CAT 2023 Question Paper Slot 1 with Solutions, Download PDF

Instructions

Faculty members in a management school can belong to one of four departments - Finance and Accounting (F&A), Marketing and Strategy (M&S), Operations and Quants (O&Q) and Behaviour and Human Resources (B&H). The numbers of faculty members in F&A, M&S, O&Q and B&H departments are 9, 7, 5 and 3 respectively.

Prof. Pakrasi, Prof. Qureshi, Prof. Ramaswamy and Prof. Samuel are four members of the school's faculty who were candidates for the post of the Dean of the school. Only one of the candidates was from O&Q.

Every faculty member, including the four candidates, voted for the post. In each department, all the faculty members who were not candidates voted for the same candidate. The rules for the election are listed below.

1. There cannot be more than two candidates from a single department.
2. A candidate cannot vote for himself/herself.
3. Faculty members cannot vote for a candidate from their own department.

After the election, it was observed that Prof. Pakrasi received 3 votes, Prof. Qureshi received 14 votes, Prof. Ramaswamy received 6 votes and Prof. Samuel received 1 vote. Prof. Pakrasi voted for Prof. Ramaswamy, Prof. Qureshi for Prof. Samuel, Prof. Ramaswamy for Prof. Qureshi and Prof. Samuel for Prof. Pakrasi.

Question 41

Which of the following can be the number of votes that Prof. Qureshi received from a single department?

Video Solution

Solution

Now, we know there is only 1 candidate from OQ, which means that the number of non-candidate voters in OQ will be 4.

We also know that the non-candidates in a particular department voted as a block, and we also know that the least number of non-candidate voters in a particular department can be 1 (BH, 3-2 faculty).

Now, we also know that R got 5 votes from non-candidates.

Now we can write 5 as

i)5

ii)4+1

iii)3+2

Considering case (i) 4+1. This is only possible when there is 1 candidate from OQ, and there are 2 candidates from BH. This implies that the number of candidates in FA and MQ is 1. Now, if we consider FA and MQ and put only 1 candidate there, it implies that there are 15 non-candidate voters between them. Now, we know this is not possible since the maximum number of non-candidate voters a candidate can get is 13. (Please note that non-candidates of a particular department vote as a block).

On similar grounds, we can eliminate Case (iii) as it also implies there is only 1 candidate in FA and MQ.

Now, considering Case (i), we know that 5+0 will happen only once when there are 5 non-candidates in a single department. This is only possible in MS (Out of 7, there will be 2 candidates and 5 non-candidates).

So we can conclude that MS has 2 candidates and that they voted Prof. R…….(i)

We also know that Prof P got 2 votes from Non-candidates. This is only possible when BH has 1 candidate.

So, we can conclude that the number of professors in FA, MS, OQ, BH is 0,2,1,1

Thus, we get the following table:

Now, if we consider Department MS, we know that there are 2 candidates from MS and R can’t be one of them as the people in that department voted for him….. (3 rd condition).

So, the possible combinations of candidates in MS are (P,Q), (Q,S), (P,S).

Now we also know that no one can vote for a candidate in their own department, so we can eliminate (P,S) and (Q,S) as we know that S voted for P and Q voted for S).

So, we can infer that P and Q are from MS.

Now, we can see that the number of votes that Prof Qureshi received from a single department can be 9 or 5 (if R is from OQ) or 4 (if R is not from OQ).

So, among the options, only Option D can be true. Therefore, Option D is the correct answer.

Question 42

If Prof. Samuel belongs to B&H, which of the following statements is/are true?
Statement A: Prof. Pakrasi belongs to M&S.
Statement B: Prof. Ramaswamy belongs to O&Q

Video Solution

Solution

Now, we know there is only 1 candidate from OQ, which means that the number of non-candidate voters in OQ will be 4.

We also know that the non-candidates in a particular department voted as a block, and we also know that the least number of non-candidate voters in a particular department can be 1 (BH, 3-2 faculty).

Now, we also know that R got 5 votes from non-candidates.

Now we can write 5 as

i)5

ii)4+1

iii)3+2

Considering case (i) 4+1. This is only possible when there is 1 candidate from OQ, and there are 2 candidates from BH. This implies that the number of candidates in FA and MQ is 1. Now, if we consider FA and MQ and put only 1 candidate there, it implies that there are 15 non-candidate voters between them. Now we know this is not possible since the maximum number of non-candidate voters a candidate can get is 13. (Please note that non-candidates of a particular department vote as a block).

On similar grounds, we can eliminate Case (iii) as it also implies there is only 1 candidate in FA and MQ.

Now, considering Case (i), we know that 5+0 will happen only one when there are 5 non-candidates in a single department. This is only possible in MS (Out of 7, there will be 2 candidates and 5 non-candidates).

So we can conclude that MS has 2 candidates and that they voted Prof. R…….(i)

We also know that Prof P got 2 votes from Non-candidates. This is only possible when BH has 1 candidate.

So, we can conclude that the number of professors in FA, MS, OQ, BH is 0,2,1,1

Thus, we get the following table:

Now, if we consider Department MS, we know that there are 2 candidates from MS and R can’t be one of them as the people in that department voted for him….. (3 rd condition).

So the possible combinations of candidates in MS are (P,Q), (Q,S), (P,S).

Now we also know that no one can vote for a candidate in their own department, so we can eliminate (P,S) and (Q,S) as we know that S voted for P and Q voted for S).

So we can infer that P and Q are from MS.

We have been told Prof Samuel belongs to B&H. So we have to consider only Case 1. In Case 1 we can see that Prof Prakash belongs to MS and Prof Ramaswamy belongs to OQ.

Therefore, both the statements are true.

So, the correct answer is Option D

Question 43

What best can be concluded about the candidate from O&Q?

Video Solution

Solution

Now, we know there is only 1 candidate from OQ, which means that the number of non-candidate voters in OQ will be 4.

We also know that the non-candidates in a particular department voted as a block, and we also know that the least number of non-candidate voters in a particular department can be 1 (BH, 3-2 faculty).

Now, we also know that R got 5 votes from non-candidates.

Now we can write 5 as

i)5

ii)4+1

iii)3+2

Considering case (i) 4+1. This is only possible when there is 1 candidate from OQ, and there are 2 candidates from BH. This implies that the number of candidates in FA and MQ is 1. Now, if we consider FA and MQ and put only 1 candidate there, it implies that there are 15 non-candidate voters between them. Now we know this is not possible since the maximum number of non-candidate voters a candidate can get is 13. (Please note that non-candidates of a particular department vote as a block).

On similar grounds, we can eliminate Case (iii) as it also implies there is only 1 candidate in FA and MQ.

Now, considering Case (i), we know that 5+0 will happen only one when there are 5 non-candidates in a single department. This is only possible in MS (Out of 7, there will be 2 candidates and 5 non-candidates).

So we can conclude that MS has 2 candidates and that they voted Prof. R…….(i)

We also know that Prof P got 2 votes from Non-candidates. This is only possible when BH has 1 candidate.

So, we can conclude that the number of professors in FA, MS, OQ, BH is 0,2,1,1

Thus, we get the following table:

Now, if we consider Department MS, we know that there are 2 candidates from MS and R can’t be one of them as the people in that department voted for him….. (3 rd condition).

So the possible combinations of candidates in MS are (P,Q), (Q,S), (P,S).

Now we also know that no one can vote for a candidate in their own department, so we can eliminate (P,S) and (Q,S) as we know that S voted for P and Q voted for S).

So we can infer that P and Q are from MS.

From the 2 cases, we can see that the candidate from OQ can either be Prof Ramaswamy or Prof Samuel.

Therefore, the correct answer is Option B.

Question 44

Which of the following statements is/are true?

Statement A: Non-candidates from M&S voted for Prof. Qureshi.
Statement B: Non-candidates from F&A voted for Prof. Qureshi.

Video Solution

Solution

Now, we know there is only 1 candidate from OQ, which means that the number of non-candidate voters in OQ will be 4.

We also know that the non-candidates in a particular department voted as a block, and we also know that the least number of non-candidate voters in a particular department can be 1 (BH, 3-2 faculty).

Now, we also know that R got 5 votes from non-candidates.

Now we can write 5 as

i)5

ii)4+1

iii)3+2

Considering case (i) 4+1. This is only possible when there is 1 candidate from OQ, and there are 2 candidates from BH. This implies that the number of candidates in FA and MQ is 1. Now, if we consider FA and MQ and put only 1 candidate there, it implies that there are 15 non-candidate voters between them. Now we know this is not possible since the maximum number of non-candidate voters a candidate can get is 13. (Please note that non-candidates of a particular department vote as a block).

On similar grounds, we can eliminate Case (iii) as it also implies there is only 1 candidate in FA and MQ.

Now, considering Case (i), we know that 5+0 will happen only one when there are 5 non-candidates in a single department. This is only possible in MS (Out of 7, there will be 2 candidates and 5 non-candidates).

So we can conclude that MS has 2 candidates and that they voted Prof. R…….(i)

We also know that Prof P got 2 votes from Non-candidates. This is only possible when BH has 1 candidate.

So, we can conclude that the number of professors in FA, MS, OQ, BH is 0,2,1,1

Thus, we get the following table:

Now, if we consider Department MS, we know that there are 2 candidates from MS and R can’t be one of them as the people in that department voted for him….. (3 rd condition).

So the possible combinations of candidates in MS are (P,Q), (Q,S), (P,S).

Now we also know that no one can vote for a candidate in their own department, so we can eliminate (P,S) and (Q,S) as we know that S voted for P and Q voted for S).

So we can infer that P and Q are from MS.

Since Prof Qureshi belongs to MS, non-candidates from MS can't vote for him. We can see that the non-candidates from FA voted for him. So, only statement B is true. Therefore, the correct answer is Option B.

Instructions

For the following questions answer them individually

Question 45

Let n be the least positive integer such that 168 is a factor of $$1134^{n}$$. If m is the least positive integer such that $$1134^{n}$$ is a factor of $$168^{m}$$, then m + n equals

Video Solution

Solution

Prime Factorising 1134, we get 1134 = $$2\times\ 3^4\times\ 7$$ and 168 = $$2^3\times\ 3\times\ 7$$

$$1134^n$$ is a factor of 168 => the factor of 2 should be atleast 3, for 168 to be a factor => n = 3.

Now, $$1134^n$$ = $$1134^3=2^3\times\ 3^{12}\times\ 7^3$$ is a factor of $$168^m=\left(2^3\times\ 3\times\ 7\right)^m$$ => m = 12 as power of 3 should be atleast 12.

=> So, m + n = 15.

Question 46

If $$x$$ and $$y$$ are positive real numbers such that $$\log_{x}(x^2 + 12) = 4$$ and $$3 \log_{y} x = 1$$, then $$x + y $$ equals

Video Solution

Solution

Given, $$\log_{x}(x^2 + 12) = 4$$

=> $$x^2+12=x^4$$

=> $$x^4-x^2-12=0$$

=> $$x^4-4x^2+3x^2-12=0$$

=> $$x^2\left(x^2-4\right)+3\left(x^2-4\right)=0$$

=> $$\left(x^2-4\right)\left(x^2+3\right)=0$$ => since, x is a positive real number (given) => x = 2.

Now, Given $$3 \log_{y} x = 1$$

=> $$\log_yx=\frac{1}{3}$$

=> $$x=y^{\frac{1}{3}}$$

=> $$y=x^3$$ => y = 8.

=> x + y = 2 + 8 = 10.

Question 47

If $$\sqrt{5x+9} + \sqrt{5x - 9} = 3(2 + \sqrt{2})$$, then $$\sqrt{10x+9}$$ is equal to

Video Solution

Solution

Given, $$\sqrt{5x+9} + \sqrt{5x - 9} = 3(2 + \sqrt{2})$$

=> $$\sqrt{\ 5x+9}+\sqrt{\ 5x-9}=6+3\sqrt{\ 2}$$

=> $$\sqrt{\ 5x+9}+\sqrt{\ 5x-9}=\sqrt{\ 36}+\sqrt{\ 18}$$

Comparing the L.H.S. and R.H.S.

=> $$5x+9=36\ $$ => $$5x=27$$ => $$x=\dfrac{27}{5}$$ (can be verified using the second term as well).

=> $$\sqrt{10x+9}$$ = $$\sqrt{\left(10\times\dfrac{27}{5}\right)+9}$$ = $$\sqrt{\ 63}=3\sqrt{\ 7}$$

Question 48

If $$x$$ and $$y$$ are real numbers such that $$x^{2} + (x - 2y - 1)^{2} = -4y(x + y)$$, then the value $$x - 2y$$ is

Video Solution

Solution

Given, $$x^{2} + (x - 2y - 1)^{2} = -4y(x + y)$$

=> $$x^2+4xy+4y^2+\left(x-2y-1\right)^2=0$$

=> $$\left(x+2y\right)^2+\left(x-2y-1\right)^2=0$$

For the L.H.S. of the equation to be 0, each of the square terms should be 0 (as squares cannot be negative)

=> x - 2y - 1 = 0 => x - 2y = 1

Question 49

The number of integer solutions of equation $$2|x|(x^{2}+1) = 5x^{2}$$ is

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Video Solution

Solution

Let us consider 3 cases:

1) x = 0, This is a solution, as both L.H.S and R.H.S will be equal (0) when x = 0. (1 solution)

2) x > 0

=> $$2x\left(x^2+1\right)=5x^2$$

=> $$2\left(x^2+1\right)=5x$$

=> $$2x^2-5x+2=0$$ => $$2x^2-4x-x-2=0$$

=> $$2x\left(x-2\right)-1\left(x-2\right)=0$$

=> $$\left(x-2\right)\left(2x-1\right)=0$$ => x = 2 or 1/2 => (1 integer solution)

3) x < 0

=> $$-2x\left(x^2+1\right)=5x^2$$

=> $$2x^2+5x+2=0$$

=> $$2x^2+4x+x+2=0$$

=> $$2x\left(x+2\right)+1\left(x+2\right)=0$$

=> $$\left(x+2\right)\left(2x+1\right)=0$$ => x = -2 or -1/2 => (1 integer solution)

So, the total number of integer solutions are 0, 2, -2 => 3.

Question 50

The equation $$x^{3} + (2r + 1)x^{2} + (4r - 1)x + 2 =0$$ has -2 as one of the roots. If the other two roots are real, then the minimum possible non-negative integer value of r is

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Video Solution

Solution

Given that -2 is a root of the given cubic equation.

=> Dividing the given equation by (x + 2), Using the Horners method of synthetic division:

coefficient of $$x^2$$ is 1, and coefficient of x is (2r+1)-2 = 2r-1 and the constant term = (4r-1)-2(2r-1) = 1.

=> The quadratic obtained by dividing the cubic = $$x^2+\left(2r-1\right)x+1=0$$, Since, this equation has 2 real roots => Discriminant should be greater than 0

=> $$\left(2r-1\right)^2>4$$ => 2r-1 > 2 or 2r-1 < -2 => r > 3/2 or r < -1/2.

=> Minimum possible non-negative integer value of r is 2.

CAT Content

CAT 2023 Question Paper Slot 1

Which of the following can be the number of votes that Prof. Qureshi received from a single department?

If Prof. Samuel belongs to B&H, which of the following statements is/are true?
Statement A: Prof. Pakrasi belongs to M&S.
Statement B: Prof. Ramaswamy belongs to O&Q

What best can be concluded about the candidate from O&Q?

Which of the following statements is/are true?

Statement A: Non-candidates from M&S voted for Prof. Qureshi.
Statement B: Non-candidates from F&A voted for Prof. Qureshi.

Let n be the least positive integer such that 168 is a factor of $$1134^{n}$$. If m is the least positive integer such that $$1134^{n}$$ is a factor of $$168^{m}$$, then m + n equals

If $$x$$ and $$y$$ are positive real numbers such that $$\log_{x}(x^2 + 12) = 4$$ and $$3 \log_{y} x = 1$$, then $$x + y $$ equals

If $$\sqrt{5x+9} + \sqrt{5x - 9} = 3(2 + \sqrt{2})$$, then $$\sqrt{10x+9}$$ is equal to

If $$x$$ and $$y$$ are real numbers such that $$x^{2} + (x - 2y - 1)^{2} = -4y(x + y)$$, then the value $$x - 2y$$ is

The number of integer solutions of equation $$2|x|(x^{2}+1) = 5x^{2}$$ is

The equation $$x^{3} + (2r + 1)x^{2} + (4r - 1)x + 2 =0$$ has -2 as one of the roots. If the other two roots are real, then the minimum possible non-negative integer value of r is

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CAT Content

CAT 2023 Question Paper Slot 1

Which of the following can be the number of votes that Prof. Qureshi received from a single department?

If Prof. Samuel belongs to B&H, which of the following statements is/are true?Statement A: Prof. Pakrasi belongs to M&S.Statement B: Prof. Ramaswamy belongs to O&Q

What best can be concluded about the candidate from O&Q?

Which of the following statements is/are true? Statement A: Non-candidates from M&S voted for Prof. Qureshi.Statement B: Non-candidates from F&A voted for Prof. Qureshi.

Let n be the least positive integer such that 168 is a factor of $$1134^{n}$$. If m is the least positive integer such that $$1134^{n}$$ is a factor of $$168^{m}$$, then m + n equals

If $$x$$ and $$y$$ are positive real numbers such that $$\log_{x}(x^2 + 12) = 4$$ and $$3 \log_{y} x = 1$$, then $$x + y $$ equals

If $$\sqrt{5x+9} + \sqrt{5x - 9} = 3(2 + \sqrt{2})$$, then $$\sqrt{10x+9}$$ is equal to

If $$x$$ and $$y$$ are real numbers such that $$x^{2} + (x - 2y - 1)^{2} = -4y(x + y)$$, then the value $$x - 2y$$ is

The number of integer solutions of equation $$2|x|(x^{2}+1) = 5x^{2}$$ is

The equation $$x^{3} + (2r + 1)x^{2} + (4r - 1)x + 2 =0$$ has -2 as one of the roots. If the other two roots are real, then the minimum possible non-negative integer value of r is

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Related Formulas With Tests

CAT Quant Questions | CAT Quantitative Ability

CAT DILR Questions | LRDI Questions For CAT

CAT Verbal Ability Questions | VARC Questions For CAT

Also Read

If Prof. Samuel belongs to B&H, which of the following statements is/are true?
Statement A: Prof. Pakrasi belongs to M&S.
Statement B: Prof. Ramaswamy belongs to O&Q

Which of the following statements is/are true?

Statement A: Non-candidates from M&S voted for Prof. Qureshi.
Statement B: Non-candidates from F&A voted for Prof. Qureshi.