# CAT Logical Reasoning Puzzle Questions PDF [Most Important]

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Puzzles are an important topic in the CAT Logical Reasoning & Data Interpretation (LRDI) Section. Usually, the Sets on Puzzles are not very tough to crack and hence one must not miss out on the LRDI Sets on Puzzles. You can check out these Puzzles questions from the CAT Previous year’s papers. Practice a good number of sums on CAT Puzzle questions so that you can answer these questions with ease in the exam. In this article, we will look into some important Puzzle Questions for CAT LRDI. These are a good source for practice; If you want to practice these questions, you can download this Important CAT Puzzle Questions PDF below, which is completely Free.

Instructions

A high security research lab requires the researchers to set a pass key sequence based on the scan of the five fingers of their left hands. When an employee first joins the lab, her fingers are scanned in an order of her choice, and then when she wants to re-enter the facility, she has to scan the five fingers in the same sequence.
The lab authorities are considering some relaxations of the scan order requirements, since it is observed that some employees often get locked-out because they forget the sequence.

Question 1: The lab has decided to allow a variation in the sequence of scans of the five fingers so that at most two scans (out of five) are out of place. For example, if the original sequence is Thumb (T), index finger (I), middle finger (M), ring finger (R) and little finger (L) then TLMRI is also allowed, but TMRLI is not.
How many different sequences of scans are allowed for any given person’s original scan?

Solution:

Let the original sequence be TIMRL

Two fingers can be out of place. This can be done if and only if two fingers interchange their position. These two can be selected in $^5C_2 = 10$ ways. In addition to these, the original sequence will also be accepted. Hence the total number of acceptable sequences = 10 + 1 = 11

Question 2: The lab has decided to allow variations of the original sequence so that input of the scanned sequence of five fingers is allowed to vary from the original sequence by one place for any of the fingers. Thus, for example, if TIMRL is the original sequence, then ITRML is also allowed, but LIMRT is not.
How many different sequences are allowed for any given person’s original scan?

a) 7

b) 5

c) 8

d) 13

Solution:

Input of the scanned sequence of five fingers is allowed to vary from the original sequence by one place for any of the fingers. This can be achieved only when two consecutive fingers are interchanged. Let the original sequence be TIMRL
Case 1: Only a set of two consecutive numbers are interchanged.
They can be selected in 5-1 = 4 ways
Case 2: Two sets of two consecutive numbers are interchanged.
(i) TI are interchanged, => (MR, RL) => 2 ways
(ii) IM are interchanged => (RL) => 1 way
Total no of ways possible = 4 + 2 + 1 = 7

Including the original sequence, we get the total number of allowed combinations as 8

Question 3: The lab has now decided to require six scans in the pass key sequence, where exactly one finger is scanned twice, and the other fingers are scanned exactly once, which can be done in any order. For example, a possible sequence is TIMTRL.
Suppose the lab allows a variation of the original sequence (of six inputs) where at most two scans (out of six) are out of place, as long as the finger originally scanned twice is scanned twice and other fingers are scanned once.
How many different sequences of scans are allowed for any given person’s original scan?

Solution:

There can be two scans out of place.

TIMTRL is the original sequence.

If T is interchanged: There will be four ways: ITMTRL, MITTRL, RIMTTL, LIMTRT

If I is interchanged: There will be four ways

If M is interchanged: There will be three ways

If T is interchanged: There will be two ways

If R is interchanged: There will be one way

Total 14.

Another sequence allowed is original, So total 15 ways.

Question 4: The lab has now decided to require six scans in the pass key sequence, where exactly one finger is scanned twice, and the other fingers are scanned exactly once, which can be done in any order. For example, a possible sequence is TIMTRL.
Suppose the lab allows a variation of the original sequence (of six inputs) so that input in the form of scanned sequence of six fingers is allowed to vary from the original sequence by one place for any of the fingers, as long as the finger originally scanned twice is scanned twice and other fingers are scanned once.
How many different sequences of scans are allowed if the original scan sequence is LRLTIM?

a) 8

b) 11

c) 13

d) 14

Solution:

1. If original sequence is given.

2. If either of LR, RL, LT, TI, IM is interchanged => 5 ways.

3. If LR and LT and IM interchanged. The sequence will look like: RLTLMI

4. If LR and LT are interchanged.

5. If LR and TI are interchanged.

6. If LR and IM are interchanged.

7. If RL and TI are interchanged.

8. If RL and IM  are interchanged.

9. If LT and IM are interchanged.

Total 13 ways possible.

Instructions

Adriana, Bandita, Chitra, and Daisy are four female students, and Amit, Barun, Chetan, and Deb are four male students. Each of them studies in one of three institutes – X, Y, and Z. Each student majors in one subject among Marketing, Operations, and Finance, and minors in a different one among these three subjects. The following facts are known about the eight students:

1. Three students are from X, three are from Y, and the remaining two students, both female, are from Z.
2. Both the male students from Y minor in Finance, while the female student from Y majors in Operations.
3. Only one male student majors in Operations, while three female students minor in Marketing.
4. One female and two male students major in Finance.
5. Adriana and Deb are from the same institute. Daisy and Amit are from the same institute.
6. Barun is from Y and majors in Operations. Chetan is from X and majors in Finance.
7. Daisy minors in Operations.

Question 5: Who are the students from the institute Z?

a) Chitra and Daisy

c) Bandita and Chitra

Solution:

There are 8 students in total – 4 male and 4 female. There are 3 institutes X, Y, and Z.
3 students are from institute X, 3 students are from institute Y, and 2 students are from institute Z. No student majors and minors in the same subject.

It has been given that both the students from institute Z are female. Also, it has been given that both the male students from institute Y minor in Finance. Therefore, the third student from institute Y should be female. Institute X should also have 2 male and 1 female student.

Both the male students from Y minor in Finance, while the female student from Y majors in Operations. Barun is from Y and majors in Operations. Chetan is from X and majors in Finance.

It has been given that one female student and 2 male students major in finance. We know that the male student from Y minors in finance. Therefore, he cannot major in finance. Therefore, both the male students from X should major in finance.

Daisy and Amit are from the same institute. Therefore, Daisy cannot be from institute Z (since Amit is a male student and both the students from Z are female). Daisy minors in operations. The girl from institute Y majors in Operations. Therefore, Daisy cannot be from institute Y as well. Daisy and Amit should be from institute X. 3 female students minor in marketing. Therefore, all girls except Daisy should minor in marketing.

Adriana and Deb are from the same institute. Therefore, both of them should be from institute Y. Bandita and Chitra should be from institute Z.

Only one male student majors in Operations. We know that Barun is the student. Two male students major in Finance. We know that Amit and Chetan major in finance. Therefore, Deb should major in Marketing.

Bandita and Chitra are from institute Z. Therefore, option C is the right answer.

Question 6: Which subject does Deb minor in?

a) Operations

b) Finance

c) Marketing

d) Cannot be determined uniquely from the given information

Solution:

There are 8 students in total – 4 male and 4 female. There are 3 institutes X, Y, and Z.
3 students are from institute X, 3 students are from institute Y, and 2 students are from institute Z. No student majors and minors in the same subject.

It has been given that both the students from institute Z are female. Also, it has been given that both the male students from institute Y minor in Finance. Therefore, the third student from institute Y should be female. Institute X should also have 2 male and 1 female student.

Both the male students from Y minor in Finance, while the female student from Y majors in Operations. Barun is from Y and majors in Operations. Chetan is from X and majors in Finance.

It has been given that one female student and 2 male students major in finance. We know that the male student from Y minors in finance. Therefore, he cannot major in finance. Therefore, both the male students from X should major in finance.

Daisy and Amit are from the same institute. Therefore, Daisy cannot be from institute Z (since Amit is a male student and both the students from Z are female). Daisy minors in operations. The girl from institute Y majors in Operations. Therefore, Daisy cannot be from institute Y as well. Daisy and Amit should be from institute X. 3 female students minor in marketing. Therefore, all girls except Daisy should minor in marketing.

Adriana and Deb are from the same institute. Therefore, both of them should be from institute Y. Bandita and Chitra should be from institute Z.

Only one male student majors in Operations. We know that Barun is the student. Two male students major in Finance. We know that Amit and Chetan major in finance. Therefore, Deb should major in Marketing.

Deb minors in Finance. Therefore, option B is the right answer.

Question 7: Which subject does Amit major in?

a) Marketing

b) Operations

c) Cannot be determined uniquely from the given information

d) Finance

Solution:

There are 8 students in total – 4 male and 4 female. There are 3 institutes X, Y, and Z.
3 students are from institute X, 3 students are from institute Y, and 2 students are from institute Z. No student majors and minors in the same subject.

It has been given that both the students from institute Z are female. Also, it has been given that both the male students from institute Y minor in Finance. Therefore, the third student from institute Y should be female. Institute X should also have 2 male and 1 female student.

Both the male students from Y minor in Finance, while the female student from Y majors in Operations. Barun is from Y and majors in Operations. Chetan is from X and majors in Finance.

It has been given that one female student and 2 male students major in finance. We know that the male student from Y minors in finance. Therefore, he cannot major in finance. Therefore, both the male students from X should major in finance.

Daisy and Amit are from the same institute. Therefore, Daisy cannot be from institute Z (since Amit is a male student and both the students from Z are female). Daisy minors in operations. The girl from institute Y majors in Operations. Therefore, Daisy cannot be from institute Y as well. Daisy and Amit should be from institute X. 3 female students minor in marketing. Therefore, all girls except Daisy should minor in marketing.

Adriana and Deb are from the same institute. Therefore, both of them should be from institute Y. Bandita and Chitra should be from institute Z.

Only one male student majors in Operations. We know that Barun is the student. Two male students major in Finance. We know that Amit and Chetan major in finance. Therefore, Deb should major in Marketing.

Amit majors in finance. Therefore, option D is the right answer.

Question 8: If Chitra majors in Finance, which subject does Bandita major in?

a) Finance

b) Cannot be determined uniquely from the given information

c) Operations

d) Marketing

Solution:

There are 8 students in total – 4 male and 4 female. There are 3 institutes X, Y, and Z.
3 students are from institute X, 3 students are from institute Y, and 2 students are from institute Z. No student majors and minors in the same subject.

It has been given that both the students from institute Z are female. Also, it has been given that both the male students from institute Y minor in Finance. Therefore, the third student from institute Y should be female. Institute X should also have 2 male and 1 female student.

Both the male students from Y minor in Finance, while the female student from Y majors in Operations. Barun is from Y and majors in Operations. Chetan is from X and majors in Finance.

It has been given that one female student and 2 male students major in finance. We know that the male student from Y minors in finance. Therefore, he cannot major in finance. Therefore, both the male students from X should major in finance.

Daisy and Amit are from the same institute. Therefore, Daisy cannot be from institute Z (since Amit is a male student and both the students from Z are female). Daisy minors in operations. The girl from institute Y majors in Operations. Therefore, Daisy cannot be from institute Y as well. Daisy and Amit should be from institute X. 3 female students minor in marketing. Therefore, all girls except Daisy should minor in marketing.

Adriana and Deb are from the same institute. Therefore, both of them should be from institute Y. Bandita and Chitra should be from institute Z.

Only one male student majors in Operations. We know that Barun is the student. Two male students major in Finance. We know that Amit and Chetan major in finance. Therefore, Deb should major in Marketing.

If Chitra majors in finance, Bandita cannot major in finance (only one female student majors in finance). She cannot major in marketing as well (since she has a minor degree in marketing). Therefore, Bandita should major in operations and hence, option C is the right answer.

Instructions

According to a coding scheme the sentence:

“Peacock is designated as the national bird of India”  is coded as 5688999 35 1135556678 56 458 13666689 1334 79 13366

This coding scheme has the following rules:
a: The scheme is case-insensitive (does not distinguish between upper case and lower case letters).
b: Each letter has a unique code which is a single digit from among 1,2,3, …, 9.
c: The digit 9 codes two letters, and every other digit codes three letters.
d: The code for a word is constructed by arranging the digits corresponding to its letters in a non-decreasing sequence.

Answer these questions on the basis of this information.

Question 9: What best can be concluded about the code for the letter L?

a) 1

b) 8

c) 1 or 8

d) 6

Solution:

We can see that India’s code is 13366 therefore we can say that I’s code is either 3 or 6.

Also, we can see that code for word “is” is 35 therefore we can say that I’s code is 3. Consequently, we can say that S’s code is 5.

Also, we can see that code of word ‘as’ is 56 therefore we can say that A’s code is 6. Consequently, we can say that S’s code is 5.

There is only one letter ‘O’ common in words ‘of’ and ‘national’. In code word as well only digit ‘9’ is common in both. Hence, we can say that letter ‘O’ is assigned numerical ‘9’. Consequently, we can say that F is assigned number 7.

It is given that ‘9’ is assigned to only two alphabets one of them is ‘O’. We can see that there are three 9’s in Peacock’s code. One of the digit ‘9’ is used for ‘O’.Remaining two 9’s must represent same letter. We can see that only letter ‘C’ has appeared twice in Peacock. Therefore, we can say that ‘C’ is assigned number ‘9’.

In word national ‘N’ has appeared twice. In code only digit ‘6’ has appeared more than once. Hence, we can say that code of letter N is ‘6’. Consequently, we can say that code for letter ‘D’ is ‘1’ because in India rest of the numerals are already taken.

In words, ‘the’ and ‘national’ only letter ‘t’ is common. In code as well only digit ‘8’ is common in two codes. Hence, we can say that letter code for letter ‘t’ is 8.

In words, ‘the’ and ‘peacock’ only letter ‘e’ is common. In code as well only digit ‘5’ is common in two codes. Hence, we can say that letter code for letter ‘e’ is 5. Consequently, we can say that leftover letter, in word “the”, ‘H’s code is 4.

We can see that code for word “NATIONAL” is 13666689. Hence, we can say that code for the letter L is ‘1’. Hence, option A is the correct answer.

Question 10: What best can be concluded about the code for the letter B?

a) 3 or 4

b) 1 or 3 or 4

c) 1

d) 3

Solution:

We can see that India’s code is 13366 therefore we can say that I’s code is either 3 or 6.

Also, we can see that code for word “is” is 35 therefore we can say that I’s code is 3. Consequently, we can say that S’s code is 5.

Also, we can see that code of word ‘as’ is 56 therefore we can say that A’s code is 6. Consequently, we can say that S’s code is 5.

There is only one letter ‘O’ common in words ‘of’ and ‘national’. In code word as well only digit ‘9’ is common in both. Hence, we can say that letter ‘O’ is assigned numerical ‘9’. Consequently, we can say that F is assigned number 7.

It is given that ‘9’ is assigned to only two alphabets one of them is ‘O’. We can see that there are three 9’s in Peacock’s code. One of the digit ‘9’ is used for ‘O’.Remaining two 9’s must represent same letter. We can see that only letter ‘C’ has appeared twice in Peacock. Therefore, we can say that ‘C’ is assigned number ‘9’.

In word national ‘N’ has appeared twice. In code only digit ‘6’ has appeared more than once. Hence, we can say that code of letter N is ‘6’. Consequently, we can say that code for letter ‘D’ is ‘1’ because in India rest of the numerals are already taken.

In words, ‘the’ and ‘national’ only letter ‘t’ is common. In code as well only digit ‘8’ is common in two codes. Hence, we can say that letter code for letter ‘t’ is 8.

In words, ‘the’ and ‘peacock’ only letter ‘e’ is common. In code as well only digit ‘5’ is common in two codes. Hence, we can say that letter code for letter ‘e’ is 5. Consequently, we can say that leftover letter, in word “the”, ‘H’s code is 4.

We can see that code for word “NATIONAL” is 13666689. Hence, we can say that code for the letter L is ‘1’.

We can see that code for word “BIRD” is 1334. 1 corresponds to D and one 3 corresponds to I. Hence, we can say that code for letters ‘R’ and ‘B’ are ‘3’ and ‘4’ in any order.

Therefore, we can say that for letter ‘B’ there are two possible numbers: 3 or 4

Hence, option A is the correct answer.

Question 11: For how many digits can the complete list of letters associated with that digit be identified?

a) 1

b) 2

c) 0

d) 3

Solution:

We can see that India’s code is 13366 therefore we can say that I’s code is either 3 or 6.

Also, we can see that code for word “is” is 35 therefore we can say that I’s code is 3. Consequently, we can say that S’s code is 5.

Also, we can see that code of word ‘as’ is 56 therefore we can say that A’s code is 6. Consequently, we can say that S’s code is 5.

There is only one letter ‘O’ common in words ‘of’ and ‘national’. In code word as well only digit ‘9’ is common in both. Hence, we can say that letter ‘O’ is assigned numerical ‘9’. Consequently, we can say that F is assigned number 7.

It is given that ‘9’ is assigned to only two alphabets one of them is ‘O’. We can see that there are three 9’s in Peacock’s code. One of the digit ‘9’ is used for ‘O’.Remaining two 9’s must represent same letter. We can see that only letter ‘C’ has appeared twice in Peacock. Therefore, we can say that ‘C’ is assigned number ‘9’.

In word national ‘N’ has appeared twice. In code only digit ‘6’ has appeared more than once. Hence, we can say that code of letter N is ‘6’. Consequently, we can say that code for letter ‘D’ is ‘1’ because in India rest of the numerals are already taken.

In words, ‘the’ and ‘national’ only letter ‘t’ is common. In code as well only digit ‘8’ is common in two codes. Hence, we can say that letter code for letter ‘t’ is 8.

In words, ‘the’ and ‘peacock’ only letter ‘e’ is common. In code as well only digit ‘5’ is common in two codes. Hence, we can say that letter code for letter ‘e’ is 5. Consequently, we can say that leftover letter, in word “the”, ‘H’s code is 4.

We can see that code for word “NATIONAL” is 13666689. Hence, we can say that code for the letter L is ‘1’.

We can see that code for word “DESIGNATED” is 1135556678. Hence, we can say that code for the letter ‘G’ is ‘7’.

We can see that code for word “PEACOCK” is 5688999. Hence, we can say that code for the letters ‘P’ and ‘K’ is ‘8’.

Digit ‘1’ is used for L and D only. We can not figure out the third letter for which digit 1 is used.

Digit ‘2’ is not used for any letter. Hence, we can not figure out all the letters for which digit 2 is correct code.

Digit ‘3’ is used for letter ‘I’ only. Hence, we can not figure out all the letters for which digit 3 is correct code.

Digit ‘4’ is used for letters ‘H’ and one of ‘B’ and ‘R’. Hence, we can not figure out all the letters for which digit 4 is correct code.

Digit ‘5’ is used for letters ‘S’ and ‘E’. We can not figure out the third letter for which digit 5 is used.

Digit ‘6’ is used for letters ‘A’ and ‘N’. We can not figure out the third letter for which digit 6 is used.

Digit ‘7’ is used for letters ‘G’ and ‘F’. We can not figure out the third letter for which digit 7 is used.

Digit ‘8’ is used for letters ‘T’, ‘P’ and K. Hence, we can say that this is one of the digit for which the complete list of letters associated is known.

Digit ‘9’ is used for letters ‘C’ and ‘O’. Hence, we can say that this is one of the digit for which the complete list of letters associated is known.

Therefore, we can say that for only two digits (8 and 9), the complete list of letters associated is known. Hence, option B is the correct answer.

Question 12: Which set of letters CANNOT be coded with the same digit?

a) S,E,Z

b) I,B,M

c) S,U,V

d) X,Y,Z

Solution:

We can see that India’s code is 13366 therefore we can say that I’s code is either 3 or 6.

Also, we can see that code for word “is” is 35 therefore we can say that I’s code is 3. Consequently, we can say that S’s code is 5.

Also, we can see that code of word ‘as’ is 56 therefore we can say that A’s code is 6. Consequently, we can say that S’s code is 5.

There is only one letter ‘O’ common in words ‘of’ and ‘national’. In code word as well only digit ‘9’ is common in both. Hence, we can say that letter ‘O’ is assigned numerical ‘9’. Consequently, we can say that F is assigned number 7.

It is given that ‘9’ is assigned to only two alphabets one of them is ‘O’. We can see that there are three 9’s in Peacock’s code. One of the digit ‘9’ is used for ‘O’.Remaining two 9’s must represent same letter. We can see that only letter ‘C’ has appeared twice in Peacock. Therefore, we can say that ‘C’ is assigned number ‘9’.

In word national ‘N’ has appeared twice. In code only digit ‘6’ has appeared more than once. Hence, we can say that code of letter N is ‘6’. Consequently, we can say that code for letter ‘D’ is ‘1’ because in India rest of the numerals are already taken.

In words, ‘the’ and ‘national’ only letter ‘t’ is common. In code as well only digit ‘8’ is common in two codes. Hence, we can say that letter code for letter ‘t’ is 8.

In words, ‘the’ and ‘peacock’ only letter ‘e’ is common. In code as well only digit ‘5’ is common in two codes. Hence, we can say that letter code for letter ‘e’ is 5. Consequently, we can say that leftover letter, in word “the”, ‘H’s code is 4.

We can see that code for word “NATIONAL” is 13666689. Hence, we can say that code for the letter L is ‘1’.

We can see that code for word “DESIGNATED” is 1135556678. Hence, we can say that code for the letter ‘G’ is ‘7’.

We can see that code for word “PEACOCK” is 5688999. Hence, we can say that code for the letters ‘P’ and ‘K’ is ‘8’.

Let us check this by options:

(A) S,E,Z: If letter ‘Z’ is assigned code ‘5’ then this case is possible.

(B) I,B,M: If letters ‘B’ and ‘M’ are assigned code ‘3’ then this case is possible.

(C) S,U,V: If letters ‘U’ and ‘V’ are assigned code ‘5’ then this case is possible. But in that case digit 5 will have 4 letters associated with it which is not possible. Hence, this is the answer.

(D) X,Y,Z: If letters ‘X’, ‘Y’ and ‘Z’ are assigned code ‘2’ then this case is possible.

Instructions

Four institutes, A, B, C, and D, had contracts with four vendors W, X, Y, and Z during the ten calendar years from 2010 to 2019. The contracts were either multi-year contracts running for several consecutive years or single-year contracts. No institute had more than one contract with the same vendor. However, in a calendar year, an institute may have had contracts with multiple vendors, and a vendor may have had contracts with multiple institutes. It is known that over the decade, the institutes each got into two contracts with two of these vendors, and each vendor got into two contracts with two of these institutes.

The following facts are also known about these contracts.
I. Vendor Z had at least one contract in every year.
II. Vendor X had one or more contracts in every year up to 2015, but no contract in any year after that.
III. Vendor Y had contracts in 2010 and 2019. Vendor W had contracts only in 2012.
IV. There were five contracts in 2012.
V. There were exactly four multi-year contracts. Institute B had a 7-year contract, D had a 4-year contract, and A and C had one 3-year contract each. The other four contracts were single-year contracts.
VI. Institute C had one or more contracts in 2012 but did not have any contract in 2011.
VII. Institutes B and D each had exactly one contract in 2012. Institute D did not have any contract in 2010.

Question 13: In which of the following years were there two or more contracts?

a) 2017

b) 2016

c) 2015

d) 2018

Solution:

From IV: A, B, C, D have one 3, 7, 3, 4-year contract respectively and all other contracts are one-year contracts.

From
I, Z has at least one contract every year, the only possible
combination is 7+3 or 7+4 year contract and that 7-year contract must be
from B.

From III, Vendor W had contracts only in 2012 and
from VII, Institutes B and D each had exactly one contract in 2012
=> W has got contracts from A and C.
From II. Vendor X had one
or more contracts in every year up to 2015, but no contract in any year
after that and from VI, VII: C and D didn’t have any contract in 2011
and 2010 respectively => A should have X as a 3-year contract from
2010-2012. Now, for 2013-2015 X can’t have B for the same. So, X must
have got contracts from either C or D in that period.

Case 1:

X has C as a 3-year contract from 2013-2015 but in this case, D can’t have any contract in 2012 so, this case is not valid.

Case 2:

X
has D for a 4-year contract from 2012-2015 and C must have Z for a
three-year contract in the period 2017-2019 such that Z has at least
one contract every year.

It is known that over the decade, the
institutes each got into two contracts with two of these vendors, and
each vendor got into two contracts with two of these institutes => A
hasn’t got any contract from 2013-2019 as it has X, W in the period
2010-2012 and similarly, C shouldn’t have any contracts in the years 2010,
2013, 2014, 2015, 2016.

From III, Vendor Y had contracts in
2010 and 2019 and in 2010 D and C hasn’t got any contract and A has
already got 2 different contracts from two different vendors => Y has
a contract from B in 2010 => B hasn’t got any contracts in 2017,
2018, 2019.

For Y the only possible contract will be from D => D has got no contracts in the years 2011, 2016, 2017, 2018.

Now, the table looks like: ‘N’ represents no contract.

Out of the given options, only 2015 has two contracts and rest have only one contract in that particular year.

Question 14: Which of the following is true?

a) B had a contract with Z in 2017

b) B had a contract with Y in 2019

c) D had a contract with X in 2011

d) D had a contract with Y in 2019

Solution:

From IV: A, B, C, D have one 3, 7, 3, 4-year contract respectively and all other contracts are one-year contracts.

From I, Z has at least one contract every year, the only possible combination is 7+3 or 7+4 year contract and that 7-year contract must be from B.

From III, Vendor W had contracts only in 2012 and from VII, Institutes B and D each had exactly one contract in 2012 => W has got contracts from A and C.
From II. Vendor X had one or more contracts in every year up to 2015, but no contract in any year after that and from VI, VII: C and D didn’t have any contract in 2011 and 2010 respectively => A should have X as a 3-year contract from 2010-2012. Now, for 2013-2015 X can’t have B for the same. So, X must have got contracts from either C or D in that period.

Case 1:

X has C as a 3-year contract from 2013-2015 but in this case, D can’t have any contract in 2012 so, this case is not valid.

Case 2:

X has D for a 4-year contract from 2012-2015 and C must have Z for a three-year contract in the period 2017-2019 such that Z has at least one contract every year.

It is known that over the decade, the institutes each got into two contracts with two of these vendors, and each vendor got into two contracts with two of these institutes => A hasn’t got any contract from 2013-2019 as it has X, W in the period 2010-2012 and similarly, C shouldn’t have any contracts in the years 2010, 2013, 2014, 2015, 2016.

From III, Vendor Y had contracts in 2010 and 2019 and in 2010 D and C hasn’t got any contract and A has already got 2 different contracts from two different vendors => Y has a contract from B in 2010 => B hasn’t got any contracts in 2017, 2018, 2019.

For Y the only possible contract will be from D => D has got no contracts in the years 2011, 2016, 2017, 2018.

Now, the table looks like: ‘N’ represents no contract.

Option D is true.

Question 15: In how many years during this period was there only one contract?

a) 3

b) 2

c) 4

d) 5

Solution:

From IV: A, B, C, D have one 3, 7, 3, 4-year contract respectively and all other contracts are one-year contracts.

From
I, Z has at least one contract every year, the only possible
combination is 7+3 or 7+4 year contract and that 7-year contract must be
from B.

From III, Vendor W had contracts only in 2012 and
from VII, Institutes B and D each had exactly one contract in 2012
=> W has got contracts from A and C.
From II. Vendor X had one
or more contracts in every year up to 2015, but no contract in any year
after that and from VI, VII: C and D didn’t have any contract in 2011
and 2010 respectively => A should have X as a 3-year contract from
2010-2012. Now, for 2013-2015 X can’t have B for the same. So, X must
have got contracts from either C or D in that period.

Case 1:

X has C as a 3-year contract from 2013-2015 but in this case, D can’t have any contract in 2012 so, this case is not valid.

Case 2:

X
has D for a 4-year contract from 2012-2015 and C must have Z for a
three-year contract in the period 2017-2019 such that Z has at least
one contract every year.

It is known that over the decade, the
institutes each got into two contracts with two of these vendors, and
each vendor got into two contracts with two of these institutes => A
hasn’t got any contract from 2013-2019 as it has X, W in the period
2010-2012 and similarly, C shouldn’t have any contracts in the years 2010,
2013, 2014, 2015, 2016.

From III, Vendor Y had contracts in
2010 and 2019 and in 2010 D and C hasn’t got any contract and A has
already got 2 different contracts from two different vendors => Y has
a contract from B in 2010 => B hasn’t got any contracts in 2017,
2018, 2019.

For Y the only possible contract will be from D => D has got no contracts in the years 2011, 2016, 2017, 2018.

Now, the table looks like: ‘N’ represents no contract.

Only during 2016, 2017 and 2018, there was only one contract.

Question 16: What BEST can be concluded about the number of contracts in 2010?

a) exactly 4

b) exactly 3

c) at least 3

d) at least 4

Solution:

From IV: A, B, C, D have one 3, 7, 3, 4-year contract respectively and all other contracts are one-year contracts.

From
I, Z has at least one contract every year, the only possible
combination is 7+3 or 7+4 year contract and that 7-year contract must be
from B.

From III, Vendor W had contracts only in 2012 and
from VII, Institutes B and D each had exactly one contract in 2012
=> W has got contracts from A and C.
From II. Vendor X had one
or more contracts in every year up to 2015, but no contract in any year
after that and from VI, VII: C and D didn’t have any contract in 2011
and 2010 respectively => A should have X as a 3-year contract from
2010-2012. Now, for 2013-2015 X can’t have B for the same. So, X must
have got contracts from either C or D in that period.

Case 1:

X has C as a 3-year contract from 2013-2015 but in this case, D can’t have any contract in 2012 so, this case is not valid.

Case 2:

X
has D for a 4-year contract from 2012-2015 and C must have Z for a
three-year contract in the period 2017-2019 such that Z has at least
one contract every year.

It is known that over the decade, the
institutes each got into two contracts with two of these vendors, and
each vendor got into two contracts with two of these institutes => A
hasn’t got any contract from 2013-2019 as it has X, W in the period
2010-2012 and similarly, C shouldn’t have any contracts in the years 2010,
2013, 2014, 2015, 2016.

From III, Vendor Y had contracts in
2010 and 2019 and in 2010 D and C hasn’t got any contract and A has
already got 2 different contracts from two different vendors => Y has
a contract from B in 2010 => B hasn’t got any contracts in 2017,
2018, 2019.

For Y the only possible contract will be from D => D has got no contracts in the years 2011, 2016, 2017, 2018.

Now, the table looks like: ‘N’ represents no contract.

The Number of contracts in 2010 is three.

Question 17: Which institutes had multiple contracts during the same year?

a) A only

b) B and C only

c) A and B only

d) B only

Solution:

From IV: A, B, C, D have one 3, 7, 3, 4-year contract respectively and all other contracts are one-year contracts.\

From
I, Z has at least one contract every year, the only possible
combination is 7+3 or 7+4 year contract and that 7-year contract must be
from B.

From III, Vendor W had contracts only in 2012 and
from VII, Institutes B and D each had exactly one contract in 2012
=> W has got contracts from A and C.
From II. Vendor X had one
or more contracts in every year up to 2015, but no contract in any year
after that and from VI, VII: C and D didn’t have any contract in 2011
and 2010 respectively => A should have X as a 3-year contract from
2010-2012. Now, for 2013-2015 X can’t have B for the same. So, X must
have got contracts from either C or D in that period.

Case 1:

X has C as a 3-year contract from 2013-2015 but in this case, D can’t have any contract in 2012 so, this case is not valid.

Case 2:

X
has D for a 4-year contract from 2012-2015 and C must have Z for a
three-year contract in the period 2017-2019 such that Z has at least
one contract every year.

It is known that over the decade, the
institutes each got into two contracts with two of these vendors, and
each vendor got into two contracts with two of these institutes => A
hasn’t got any contract from 2013-2019 as it has X, W in the period
2010-2012 and similarly, C shouldn’t have any contracts in years 2010,
2013, 2014, 2015, 2016.

From III, Vendor Y had contracts in
2010 and 2019 and in 2010 D and C hasn’t got any contract and A has
already got 2 different contracts from two different vendors => Y has
a contract from B in 2010 => B hasn’t got any contracts in 2017,
2018, 2019.

For Y the only possible contract will be from D => D has got no contracts in the years 2011, 2016, 2017, 2018.

Now, the table looks like: ‘N’ represents no contract.

B and A have multiple contracts in a single year.

Question 18: Which institutes and vendors had more than one contracts in any year?

a) B, W, X, and Z

b) A, B, W, and X

c) A, D, W, and Z

d) B, D, W, and X

Solution:

From IV: A, B, C, D have one 3, 7, 3, 4-year contract respectively and all other contracts are one-year contracts.\

From
I, Z has at least one contract every year, the only possible
combination is 7+3 or 7+4 year contract and that 7-year contract must be
from B.

From III, Vendor W had contracts only in 2012 and
from VII, Institutes B and D each had exactly one contract in 2012
=> W has got contracts from A and C.
From II. Vendor X had one
or more contracts in every year up to 2015, but no contract in any year
after that and from VI, VII: C and D didn’t have any contract in 2011
and 2010 respectively => A should have X as a 3-year contract from
2010-2012. Now, for 2013-2015 X can’t have B for the same. So, X must
have got contracts from either C or D in that period.

Case 1:

X has C as a 3-year contract from 2013-2015 but in this case, D can’t have any contract in 2012 so, this case is not valid.

Case 2:

X
has D for a 4-year contract from 2012-2015 and C must have Z for a
three-year contract in the period 2017-2019 such that Z has at least
one contract every year.

It is known that over the decade, the
institutes each got into two contracts with two of these vendors, and
each vendor got into two contracts with two of these institutes => A
hasn’t got any contract from 2013-2019 as it has X, W in the period
2010-2012 and similarly, C shouldn’t have any contracts in years 2010,
2013, 2014, 2015, 2016.

From III, Vendor Y had contracts in
2010 and 2019 and in 2010 D and C hasn’t got any contract and A has
already got 2 different contracts from two different vendors => Y has
a contract from B in 2010 => B hasn’t got any contracts in 2017,
2018, 2019.

For Y the only possible contract will be from D => D has got no contracts in the years 2011, 2016, 2017, 2018.

Now, the table looks like: ‘N’ represents no contract.

A, B, W, X had more than one contracts in a single year

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