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Practice Question - 64 (Tables) | 100 DILR Questions for CAT Preparation PDF Download

A company administers a written test comprising of three sections of 20 marks each - Data Interpretation (DI), Written English (WE) and General Awareness (GA), for recruitment. A composite score for a candidate (out of 80) is calculated by doubling her marks in DI and adding it to the sum of her marks in the other two sections. Candidates who score less than 70% marks in two or more sections are disqualified. From among the rest, the four with the highest composite scores are recruited. If four or less candidates qualify, all who qualify are recruited.

Ten candidates appeared for the written test. Their marks in the test are given in the table below. Some marks in the table are missing, but the following facts are known:

  1. No two candidates had the same composite score.
  2. Ajay was the unique highest scorer in WE.
  3. Among the four recruited, Geeta had the lowest composite score.
  4. Indu was recruited.
  5. Danish, Harini, and Indu had scored the same marks the in GA.
  6. Indu and Jatin both scored 100% in exactly one section and Jatin’s composite score was 10 more than Indu’s.

Practice Question - 64 (Tables) | 100 DILR Questions for CAT Preparation

Q1: Which of the following statements MUST be true?

  1. Jatin's composite score was more than that of Danish.
  2. Indu scored less than Chetna in DI.
  3. Jatin scored more than Indu in GA.

(a) Both 2 and 3
(b) Only 1
(c) Only 2
(d) Both 1 and 2

Practice Question - 64 (Tables) | 100 DILR Questions for CAT PreparationView Answer  Practice Question - 64 (Tables) | 100 DILR Questions for CAT Preparation

Ans: (d) 
It is given that Indu and Jatin both scored 100% in exactly one section. We can say that Jatin scored 100% marks in DI. Therefore, Jatin's composite score = 2*20+16+14 = 70
It is given that Jatin’s composite score was 10 more than Indu’s. Therefore, we can say that Indu's composite score = 70 - 10 = 60.
Indu also scored 100% in exactly one section. 

Case 1: Indu scored 100% marks in DI.
If Indu scored 100% marks in DI, then Indu's score in GA = 60 - 2*20 - 8 = 12 which is less than 70% of maximum possible marks. Indu already has less than 70% in WE, therefore we Indu can't be recruited . Hence, we can reject this case. 
Consequently, we can say that Indu scored 100% marks in WE. Therefore, Indu's score in Practice Question - 64 (Tables) | 100 DILR Questions for CAT Preparation

It is also given that Danish, Harini, and Indu had scored the same marks the in GA. 

Practice Question - 64 (Tables) | 100 DILR Questions for CAT Preparation

We are given that, among the four recruited, Geeta had the lowest composite score. 
Maximum composite score that Geeta can get = 2*14 + 6 + 20  = 54 {Assuming 100% marks in WE}. Since, Geeta was recruitedat a composite score of 54 or less we can say that Ester was definitely recruited. 
It is given that no two candidates had the same composite score. We can see that Chetna's composite score is 54. Hence, Geeta can't have a composite score of 54. Therefore, we can say that Geeta's composite score is 53 or less. 
We already know the four people(Jatin, Indu, Geeta, Ester) which were recruited. Hence, we cab say that Danish was rejected at a composite score of 51. Hence, we can say that Geeta's composite score in 52 or more. 
Consequently, we can say that Geeta's composite score if either 52 or 53. Therefore we can say that Geeta scored either 18 {52-(2*14+6)} or 19 {53-(2*14+6)} marks in WE. 

Practice Question - 64 (Tables) | 100 DILR Questions for CAT Preparation

Ajay was the unique highest scorer in WE. 

Case 1: Geeta scored 19 marks in WE.
We can say that if Geeta scored 19 marks in WE, then Ajay scored 20 marks in DI. In that case Ajay's composite score = 2*8 + 20 + 16 = 52. Which is a possible case. 

Case 1: Geeta scored 18 marks in WE.
We can say that if Geeta scored 18 marks in WE, then Ajay can score either 19 or 20 marks in DI. 
If Ajay scored 20 marks in DI then in that case Ajay's composite score = 2*8 + 20 + 16 = 52 which will be same as Geeta's composite score. Hence, we can say that in this case Ajay can't score 20 marks.
If Ajay scored 19 marks in DI then in that case Ajay's composite score = 2*8 + 19 + 16 = 51 which will be same as Danish's composite score. Hence, we can say that in this case Ajay can't score 19 marks.
Therefore, we can say that case 2 is not possible at all.

Practice Question - 64 (Tables) | 100 DILR Questions for CAT Preparation

Let us check all the statement one by one. 
Statement 1: Jatin's composite score was more than that of Danish. We can see that this statement is correct. 
Statement 2:  Indu scored less than Chetna in DI. We can see that Indu scored 16 marks in DI whereas Chetna scored 19 marks in DI. Hence, we can say that this statement is also correct. 
Statement 3: Jatin scored more than Indu in GA. We can see that Jatin scored 14 marks in GA whereas Indu scored 20 marks in GA. Hence, we can say that this statement is incorrect. 
Hence, we can say that option (d) is the correct answer. 

Q2: Which of the following statements MUST be FALSE?
(a) Bala scored same as Jatin in DI
(b) Harini’s composite score was less than that of Falak
(c) Bala’s composite score was less than that of Ester
(d) Chetna scored more than Bala in DI

Practice Question - 64 (Tables) | 100 DILR Questions for CAT PreparationView Answer  Practice Question - 64 (Tables) | 100 DILR Questions for CAT Preparation

Ans: (a)
It is given that Indu and Jatin both scored 100% in exactly one section. We can say that Jatin scored 100% marks in DI. Therefore, Jatin's composite score = 2*20+16+14 = 70
It is given that Jatin’s composite score was 10 more than Indu’s. Therefore, we can say that Indu's composite score = 70 - 10 = 60.
Indu also scored 100% in exactly one section. 

Case 1: Indu scored 100% marks in DI.
If Indu scored 100% marks in DI, then Indu's score in GA = 60 - 2*20 - 8 = 12 which is less than 70% of maximum possible marks. Indu already has less than 70% in WE, therefore we Indu can't be recruited . Hence, we can reject this case. 

Consequently, we can say that Indu scored 100% marks in WE. Therefore, Indu's score in Practice Question - 64 (Tables) | 100 DILR Questions for CAT Preparation

It is also given that Danish, Harini, and Indu had scored the same marks the in GA. 

Practice Question - 64 (Tables) | 100 DILR Questions for CAT Preparation

We are given that, among the four recruited, Geeta had the lowest composite score. 
Maximum composite score that Geeta can get = 2*14 + 6 + 20  = 54 {Assuming 100% marks in WE}. Since, Geeta was recruitedat a composite score of 54 or less we can say that Ester was definitely recruited. 
It is given that no two candidates had the same composite score. We can see that Chetna's composite score is 54. Hence, Geeta can't have a composite score of 54. Therefore, we can say that Geeta's composite score is 53 or less. 
We already know the four people(Jatin, Indu, Geeta, Ester) which were recruited. Hence, we cab say that Danish was rejected at a composite score of 51. Hence, we can say that Geeta's composite score in 52 or more. 
Consequently, we can say that Geeta's composite score if either 52 or 53. Therefore we can say that Geeta scored either 18 {52-(2*14+6)} or 19 {53-(2*14+6)} marks in WE. 

Practice Question - 64 (Tables) | 100 DILR Questions for CAT Preparation

Ajay was the unique highest scorer in WE. 

Case 1: Geeta scored 19 marks in WE.
We can say that if Geeta scored 19 marks in WE, then Ajay scored 20 marks in DI. In that case Ajay's composite score = 2*8 + 20 + 16 = 52. Which is a possible case. 

Case 1: Geeta scored 18 marks in WE.
We can say that if Geeta scored 18 marks in WE, then Ajay can score either 19 or 20 marks in DI. 
If Ajay scored 20 marks in DI then in that case Ajay's composite score = 2*8 + 20 + 16 = 52 which will be same as Geeta's composite score. Hence, we can say that in this case Ajay can't score 20 marks.
If Ajay scored 19 marks in DI then in that case Ajay's composite score = 2*8 + 19 + 16 = 51 which will be same as Danish's composite score. Hence, we can say that in this case Ajay can't score 19 marks.
Therefore, we can say that case 2 is not possible at all.

Practice Question - 64 (Tables) | 100 DILR Questions for CAT Preparation

Let us check all the statement one by one.
Option A: Bala scored same as Jatin in DI. We can say that Bala scored 20 marks in DI. In that Bala's composite score = 2*20 + 9 + 11 = 60 which is sam as Indu's composite score. Therefore, we can say that this is a false statement. Hence, option (a) is the correct answer. 

Q3: If all the candidates except Ajay and Danish had different marks in DI, and Bala's composite score was less than Chetna's composite score, then what is the maximum marks that Bala could have scored in DI?
Practice Question - 64 (Tables) | 100 DILR Questions for CAT PreparationView Answer  Practice Question - 64 (Tables) | 100 DILR Questions for CAT Preparation

Ans: 13
It is given that Indu and Jatin both scored 100% in exactly one section. We can say that Jatin scored 100% marks in DI. Therefore, Jatin's composite score = 2*20+16+14 = 70
It is given that Jatin’s composite score was 10 more than Indu’s. Therefore, we can say that Indu's composite score = 70 - 10 = 60.
Indu also scored 100% in exactly one section. 

Case 1: Indu scored 100% marks in DI.
If Indu scored 100% marks in DI, then Indu's score in GA = 60 - 2*20 - 8 = 12 which is less than 70% of maximum possible marks. Indu already has less than 70% in WE, therefore we Indu can't be recruited . Hence, we can reject this case. 

Consequently, we can say that Indu scored 100% marks in WE. Therefore, Indu's score in Practice Question - 64 (Tables) | 100 DILR Questions for CAT Preparation

It is also given that Danish, Harini, and Indu had scored the same marks the in GA. 

Practice Question - 64 (Tables) | 100 DILR Questions for CAT Preparation

We are given that, among the four recruited, Geeta had the lowest composite score. 
Maximum composite score that Geeta can get = 2*14 + 6 + 20  = 54 {Assuming 100% marks in WE}. Since, Geeta was recruitedat a composite score of 54 or less we can say that Ester was definitely recruited. 
It is given that no two candidates had the same composite score. We can see that Chetna's composite score is 54. Hence, Geeta can't have a composite score of 54. Therefore, we can say that Geeta's composite score is 53 or less. 
We already know the four people(Jatin, Indu, Geeta, Ester) which were recruited. Hence, we cab say that Danish was rejected at a composite score of 51. Hence, we can say that Geeta's composite score in 52 or more. 
Consequently, we can say that Geeta's composite score if either 52 or 53. Therefore we can say that Geeta scored either 18 {52-(2*14+6)} or 19 {53-(2*14+6)} marks in WE. 

Practice Question - 64 (Tables) | 100 DILR Questions for CAT Preparation

Ajay was the unique highest scorer in WE. 
Case 1: Geeta scored 19 marks in WE.
We can say that if Geeta scored 19 marks in WE, then Ajay scored 20 marks in DI. In that case Ajay's composite score = 2*8 + 20 + 16 = 52. Which is a possible case. 

Case 1: Geeta scored 18 marks in WE.
We can say that if Geeta scored 18 marks in WE, then Ajay can score either 19 or 20 marks in DI. 
If Ajay scored 20 marks in DI then in that case Ajay's composite score = 2*8 + 20 + 16 = 52 which will be same as Geeta's composite score. Hence, we can say that in this case Ajay can't score 20 marks.
If Ajay scored 19 marks in DI then in that case Ajay's composite score = 2*8 + 19 + 16 = 51 which will be same as Danish's composite score. Hence, we can say that in this case Ajay can't score 19 marks.
Therefore, we can say that case 2 is not possible at all.

Practice Question - 64 (Tables) | 100 DILR Questions for CAT Preparation

It is given that all the candidates except Ajay and Danish had different marks in DI and Bala's composite score was less than Chetna's composite score. 
Let us assume that Bala scored 'x' marks in DI.
⇒ 2*x + 9 + 11 < 54
⇒ x < 17
We can see that Bala's score will be less than 17. Bala's maximum score in DI will be the largest possibel number less than 17 which is not same as any other candidate's score in DI. From, the table we can see that 16, 15 and 14 are already taken by Indu, Falak and Geeta respectively. 
Therefore, we can say that Bala can score a maximum of 13 marks in DI.

Q4: If all the candidates scored different marks in WE then what is the maximum marks that Harini could have scored in WE?
Practice Question - 64 (Tables) | 100 DILR Questions for CAT PreparationView Answer  Practice Question - 64 (Tables) | 100 DILR Questions for CAT Preparation
Ans: 14
It is given that Indu and Jatin both scored 100% in exactly one section. We can say that Jatin scored 100% marks in DI. Therefore, Jatin's composite score = 2*20+16+14 = 70
It is given that Jatin’s composite score was 10 more than Indu’s. Therefore, we can say that Indu's composite score = 70 - 10 = 60.
Indu also scored 100% in exactly one section. 

Case 1: Indu scored 100% marks in DI.
If Indu scored 100% marks in DI, then Indu's score in GA = 60 - 2*20 - 8 = 12 which is less than 70% of maximum possible marks. Indu already has less than 70% in WE, therefore we Indu can't be recruited . Hence, we can reject this case. 

Consequently, we can say that Indu scored 100% marks in WE. Therefore, Indu's score in Practice Question - 64 (Tables) | 100 DILR Questions for CAT Preparation

It is also given that Danish, Harini, and Indu had scored the same marks the in GA. 

Practice Question - 64 (Tables) | 100 DILR Questions for CAT Preparation

We are given that, among the four recruited, Geeta had the lowest composite score. 
Maximum composite score that Geeta can get = 2*14 + 6 + 20  = 54 {Assuming 100% marks in WE}. Since, Geeta was recruitedat a composite score of 54 or less we can say that Ester was definitely recruited. 
It is given that no two candidates had the same composite score. We can see that Chetna's composite score is 54. Hence, Geeta can't have a composite score of 54. Therefore, we can say that Geeta's composite score is 53 or less. 
We already know the four people(Jatin, Indu, Geeta, Ester) which were recruited. Hence, we cab say that Danish was rejected at a composite score of 51. Hence, we can say that Geeta's composite score in 52 or more. 
Consequently, we can say that Geeta's composite score if either 52 or 53. Therefore we can say that Geeta scored either 18 {52-(2*14+6)} or 19 {53-(2*14+6)} marks in WE. 

Practice Question - 64 (Tables) | 100 DILR Questions for CAT Preparation

Ajay was the unique highest scorer in WE. 
Case 1: Geeta scored 19 marks in WE.
We can say that if Geeta scored 19 marks in WE, then Ajay scored 20 marks in DI. In that case Ajay's composite score = 2*8 + 20 + 16 = 52. Which is a possible case. 

Case 1: Geeta scored 18 marks in WE.
We can say that if Geeta scored 18 marks in WE, then Ajay can score either 19 or 20 marks in DI. 
If Ajay scored 20 marks in DI then in that case Ajay's composite score = 2*8 + 20 + 16 = 52 which will be same as Geeta's composite score. Hence, we can say that in this case Ajay can't score 20 marks.
If Ajay scored 19 marks in DI then in that case Ajay's composite score = 2*8 + 19 + 16 = 51 which will be same as Danish's composite score. Hence, we can say that in this case Ajay can't score 19 marks.
Therefore, we can say that case 2 is not possible at all.

Practice Question - 64 (Tables) | 100 DILR Questions for CAT Preparation

It is given that all the candidates scored different marks in WE.
We can see that Ajay, Geeta and Ester has already scored 20, 19 and 18 marks in WE. Therefore, Harini can score a maximum of 17 marks in WE. If Harini's score in WE is 17, then Harini's composite score = 2*5+17+20 = 47 which is same as Falak's composite score. Hence, we can say that Harini can't score 17 marks in WE. Jatin and Danish have already scored 16 and 15 marks respectively. 
Therefore, we can say that the maximum marks that Harini could have scored in WE = 14. 

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FAQs on Practice Question - 64 (Tables) - 100 DILR Questions for CAT Preparation

1. What are the key components of tables used in data representation for exams?
Ans. Tables are structured formats that present data in rows and columns, making it easier to read and analyze. The key components include headings or titles that identify the content of each column, rows that hold individual data entries, and sometimes footnotes or legends that provide additional context or explanations.
2. How can I effectively analyze data presented in tables during the exam?
Ans. To effectively analyze data in tables, start by reviewing the headings to understand what each column represents. Look for patterns, trends, or anomalies in the data. Compare values across rows and columns to draw conclusions or make calculations. Summarizing key findings in your own words can also help reinforce understanding.
3. What strategies can I use to memorize data from tables for the exam?
Ans. To memorize data from tables, use techniques such as visualization, where you picture the table in your mind. Create mnemonic devices or associations with the data. Practice recalling the information by covering the table and testing yourself. Repetition and summarizing the content in your own words can also enhance retention.
4. Are there common mistakes to avoid when interpreting tables in exams?
Ans. Yes, common mistakes include misreading headings, overlooking footnotes that provide important context, and jumping to conclusions without analyzing the data thoroughly. It's also important to avoid comparing data from different tables without ensuring they are relevant and comparable.
5. How can tables help in organizing information for essay writing in exams?
Ans. Tables help organize information by breaking down complex data into manageable parts, allowing for clearer comparisons and contrasts. When writing an essay, referring to tables can provide concrete evidence to support arguments and enhance clarity, making it easier for readers to follow the logic and flow of the writing.
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