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Test Level 3: Caselets - 1 - CAT MCQ


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10 Questions MCQ Test Level-wise Tests for CAT - Test Level 3: Caselets - 1

Test Level 3: Caselets - 1 for CAT 2024 is part of Level-wise Tests for CAT preparation. The Test Level 3: Caselets - 1 questions and answers have been prepared according to the CAT exam syllabus.The Test Level 3: Caselets - 1 MCQs are made for CAT 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test Level 3: Caselets - 1 below.
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Test Level 3: Caselets - 1 - Question 1
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Directions: Refer to the data below and answer the question that follows.

The table below gives the number of times 8 colleges viz. P, Q, R, S, T, U, V and W won gold, silver or bronze medals and obtained 4th, 5th and 6th places in intercollege competitions held this year.

The number of points the college scored during the competitions is calculated using the following formula:
Number of points = (Gold medals) × 3 + (Silver medals) × 2 + (Bronze medals) +[(1 - 0.1 × place) × (places)
[Note: (places) or (medals) refers to the number of times the college won that place or that medal]
Some additional facts that are known are as follows:
(i) The number of bronze medals won by college P is the same as the number of silver medals won by college Q.
(ii) The number of times college U was at the 4th place is the same as the number of bronze medals won by college T.
(iii) No two colleges are assigned the same place in a game.

Q. How many times did college P win the bronze medal?

Detailed Solution for Test Level 3: Caselets - 1 - Question 1

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 4th 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 5th place x times and the 6th place y times and school U win the 5th place z times and the 6th 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 7th and 8th places won by the college = 36 – (gold, silver, bronze, 4th, 5th, 6th places won by the college)
Hence, we can complete the table as follows:

Number of times college P won the bronze medal = 6

Test Level 3: Caselets - 1 - Question 2
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Directions: Refer to the data below and answer the question that follows.

The table below gives the number of times 8 colleges viz. P, Q, R, S, T, U, V and W won gold, silver or bronze medals and obtained 4th, 5th and 6th places in intercollege competitions held this year.

The number of points the college scored during the competitions is calculated using the following formula:
Number of points = (Gold medals) × 3 + (Silver medals) × 2 + (Bronze medals) +[(1 - 0.1 × place) × (places)
[Note: (places) or (medals) refers to the number of times the college won that place or that medal]
Some additional facts that are known are as follows:
(i) The number of bronze medals won by college P is the same as the number of silver medals won by college Q.
(ii) The number of times college U was at the 4th place is the same as the number of bronze medals won by college T.
(iii) No two colleges are assigned the same place in a game.

Q. How many colleges scored more than 40 points?

Detailed Solution for Test Level 3: Caselets - 1 - Question 2

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 4th 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 5th place x times and the 6th place y times and school U win the 5th place z times and the 6th 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 7th and 8th places won by the college = 36 – (gold, silver, bronze, 4th, 5th, 6th places won by the college)
Hence, we can complete the table as follows:

Number of colleges that scored more than 40 points = 3

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Test Level 3: Caselets - 1 - Question 3
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Directions: Refer to the data below and answer the question that follows:
The daily demand (in number of customers) for air travel from Delhi to Bangalore and from Bangalore to Delhi, each is 1200 customers. The breakdown of demand by time is given in the following table:

The demand for travel from Bangalore to Delhi is identical to that from Delhi to Bangalore. Any unmet demand in any time period does not spill over to the next time period.
Space jet, an airline, which flies only between Delhi and Bangalore has the following costs:
1. A surcharge A per passenger of Rs. 100 to be paid to the Airlines Regulatory Board.
2. A fixed cost B of Rs. 72000 per trip made.
The airline owns 2 planes, each with a maximum capacity of 150 passengers. One plane is parked at Delhi and the other at Bangalore at the beginning of the day. One trip (from Delhi to Bangalore or vice versa) takes 1.5 hours. The airline charges a uniform fare of Rs. 1000 per passenger per trip.

Customer satisfaction index (CSI) is defined as: 

CSI = 1 - (Unmet demand)/(Total number of passengers that the airline carries on that day)

Q. Let P be the maximum profit possible in the first time period of 6 am to 8:59 am. Based on new information, it is known that 60% of the demand in the first time period consists of business travellers who will pay Rs. 2500 per trip, but will not wait beyond 7 am. Hence, the airline company decided to purchase two more planes with a capacity of 150 each. If it has to maintain the same profit P, what is the fixed cost per trip that the airline has to pay for the new plane? Assume the same surcharge of Rs. 100 per passenger.

Detailed Solution for Test Level 3: Caselets - 1 - Question 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

Test Level 3: Caselets - 1 - Question 4
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Directions: Refer to the data below and answer the question that follows:
The daily demand (in number of customers) for air travel from Delhi to Bangalore and from Bangalore to Delhi, each is 1200 customers. The breakdown of demand by time is given in the following table:

The demand for travel from Bangalore to Delhi is identical to that from Delhi to Bangalore. Any unmet demand in any time period does not spill over to the next time period.
Space jet, an airline, which flies only between Delhi and Bangalore has the following costs:
1. A surcharge A per passenger of Rs. 100 to be paid to the Airlines Regulatory Board.
2. A fixed cost B of Rs. 72000 per trip made.
The airline owns 2 planes, each with a maximum capacity of 150 passengers. One plane is parked at Delhi and the other at Bangalore at the beginning of the day. One trip (from Delhi to Bangalore or vice versa) takes 1.5 hours. The airline charges a uniform fare of Rs. 1000 per passenger per trip.

Customer satisfaction index (CSI) is defined as: 

CSI = 1 - (Unmet demand)/(Total number of passengers that the airline carries on that day)

Q. If the airline wants to make exactly 10 trips (5 each way) per day in such a way that maximises the profit for the airline, which of the following cannot be the start time for the third flight from Delhi to Bangalore?
Detailed Solution for Test Level 3: Caselets - 1 - Question 4

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.

Test Level 3: Caselets - 1 - Question 5
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Directions: Answer the question on the basis of the information given below.
The following table provides information about the marks obtained by following sixteen students in Quantitative Aptitude, Logical Reasoning and Data Interpretation and Verbal Ability in an MBA entrance test for an MBA Institute.
(i) The total marks obtained by the student in the test is the sum of the marks obtained by the student in the mentioned three sections. It also provides information about the centre at which the student is enroled.
(ii) The students belong to either one of the three centres namely Jaipur, Delhi and Punjab. Each student is enroled at only one centre.
(iii) Each student is given only one rank from 1 to 16 based on the marks obtained by him/her in the entrance test. This rank is called 'overall rank'.
Note: 
A student A (assume) is given a numerically lesser rank than the other student B (assume) if the total marks obtained by A is greater than the total marks obtained by B.

  • If the total marks obtained by A is same as those obtained by B, then the student having obtained more marks in Verbal Ability section is given a numerically lesser rank.
  • If marks obtained by two students in Verbal Ability section are also same, then the student having obtained more marks in Logical Reasoning and Data Interpretation section is given a numerically lesser rank.

Q. From which of the following centres did the maximum possible number of students obtain a total of at least 33 marks and a total of at most 57 marks?
Detailed Solution for Test Level 3: Caselets - 1 - Question 5

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.

Test Level 3: Caselets - 1 - Question 6
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Directions: Answer the question on the basis of the information given below.
The following table provides information about the marks obtained by following sixteen students in Quantitative Aptitude, Logical Reasoning and Data Interpretation and Verbal Ability in an MBA entrance test for an MBA Institute.
(i) The total marks obtained by the student in the test is the sum of the marks obtained by the student in the mentioned three sections. It also provides information about the centre at which the student is enroled.
(ii) The students belong to either one of the three centres namely Jaipur, Delhi and Punjab. Each student is enroled at only one centre.
(iii) Each student is given only one rank from 1 to 16 based on the marks obtained by him/her in the entrance test. This rank is called 'overall rank'.
Note: 
A student A (assume) is given a numerically lesser rank than the other student B (assume) if the total marks obtained by A is greater than the total marks obtained by B.

  • If the total marks obtained by A is same as those obtained by B, then the student having obtained more marks in Verbal Ability section is given a numerically lesser rank.
  • If marks obtained by two students in Verbal Ability section are also same, then the student having obtained more marks in Logical Reasoning and Data Interpretation section is given a numerically lesser rank.

Q. How many male student(s) has/have 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?
Detailed Solution for Test Level 3: Caselets - 1 - Question 6

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.

Test Level 3: Caselets - 1 - Question 7
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Directions: Answer the following question based on the information given below.
Tina, a blast furnace expert, who works as a technology trouble shooter stays in Jamshedpur. She has got an important assignment in Delhi, which requires 6 hours to complete. The work is so critical that she has to start working the moment she reaches the client's premises.
She is considering various options for her onward and return journey between Jamshedpur and Delhi.
A quick search revealed that ticket from Jamshedpur to Delhi is available on 2 trains. Trains 12801 and 12443 depart from Jamshedpur station at 06:45 hrs and 15:55 hrs and reach Delhi next day at 04:50 hrs and 10:35 hrs, respectively. Trains 12444 and 12802 start from Delhi at 17:20 hrs and 22:20 hrs and reach Jamshedpur next day at 10:35 hrs and 20:05 hrs, respectively.
Another option is to reach Ranchi by a 3-hour road trip and take a flight to Delhi from Ranchi. The distance between Delhi and Ranchi is covered in 105 minutes both ways by any of the scheduled flights. Air India operates 2 flights AI 9810 and AI 810, which depart Ranchi at 08:00 hours and 15:25 hrs, respectively. Flight number IT - 3348 operated by Kingfisher Airlines departs Ranchi at 19:20 hrs. Return flights operated by Air India - AI 9809 and AI809 depart Delhi at 05:50 hrs and 11:00 hrs, respectively. Flight number IT - 3347 operated by Kingfisher Airlines departs Delhi at 17:10 hrs.
From Tina's home, Jamshedpur railway station is 5 minutes drive, and her destination at Delhi is 90 minutes and 30 minutes drive from airport and railway station, respectively. One has to reach the airport at least 1 hour before the scheduled departure to complete the boarding procedure. At every railway station, she loses 5 minutes in navigating through the crowd.

Q. If Tina decides to minimise the in-between waiting period, the option she should choose from the ones given below will be:
Detailed Solution for Test Level 3: Caselets - 1 - Question 7

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.

Test Level 3: Caselets - 1 - Question 8
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Directions: Answer the following question based on the information given below.
Tina, a blast furnace expert, who works as a technology trouble shooter stays in Jamshedpur. She has got an important assignment in Delhi, which requires 6 hours to complete. The work is so critical that she has to start working the moment she reaches the client's premises.
She is considering various options for her onward and return journey between Jamshedpur and Delhi.
A quick search revealed that ticket from Jamshedpur to Delhi is available on 2 trains. Trains 12801 and 12443 depart from Jamshedpur station at 06:45 hrs and 15:55 hrs and reach Delhi next day at 04:50 hrs and 10:35 hrs, respectively. Trains 12444 and 12802 start from Delhi at 17:20 hrs and 22:20 hrs and reach Jamshedpur next day at 10:35 hrs and 20:05 hrs, respectively.
Another option is to reach Ranchi by a 3-hour road trip and take a flight to Delhi from Ranchi. The distance between Delhi and Ranchi is covered in 105 minutes both ways by any of the scheduled flights. Air India operates 2 flights AI 9810 and AI 810, which depart Ranchi at 08:00 hours and 15:25 hrs, respectively. Flight number IT - 3348 operated by Kingfisher Airlines departs Ranchi at 19:20 hrs. Return flights operated by Air India - AI 9809 and AI809 depart Delhi at 05:50 hrs and 11:00 hrs, respectively. Flight number IT - 3347 operated by Kingfisher Airlines departs Delhi at 17:10 hrs.
From Tina's home, Jamshedpur railway station is 5 minutes drive, and her destination at Delhi is 90 minutes and 30 minutes drive from airport and railway station, respectively. One has to reach the airport at least 1 hour before the scheduled departure to complete the boarding procedure. At every railway station, she loses 5 minutes in navigating through the crowd.

Q. Tina gets a message that her work has to be completed between 09:00 hrs and 17:00 hrs. If she wants to minimise the total time out of Jasmshedpur, the best option, from among the following, is to go by:
Detailed Solution for Test Level 3: Caselets - 1 - Question 8

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.

Test Level 3: Caselets - 1 - Question 9
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Directions: Read the information carefully to answer the question that follows.
Dr. Ravi Patel teaches financial markets course at Indian Management Institute, which is being taught in two parts - one part in first trimester and the other part in second trimester. Initially, the professor divides the students randomly in five classes in first trimester and based on the performance of students in first trimester, he further divides the students of each class into two parts as below average and above average performers. It is observed that in each class, the ratio of below average and above average performers is at least 13/10.
Initially, the classes were named as classes P, Q, R, S and T and after further division, the classes were renamed as classes 1, 2, 3, 4 ,....,10 and the classes in trimester 2 are named randomly i.e. it is not necessary that two parts of class A are numbered one after the other.
It is also given that 1/5 of total students in first trimester were in class R, 23% of total students were in class Q and 17% of total students were in class T. The ratio of students in classes P and S was 11 : 9. In the second trimester, there were 20 students in class 1. The number of students in classes 3, 10 and 9 were twice, thrice and four times the number of students in class 1. The number of students in classes 4 and 5 are in ratio 1 : 3. The number of students in class 2 were 5 more than the number of students in class 9. The number of students in class 8 was 40% more than the number of students in class 4. The number of students in classes 6 and 7 is 50 and 30 and the total number of students in classes 4, 5 and 8 is 125% more than the number of students in class 10. Each student in first trimester was classified as either below average or above average performer.

Q. In how many classes was the number of students who performed below average at least 90% more than the number of students who performed above average?
Detailed Solution for Test Level 3: Caselets - 1 - Question 9

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.

Test Level 3: Caselets - 1 - Question 10
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Directions: Read the information carefully to answer the question that follows.
Dr. Ravi Patel teaches financial markets course at Indian Management Institute, which is being taught in two parts - one part in first trimester and the other part in second trimester. Initially, the professor divides the students randomly in five classes in first trimester and based on the performance of students in first trimester, he further divides the students of each class into two parts as below average and above average performers. It is observed that in each class, the ratio of below average and above average performers is at least 13/10.
Initially, the classes were named as classes P, Q, R, S and T and after further division, the classes were renamed as classes 1, 2, 3, 4 ,....,10 and the classes in trimester 2 are named randomly i.e. it is not necessary that two parts of class A are numbered one after the other.
It is also given that 1/5 of total students in first trimester were in class R, 23% of total students were in class Q and 17% of total students were in class T. The ratio of students in classes P and S was 11 : 9. In the second trimester, there were 20 students in class 1. The number of students in classes 3, 10 and 9 were twice, thrice and four times the number of students in class 1. The number of students in classes 4 and 5 are in ratio 1 : 3. The number of students in class 2 were 5 more than the number of students in class 9. The number of students in class 8 was 40% more than the number of students in class 4. The number of students in classes 6 and 7 is 50 and 30 and the total number of students in classes 4, 5 and 8 is 125% more than the number of students in class 10. Each student in first trimester was classified as either below average or above average performer.

Q. Out of total students in class S, what percent performed below average?
Detailed Solution for Test Level 3: Caselets - 1 - Question 10

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 = $\frac{60}{90}\times 100 = 66.67$%

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