Caselets Questions for CAT with Answers PDF

Since the total number of gold medals won is 36, the total number of games played should be 36, i.e. total number of silver medals, bronze medals and the places must be 36.
Therefore, silver medals won by college Q = 36 – (5 + 4 + 8 + 3 + 7 + 2 + 1) = 36 – 30 = 6
Then, from (i), bronze medals won by college P = 6
Number of times college U won the 4^th place = 36 – (0 + 4 + 5 + 8 + 8 + 4 + 4) = 36 – 33 = 3
Then, from (ii), bronze medals won by college T = 3
So, bronze medals won by college S = 36 – (6 + 3 + 2 + 3 + 10 + 5 + 5) = 36 – 34 = 2
Let college R win the 5^th place x times and the 6^th place y times and school U win the 5^th place z times and the 6^th place w times. Then, we get
x + z = 3, y + w = 14 and 0.5x + 0.4y = 4.4
i.e. x + z = 3, 5x + 4y = 44
Now, consider the cases for x + z = 3
For all other values of x (i.e. 1, 2, 3), we do not get integer values for y and w.
So, x = 0, y = 11, w = 3 and z = 3.
We know that every college played 36 games. Therefore, the total number of medals and places won by each college must add up to 36.
Therefore, the number of 7^th and 8^th places won by the college = 36 – (gold, silver, bronze, 4^th, 5^th, 6^th places won by the college)
Hence, we can complete the table as follows:

Number of times college P won the bronze medal = 6

Since the total number of gold medals won is 36, the total number of games played should be 36, i.e. total number of silver medals, bronze medals and the places must be 36.
Therefore, silver medals won by college Q = 36 – (5 + 4 + 8 + 3 + 7 + 2 + 1) = 36 – 30 = 6
Then, from (i), bronze medals won by college P = 6
Number of times college U won the 4^th place = 36 – (0 + 4 + 5 + 8 + 8 + 4 + 4) = 36 – 33 = 3
Then, from (ii), bronze medals won by college T = 3
So, bronze medals won by college S = 36 – (6 + 3 + 2 + 3 + 10 + 5 + 5) = 36 – 34 = 2
Let college R win the 5^th place x times and the 6^th place y times and school U win the 5^th place z times and the 6^th place w times. Then, we get
x + z = 3, y + w = 14 and 0.5x + 0.4y = 4.4
i.e. x + z = 3, 5x + 4y = 44
Now, consider the cases for x + z = 3
For all other values of x (i.e. 1, 2, 3), we do not get integer values for y and w.
So, x = 0, y = 11, w = 3 and z = 3.
We know that every college played 36 games. Therefore, the total number of medals and places won by each college must add up to 36.
Therefore, the number of 7^th and 8^th places won by the college = 36 – (gold, silver, bronze, 4^th, 5^th, 6^th places won by the college)
Hence, we can complete the table as follows:

Number of colleges that scored more than 40 points = 3

The implication of unmet demand not spilling over is equivalent to each time period's demand, independent of all other time periods. Hence, the maximum profit is obtained when the number of profitable passengers is maximised for each time period. More than 80 passengers are needed for a profitable trip as for 80 passengers, the fare is Rs. 80000 and the cost is Rs. 72000 + Rs. 8000 = Rs. 80000. For each additional passenger over and above 80 passengers in a trip, the airline makes a profit of Rs. 1000 – Rs. 100 = Rs. 900.
Note that all the demand is now at the beginning of the time period. In a 2 hours and 59 minutes span, a plane can start a maximum of two trips since it takes 1.5 hours for a trip. Consider the first time period: 6 am - 8:59 am. Since there 360 passengers waiting for the trip, starting at 6 am, at each destination, the airline can make two trips (of each plane) with the maximum capacity of 150 passengers at a time from Delhi to Bangalore and Bangalore to Delhi. Similarly, a plane at Bangalore will carry 150 passengers at a time from Bangalore to Delhi and Delhi to Bangalore. Consider the time period 9 am - 11:59 am. Since there are 240 passengers waiting for the trip at each destination, two trips (of each plane) cannot be made with 150 passengers and another with 90 passengers or 2 trips with 120 passengers in each trip.

Applying similar arguments to each time period, we obtain the following table for trips from Delhi to Bangalore to maximise the profit:

Plane at Delhi has 470 profitable passengers.
Therefore, profit for plane at Delhi = 470 x Rs. 900 = Rs. 423000
The table for trips from Bangalore to Delhi is identical.
Total profit (considering the plane at Bangalore also) is Rs. 423000 x 2 = Rs. 846000. From the table given above, maximum profit possible in the first period without any new information = P = 140 x Rs. 900 = Rs. 126000.
Normally, only 2 trips each way would have been possible. However, since there are 2 more planes, 4 trips each way are possible between 6 am and 8:59 am. This implies a total capacity of 150 x 4 = 600 passengers.
Hence, all 360 people can be flown.

Number of business travellers = 60% x 360 = 216
Revenue from business travellers = 216 x Rs. 2500 = Rs. 540000
Number of normal travellers = 40% x 360 = 144 
Revenue from normal travellers = 144 x Rs. 1000 = Rs. 144000 
Total revenue = Rs. 684000
Surcharge = 360 x Rs. 100 = Rs. 36000
Fixed cost of old plane = Rs. 72000 x 2
Let the fixed cost of the new plane be F.
Profit = Rs. 684000 – Rs. 144000 – Rs. 36000 – F x 2
Equating this to P, we get
Rs. 126000 = Rs. 684000 – Rs. 180000 – 2F
F = (Rs. 378000)/2 = Rs. 189000

The implication of unmet demand not spilling over is equivalent to each time period's demand, independent of all other time periods. Hence, the maximum profit is obtained when the number of profitable passengers is maximised for each time period. More than 80 passengers are needed for a profitable trip as for 80 passengers, the fare is Rs. 80000 and the cost is Rs. 72000 + Rs. 8000 = Rs. 80000. For each additional passenger over and above 80 passengers in a trip, the airline makes a profit of Rs. 1000 – Rs. 100 = Rs. 900.
Note that all the demand is now at the beginning of the time period. In a 2 hours and 59 minutes span, a plane can start a maximum of two trips since it takes 1.5 hours for a trip. Consider the first time period: 6 am - 8:59 am. Since there 360 passengers waiting for the trip, starting at 6 am, at each destination, the airline can make two trips (of each plane) with the maximum capacity of 150 passengers at a time from Delhi to Bangalore and Bangalore to Delhi. Similarly, a plane at Bangalore will carry 150 passengers at a time from Bangalore to Delhi and Delhi to Bangalore. Consider the time period 9 am - 11:59 am. Since there are 240 passengers waiting for the trip at each destination, two trips (of each plane) cannot be made with 150 passengers and another with 90 passengers or 2 trips with 120 passengers in each trip.

Applying similar arguments to each time period, we obtain the following table for trips from Delhi to Bangalore to maximise the profit:

Since the profit is to be maximised, the trip should be a full capacity flight, i.e. of 150 passengers.
If the first two trips start between 6 am and 8:59 am, then the third trip can start between 9 am and 11:59 am or 3 pm and 5:59 pm as the demand in these periods is more than 150, which is not the same as between 12 noon and 2:59 pm.
Note: The third trip cannot start between 6 pm and 8:59 pm, as then, the next 2 trips will not be possible on the same day.

The total marks obtained by the students and their overall ranks are tabulated in the following table:

3 students each from Punjab and Delhi, obtained a total of at least 33 marks and a total of at most 57 marks.

The total marks obtained by the students and their overall ranks are tabulated in the following table:

Here, only Manish has obtained more marks in Logical Reasoning and Data Interpretation than the marks obtained by Deepika in Logical Reasoning and Data Interpretation, but less marks in Verbal Ability than the marks obtained by Dimple in Verbal Ability.

Let us evaluate this option wise.
Let us evaluate option 1 first.
According to this, she goes by train 12801 and returns by IT 3347.
Train 12801 departs from Jamshedpur at 06:45 and arrives in Delhi at 04:50.
She departs for client site at 4:55 and  reaches the client's site at 05:25 and her work there is finished by 11:25.
She then reaches the airport at 12:55 and checks in by 13:55.
IT 3347 departs at 17:10.
So, she has a waiting period of 3 hours 15 minutes.
Let us now evaluate option 2.
According to this, she goes by train 12443 and comes back by train 12802.
Train 12443 departs from Jamshedpur at 15:55 and arrives in at Delhi at 10:35 the next day.
Thereafter she goes to the client's site at 10:40.
She reaches there at 11:10, and works till 17:10.
She then travels to the railway station at 17:40 and boards the train at 17:45.
The train however departs at 22:20.
Thus, she has to wait for 4 hours and 35 minutes.
Let us now evaluate option 3.
According to this, she goes by AI 9810 and returns by train 12802.
AI 9810 departs from Ranchi at 08:00 hrs and reaches Delhi at 09:45.
Then, she travels to the client's site and reaches there at 11:15.
Then, she works there till 17:15.
She then reaches the station at 17:45 and boards the train at 17:50.
However, the train departs at 22:20.
Thus, she has to wait for 4 hours and 30 minutes.
Let us now evaluate option 4.
According to this, she goes by flight AI 810 and returns by flight AI 9809.
Flight AI 810 departs from Ranchi at 15:25 and arrives in Delhi at 17:10.
She then travels to the client's site and reaches there at 18:40.
She works there till 00:40 and then travels to the airport where she reaches at 02:10.
She then boards the flight at 03:10.
The flight departs at 05: 50 hours.
Thus, she has to wait for 2 hours and 40 minutes.
It is thus evident that the waiting time is the least for option 4.
Hence, answer option 4 is correct.

Let us evaluate this option wise.
Evaluating option 1, train 12443 departs from Jamshedpur for Delhi at 15:55 and reaches Delhi the next day at 10:35.
She departs for the client's site at 10:40 and reaches there at 11:10 where she works till 17:10.
As by doing so, she overshoots the feasible work interval by 10 minutes, this scenario and hence travel option is not a feasible one.
Let us now evaluate option 2.
She departs in train 12801 from Jamshedpur for Delhi at 06:45 and reaches Delhi the next day at 04:50 hrs.
She leaves the station at 04:55 and reaches the client's site at 05:25.
She begins her work at 09:00 and finishes it by 15:00.
She reaches back Delhi railway station at 15:30 and can board a train by 15:35.
She boards the 12802 at its designated time for Jamshedpur where she reaches at 20:05 the next day.
Thus, she is away from Jamshedpur from 6:45 to 20:05 for a period of 37 hours and 20 minutes.
Let us now evaluate option 3.
She leaves Jamshedpur for Ranchi airport at 04:00 where she boards the flight AI 9810 at 08:00.
She reaches Delhi at 09:45 and reaches the client's site at 11:15.
Then, as she needs 6 hours to work between 9:00 and 17:00, this is not a feasible option.
Hence, this option is ruled out.
Let us now evaluate option 4.
She departs from Jamshedpur for Ranchi airport at 11:25 and boards the flight AI 810 for Delhi at 15:25.
She reaches Delhi at 16:55.
Next day, she goes to the client's site and works there from 09:00 to 15:00.
Next day she boards the flight AI 9809 for Ranchi at 05:50 which reaches Ranchi at 07:35.
She then goes by road to Jamshedpur, and finally reaches there at 10:35.
Thus, she is away from Jamshedpur for 47 hours and 10 minutes.
Hence, out of all the options, option 2 is the option by means of which she has to spend the least time out of Jamshedpur.
Thus, answer option 2 is correct.

Number of students in class 1 = 20
Number of students in class 3 = 40
Number of students in class 9 = 80
Number of students in class 10 = 60
Number of students in class 2 = 85
Number of students in class 6 = 50
Number of students in class 7 = 30
Let the number of students in class 4 = x
Number of students in class 5 = 3x
Number of students in class 8 = 1.4x
Also,
x + 1.4x + 3x = 135
x = 25
Number of students in class 4 = 25
Number of students in class 5 = 75
Number of students in class 8 = 35
So, total students = 20 + 40 + 80 + 60 + 85 + 50 + 30 + 135 = 500
Number of students in class R = 100
Number of students in class Q = 0.23 × 500 = 115
Number of students in class T = 0.17 × 500 = 85
Remaining students = 200
Number of students in class P = 11 × 200/20 = 110
Number of students in class S = 200 - 110 = 90
Since each of classes P, Q, R, S and T is divided into two parts, so possible choices are
For 85, its
(50, 35) and (60, 25)
For 90, its
(60, 30) and (50, 40)
Similarly, for 100 as sum, 3 cases are possible and for 110 as sum, 4 cases are possible.
It is also mentioned that in each class, the ratio of below average and above average performers is at least 13/10.
So, following table can be formed from given information after considering ratio case.
Class S is divided into class 7 (above average students) and 10 (below average students)
Class T is divided into class 6 (below average students) and class 8 (above average students)
Class R is divided into class 1 (above average students) and class 9 (below average students)
Class P is divided into class 4 (above average students) and class 2 (below average students)
Class Q is divided into class 3 (above average students) and class 5 (below average students)
The ratio of number of students who performed below average to the number of students who performed above average in each class is as follows.
Class P = 85/25 = 3.4 : 1
Class Q = 75/40 = 1.875 : 1
Class R = 80/20 = 4 : 1
Class S = 60/30 = 2 : 1
Class T = 50/35 = 1.4285 : 1
So, the number of students who performed below average were at least 90% more than the number of students who performed above average in 3 classes.

Number of students in class 1 = 20
Number of students in class 3 = 40
Number of students in class 9 = 80
Number of students in class 10 = 60
Number of students in class 2 = 85
Number of students in class 6 = 50
Number of students in class 7 = 30
Let the number of students in class 4 = x
Number of students in class 5 = 3x
Number of students in class 8 = 1.4x
Also,
x + 1.4x + 3x = 135
x = 25
Number of students in class 4 = 25
Number of students in class 5 = 75
Number of students in class 8 = 35
So, total students = 20 + 40 + 80 + 60 + 85 + 50 + 30 + 135 = 500
Number of students in class R = 100
Number of students in class Q = 0.23 × 500 = 115
Number of students in class T = 0.17 × 500 = 85
Remaining students = 200
Number of students in class P = 11 × 200/20 = 110
Number of students in class S = 200 - 110 = 90
Since each of classes P, Q, R, S and T is divided into two parts, so possible choices are
For 85, its
(50, 35) and (60, 25)
For 90, its
(60, 30) and (50, 40)
Similarly, for 100 as sum, 3 cases are possible and for 110 as sum, 4 cases are possible.
It is also mentioned that in each class, the ratio of below average and above average performers is at least 13/10.
So, following table can be formed from given information after considering ratio case.
Class S is divided into class 7 (above average students) and 10 (below average students)
Class T is divided into class 6 (below average students) and class 8 (above average students)
Class R is divided into class 1 (above average students) and class 9 (below average students)
Class P is divided into class 4 (above average students) and class 2 (below average students)
Class Q is divided into class 3 (above average students) and class 5 (below average students)
Below average students of class S are 60 and total students are 90.
Percentage = %