XAT Puzzles Questions

XAT LR Puzzles questions are the similar type of CAT DILR Puzzles that gets asked each year in the exam. To solve these sets, some of the fundas of Arithmetic, Algebra, Number System and Probability also become handy. To check some other common important sets that appear in XAT QA & DI section, checking with the XAT syllabus will help you further. To ace these kind of sets basic arithmetic skills are necessary and some prior experience to similar sets will help.

We have compiled all the previously repeated questions with detailed solutions. You could also download them as PDF. We advise aspirants to take XAT mock tests and assessing where to focus further. These tests will help you in familiarizing with the kind of questions that are asked in the exam.

XAT Puzzles Questions

Instructions

Read the following scenario and answer the THREE questions that follow.

A pencil maker ships pencils in boxes of size 50, 100 and 200. Due to packaging issues, some pencils break. About the 20 boxes he has supplied to a shop, the following information is available:
* Box no. 1 through 6 have 50 pencils, Box no. 7 through 16 have 100 pencils and Box no. 17 through 20 have 200 pencils.
* No box has less than 5% or more than 20% broken pencils.
Following is the frequency table of the number of broken pencils for the twenty boxes:

Question 1

Which of the following can possibly be the sequence of the number of brokenpencils in Boxes 7-16?

6,7,9, 11,15,19,20,20,20,29

5,6,6,6,11,15,15,20,20,20

7,7,7,7,11,15,15,19,20,20

7,7,9,9,11,13,15,19,20,20

5,7,7,7,9,11,15,20,20,20

Video Solution

Solution

Boxes 7 to 16 contain a total of 100 pencils each. The minimum number of broken pencils the box can hold is 5 percent of the total pencils and a maximum of 20 percent of the total pencils.

5 percent of 100 = 5 and 20 percent of 100 = 20 pencils.

Hence the number of broken pencils must be in the range of 5 to 20.

The frequency of the different number of broken pencils is :

5 - 1

6 - 2

7 - 4

9 - 3

11 - 1

15 - 2

19 - 1

20 - 3

29 - 1

31 - 1

33 - 1

The boxes cannot contain 29, 31, 33 to be the number of broken pencils because they are beyond 20 percent.

Since boxes 1- 6 can contain only between 2.5 to 10 pencils. The remaining boxes which include broken pencils of numbers less than 10 must be a part of 7 - 16. Because boxes 17 - 20 cannot contain broken pencils of numbers less than 10.

Hence 7 - 16 must have 4 boxes that contain less than 10 broken pencils.

Going through the options :

Option A fails because this includes only 3 boxes with less than 10 pencils.

Option B fails because we only have 2 boxes with 6 broken pencils but this includes 3.

Option D fails because it does not include a box of 15 and a box of 20 pencils which can only be a part of boxes with 100 or boxes with 200 pencils. Since boxes 17 - 20 can include only one among 15 or 20 because 29, 31, 33 are a part of this group. Hence this case fails.

Option E fails because this includes 5 boxes with broken pencils less than 10 but this is not possible because this must exactly contain 4 boxes with less than 10 pencils.

Option C is a feasible case containing :

1-6 ( 5, 6, 6, 9, 9, 9)

7 - 16 ( 7, 7, 7, 7, 11, 15, 15, 19, 20, 20 )

17 - 20 (20, 29, 31, 33)

Instructions

Read the following scenario and answer the THREE questions that follow.

Question 2

Which of the following cannot be inferred conclusively from the given information?

No box numbered 1-6 has more broken pencils than any box numbered 17-20.

A box with the highest percentage of broken pencils has 100 pencils.

Four among the boxes numbered 7 to 16 have less than 10 broken pencils.

Exactly three boxes have 20% broken pencils.

Three among the boxes numbered 17 to 20 have 29, 31 and 33 broken pencils in someorder.

Video Solution

Solution

Going by the given options :

Option A: The boxes numbered 1- 6 have a capacity of 50 pencils. The maximum number of broken pencils they can contain is 20 percent of 20 pencils = 10 pencils. For boxes numbered 17 to 20 they must contain a minimum of 5 percent which is equal to 10 pencils out of 200. Hence this is true.

Option B: Boxes with broken pencils of 29, 31, 33 must be a part of 17 - 20. There must be one box containing broken pencils in the range of 10 to 20. There are three boxes in total containing exactly 20 pencils. A maximum of only one of the three can be a part of 17 - 20. The remaining must be a part of boxes 7- 16 because they cannot be a part of 1-6. Hence at least one box among 7 - 16 contains 20 percent of broken pencils which is the highest.

Option C: There are a total of ten boxes with less than 10 broken pencils. They can either be a part of boxes 1-6 or 7 - 16. Since boxes 1-6 can only take broken pencils with less than 10 in number. Hence of the 10 six must be a part of 1-6 and the remaining 4 must be a part of 7 - 16.

Option D: The only possibility for containing 20 percent of the broken pencils is only possible for 20 broken pencils which is 20 percent of 100. There must be at least 2 boxes in the range of 7 - 16 which contain 20 broken pencils which is equal to 20 percent. The third box can either be a part of (7-16) to (17-20). If this belongs to 17 - 20 then the case is not possible and hence cannot be concluded.

Option E: Boxes containing 29, 31, and 33 broken pencils must be a part of boxes 17 -20 because they are higher than the 20 percent range of boxes (1-6) and (7-16). Hence this can be concluded.

Instructions

Read the following scenario and answer the THREE questions that follow.

Question 3

Suppose that additionally it is known that the number of broken pencils in Boxes 17-20 are in increasing order. Which among the following additional information, if true, is not sufficient to uniquely know the number of defective pencils in each of the boxes numbered 17-20?

Boxes no. 7-16 contains a total of 124 defective pencils.

Boxes no. 17-20 contain a total of 108 defective pencils.

Boxes no. 11-16 contain a total of 101 defective pencils.

Box no. 17 contains more defective pencils than any box from among boxes no. 1-14.

Boxes no. 7-16 contain a total of 133 defective pencils.

Video Solution

Solution

Going by the options :

Option A: Boxes 7- 16 contain a total of 124 pencils. Boxes (1-6) has 6 boxes with broken pencils which can be included from :

(5, 6, 6, 7, 7, 7, 7, 9, 9, 9). The minimum possible sum of the 6 pencils is : (5+6+6+7) = 24 and the maximum possible sum is (7+9+9+9) = 34.

Boxes 7 - 16 contains all the boxes with broken pencils except one among the boxes with broken pencils among 11 - 20 and hence ;

This can contain : (11+15+15+19+20+20) or (15+15+19+20+20+20) or (11+15+15+20+20+20) or (11+15+19+20+20+20 ) =100/109/101/105.

THe only possible case to contain 124 pencils is by considering the case : (24+100 ) = (5, 6,6,7, 11+15+15+19+20+20) = 124 .

Hence box 17 - 20 must contain (20, 29, 31, 33).

Option B: Boxes 17 - 20 contain a total of 108 pencils. Since 29, 31, 33 pencils must be a part of 17 - 20 boxes. The remaining box must contain 108 - (29 +31+33) = 15 pencils. Hence the order is (15, 29, 31, 33).

Option C: Boxes 11- 16 contain a total of 101 defective pencils. This is only possible if the boxes here contain : ( 20, 20, 20, 15, 15, 11) pencils. Hence the box containing 19 pencils must be a part of boxes 17 - 20 and the remaining three contain 29, 31, 33. (19, 29, 31, 33)

Option D : Box number 17 containing more pencils than any box from box number 1- 14. Hence this only possible if Box 15, 16, 17 contains 20 pencils each and 18, 19, 20 contain (29, 31, 33).

Option E : Box 7-16 containing 133 broken pencils :

Boxes (1-6) has 6 boxes with broken pencils which can be included from :

(5, 6, 6, 7, 7, 7, 7, 9, 9, 9). The minimum possible sum of the 6 pencils is : (5+6+6+7) = 24 and the maximum possible sum is (7+9+9+9) = 34.

Boxes 7 - 16 contains all the boxes with broken pencils except one among the boxes with broken pencils among 11 - 20 and hence ;

This can contain : (11+15+15+19+20+20) or (15+15+19+20+20+20) or (11+15+15+20+20+20) or (11+15+19+20+20+20 ) =100/109/101/105.

This has multiple possibilities which include (109 + 24) or (101+32) or (105+28). Hence cannot be uniquely determined

Instructions

Read the following scenario and answer the THREE questions that follow.

An examination had ten multiple choice questions; labelled Q1 to Q10 respectively. Each question had four answer options — A, B, C and D — of which one and only one was the correct answer. For each correct answer, the candidate obtained 1 mark. There were no negative marks for wrong answers. The answers chosen by six candidates named Om, Pavan, Qadir, Rakesh, Simranjeet and Tracey to each of the ten questions and the total marks obtained by each of them are shown in the table.

Question 4

What is the correct answer for Q5?

Not possible to determine uniquely

Video Solution

Solution

As per the given conditions :

Qadir and Simarjeet both of them scored a score of 7 each.

In the table, it can be found that both Qadir and Simarjeet marked the same answer for exactly four questions. They are :

Q3, Q5, Q8, and Q10.

Both of them had 3 more questions correctly. Hence for the rest 6 questions 3 of them must have been answered by one of them correctly and the other three by another one of them.

Hence the answers must have been :

Now looking at the score of Rakesh he scored a total of 3 marks.

All the questions for which both Qadir and Simarjeet have answered correctly he has marked them wrong. He for whatever question he must have answered correctly exactly one among Qadir and Simarjeet must answer the same. Because at least one of them answered each question correctly.

Hence he must have answered correctly 3 questions among ( 1, 6, 7, 9)

Of the different possibilities are ( 1, 6, 7), (1, 6, 9), (6, 7, 9), (1, 7, 9).

But the cases 6, 7, 9 fails because if he answered all three of them correctly then he must have answered 1 wrongly but since if 1 is answered wrong then all of 1, 6, 7, 9 are answered correctly by Simarjeet when Simarjeet can answer actually a total of 3 questions only correctly hence this case fails.

Similarly if considered the cases ( 1, 6, 7) and (1, 6, 9) as the questions which were answered correctly. The cases fail because if they are answered correctly we cannot possibly have Pavan answering 5 questions correctly.

Hence the only possibility is he must have answered the questions :

( 1, 7, 9) correctly.

Hence the correct answers are :

Now drawing the table based on the answers marked by them :

Simarjeet must have answered 1 question of the Q2, Q4, and Q6 correctly. Qadir must have answered the remaining two of them correctly.

If Simarjeet answered Q4 correctly and Q2, Q6 wrong the answers for Q2, Q4 and Q6 will be: B, B, and D

But if Q2 and Q4 are answered as B then Om must have scored 3 marks instead of 2 and hence the case fails.

If Simarjeet answered Q6 correctly and Q2, Q4 wrong the answers for Q2, Q4 and Q6 will be: A, C, and C.

But Pavan cannot score 5 marks.

Hence Simarjeet must have answered Q2 correctly and answers for Q2, Q4, and Q6 will be :

(B, C, D).

The correct answer for Q5 is C.

Instructions

Read the following scenario and answer the THREE questions that follow.

Question 5

For which of these questions is D the correct answer?

Both Q1 and Q9

Both Q1 and Q8

Video Solution

Solution

As per the given conditions :

Qadir and Simarjeet both of them scored a score of 7 each.

In the table, it can be found that both Qadir and Simarjeet marked the same answer for exactly four questions. They are :

Q3, Q5, Q8, and Q10.

Both of them had 3 more questions correctly. Hence for the rest 6 questions 3 of them must have been answered by one of them correctly and the other three by another one of them.

Hence the answers must have been :

Now looking at the score of Rakesh he scored a total of 3 marks.

Hence he must have answered correctly 3 questions among ( 1, 6, 7, 9)

Of the different possibilities are ( 1, 6, 7), (1, 6, 9), (6, 7, 9), (1, 7, 9).

Hence the only possibility is he must have answered the questions :

( 1, 7, 9) correctly.

Hence the correct answers are :

Now drawing the table based on the answers marked by them :

Simarjeet must have answered 1 question of the Q2, Q4, and Q6 correctly. Qadir must have answered the remaining two of them correctly.

If Simarjeet answered Q4 correctly and Q2, Q6 wrong the answers for Q2, Q4 and Q6 will be: B, B, and D

But if Q2 and Q4 are answered as B then Om must have scored 3 marks instead of2 and hence the case fails.

If Simarjeet answered Q6 correctly and Q2, Q4 wrong the answers for Q2, Q4 and Q6 will be: A, C, and C.

But Pavan cannot score 5 marks.

Hence Simarjeet must have answered Q2 correctly and answers for Q2, Q4, and Q6 will be :

(B, C, D).

Q6 and Q8 have D as the correct answer.

Instructions

Read the following scenario and answer the THREE questions that follow.

Question 6

Which of these questions witnessed the least number of the students answering correctly?

Both Q3 and Q4

Q10

Video Solution

Solution

As per the given conditions :

Qadir and Simarjeet both of them scored a score of 7 each.

In the table, it can be found that both Qadir and Simarjeet marked the same answer for exactly four questions. They are :

Q3, Q5, Q8, and Q10.

Both of them had 3 more questions correctly. Hence for the rest 6 questions 3 of them must have been answered by one of them correctly and the other three by another one of them.

Hence the answers must have been :

Now looking at the score of Rakesh he scored a total of 3 marks.

Hence he must have answered correctly 3 questions among ( 1, 6, 7, 9)

Of the different possibilities are ( 1, 6, 7), (1, 6, 9), (6, 7, 9), (1, 7, 9).

Hence the only possibility is he must have answered the questions :

( 1, 7, 9) correctly.

Hence the correct answers are :

Now drawing the table based on the answers marked by them :

Simarjeet must have answered 1 question of the Q2, Q4, and Q6 correctly. Qadir must have answered the remaining two of them correctly.

If Simarjeet answered Q4 correctly and Q2, Q6 wrong the answers for Q2, Q4 and Q6 will be: B, B, and D

But if Q2 and Q4 are answered as B then Om must have scored 3 marks instead of2 and hence the case fails.

If Simarjeet answered Q6 correctly and Q2, Q4 wrong the answers for Q2, Q4 and Q6 will be: A, C, and C.

But Pavan cannot score 5 marks.

Hence Simarjeet must have answered Q2 correctly and answers for Q2, Q4, and Q6 will be :

(B, C, D).

Q4 was answered wrong by 5 members.

Instructions

Questions are based on a set of conditions. In answering some of the questions, it may be useful to draw a rough diagram. Choose the response that most accurately and completely answers each question.

A BPO has assigned duty to nine operators - Abdulla, Ballal, Chandan, Dogra, Eshita, Falguni, Ganguli, Henri and Indra - on Monday, January 05, 2009 from 00:00 hours. Each operator commences duty at any of the following hours: 00:00 hrs, 04:00 hrs, 08:00 hrs, 12:00 hrs, 16:00 hrs and 20:00 hrs. At any point in time, at least one operator is required, to take clients' calls. Each operator works continuously for eight hours. All operators located at any single location start work simultaneously. The operators took training in five different colleges -Abhiman College, Sutanama College, Gutakal College, Barala College and Khatanama College. These colleges are located in the cities Jamshedpur, Pune, Noida, Hyderabad and Mangalore, not necessarily in that order. The operators operate from the cities where their respective colleges are located.

Indra operates alone from a city other than Mangalore and Jamshedpur. Operator(s) trained in Abhiman College will start working at 12:00 hrs. Only Dogra and Falguni operate from Pune, but they are not trained in Gutakal College. Three of the operators took training from Sutanama College, and they operate from Noida. The operator(s) from Jamshedpur will start working at 0:00 hrs. Abdulla and Henri operate together as a two member team from a single location. They do not operate from Mangalore. No operator(s) will join at 20:00 hrs. Ballal, who alone operates from his location, was not trained in Barala College, and will commence his duty four hours after the operator(s) trained in Gutakal College. The operator(s) trained in Barala College operate from Hyderabad. The number of operator(s) trained in Khatanama College is same as the number of operator(s) trained in Barala College.

Question 1

Which of the following statements must be true?

Dogra and Henri took training from Khatanama College

Indra took training from Barala College

Dogra and Falguni took training from Barala College

Indra took training from Ahiman College

Ballal took training from Abhiman College

Video Solution

Solution

Let A, B, C, D, E, F, G, H and I represent Abdulla, Ballal, Chandan, Dogra, Eshita, Falguni, Ganguli, Henri and Indra.
Indira operates alone from Pune, Noida or Hyderabad and only Dogra and Falguni operate from Pune, while there are three operators from Noida
who have studies in Sutanama College. Since Indira works alone, she cannot be from Pune or Noida.
Thus, Indira works alone from Hyderabad.
Thus, D, F- Pune - 2
I- Hyderabad - 1
-Noida- 3
We are given, Abdulla and Henri operate together as a two member team from a single location and they do not operate from Mangalore.
Thus, A and H work from Jamshedpur and Only 1 person work from Mangalore.

D, F- Pune - 2
I- Hyderabad - 1
- Noida- 3
A, H- Jamshedpur
- Mangalore- 1
Ballal operates alone from his location.
Thus, we get:-

D, F- Pune - 2
I- Hyderabad - 1
C, E, G - Noida- 3
A, H- Jamshedpur
B - Mangalore- 1
Three of the operators took training from Sutanama College, and they operate from Noida
Thus, C, E and G took training from Sutanama College.
The operator(s) trained in Barala College operate from Hyderabad.
Thus, I is trained in Barala College.
The number of operator(s) trained in Khatanama College is same as the number of operator(s) trained in Barala College.
Thus, only 1 person is trained from Khatanama College and that person is B.
Only Dogra and Falguni operate from Pune, but they are not trained in Gutakal College.
Thus, D and F are trained in Abhiman college.
Thus, A and H are trained in Gutakal College.
The operator(s) from Jamshedpur will start working at 0:00 hrs
Hence, A and H will start working from 0.
Operator(s) trained in Abhiman College will start working at 12:00 hrs.
Hence, D and F will start working from 12

Ballal will commence his duty four hours after the operator(s) trained in Gutakal College
Hence, B will commence his duty from 4.
Thus, the final table is:-

From the table, we can see that only option B is true.
Hence, option B is the correct answer.

Instructions

Question 2

Which of the following can be true for the operators operating from 20:00 hrs (of January 05, 2009) to 00:00 hrs (of January 06, 2009)?

Operators took training from Khatanama College and operate from Mangalore.

Operators took training from Barala College and operate from Pune.

Operators took training from Sutanama College and operate from Noida.

Operators took training from Gutakal College and operate from Mangalore.

Operators took training from Abhiman College and operate from Pune.

Video Solution

Solution

D, F- Pune - 2
I- Hyderabad - 1
- Noida- 3
A, H- Jamshedpur
- Mangalore- 1
Ballal operates alone from his location.
Thus, we get:-

Ballal will commence his duty four hours after the operator(s) trained in Gutakal College
Hence, B will commence his duty from 4.
Thus, the final table is:-

From the table, we can see that only option C can be true.
Hence, option C is the correct answer.

Instructions

Question 3

Which of the following statements must be true for the operator(s) trained in Gutakal College?

They are Abdulla and Henri, and work from Jamshedpur.

They are Dogra and Falguni, and work from Pune.

She is Eshita, and works from Mangalore.

She is Indra, and works from Pune.

They are Chandan and Ganguli, and work from Jamshedpur.

Video Solution

Solution

D, F- Pune - 2
I- Hyderabad - 1
- Noida- 3
A, H- Jamshedpur
- Mangalore- 1
Ballal operates alone from his location.
Thus, we get:-

Ballal will commence his duty four hours after the operator(s) trained in Gutakal College
Hence, B will commence his duty from 4.
Thus, the final table is:-

From the table, we can see that option A is the correct answer.

Instructions

Question 4

Which of the following is not definitely true?

At least three operators will be working between 04:00 hrs to 08:00 hrs.

At most five operators will be working between 04:00 hrs to 08:00 hrs.

At most five operators will be working between 12:00 hrs to 16:00 hrs.

At most six operators will be working between 16:00 hrs to 20:00 hrs.

At least three operators will be working between 16:00 hrs to 20:00 hrs.

Video Solution

Solution

D, F- Pune - 2
I- Hyderabad - 1
- Noida- 3
A, H- Jamshedpur
- Mangalore- 1
Ballal operates alone from his location.
Thus, we get:-

Ballal will commence his duty four hours after the operator(s) trained in Gutakal College
Hence, B will commence his duty from 4.
Thus, the final table is:-

From the table, if Chandan, Eshita and Ganguli start working at 4:00, then between 4:00 to 8:00, there will be 6 people working. Thus, option B is incorrect.
Hence, option B is the correct answer.

Instructions

Question 5

If five operators are working between 16:00 hrs and 20:00 hrs, which of the following must be true?

The only operator working between 8:00 hrs and 12:00 hrs is Ballal.

The operators working between 12:00 hrs and 16:00 hrs are Ballal, Dogra and Falguni.

The operators working between 12:00 hrs and 16:00 hrs are Dogra, Henri and Falguni.

The operators working between 16:00 hrs and 0:00 hrs are Indra, Dogra, Falguni, Chandan and Ganguli.

The operators working between 20:00 hrs and 0:00 hrs are Chandan Ganguli and Eshita.

Video Solution

Solution

D, F- Pune - 2
I- Hyderabad - 1
- Noida- 3
A, H- Jamshedpur
- Mangalore- 1
Ballal operates alone from his location.
Thus, we get:-

Ballal will commence his duty four hours after the operator(s) trained in Gutakal College
Hence, B will commence his duty from 4.
Thus, the final table is:-

Five operators are working between 16:00 hrs and 20:00 hrs. This is only possible if Chandan, Ishita and Ganguli start working at 16:00. Hence, the operators working between 20:00 hrs and 0:00 hrs are Chandan Ganguli and Eshita.
Hence, option E is the correct answer.

Instructions

Questions are based on a set of conditions. In answering some of the questions, it may be useful to draw a rough diagram. Choose the response that most accurately and completely answers each question.

Five colleagues pooled their efforts during the office lunch-hour to solve the crossword in the daily paper.

Colleagues: Mr. Bineet, Mr. Easwar, Ms. Elsie, Ms. Sheela, Ms. Titli.
Answers: Burden, Barely, Baadshah, Rosebud, Silence.
Numbers: 4 down, 8 across, 15 across, 15 down, 21 across.
Order: First, second, third, fourth, fifth.

1. Titli produced the answer to 8 across, which had the same number of letters as the previous answer to be inserted, and one more than the subsequent answer which was produced by one of the men.
2. It was not Bineet who solved the clue to ‘Burden’, and Easwar did not solve 4 down.
3. The answers to 15 across and 15 down did not have the same number of letters.
4. 'Silence', which was not the third word to be inserted, was the answer to an across clue.
5. 'Barely' was the first word to be entered in the grid, but 'Baadshah' was not the second answer to be found.
6. Elsie's word was longer than Bineet's; Sheela was neither the first nor the last to come up with an answer.
7. Fifth one to be worked out was an answer to an across clue

Question 6

What was Sheela’s word?

Baadshah

Silence

Rosebud

Barely

Burden

Video Solution

Solution

Titli produced the answer to 8 across, which had the same number of letters as the previous answer to be inserted, and one more than the subsequent answer which was produced by one of the men. So, Titli must have produced a 7-letter word i.e either Silence or Rosebud. The word next to Titli's word must be a 6-letter word which is Burden. It cannot be Barely because it has been given that Barely is the first word to be inserted. Also, the word preceding Titli's word must be a 7-letter word.

Thus, both the 7-letter word and Burden must be at he consecutive places. They can be either at second, third and fourth position or third, fourth and fifth position. But, in the second case, Baadshah would be at the second position which is not possible as given in the question. So, both the 7-letter words and Burden must be at the second, third and the fourth position respectively and Baadshah would be at the fifth position. Also, Silence is not at the third position. So, it must be at the second position and Rosebus must be at the third position.

15 across and 15 down must start from the same alphabet and their length must be unequal. Also, Silence is an answer to across clue which can be either 15 across or 21 across. However, if Silence is an answer to 15 across, the answer to 15 down must also start with 'S'. But, there is no other word starting with 'S'. So, Silence must be the answer to 21 across clue. Fifth one to be worked out was an answer to an across clue. So, Baadshah should be an answer to across clue.

Elsie's word was longer than Bineet's. So, Elsie's word must be either Baadshah or Silence and Bineet's word should be either Barely or Silence. In both the cases, Burden would be Easwar's word and it would be the answer to 15 down clue.
Sheela was neither the first nor the last to come up with an answer. So, her answer must be Silence. Bineet's word must be Barely and Elsie's word must be Baadshah.

From the table we can see that Sheela's word was Silence.
Hence, option B is the correct answer.

Instructions

Questions are based on a set of conditions. In answering some of the questions, it may be useful to draw a rough diagram. Choose the response that most accurately and completely answers each question.

Five colleagues pooled their efforts during the office lunch-hour to solve the crossword in the daily paper.

Colleagues: Mr. Bineet, Mr. Easwar, Ms. Elsie, Ms. Sheela, Ms. Titli.
Answers: Burden, Barely, Baadshah, Rosebud, Silence.
Numbers: 4 down, 8 across, 15 across, 15 down, 21 across.
Order: First, second, third, fourth, fifth.

1. Titli produced the answer to 8 across, which had the same number of letters as the previous answer to be inserted, and one more than the subsequent answer which was produced by one of the men.
2. It was not Bineet who solved the clue to ‘Burden’, and Easwar did not solve 4 down.
3. The answers to 15 across and 15 down did not have the same number of letters.
4. 'Silence', which was not the third word to be inserted, was the answer to an across clue.
5. 'Barely' was the first word to be entered in the grid, but 'Baadshah' was not the second answer to be found.
6. Elsie's word was longer than Bineet's; Sheela was neither the first nor the last to come up with an answer.
7. Fifth one to be worked out was an answer to an across clue

Question 7

What could be Titli’s answer?

Baadshah

Silence

Rosebud

Barely

Burden

Video Solution

Solution

Elsie's word was longer than Bineet's. So, Elsie's word must be either Baadshah or Silence and Bineet's word should be either Barely or Silence. In both the cases, Burden would be Easwar's word and it would be the answer to 15 down clue. Sheela was neither the first nor the last to come up with an answer. So, her answer must be Silence. Bineet's word must be Barely and Elsie's word must be Baadshah.

From the table we can see that Titli's answer is Rosebud.
Hence, option C is the correct answer.

Instructions

Questions are based on a set of conditions. In answering some of the questions, it may be useful to draw a rough diagram. Choose the response that most accurately and completely answers each question.

Five colleagues pooled their efforts during the office lunch-hour to solve the crossword in the daily paper.

Colleagues: Mr. Bineet, Mr. Easwar, Ms. Elsie, Ms. Sheela, Ms. Titli.
Answers: Burden, Barely, Baadshah, Rosebud, Silence.
Numbers: 4 down, 8 across, 15 across, 15 down, 21 across.
Order: First, second, third, fourth, fifth.

1. Titli produced the answer to 8 across, which had the same number of letters as the previous answer to be inserted, and one more than the subsequent answer which was produced by one of the men.
2. It was not Bineet who solved the clue to ‘Burden’, and Easwar did not solve 4 down.
3. The answers to 15 across and 15 down did not have the same number of letters.
4. 'Silence', which was not the third word to be inserted, was the answer to an across clue.
5. 'Barely' was the first word to be entered in the grid, but 'Baadshah' was not the second answer to be found.
6. Elsie's word was longer than Bineet's; Sheela was neither the first nor the last to come up with an answer.
7. Fifth one to be worked out was an answer to an across clue

Question 8

What was Titli’s order?

First

Second

Third

Fourth

Fifth

Video Solution

Solution

Elsie's word was longer than Bineet's. So, Elsie's word must be either Baadshah or Silence and Bineet's word should be either Barely or Silence. In both the cases, Burden would be Easwar's word and it would be the answer to 15 down clue. Sheela was neither the first nor the last to come up with an answer. So, her answer must be Silence. Bineet's word must be Barely and Elsie's word must be Baadshah.

From the table we can see that Titli's order is third.
Hence, option C is the correct answer.

Instructions

Questions are based on a set of conditions. In answering some of the questions, it may be useful to draw a rough diagram. Choose the response that most accurately and completely answers each question.

Five colleagues pooled their efforts during the office lunch-hour to solve the crossword in the daily paper.

Colleagues: Mr. Bineet, Mr. Easwar, Ms. Elsie, Ms. Sheela, Ms. Titli.
Answers: Burden, Barely, Baadshah, Rosebud, Silence.
Numbers: 4 down, 8 across, 15 across, 15 down, 21 across.
Order: First, second, third, fourth, fifth.

1. Titli produced the answer to 8 across, which had the same number of letters as the previous answer to be inserted, and one more than the subsequent answer which was produced by one of the men.
2. It was not Bineet who solved the clue to ‘Burden’, and Easwar did not solve 4 down.
3. The answers to 15 across and 15 down did not have the same number of letters.
4. 'Silence', which was not the third word to be inserted, was the answer to an across clue.
5. 'Barely' was the first word to be entered in the grid, but 'Baadshah' was not the second answer to be found.
6. Elsie's word was longer than Bineet's; Sheela was neither the first nor the last to come up with an answer.
7. Fifth one to be worked out was an answer to an across clue

Question 9

What was Easwar’s number?

4 down

21 across

8 across

15 down

15 across

Video Solution

Solution

Elsie's word was longer than Bineet's. So, Elsie's word must be either Baadshah or Silence and Bineet's word should be either Barely or Silence. In both the cases, Burden would be Easwar's word and it would be the answer to 15 down clue. Sheela was neither the first nor the last to come up with an answer. So, her answer must be Silence. Bineet's word must be Barely and Elsie's word must be Baadshah.

From the table we can see that Easwar's number is 15 down.
Hence, option D is the correct answer.

Instructions

Questions are based on a set of conditions. In answering some of the questions, it may be useful to draw a rough diagram. Choose the response that most accurately and completely answers each question.

Five colleagues pooled their efforts during the office lunch-hour to solve the crossword in the daily paper.

Colleagues: Mr. Bineet, Mr. Easwar, Ms. Elsie, Ms. Sheela, Ms. Titli.
Answers: Burden, Barely, Baadshah, Rosebud, Silence.
Numbers: 4 down, 8 across, 15 across, 15 down, 21 across.
Order: First, second, third, fourth, fifth.

1. Titli produced the answer to 8 across, which had the same number of letters as the previous answer to be inserted, and one more than the subsequent answer which was produced by one of the men.
2. It was not Bineet who solved the clue to ‘Burden’, and Easwar did not solve 4 down.
3. The answers to 15 across and 15 down did not have the same number of letters.
4. 'Silence', which was not the third word to be inserted, was the answer to an across clue.
5. 'Barely' was the first word to be entered in the grid, but 'Baadshah' was not the second answer to be found.
6. Elsie's word was longer than Bineet's; Sheela was neither the first nor the last to come up with an answer.
7. Fifth one to be worked out was an answer to an across clue

Question 10

What was Bineet’s word?

Barely

Burden

Silence

Rosebud

Baadshah

Solution

Elsie's word was longer than Bineet's. So, Elsie's word must be either Baadshah or Silence and Bineet's word should be either Barely or Silence. In both the cases, Burden would be Easwar's word and it would be the answer to 15 down clue. Sheela was neither the first nor the last to come up with an answer. So, her answer must be Silence. Bineet's word must be Barely and Elsie's word must be Baadshah.

From the table, we can see that Bineet's word is Barely.
Hence, option A is the correct answer.

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