in the month of February = Rs. 1,25,000