Section-Wise Breakdown of CAT 2023 Question Paper
The CAT 2023 question paper had a lot of differences when compared with the previous year’s CAT question papers. The proper approach to understanding the CAT 2023 question paper is through proper CAT 2023 exam analysis through a section-wise breakdown. Let us now discuss the CAT 2023 exam paper section-wise breakdown.
VARC (Verbal Ability and Reading Comprehension):
Number of Questions: 24
Time Allotted: 40 minutes
Difficulty Level: Easy to Moderate and slightly easier than last year
The Verbal Ability and Reading Comprehension section of the CAT 2023 question paper had 24 questions with a balance of CAT questions testing both verbal skills and reading ability questions of the candidates. The presence of the odd one-out questions was one deviation observed in the CAT 2023 question paper when compared with the CAT previous year’s question papers. The various questions and their difficulty level in the VARC section of the CAT 2023 question paper are given below:
CAT SLOT 1 |
Section | No of Qs. | Difficulty Level |
RC-1: Return of wolves' predation in a French region | 4 | Easy |
RC-2: Current Economies of North and South Korea | 4 | Easy |
RC-3: Indian Ocean Literature | 4 | Medium |
RC-4: Modern materialism versus ancient societies | 4 | Difficult |
CAT SLOT 2 |
RC - 1: Second-hand Shopping and Fast Fashion | 4 | Easy |
RC-2: Translated Streaming in Europe | 4 | Easy |
RC - 3: Why Liberalism Failed (Book Review) | 4 | Medium |
RC - 4: What is a Historical Fact | 4 | Difficult |
CAT SLOT 3 |
RC Passage | Questions | Difficulty |
RC - 1: Colonialism and Global Warming (The Nutmeg’s Curse) | 4 | Easy |
RC - 2: Understanding Romantic Aesthetics | 4 | Easy |
RC - 3: Cultural Patrimony Laws | 4 | Medium |
RC - 4: Rationality by Steven Pinker | 4 | Difficult |
CAT Question Paper 2023: VARC Questions
DIRECTIONS for the question: The passage below is accompanied by a set of questions. Choose the best answer to each question.
The Positivists, anxious to stake out their claim for history as a science, contributed the weight of their influence to the cult of facts. First, ascertain the facts, said the positivists, then draw your conclusions from them…..This is what may [be] called the common-sense view of history. History consists of a corpus of ascertained facts. The facts are available to the historian in documents, inscriptions, and so on…[Sir George Clark] contrasted the "hard core of facts" in history with the surrounding pulp of disputable interpretation forgetting perhaps that the pulpy part of the fruit is more rewarding than the hard core.…. It recalls the favourite dictum of the great liberal journalist C. P. Scott: "Facts are sacred, opinion is free.". . .
What is a historical fact?..... According to the common-sense view, there are certain basic facts which are the same for all historians and which form, so to speak, the backbone of history—the fact, for example, that the Battle of Hastings was fought in 1066. But this view calls for two observations. In the first place, it is not with facts like these that the historian is primarily concerned. It is no doubt important to know that the great battle was fought in 1066 and not in 1065 or 1067, and that it was fought at Hastings and not at Eastbourne or Brighton. The historian must not get these things wrong. But [to] praise a historian for his accuracy is like praising an architect for using well-seasoned timber or properly mixed concrete in his building. It is a necessary condition of his work, but not his essential function. It is precisely for matters of this kind that the historian is entitled to rely on what have been called the "auxiliary sciences" of history—archaeology, epigraphy, numismatics, chronology, and so forth….
The second observation is that the necessity to establish these basic facts rests not on any quality in the facts themselves, but on an apriori decision of the historian. In spite of C. P. Scott's motto, every journalist knows today that the most effective way to influence opinion is by the selection and arrangement of the appropriate facts. It used to be said that facts speak for themselves. This is, of course, untrue. The facts speak only when the historian calls on them: it is he who decides to which facts to give the floor, and in what order or context. The only reason why we are interested to know that the battle was fought at
Hastings in 1066 is that historians regard it as a major historical event… Professor Talcott Parsons once called [science] "a
selective system of cognitive orientations to reality." It might perhaps have been put more simply. But history is, among other things, that. The historian is necessarily selective. The belief in a hard core of historical facts existing objectively and independently of the interpretation of the historian is a preposterous fallacy, but one which it is very hard to eradicate.
Question No. : 1
According to this passage, which one of the following statements best describes the significance of archaeology for historians?
A) Archaeology helps historians to interpret historical facts.
B) Archaeology helps historians to ascertain factual accuracy.
C) Archaeology helps historians to carry out their primary duty.
D) Archaeology helps historians to locate the oldest civilisations in history.
Solution:
Answer 2: The author in the passage explains that while historians depend on auxiliary sciences like archaeology to establish fundamental facts (such as the date and location of historical events), this accuracy is a prerequisite rather than the essence of their work. Praising historians for accuracy is akin to praising an architect merely for using good materials, which is necessary but not the core of their role. Thus, option 2 best aligns with this view.
Question No. : 2
All of the following, if true, can weaken the passage’s claim that facts do not speak for themselves, EXCEPT:
A) the truth value of a fact is independent of the historian who expresses it.
B) facts, like truth, can be relative: what is fact for person X may not be so for person Y.
C) a fact, by its very nature, is objective and universal, irrespective of the context in which it is placed.
D) the order in which a series of facts is presented does not have any bearing on the production of meaning.
Solution:
Answer 2: Option 2 is correct because it aligns with the passage's argument that the interpretation of facts can be influenced by different perspectives, suggesting that facts, much like truth, can be seen as relative.
Question No. : 3
If the author of the passage were to write a book on the Battle of Hastings along the lines of his/her own reasoning, the focus of the historical account would be on:
A) providing a nuanced interpretation by relying on the auxiliary sciences.
B) producing a detailed timeline of the various events that led to the Battle.
C) exploring the socio-political and economic factors that led to the Battle.
D) deriving historical facts from the relevant documents and inscriptions.
Solution:
Answer 3: The passage asserts that historians should go beyond mere fact-finding to understand the context and motivations behind historical events. This comprehensive and interpretive approach is reflected in Option 3, which emphasizes understanding the underlying causes and influences.
Question No. : 4
All of the following describe the “common-sense view” of history, EXCEPT:
A) history can be objective like the sciences if it is derived from historical facts.
B) real history can be found in ancient engravings and archival documents.
C) history is like science: a selective system of cognitive orientations to reality.
D) only the positivist methods can lead to credible historical knowledge.
Solution:
Answer 3: Option 3 is the best choice because it highlights the fallacious belief in the objectivity of historical facts, which contrasts with the passage's critique of the oversimplified view of historical methodology as purely positivist.
DIRECTIONS for the question: The four sentences (labelled 1, 2, 3, and 4) given in this question, when properly sequenced, form a coherent paragraph. Decide on the proper order for the sentence and key in this sequence of four numbers as your answer.
Sentences:
Like the ants that make up a colony, no single neuron holds complex information like self-awareness, hope, or pride.
Although the human brain is not yet understood enough to identify the mechanism by which emergence functions, most neurobiologists agree that complex interconnections among the parts give rise to qualities that belong only to the whole.
Nonetheless, the sum of all neurons in the nervous system generates complex human emotions like fear and joy, none of which can be attributed to a single neuron.
Human consciousness is often called an emergent property of the human brain.
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Solution:
Answer 4132: Sequence 4132 logically progresses from introducing the concept of human consciousness as an emergent property, elaborating on neuron organisation, and discussing the collective contribution of neurons to complex mental processes.
DILR (Data Interpretation and Logical Reasoning):
Number of Questions: 20
Time Allotted: 40 minutes
Difficulty Level: Slightly more difficult than 2022
The Data Interpretation and Logical Reasoning (DILR) section of the CAT 2023 question paper followed the same format as the CAT 2022 question paper. The number of questions under this section was set to be 20. The difficultly level of the DILR section however was observed to be slightly more difficult than the CAT 2022 question paper. There were 4 sets of questions with each set containing 5 questions. The analytical skills of the candidates were put to the test in this section. The various CAT 2023 DILR questions are given below:
CAT 2023 SLOT 1 |
Set No. | Area | Set Descriptions | No. of Qs | LOD | Q. Type |
1 | LR | (Arrangements) Professors in depart- ments | 5 | Easy to Medium | 5 MCQs |
2 | DI | (Reasoning based DI) Gig workers and Restaurants, Mathematical Reasoning Involved Measures of central tendency | 5 | Medium to Difficult | 1MCQ+4 TITA |
3 | LR | Houses on sale | 5 | Medium to Difficult | 3MCQ+2 TITA |
4 | DI | Visa Office Special Equation handling | 5 | Medium to Difficult | 3MCQ+2 TITA |
CAT 2023 SLOT 2 |
1 | LR | 5 firms - Funds raised in consecutive years | 5 | 4 MCQ, 2 TITA | Easy to Medium |
2 | DI | Draw competition | 5 | 1 MCQ, 4 TITA | Medium to Difficult |
3 | LR | 3 friends visited a park. Many rides. Scheduling questions. | 5 | 2 MCQ, 3 TITA | Medium to Difficult |
4 | DI | Nine Boxes - sudoku-type puzzle | 5 | 3 MCQ, 2 TITA | Medium to Difficult |
CAT 2023 SLOT 3 |
1 | DI | Number of enrollments in offline and online courses | 5 | 3MCQ, 2 TITA | Easy to Medium |
2 | DI | Six Police Stations with map (route based) | 5 | 3 MCQ, 2 TITA | Medium |
3 | DI | Projects and scores of 3 boys and 3 girls | 5 | 3MCQ, 2 TITA | Medium to Difficult |
4 | LR | AC Company - 4 Dealers | 5 | 3MCQ, 2 TITA | Medium to Difficult |
CAT Question Paper 2023: DILR Questions
Q1- Q5: 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 1. Which two candidates can belong to the same department?
Prof. Pakrasi and Prof. Samuel
Prof. Pakrasi and Prof. Qureshi
Prof. Qureshi and Prof. Ramaswamy
Prof. Ramaswamy and Prof. Samuel
Correct Answer: B
Solution:
Here Take the first letter of the name of professor to represent the professor.
Given,
Department | No of Faculty members | Candidate for Dean | No of non-candidate voters |
FA | 9 | 1 or 2 | 8 or 7 |
MS | 7 | 1 or 2 | 6 or 5 |
OQ | 5 | 1 | 4 |
BH | 3 | 1 or 2 | 2 or 1 |
We know that the non-candidates in a particular department voted for the same candidate, and we also know that the least number of non-candidate voters in a particular department can be 1.
R got total 6 votes, out of which 1 is from the dean candidate P and 5 from non-candidates.
from above table we can see that 5 can be formed by (5 + 0) or (4 +1).
If we take 2nd case: It is only possible when there is 1 candidate from OQ, and there are 2 candidates from BH, which implies that number of candidates in FA and MQ is 1. But P received 2 non candidate votes only which is possible only if BH has 2 non candidate faculty and 1 candidate. So, this case is not possible.
Now, considering 1st case (a), we get that 5+0 will happen only one when there are 5 non-candidates in a single department.
This can be possible only in MS (where out of 7, there will be 2 candidates and 5 non-candidates)
So, we get the following table:
Department | No of Faculty members | Candidate for Dean | No of non-candidate voters |
FA | 9 | 0 | 9 |
MS | 7 | 2 | 5 |
OQ | 5 | 1 | 4 |
BH | 3 | 1 | 2 |
We also can draw
Candidate | Total Votes | Candidate Votes | Non-Candidate Vote |
P | 3 | 1(S) | 2 (BH) |
Q | 14 | 1(R) | 13 (FA and OQ) |
R | 6 | 1(P) | 5 (MS) |
S | 1 | 1(Q) | 0 |
Let's 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)
Now, the possible combinations of candidates in MS are (P,Q), (Q,S) and (P,S).
Now , we can get that P and Q are from MS.
Question 2. Which of the following can be the number of votes that Prof. Qureshi received from a single department?
8
7
9
6
Correct Answer: C
Solution:
Here Take the first letter of the name of professor to represent the professor.
Given,
Department | No of Faculty members | Candidate for Dean | No of non-candidate voters |
FA | 9 | 1 or 2 | 8 or 7 |
MS | 7 | 1 or 2 | 6 or 5 |
OQ | 5 | 1 | 4 |
BH | 3 | 1 or 2 | 2 or 1 |
We know that the non-candidates in a particular department voted for the same candidate, and we also know that the least number of non-candidate voters in a particular department can be 1.
R got total 6 votes, out of which 1 is from the dean candidate P and 5 from non-candidates.
from above table we can see that 5 can be formed by (5 + 0) or (4 +1).
If we take 2nd case: It is only possible when there is 1 candidate from OQ, and there are 2 candidates from BH, which implies that number of candidates in FA and MQ is 1. But P received 2 non candidate votes only which is possible only if BH has 2 non candidate faculty and 1 candidate. So, this case is not possible.
Now, considering 1st case (a), we get that 5+0 will happen only one when there are 5 non-candidates in a single department.
This can be possible only in MS (where out of 7, there will be 2 candidates and 5 non-candidates)
So, we get the following table:
Department | No of Faculty members | Candidate for Dean | No of non-candidate voters |
FA | 9 | 0 | 9 |
MS | 7 | 2 | 5 |
OQ | 5 | 1 | 4 |
BH | 3 | 1 | 2 |
Question 3. 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.
Both statements A and B
Only statement B
Only statement A
Neither statement A nor statement B
Correct Answer: A
Solution:
Here Take the first letter of the name of professor to represent the professor.
Given,
Department | No of Faculty members | Candidate for Dean | No of non-candidate voters |
FA | 9 | 1 or 2 | 8 or 7 |
MS | 7 | 1 or 2 | 6 or 5 |
OQ | 5 | 1 | 4 |
BH | 3 | 1 or 2 | 2 or 1 |
We know that the non-candidates in a particular department voted for the same candidate, and we also know that the least number of non-candidate voters in a particular department can be 1.
R got total 6 votes, out of which 1 is from the dean candidate P and 5 from non-candidates.
from above table we can see that 5 can be formed by (5 + 0) or (4 +1).
If we take 2nd case: It is only possible when there is 1 candidate from OQ, and there are 2 candidates from BH, which implies that number of candidates in FA and MQ is 1. But P received 2 non candidate votes only which is possible only if BH has 2 non candidate faculty and 1 candidate. So, this case is not possible.
Now, considering 1st case (a), we get that 5+0 will happen only one when there are 5 non-candidates in a single department.
This can be possible only in MS (where out of 7, there will be 2 candidates and 5 non-candidates)
So, we get the following table:
Department | No of Faculty members | Candidate for Dean | No of non-candidate voters |
FA | 9 | 0 | 9 |
MS | 7 | 2 | 5 |
OQ | 5 | 1 | 4 |
BH | 3 | 1 | 2 |
We also can draw
Candidate | Total Votes | Candidate Votes | Non-Candidate Vote |
P | 3 | 1(S) | 2 (BH) |
Q | 14 | 1(R) | 13 (FA and OQ) |
R | 6 | 1(P) | 5 (MS) |
S | 1 | 1(Q) | 0 |
Let's 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)
Now, the possible combinations of candidates in MS are (P,Q), (Q,S) and (P,S).
Now , we can get that P and Q are from MS.
Question 4. What best can be concluded about the candidate from O&Q?
It was either Prof. Pakrasi or Prof. Qureshi.
It was Prof. Samuel.
It was Prof. Ramaswamy.
It was either Prof. Ramaswamy or Prof. Samuel.
Correct Answer: D
Solution:
Here Take the first letter of the name of professor to represent the professor.
Given,
Department | No of Faculty members | Candidate for Dean | No of non-candidate voters |
FA | 9 | 1 or 2 | 8 or 7 |
MS | 7 | 1 or 2 | 6 or 5 |
OQ | 5 | 1 | 4 |
BH | 3 | 1 or 2 | 2 or 1 |
We know that the non-candidates in a particular department voted for the same candidate, and we also know that the least number of non-candidate voters in a particular department can be 1.
R got total 6 votes, out of which 1 is from the dean candidate P and 5 from non-candidates.
from above table we can see that 5 can be formed by (5 + 0) or (4 +1).
If we take 2nd case: It is only possible when there is 1 candidate from OQ, and there are 2 candidates from BH, which implies that number of candidates in FA and MQ is 1. But P received 2 non candidate votes only which is possible only if BH has 2 non candidate faculty and 1 candidate. So, this case is not possible.
Now, considering 1st case (a), we get that 5+0 will happen only one when there are 5 non-candidates in a single department.
This can be possible only in MS (where out of 7, there will be 2 candidates and 5 non-candidates)
So, we get the following table:
Department | No of Faculty members | Candidate for Dean | No of non-candidate voters |
FA | 9 | 0 | 9 |
MS | 7 | 2 | 5 |
OQ | 5 | 1 | 4 |
BH | 3 | 1 | 2 |
We also can draw
Candidate | Total Votes | Candidate Votes | Non-Candidate Vote |
P | 3 | 1(S) | 2 (BH) |
Q | 14 | 1(R) | 13 (FA and OQ) |
R | 6 | 1(P) | 5 (MS) |
S | 1 | 1(Q) | 0 |
Let's 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)
Now, the possible combinations of candidates in MS are (P,Q), (Q,S) and (P,S).
Now , we can get that P and Q are from MS.
Question 5. 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.
Only statement B
Only statement A
Both statements A and B
Neither statement A nor statement B
Correct Answer: A
Solution:
Here Take the first letter of the name of professor to represent the professor.
Given,
Department | No of Faculty members | Candidate for Dean | No of non-candidate voters |
FA | 9 | 1 or 2 | 8 or 7 |
MS | 7 | 1 or 2 | 6 or 5 |
OQ | 5 | 1 | 4 |
BH | 3 | 1 or 2 | 2 or 1 |
We know that the non-candidates in a particular department voted for the same candidate, and we also know that the least number of non-candidate voters in a particular department can be 1.
R got total 6 votes, out of which 1 is from the dean candidate P and 5 from non-candidates.
from above table we can see that 5 can be formed by (5 + 0) or (4 +1).
If we take 2nd case: It is only possible when there is 1 candidate from OQ, and there are 2 candidates from BH, which implies that number of candidates in FA and MQ is 1. But P received 2 non candidate votes only which is possible only if BH has 2 non candidate faculty and 1 candidate. So, this case is not possible.
Now, considering 1st case (a), we get that 5+0 will happen only one when there are 5 non-candidates in a single department.
This can be possible only in MS (where out of 7, there will be 2 candidates and 5 non-candidates)
So, we get the following table:
Department | No of Faculty members | Candidate for Dean | No of non-candidate voters |
FA | 9 | 0 | 9 |
MS | 7 | 2 | 5 |
OQ | 5 | 1 | 4 |
BH | 3 | 1 | 2 |
We also can draw
Candidate | Total Votes | Candidate Votes | Non-Candidate Vote |
P | 3 | 1(S) | 2 (BH) |
Q | 14 | 1(R) | 13 (FA and OQ) |
R | 6 | 1(P) | 5 (MS) |
S | 1 | 1(Q) | 0 |
Let's 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)
Now, the possible combinations of candidates in MS are (P,Q), (Q,S) and (P,S).
Now , we can get that P and Q are from MS.
CAT Question Paper 2023: QA (Quantitative Aptitude)
Number of Questions: 22
Time Allotted: 40 minutes
Difficulty Level: Lengthier than the 2022 paper, Medium to Difficult
Out of all the sections in the CAT 2023 question paper, the quantitative aptitude section posed the most difficulty for the candidates. The paper was lengthy and more difficult compared to the previous year's CAT question papers. The questions covered a wide range of mathematical concepts, requiring candidates to display their quantitative aptitude. Another interesting observation is that the algebraic questions outweighed the arithmetic questions in the CAT 2023 question paper on quantitative aptitude. The division of questions in the CAT 2023 question papers across all the slots is given below. Having a idea of the important CAT quantitative aptitude formulas can be really beneficial for the candidates while preparing for this section.
Area | Slot 1 (Total: 22) | Slot 2 (Total: 22) | Slot 3 (Total: 14) |
Arithmetic | 7 | 7 | 8 |
Algebra | 8 | 8 | 4 |
Geometry | 4 | 3 | 3 |
Modern Maths | 2 | 2 | 4 |
Numbers | 1 | 2 | 3 |
MCQs | 15 | 14 | 14 |
TITA | 7 | 8 | 8 |
CAT Question Paper 2023: Quantitative Aptitude Questions
Q. 1) If x is a positive real number such that x8 + (1/x)8 = 47, then the value of x9 + (1/x)9 is
A. 34√5
B. 40√5
C. 36√5
D. 30√5
Ans: A
Solution:
Given:
x⁸ + (1/x)⁸ = 47
⇒ (x⁴ + 1/x⁴)² - 2 = 47
⇒ (x⁴ + 1/x⁴)² = 49
⇒ x⁴ + 1/x⁴ = 7
Next,
(x² + 1/x²)² - 2 = 7
⇒ (x² + 1/x²)² = 9
⇒ x² + 1/x² = 3
Now,
(x + 1/x)² - 2 = 3
⇒ x + 1/x = √5
Cubing both sides:
(x + 1/x)³ = 5√5
⇒ x³ + 1/x³ + 3 × √5 = 5√5
⇒ x³ + 1/x³ = 2√5
Cubing again:
x⁹ + 1/x⁹ + 3 × 2√5 = 40√5
⇒ x⁹ + 1/x⁹ = 34√5
Q. 2) Let n and m be two positive integers such that there are exactly 41 integers greater than 8m and less than 8n, which can be expressed as powers of 2. Then, the smallest possible value of n + m is
A. 44
B. 16
C. 42
D. 14
Ans: B
Solution:
Smallest value of 8m is 8 i.e. 23 (for m = 1)
Next integer in the form of 2x will be 24.
And we have 41 integers in the form of 2x greater than 8m and less than 8n.
So, last integer in the form of 2x and less than 8n is 244.
So, 8n = 245 which implies that 3n = 45 and n = 15.
So, smallest possible value of n + m = 15 + 1 = 16.
Q. 3) For some real numbers a and b, the system of equations x +y = 4 and (a+5)x + (b2 -15)y = 8b has infinitely many solutions for x and y. Then, the maximum possible value of ab is
A. 15
B. 55
C. 33
D. 25
Ans: C
Solution:
For the given equations:
x + y = 4 and (a + 5)x + (b² - 15)y = 8b
Condition for infinite many solutions:
1/(a + 5) = 1/(b² - 15) = 4/(8b)
Solving 1/(b² - 15) = 4/(8b),
we get 8b = 4(b² - 15)
On solving this quadratic equation, we get b = -3, 5.
Next, solving 1/(a + 5) = 4/(8b),
we get 8b = 4(a + 5).
For b = -3, a = -11, and hence ab = (-3) × (-11) = 33.
For b = 5, a = 5, and hence ab = 5 × 5 = 25.
So, the maximum value of ab is 33.
