The CAT Exam Roadmap: Navigating Success through Practice Set Questions- Day 20

CAT 2025 Sample Paper with solutions PDF

A CAT sample paper with solutions PDF is a mock version of the actual exam, designed to simulate the CAT question pattern and difficulty level. It helps aspirants understand the exam structure, improve time management, and assess their preparation level. Practising sample papers regularly also boosts confidence and highlights weak areas that need focused revision.

This CAT Sample paper consists of three Sections:

• Part- A consists of 9 questions.

• Part- B consists of 7 questions.

• Part- C consists of 9 questions.

Quantitative Aptitude

Q1. For the given pair (x, y) of positive integers, such that (-x + 3y/2) = 7 and x < 210. How many integer values of y satisfy the given conditions? (CAT previous year questions)

1. 340

2. 341

3. 342

4. 1024

Q2. A quadratic function f(x) attains a maximum of 12 at x = 1. The value of the function at x = 0 is 7. What is the value of f (x) at x = -11? (CAT previous year questions)

1. -805

2. -719

3. -859

4. -883

Q3. If the roots of the equation in y i.e. y3 – ay2 + by – c = 0 are three consecutive integers, then the smallest possible value of b (nearest to its integer value) is ______________ [TITA]

Q4. The sum of the integers in the solution set of |x2 - 7x| > 5 is: (CAT exam model paper)

1. 10

2. 15

3. 20

4. 0

Q5. The equation x2 + ax + (3-b) = 0 has real roots. What is the minimum value of a2 + b2? (CAT previous year questions)

1. 0

2. 2

3. 4

4. 8

Q6. X is p% less than Y and Z is p% more than Y. If X is 2.5p% more than Z, then find p.
(Note: p ≠ 0) - (CAT previous year questions)

1. 20

2. 40

3. 60

4. 50

Q7. The value of the base of a triangle for which area is minimum if the relation between the perpendicular distance from the vertex and the base is p = 7b2 - 84 is _______________ (where b represents base and p represents perpendicular distance) [TITA]

Q8. Let f (x) = max (2x + 1, 3 − 4x), where x is any real number. Value of f(x) at x = 7.5 is: (CAT exam model paper)

1. -11

2. - 27

3. 16

4. 43

Q9. Let g (x) be a function such that g (x + 1) g (x) = 12 for every real x. Then what is the value of g (11) g (0)? (CAT exam model paper)

1. 1

2. 12

3. 1/12

4. 6

Logical Reasoning

Direction (Q1-Q3):

Four friends Arun, Bittu, Cirag and Dheeraj went for an excursion with their wives Priya, Renu, Sharita and Varti, not necessarily in the same order. Each couple hails from a different city, including Madurai, Cutnee, Kurnool, and Hapur, but they are not necessarily in that order. They went to Delhi to visit the Pinkfort, where they sat in a row. Each wife always sat to the immediate right of her husband.

(CAT practice questions)

(i) Bittu sat to the immediate right of Priya.

(ii) Dheeraj is from Hapur and Priya is not from Madurai.

(iii) Renu and her husband were sitting to the immediate right of the couple that hailed from Cutnee.

(iv) Chirag and his wife were sitting to the immediate left of the couple from Kurnool and Chirag was sitting to the immediate right of Sharita.

(v) Arun sat to the immediate right of Renu, who is not from Madurai.

Q1. Who is Dheeraj’s wife?

1. Sharita

2. Priya

3. Varti

4. Renu

Q2. Which couple is from Cutnee?

1. Varti and Bittu

2. Dheeraj and Renu

3. Chirag and Priya

4. Arun and Sharita

Q3. Who is the husband in the couple, who is seated second in the row from left to right?

1. Arun

2. Bittu

3. Chirag

4. Dheeraj

Direction (Q4- Q7) (CAT practice questions)

A cube is formed by joining 216 smaller and identical cubes. The bigger cube is painted black colour on all its faces. From the bigger cube, 64 smaller cubes from a corner were taken out and the cube formed by joining these 64 smaller cubes is painted pink in all its faces. Again, this cube is fitted back into its usual position in the large cube.

Q4. How many smaller cubes have no faces painted?

Q5. How many smaller cubes have three faces painted?

Q6. How many smaller cubes have only two faces painted?

Q7. How many smaller cubes have only one face painted?

VARC (CAT solved papers)

In the questions below, find the phrase with an error.

Q1: If I had known, I would lend you my car.

1. I will lend you my car

2. I would lent you my car

3. I would have lent you my car

4. No improvement needed

Q2. You have no trouble at school, if you had done your homework.

1. You would have had no trouble at school.

2. You will have no trouble at school.

3. You would had no trouble at school

4. No improvement needed

Q3. If you had done better at your interview, you get that job.

1. You will get that job.

2. You would get that job.

3. You would have got that job.

4. No improvement needed

Choose appropriate sentence

Q4. While the guests danced, the thieves broke into the house.

1. While the guests have danced

2. While the guests were dancing

3. While the guests had danced

4. No improvement needed

Error detection - (CAT exam paper sample)

Q5. Take a shower (a)/ you will (b)/ feel better. (c) / No error (d)

1. (a)

2. (b)

3. (c)

4. (d)

Q6. Whenever I look (a)/ at the moon, my heart (b)/ fills with the pleasure. (c)/ No error(d)

1. (a)

2. (b)

3. (c)

4. (d)

Q7. The whole block of flats (a)/ including two shops were (b)/ destroyed in fire. (c)/ No error (d)

1. (a)

2. (b)

3. (c)

4. (d)

Q8. The man who cannot (a)/ believe his senses and the man who cannot (b)/ believe anything else are insane. (c)/ No error(d)

1. (a)

2. (b)

3. (c)

4. (d)

Fill the blank with the correct Article:

Q9. If you go by __ train you can have quite__ comfortable journey, but make sure you get ___ express, not __ train that stops at all the stations.

1. a, an, a, the

2. an, a, the, a

3. No article, an, a, the

4. No article, a, an, a

CAT 2025 Quantitative Aptitude Sample Papers Solutions

CAT Quantitative Aptitude sample papers with detailed answers and solutions help in effective preparation. These papers resemble the actual exam pattern and include questions from all three sections, Quantitative Aptitude. Each question is accompanied by a step-by-step solution, helping aspirants understand the correct approach.

Answer 1: (B)

3y/2 = 7 + x

⇒ y = (14 + 2x)/3; for y to be integer (14 + 2x) should be divisible by 3

So, x = 2; y = 6

When x= 5, y = 8

When x = 8, y = 10

When x = 11, y = 12 and so on…

So, there is an AP in x with the first term being 2 and the common difference is 3.

nth term in x must be less than or equal to 210 i.e., 1024.

Therefore, 2 + (n-1)3 ≤ 1024

⇒ n ≤ (1022/3) +1

Therefore, n must be 341.

Hence, there can be 341 solutions.

Answer 2: (D)

If the function gets the maximum of 12 at x = 1

f(x) = a(x – 1)2 + 12

So, f (0) = a + 12 = 7 [Given]

⇒ a = –5

f(x) = –5(x – 1)2 + 12

So, f(-11) = –5(-12 – 1)2 + 12 = –845 + 12 = –833

Hence, the correct answer is –833.

Answer 3:

Let roots be (n - 1), n and (n + 1).

In a cubic equation, the coefficient of y divided by the coefficient of y3 represents the sum of the product of roots taken two at a time.

So, b = the sum of the product of roots taken two at a time

n(n - 1) + (n + 1)n + (n - 1) (n + 1) = b

⇒ n2 - n + n2 + n + n2 - 1 = b

⇒ 3n2 - 1 = b

The value of b will be minimal when the value of n2 is minimum.

i.e., n2 = 0

So, 3 × 0 - 1 = b

∴ b = -1

Hence, the correct answer is -1.

Answer 4: (D)

|x2 - 7x| > 5

⇒ x2 - 7x > 5 or x2 - 7x < -5

Case 1:

⇒ x2-7x > 5

⇒ x2 - 7x – 5 > 0

⇒ x > [(7±√69)/2], which gives two solutions x > 7 or x < 0 (for x to be integer).

Case 2:

⇒ x2 - 7x < -5

⇒ x2 - 7x + 5 < 0

⇒ x < [(7±√29)/2], which gives two solutions x < 7 or x > 0 (for x to be integer).

So, x = [1, 2, 3, 4, 5, 6]

There is no common solution in both cases.

Hence, the correct answer is 0.

Answer 5: (D)

For real roots; a2 – 4(3 – b) ≥ 0

⇒ a2 + 4b – 12 ≥ 0

⇒ a2 + 4b + b2 – b2 – 12 ≥ 0

⇒ a2 + b2 ≥ b2 – 4b + 12

⇒ a2 + b2 ≥ b2 – 4b + 4 + 8

⇒ a2 + b2 ≥ (b – 2)2 + 8

At b = 2, a2 + b2 has the minimum value. i.e., 8.

Hence, the correct answer is 8.

Answer 6: (A)

X = (100 – p) Y/100

Z = (100 + p) Y/100

X = (100 + 2.5p) Z/100

From 1st and 2nd relation, we get X = Z (100 – p) / (100 + p)

So, Z (100 – p) / (100 + p) = (100 + 2.5p) Z/100

⇒ 100(100 – p) = (100 + p) (100 + 2.5p)

⇒ – 100p = 100p – 250p + 2.5p2

⇒ 50p = 2.5p2

⇒ p = 20

Hence, the correct answer is 20.

Answer 7:

Area = ½ (base) (height) = ½ b (7b2 - 84) = 7b3/2 - 84b1/2

For Area to be minimum, find dA/db = 21/2 b2 – 42 = 0

⇒ b2 = 4

⇒ b = 2 or b = -2

Now find d2A/db2 = 21b

At b = 2; d2A/db2 = 42, which is greater than 0.

So, at b = 2, we get the minimum value of Area.

Hence, the correct answer is 2.

Answer 8: (C)

f(x) = max (2 × 7.5 + 1, 3 – 4 × 7.5) at x = 7.5

⇒ f(x) = max (16, -27) at x = 7.5

Hence, the correct answer is 16.

Answer 9: (B)

For x = 0; g(1) = 12/g(0)

Similarly g(2) = 12/g(1), which gives g(2) = g(0)

g(3) = g(1) and so on.

Therefore, g(11) = g(9) = g(7) = ……= g(1) = 12/g(0)

⇒ g (11) g (0) = 12

Hence, the correct answer is 12.

CAT Logical Reasoning Sample Paper Solutions

CAT Logical Reasoning sample paper solutions provide in-depth explanations for a wide range of LR questions, including puzzles, arrangements, blood relations, and logical sequences. These step-by-step solutions help candidates understand the reasoning behind each answer, improve their problem-solving techniques.

Solution 1-3:

Males: Arun, Bittu, Cirag and Dheeraj

Females: Priya, Renu, Sharita and Varti

Cities: Madurai, Cutnee, Kurnool and Hapur

Regarding the couples, the wife always sits to the immediate right of her husband.

From (i), Bittu is not the husband of Priya.

From (ii), Dheeraj is from Hapur.

From (iii), Renu is not from Cutnee.

From (iv), Cirag is not from Kurnool and is not the husband of Sharita.

From (v), Arun is not married to Renu and Renu is not from Madurai.

Possibility Table

So, according to the table, there were two possibilities for Chirag's wife either Priya or Varita so we will consider these cases one by one.

Case I: If we consider Priya as Chirag’s wife and Priya can’t be from Madurai so they must be from Cutnee.

So, the arrangement could look like this:

------ Sharita Chirag --- Priya Bittu---Renu Arun-------

Now after this arrangement only one man and one woman is left so we can consider the woman as Arun’s wife and the man as Sharita’s Husband

Dheeraj ------ Sharita Chirag --- Priya Bittu---Renu Arun-------Varti

This arrangement fulfils all the conditions, so it is not required to consider any other case.

Answer 1: Dheeraj is married to Sharita. [Which is option (A)]

Answer 2: Chirag and Priya are from Cutnee. [Which is option (C)]

Answer 3: Chirag and his wife are seated second in the row. [Which is option (C)]

Solution 4 to 7:

Answer 4: By assuming a 4 × 4 × 4 cube is also not painted with pink colour, the number of cubes with no faces painted is = (6 – 2)3 = 64. Of these 64 smaller cubes now, in the second, third and fourth layers leftmost 9 cubes are painted pink, on the bottom surface four smaller cubes are painted pink and on the fourth layer back six smaller cubes are painted.

Hence, the total number of cubes with no face painted = 64 – 19 = 45

Answer 5: The number of cubes which are three faces painted before removing 64 smaller cubes = 8 and after painting with pink colour to those 64 smaller cubes seven new cubes will have three faces painted.

Hence, the total number of cubes with three faces painted = 8 + 7 = 15

Answer 6: The number of smaller cubes painted on two faces in the larger cube after removing 4 × 4 × 4 is illustrated as shown below.

9 edges with four smaller cubes with two faces painted = 9 × 4 = 36; 3 edges with one cube with two faces painted = 3 × 1 = 3

In the pink cube, there are 24 cubes on 12 edges which are painted on two faces.

Hence, the number of cubes with two faces painted = 36 + 3 + 24 = 63

Answer 7: The number of cubes with one face painted in the larger cube = 3 × 16 + 3 × 7 = 69. Number of cubes with one face painted in the 4 × 4 cube = 6 × 4 = 24

Hence, the total number of smaller cubes with one face painted = 69 + 24 = 93

CAT VARC Sample Papers Solutions

CAT VARC sample paper solutions offer detailed explanations for reading passages, para jumbles, sentence completion, and odd sentence questions. These solutions guide candidates on how to approach each question type, identify key ideas, and eliminate incorrect choices.

Answer 1: (C)

if clause: had known (past perfect), Main clause: would/ could have lent.

Answer 2: (A)

if clause: had done (past perfect), the main clause: would/ could/ might have had.

Answer 3: (C)

if clause: had done (past perfect), the main clause: would/ could/ might have had.

Answer 4: (B)

The previous event is simple past, then the next event must be past continuous for continuous action.

Answer 5: D

“take a shower” is an idiomatic phrase. Hence, No error.

Answer 6: (C)

“with pleasure” is an idiomatic phrase. With pleasure replaced with “pleasure”.

Answer 7: (B)

“was” is used instead of “were”. The whole block is considered a singular subject and the singular subject takes a singular verb.

Answer 8: (D)

when two nouns get added with the use of ‘and’ and a possessive adjective article is used before both nouns then we use a plural verb but when the possessive adjective article is used with only one noun then it is considered to be one and hence singular verb is used.

Answer 9: (D)

No article is used between By and Means of transport.
Before comfortable “a” should be used.
Before “express”, “an” will be used as there is a vowel at the start.
Before “train”, “a” will be used.

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