CAT '21DI-LR

FUNDAS

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Table of Contents

Understanding DILR Section of CAT .............................................................................................................................. 2

Differentiating Between DI and LR Questions ........................................................................................................... 3

Strengthen Your Foundation ............................................................................................................................................. 3

Calculations and Approximation .................................................................................................................................. 3

Skill-sets Required ............................................................................................................................................................. 4

Percentage-Ratio Equivalence ...................................................................................................................................... 5

Fundas To Crack DILR Questions in CAT ..................................................................................................................... 6

Solved Practice Questions.................................................................................................................................................. 6

DI + LR .................................................................................................................................................................................... 6

DI + LR .................................................................................................................................................................................... 8

LR ........................................................................................................................................................................................... 10

DI + LR .................................................................................................................................................................................. 11

DI ............................................................................................................................................................................................ 13

Detailed Explanations ........................................................................................................................................................ 14

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Aspirants taking CAT and other Management Entrance tests find Data Interpretation to be one of the toughest subject areas in the test. The time pressure adds anxiety to this toughness. With less than two minutes to answer each question (if you attempt them all), candidates also need to decipher the data given in the form of tables, pie chart, bar graph, scatter plots, line graphs, and puzzles. However, this doesn’t mean that you can’t score 95+ percentile in DILR!

In this book we’ll be covering tips to improve Data Interpretation for CAT, ways to practice smarter—and raise your scores accordingly.

Understanding DILR Section of CAT

Data Interpretation and Logical Reasoning section of CAT exam clubs two areas in one. While Data Interpretation includes questions on interpreting data from tables, charts, graphs etc., Logical Reasoning is about solving complex puzzles and caselets which are analytical in nature.

The 3 type of question that aspirants can expect are

Only calculation driven based on Data represented in form of Table, Line Graph, Bar Graph, Pie Chart, etc. Only reasoning driven based on Puzzles, Family Tree, Arrangements, etc. Mixed questions that would involve both Logical Reasoning as well as Data Interpretation. These questions can be calculative as well as reasoning driven wherein candidates need to infer the data from the given statements and then answer the relevant questions accordingly.”

Particulars Details

Section Name Data Interpretation Logical Reasoning

Total Sub-questions 12 12

Number of Problems 3 3

Type of Questions Structured Date – Table, Graph, Caselet

Unstructured Data – Puzzles, Arrangement,

Non-MCQs Questions 4 2

Total time to attempt DILR Section

40 minutes

Best Data Interpretation for CAT Books

How to prepare for Data Interpretation for CAT by Arun Sharma Logical Reasoning and Data Interpretation for the CAT by Nishit K. Sinha

Note: If duration of CAT’21 is of 3 hrs, total number of questions in DILR will be 32 (16+16) to be attempted in 60 min.

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Differentiating Between DI and LR Questions

The DILR for CAT has an undefined syllabus when compared to the other section in CAT. The two sub-section of the DILR Syllabus are:

Data Interpretation: Data is visualised in form of a graph, chart or another form and aspirants need to solve a set of questions based on the information given. Logical Reasoning: Information is given in a non-structured manner and aspirants are required to structure it to solve the set of question that follows.

However, the CAT DILR Section has been shifting towards questions which are a combination of DI and LR rather than separate DI and LR questions which were common in previous CATs.

DI LR

Tables Arrangements Data Charts Games, Tournament, Results Bar Graphs Mathematical Reasoning Pie Charts Logical Reasoning Puzzles Line Charts, Radar Diagrams Matrix-based Arrangement Caselets Mixed / Miscellaneous

Strengthen Your Foundation

Calculations and Approximation

DILR questions in CAT have moved from east easy tables to more logical caselets and puzzles with easy calculations. But you can’t skip being really good at quick calculations and approximations. Studies conducted on previous CAT toppers suggest that if you double your calculation speed, you would have 3-4 minutes of extra time during the exam. So if you can attempt even 2 more questions in these 3-4 minutes can give you an edge of at least 2 percentiles.

Let’s have a look at the most frequent mathematical operations and calculations done while solving DILR problems in CAT

1. Addition2. Subtraction 3. Multiplication

a. Base Method b. Place value Method

4. Division a. Approximation

5. Ratio Comparison a. Cross Multiplication Methodb. Percentage Comparison Method

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Percentage

Percentage is of immense importance when it comes to problem solving in quantitative aptitude (QA) and data interpretation (DI). Despite the fact that the number of questions asked from this concept is not very high in the CAT, a sound understanding and good practice of percentage calculations comes handy in almost all the topics of QA and DI. However, percentage is an important concept for all the other B-school and aptitude examinations.

Let us first have some standard meaning of percentage:

Percentage as a common platform

For example,

Name Marks obtained Total marks Percentage

Amit 100 1000 10% Binay 25 25 100%

Chandan 45 300 15%

Deepak 50 100 50%

Percentage as a rate measurer

For example,

1. What is x% of y?2. What percentage of x is y?3. Change in value

o Percentage change = (Change/Initial value) *100o Percentage point change - The numerical difference between the values for

which we have to calculate change.

Skill-sets Required

Let’s have a basic list of acquired skillsets required to easily solve DILR questions. DILR problems require you to understand data and retain the information, time spent observing various data presentation and visualisations and attention towards restrictions and conditions given in presentation. We have assumed that the learner has a basic level of understanding of DILR and has just started.

1. Knowing the Terminology - There can be various economic or non-economic terms like CAGR,inflation etc., involved in the data-set, and it is a prerequisite to have a good understanding ofthese terms.

2. Knowing the Different Formats of Presenting the Data - Having discussed earlier, thesignificance of organized data, we can automatically conclude that the same data whenpresented in an unorganized way makes difficult reading and is not of much use. The same datacan be presented differently. In one format, it might make easy reading; in another format, itmight make difficult reading.

3. Knowing What is Asked - Many a times aspirants are not able to solve a question because theyintimidated with the data. And as a result, they lose track of what is to be found out. You can’t

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get 'emotionally attached' to a question. Just focus on extracting information from the data-set and build an algorithm to solve similar problems whenever they occur.

Percentage-Ratio Equivalence

Most of the percentage calculations like 25%, 37.5%, 33.33% etc. lie around some particular ratios. Having a good command over these ratios help in achieving good percentage calculation speed essential for DI calculations. (See more such calculation in ratio, proportion and variation chapter.)

1 2 3 4 5 6 7 8 9 10

1 100% 200 300 400 500 600 700 800 900 1000

2 50 100 150 200 250

3 33.33 60.66 100

4 25 50 75 100

5 20 40 60 80 100

6 16.66 33.33 50 66.66 83.33 100

7 14.28 28.56 42.85 57. 14 71.42 85.71 100

8 12.5 25 37.5 50 62.5 75 87.5 100

9 11.11 22.22 33.33 44.44 55.55 66.66 77.77 88.88 100

10 10 20 30 40 50 60 70 80 90 100

11 9.09 18.1 27.27 36.36 45.45 54.54 63.63 72.72 81.81 90.90

12 8.33 16.66 25 33.33 41.66 50 58.33 66.66 75 83.33

1 3 7.69 15.38 23.07 30.7 38.46 46.15 53.84 61.53 69.23 76.92

14 7. 14 14.28 21.42 28.57 35.71 42.85 50.00 57.14 64.28 71.42

15 6.66 13 .33 20 26.66 33.33 40 46.66 53.33 60 66.66

16 6.25 12.5 18.75 25 31.25 37 .5 43.75 50 56.25 62.5

How to read the table?

The number that lies at the intersection of a row and a column is the percentage equivalence of their ratio (entries on the first row as numerator).

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Fundas To Crack DILR Questions in CAT

Here are few of the DILR fundas to crack the CAT DATA INTERPRETATION

Develop the understanding for voluminous data Familiarize yourself with different kinds of charts Enhance mental math skills Calculate the increase or decrease of percentage or comparing ratio to save time Regularly solve DI sets that require reasoning skills and logic and increasing your familiarity with such sets Don’t reject DI sets based on the volume of data given. More data does not necessarily mean a more difficult set and vice versa. Solve a bulk of easy questions with some difficult ones at a time. Attempt questions that you can confidently answer in 2 to 2.5 minutes Ration your time between the Quant and DI section would help you neither neglect nor spend excessive time on any of these areas And last but not the least, Practice, Practice and Practice DI sets regularly!!

Solved Practice Questions

Data Interpretation assesses aspirants’ ability to analyse the given data, derive a meaningful conclusions and to make appropriate decisions. This makes sense as the situations mentioned in these questions often reflect a business situation; where one is required to take a decision on the basis of available information and future trends.

First thing aspirants need to consider while solving DI question is to understand what information is given and what is asked in the questions that follow. Management students must perform solving problems under strict time restrictions when they head towards their professional career.

Each DI question contains three separate parts –

1. Data-set2. Explanations and restrictions given for the data and3. Questions

Now let’s have a look at some of the most challenges DILR Questions asked in recent years and develop a clear understanding of the process to solve them.

DI + LR

Instruction Q. 1-4

There were seven elective courses – E1 to E7 - running in a specific term in a college. Each of the 300 students enrolled had chosen just one elective from among these seven. However, before the start of the term, E7 was withdrawn as the instructor concerned had left the college. The students who had opted for E7 were allowed to join any of the remaining electives. Also, the students who had chosen other electives were given one chance to change their choice. The table below captures the movement of the students from one elective to another during this process. Movement from one elective to the

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same elective simply means no movement. Some numbers in the table got accidentally erased; however, it is known that these were either 0 or 1.

Further, the following are known: a) Before the change process there were 6 more students in E1 than in E4, but after the reshuffle,

the number of students in E4 was 3 more than that in E1.b) The number of students in E2 increased by 30 after the change process.c) Before the change process, E4 had 2 more students than E6, while E2 had 10 more students

than E3.

The table is given below –

To Elective

E1 E2 E3 E4 E5 E6

From Elective

E1 9 5 10 1 4 2 E2 34 8 2 2 E3 2 6 25 2 E4 3 2 14 4 E5 5 30 E6 7 3 2 9 E7 4 16 30 5 5 41

1. How many elective courses among E1 to E6 had a decrease in their enrolments after the changeprocess?

A. 4B. 1C. 2D. 3

Answer key: C

Solution: Only E1 and E4 had a decrease in their enrolments after the change process.

2. After the change process, which of the following is the correct sequence of number of students inthe six electives E1 to E6?

A. 19, 76, 79, 21, 45, 60B. 19, 76, 78, 22, 45, 60C. 18, 76, 79, 23, 43, 61D. 18, 76, 79, 21, 45, 61

Answer key: D

Solution: The correct sequence of number of students in E1 to E6 is given in option D.

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3. After the change process, which course among E1 to E6 had the largest change in its enrolment asa percentage of its original enrolment?

A. E1B. E2C. E3D. E6

Answer key: D Solution: Required percentage change for E1 = (31 – 18)/31 * 100 = 41.94% Required percentage change for E2 = (76 – 46)/46 * 100 = 65.22% Required percentage change for E3 = (79 – 36)/36 * 100 = 119.44% Required percentage change for E6 = (61 – 23)/23 * 100 = 165.22%

4. Later, the college imposed a condition that if after the change of electives, the enrolment in anyelective (other than E7) dropped to less than 20 students, all the students who had left that coursewill be required to re-enrol for that elective.Which of the following is a correct sequence of electives in decreasing order of their final enrolments?

A. E2, E3, E6, E5, E1, E4 B. E3, E2, E6, E5, E4, E1 C. E2, E5, E3, E1, E4, E6 D. E2, E3, E5, E6, E1, E3

Answer key: A Solution: Only in E1, the enrolment has dropped to less than 20 students. So, final enrolments in E1 = 18 + no. of students who left E1 = 18 + (31 – 9) = 40 After calculating the final enrolments in all the other electives after subtracting no. of students who moved from E1 to these electives and then comparing, we get E2 > E3 > E6 > E5 > E1 > E4

DI + LR

Instructions Q. 5-10

In a certain board examination, students were to appear for examination in five subjects: English, Hindi, Mathematics, Science and Social Science. Due to a certain emergency situation, a few of the examinations could not be conducted for some students. Hence, some students missed one examination and some others missed two examinations. Nobody missed more than two examinations. The board adopted the following policy for awarding marks to students. If a student appeared in all five examinations, then the marks awarded in each of the examinations were on the basis of the scores obtained by them in those examinations.

If a student missed only one examination, then the marks awarded in that examination was the average of the best three among the four scores in the examinations they appeared for. If a student missed two examinations, then the marks awarded in each of these examinations was the average of the best two among the three scores in the examinations they appeared for.

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The marks obtained by six students in the examination are given in the table below. Each of them missed either one or two examinations.

English Hindi Mathematics Science Social Science Alva 80 75 70 75 60 Bithi 90 80 55 85 85 Carl 75 80 90 100 90 Deep 70 90 100 90 80 Esha 80 85 95 60 55 Foni 83 72 78 88 83

The following facts are also known. a) Four of these students appeared in each of the English, Hindi, Science, and Social Science

examinations.b) The student who missed the Mathematics examination did not miss any other examination.c) One of the students who missed the Hindi examination did not miss any other examination.

The other student who missed the Hindi examination also missed the Science examination.

5. Who among the following did not appear for the Mathematics examination?A. CarlB. AlvaC. EshaD. Foni

Answer key: A Solution: Here, Carl did not appear for Mathematics examination.

6. Which students did not appear for the English examination?A. Cannot be determined B. Alva and BithiC. Carl and DeepD. Esha and FoniAnswer key: D Solution:Here, Esha and Foni did not appear for the English examination.

7. What BEST can be concluded about the students who did not appear for the Hindi examination?A. Two among Alva, Deep and EshaB. Alva and EshaC. Alva and DeepD. Deep and Esha

Answer key: C Solution: Only Alva and Deep did not appear for the Hindi examination.

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8. What BEST can be concluded about the students who missed the Science examination?A. Deep and BithiB. Alva and BithiC. Alva and DeepD. Bithi and one out of Alva and Deep

Answer key: D Solution: Here, Bithi and one out of Alva and Deep missed the Science examination.

9. How many out of these six students missed exactly one examination?“Type in The Answer”

Answer: 3 Solution: Esha, Carl and one among Alva and Deep missed exactly one examination.

10. For how many students can we be definite about which examinations they missed?“Type in The Answer”

Answer: 4 Solution: For four students, we can be definite about which examinations they missed.

LR

Instruction Q. 11-14

Twenty-five coloured beads are to be arranged in a grid comprising of five rows and five columns. Each cell in the grid must contain exactly one bead. Each bead is coloured either Red, Blue or Green. While arranging the beads along any of the five rows or along any of the five columns, the rules given below are to be followed:

a) Two adjacent beads along the same row or column are always of different colours.b) There is at least one Green bead between any two Blue beads along the same row or

column.c) There is at least one Blue and at least one Green bead between any two Red beads along

the same row or column.

Every unique, complete arrangement of twenty-five beads is called a configuration.

11. The total number of possible configurations using beads of only two colours is:“Type in The Answer”

Answer: 2

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12. What is the maximum possible number of Red beads that can appear in any configuration?“Type in The Answer”

Answer: 9

13. What is the minimum number of Blue beads in any configuration?“Type in The Answer”

Answer: 6

14. Two Red beads have been placed in ‘second row, third column’ and ‘third row, second column’.How many more Red beads can be placed so as to maximize the number of Red beads used in theconfiguration?

“Type in The Answer” Answer: 6

DI + LR

Instruction Q. 15-18

Sixteen patients in a hospital must undergo a blood test for a disease. It is known that exactly one of them has the disease. The hospital has only eight testing kits and has decided to pool blood samples of patients into eight vials for the tests. The patients are numbered 1 through 16, and the vials are labelled A, B, C, D, E, F, G, and H. The following table shows the vials into which each patient’s blood sample is distributed.

Patient Vials Patient Vials 1 B,D,F,H 9 A,D,F,H 2 B,D,F,G 10 A,D,F,G 3 B,D,E,H 11 A,D,E,H 4 B,D,E,G 12 A,D,E,G 5 B,C,F,H 13 A,C,F,H 6 B,C,F,G 14 A,C,F,G 7 B,C,E,H 15 A,C,E,H 8 B,C,E,G 16 A,C,E,G

If a patient has the disease, then each vial containing his/her blood sample will test positive. If a vial tests positive, one of the patients whose blood samples were mixed in the vial has the disease. If a vial tests negative, then none of the patients whose blood samples were mixed in the vial has the disease.

15. Suppose vial C tests positive and vials A, E and H test negative. Which patient has the disease?A. Patient 8 B. Patient 14 C. Patient 6 D. Patient 2

Answer key: C Solution:

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If vial C tests positive, then one of patients 5, 6, 7, 8, 13, 14, 15 and 16 has the disease. If vials A, E and H test negative then except patient 6, none of the above has the disease. So, patient 6 has the disease.

16. Suppose vial A tests positive and vials D and G test negative. Which of the following vials shouldwe test next to identify the patient with the disease?

A. Vial HB. Vial EC. Vial BD. Vial C

Answer key: B Solution: If vial A tests positive and vials D and G test negative, then one among patients 13 and 15 has the disease. Vials common for patients 13 and 15 are vials A, C and H. So, we need to test vial E or F.

17. Which of the following combinations of test results is NOT possible?A. Vial B positive, vials C, F and H negative B. Vials A and G positive, vials D and E negative C. Vials B and D positive, vials F and H negative D. Vials A and E positive, vials C and D negative

Answer key: D Solution: If vials A and E test positive, then one of patients 11, 12, 15 and 16 has the disease. So, at least one of the vials C and D must test positive.

18. Suppose one of the lab assistants accidentally mixed two patients' blood samples before theywere distributed to the vials. Which of the following correctly represents the set of all possiblenumbers of positive test results out of the eight vials?

A. {4,5,6,7}B. {4,5,6,7,8}C. {4,5}D. {5,6,7,8}

Answer key: B

Solution: Suppose blood samples of patients 5 and 12 are mixed and one of them has the disease, then all 8 vials will test positive. Suppose blood samples of patients 2 and 14 are mixed and one of them has the disease, then 6 vials will test positive. If any one patient let’s say patient 6 has the disease and his blood sample is not mixed, then 4 vials will definitely test positive. So, answer set must contain 4, 6 and 8.

3 MONTHSPREPARATIONSTRATEGYFOR CATDI-LR, QUANT & VARC

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DI

Instructions Q. 19-22 Five vendors are being considered for a service. The evaluation committee evaluated each vendor on six aspects – Cost, Customer Service, Features, Quality, Reach, and Reliability. Each of these evaluations are on a scale of 0 (worst) to 100 (perfect). The evaluation scores on these aspects are shown in the radar chart. For example, Vendor 1 obtains a score of 52 on Reliability, Vendor 2 obtains a score of 45 on Features and Vendor 3 obtains a score of 90 on Cost.

19. On which aspect is the median score of the five vendors the least?A. CostB. Quality C. Reliability D. Customer Service

Answer key: D Solution: It can be found that median score is least on Customer Service.

20. A vendor's final score is the average of their scores on all six aspects. Which vendor has thehighest final score?

A. Vendor 1 B. Vendor 3 C. Vendor 4 D. Vendor 2

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Answer key: B Solution: Final score is directly proportional to total scores in all 6 aspects Total score of vendor 1 = 75 + 55 + 80 + 75 + 40 + 55 = 380 Similarly, Total score of vendor 3 = 400 Total score of vendor 4 = 340 Total score of vendor 2 = 330

21. List of all the vendors who are among the top two scorers on the maximum number of aspects is:A. Vendor 1 and Vendor 2 B. Vendor 2, Vendor 3 and Vendor 4 C. Vendor 2 and Vendor 5 D. Vendor 1 and Vendor 5

Answer key: D Solution: Vendor 1 is among top two scorers on 3 aspects – Reach, Quality and Customer Service Vendor 2 is among top two scorers on 2 aspects – Cost and Quality Vendor 3 is among top two scorers on 2 aspects – Cost and Reliability Vendor 4 is among top two scorers on 2 aspects – Features and Customer Service Vendor 5 is among top two scorers on 3 aspects – Reliability, Reach and Features

22. List of all the vendors who are among the top three vendors on all six aspects is: A. None of the VendorsB. Vendor 1 C. Vendor 1 and Vendor 3 D. Vendor 3 Answer key: D Solution:Only vendor 3 is among top three vendors on all six aspects.

Detailed Explanations

Explanations Question 1-4

No. of students who had opted for E1 before change process = Sum of no. of students who didn’t move from E1 and the no. of students who moved from E1 to other electives = 9 + 5 + 10 + 1 + 4 + 2 = 31

Before the change process there were 6 more students in E1 than in E4.

So, no. of students in E4 before the change process = 31 – 6 = 25

No. of students who had opted for E7 before change process = No. of students who moved from E7 to other electives = 4 + 16 + 30 + 5 + 5 + 41 = 101

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Elective Number of students

Before change process

After change process

E1 31 E2 E3 E4 25 E5 E6 E7 101 Not applicable

No. of students who had opted for E2 after change process = Sum of no. of students who didn’t move from E2 and the no. of students who moved to E2 = 5 + 34 + 6 + 3 + 5 + 7 + 16 = 76 The number of students in E2 increased by 30 after the change process No. of students who had opted for E2 before change process = 76 – 30 = 46 Before the change process, E4 had 2 more students than E6, while E2 had 10 more students than E3. No. of students who had opted for E6 before change process = 25 – 2 = 23 No. of students who had opted for E3 before change process = 46 – 10 = 36 No. of students who had opted for E5 before change process = Total no. of students enrolled in the college – Total no. of students who opted for all the electives except E5 before change process = 300 – (31 + 46 + 36 + 25 + 23 + 101) = 38

Elective Number of students

Before change process

After change process

E1 31 E2 46 76 E3 36 E4 25 E5 38 E6 23 E7 101 Not applicable

Sum of no. of students who moved from E4 to E1 and that who moved from E4 to E5 = 25 – (3 + 2 + 14 + 4) = 2 Since maximum entry for a missing value = 1,So, no. of students who moved from E4 to E1 = 1 = no. of students who moved from E4 to E5.Similarly,No. of students who moved from E2 to E1 = 0 = no. of students who moved from E2 to E4.No. of students who moved from E6 to E1 = 1 = no. of students who moved from E6 to E4.

To Elective

E1 E2 E3 E4 E5 E6

From Elective

E1 9 5 10 1 4 2 E2 0 34 8 0 2 2 E3 2 6 25 2 E4 1 3 2 14 1 4 E5 5 30

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E6 1 7 3 1 2 9 E7 4 16 30 5 5 41

After the reshuffle, the number of students in E4 was 3 more than that in E1. Minimum no. of students in E4 after change process = 1 + 0 + 0 + 14 + 0 + 1 + 5 = 21 Maximum no. of students in E1 after change process = 9 + 0 + 2 + 1 + 1 + 1 + 4 = 18 So, minimum difference between no. of students in E4 after change process and no. of students in E1 after change process = 3

To Elective

E1 E2 E3 E4 E5 E6

From Elective

E1 9 5 10 1 4 2 E2 0 34 8 0 2 2 E3 2 6 25 0 1 2 E4 1 3 2 14 1 4 E5 1 5 0 30 E6 1 7 3 1 2 9 E7 4 16 30 5 5 41

Elective Number of students

Before change process

After change process

E1 31 18 E2 46 76 E3 36 E4 25 21 E5 38 45 E6 23 E7 101 Not applicable

Sum of no. of students who moved from E5 to E3 and that who moved from E5 to E6 = 38 – (1 + 5 + 0 + 30) = 2So, no. of students who moved from E5 to E3 = 1 = no. of students who moved from E5 to E6.

To Elective

E1 E2 E3 E4 E5 E6

From Elective

E1 9 5 10 1 4 2 E2 0 34 8 0 2 2 E3 2 6 25 0 1 2 E4 1 3 2 14 1 4 E5 1 5 1 0 30 1 E6 1 7 3 1 2 9 E7 4 16 30 5 5 41

Finally,

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Elective Number of students

Before change process After change process E1 31 18 E2 46 76 E3 36 79 E4 25 21 E5 38 45 E6 23 61 E7 101 Not applicable

Explanations Question 5-10

One of the students who missed the Hindi examination did not miss any other examination. We check for average of top three scores in examinations other than Hindi examination: Average score of Alva = (80 + 75 + 70)/3 = 75 = score of Alva in Hindi

Similarly, checking for others, we get that only Alva and Deep have got same score in Hindi as the average of top three scores in other examinations.

The other student who missed the Hindi examination also missed the Science examination. For Alva, average of top two scores in examinations other than Hindi and Science = (80 + 70)/2 = 75 = Score obtained by Alva in each of Hindi and Science examinations.

For Deep, average of top two scores in examinations other than Hindi and Science = (100 + 80)/2 = 90 = Score obtained by Deep in each of Hindi and Science examinations. So, one among Alva and Deep missed only Hindi examination and the other missed Hindi as well as Science examinations.

The student who missed the Mathematics examination did not miss any other examination. We can verify that only Carl is the student who missed only Mathematics examination by following the same procedure as in the first few steps of this solution. Four of these students appeared in each of the English, Hindi, Science, and Social Science examinations.

This means that there are two students who missed each of these four examinations. Alva and Deep missed Hindi examination. Bithi can’t miss English examination (average score can’t be equal to highest score)

So, Esha and Foni missed English examination. Esha and Foni can’t miss Science examination (average score can’t be less than top 3 scores)

So, Bithi missed Science Examination apart from one among Alva and Deep. Esha can’t miss Social Science examination (average score can’t be equal to least score).

So, Bithi and Foni missed Social Science examination.

So, listing examination(s) missed by each student: Alva – Hindi or Hindi and Science Bithi – Science and Social Science

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Carl – Mathematics Deep – Hindi or Hindi and Science Esha – English Foni – English and Social Science

Explanations Question 11-14

11. Solution:Only Blue and Green beads can be used as including one colour as Red will contradict rule c. Also,there is only one configuration possible in each case of arrangement starting with Blue bead andGreen bead:

Case 1: G B G B G B G B G B G B G B G B G B G B G B G B G

Case 2: B G B G B G B G B G B G B G B G B G B G B G B G B

12. Solution:For maximum possible number of Red beads to appear in any configuration, there must be a maximumof 2 gaps between two Red beads.

R R R R

R R R

R R So, maximum 9 Red beads can appear.

13. Solution:For minimum number of Blue beads in a configuration, the number of Red beads and that of Greenbeads need to be maximized. If we maximize the number of Red beads, then one such configurationwill be

R G B R G G R G B R B G R G B R B G R G G R B G R

So, there will be minimum 6 Blue beads.

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14. Solution:R R

R R R

R R R

So, 6 more Red beads can be placed.

Explanations Question 15-18

15. Solution:If vial C tests positive, then one of patients 5, 6, 7, 8, 13, 14, 15 and 16 has the disease.If vials A, E and H test negative then except patient 6, none of the above has the disease.So, patient 6 has the disease.

16. Solution:If vial A tests positive and vials D and G test negative, then one among patients 13 and 15 has thedisease.Vials common for patients 13 and 15 are vials A, C and H.So, we need to test vial E or F.

17. Solution:If vials A and E test positive, then one of patients 11, 12, 15 and 16 has the disease.So, at least one of the vials C and D must test positive.

18. Solution:Suppose blood samples of patients 5 and 12 are mixed and one of them has the disease, then all 8vials will test positive.Suppose blood samples of patients 2 and 14 are mixed and one of them has the disease, then 6vials will test positive.If any one patient let’s say patient 6 has the disease and his blood sample is not mixed, then 4 vialswill definitely test positive.So, answer set must contain 4, 6 and 8.

Explanations Question 19-22

Tabulating the data from the given radar graph,

Vendor Score on various aspects Cost Reliability Reach Quality Features Customer

Service 1 75 55 80 75 40 55 2 80 40 55 70 45 40 3 90 75 65 65 55 50 4 70 25 45 40 90 70 5 60 60 70 50 75 30

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