Q: CAT 2025 Slot 1: The number of distinct pairs of integers (x, y) satisfying the inequalities x > y ≥ 3 and x + y < 14 is ____.

Answer: 16. We fix y and find all possible x values that satisfy both conditions: x > y x + y < 14 y Allowed x values Count 3 4, 5, 6, 7, 8, 9, 10 7 4 5, 6, 7, 8, 9 5 5 6, 7, 8 3 6 7 1 Total 16

Q: CAT 2024 Slot 1: In the XY-plane, the area, in sq. units, of the region defined by the inequalities $ y \geq x + 4 $ and $ -4 \leq x^2 + y^2 + 4(x - y) \leq 0 $ is:

Answer: B · $ 2\pi $. Area of Region Defined by Inequalities Finding the Area Defined by Given Inequalities Given inequalities: $ y \geq x + 4 $ $ -4 \leq x^2 + y^2 + 4(x - y) \leq 0 $ Step 1: Simplify the second inequality \[ x^2 + y^2 + 4x - 4y = (x + 2)^2 - 4 + (y - 2)^2 - 4 = (x + 2)^2 + (y - 2)^2 - 8 \] So the inequality becomes: \[ -4 \leq (x + 2)^2 + (y - 2)^2 - 8 \leq 0 \Rightarrow 4 \leq (x + 2)^2 + (y - 2)^2 \leq 8 \] This represents an annular region (ring) centered at $ (-2, 2) $, with: Inner radius: $ \sqrt{4} = 2 $ Outer radius: $ \sqrt{8} = 2\sqrt{2} $ Step 2: Apply the line constraint \( y

Q: CAT 2021 Slot 3: The number of distinct pairs of integers (m, n) satisfying |1+mn| < |m+n| < 5 is

Answer: 12. |1 + mn| 1, or |n| > 1 and m = 0 Now for the case when m = 0 and |n| > 1 |m + n| 1 |m + n| < 5 |0 + m| < 5 So m can be ±±2, ±±3, ±±4 Again we have 6 cases. Hence the answer is 12.

Q: CAT 2021 Slot 2: For all possible integers n satisfying 2.25 ≤ 2 + 2^n + 2 ≤ 202, the number of integer values of 3 + 3^n + 1 is

Answer: 7. 2.25 ≤ 2 + 2^n + 2 ≤ 202 So, 2 + 2 ^n+2 has to be either lesser than or equal to 202. 2^7 = 128 and so will satisfy the inequality, but 2^8 wont. Hence, the value of n can be 5 at most. Further, subtracting 2 from the first two parts of the inequality, 0.25 ≤ 2^n+2 Or, 2^-2 ≤ 2^n+2 Hence, n ≥ -4 So from the inequality we can say that the integral value of n can be [-4, 5]. For 3 + 3^n+1 to have integral values, n but be greater than or equal to -1 and go up till 5. Hence the number of values of n can be -1, 0, 1, 2, 3, 4, 5. 7 values

Q: CAT 2021 Slot 1: The number of integers n that satisfy the inequalities |n - 60| < |n - 100| < |n - 20| is

Answer: D · 19. |n - 60| < |n - 100| < |n - 20| To solve this question, we need to know the basic concept that ‘if x and y are two points on the number line, then the distance between these two points can be represented by |x - y| or |y - x|.’ Let's take the first part of the inequality |n - 60| < |n - 100| This inequality holds good for the values of n below 80. So this is the turning point. Hence ‘n’ should be less than 80. Now Let's check for the later part of the inequality. |n - 100| < |n - 20| This inequality holds good for the values of n above 60. So, 60 is the turning point here. Hence ‘n’ should be

Q: CAT 2020 Slot 2: If x and y are non-negative integers such that x + 9 = z, y + 1 = z and x + y < z + 5, then the maximum possible value of 2x + y equals

Answer: 23. x + x + 8 < x + 9 + 5 2x + 8 < x + 14 x < 6 Maximum possible of x = 5, then y = 13 2x + y = 2(5) + 13 = 23

Q: CAT 2018 Slot 2: If a and b are integers such that 2x^2 - ax + 2 ＞ 0 and x^2 - bx + 8 ≥ 0 for all real numbers x, then the largest possible value of 2a - 6b is

Answer: 36. Given that a and b are integers such that 2x^2 − ax + 2 > 0 and x^2 − bx + 8 ≥ 0 for all real numbers x We have to find the largest possible value of 2a – 6b Using the completion of squares method 2x^2 − ax + 2 > 0 can be written as x^2 - a/2x + 1 > 0 We know that (x – k )^2 = x^2 – 2kx + k^2 Here we have a/2 = 2k or k = a/4 This a^2/16is added on both sides ⟹ x^2 - a^2/16 + a^2/16 > a^2/16 - 1 ⟹ (x - a/4)^2 > a^2/16 - 1 [ this a^2/16 - 1 always holds good if it is negative or zero] a^2/16 - 1 < 0 a^2 < 16 a ∈(−4 , 4) Since we have to find the largest value as possible and it cannot be 4 so a

Question 1

CAT 2025 Slot 3: For real values of x, the range of the function f(x) = (2x - 3)/(2x² + 4x - 6)

Accepted Answer

Answer: C · (-oo , 1/8] U [1/2 , oo). We can factor the denominator: 2x² + 4x - 6 = 2(x + 3)(x - 1) So, the function becomes: f(x) = (2x - 3) / (2(x + 3)(x - 1)) , thus x ≠ -3 , 1 Range of f(x) f(0) = 1/2 f(1) = not possible f(2) = 1/10 f(3) = 1/8 Only option 3 includes f(0) and f(3) Thus option 3 is the answer

Question 2

CAT 2025 Slot 2: The set of all real values of x for which (x² − |x + 9| + x) > 0, is

Accepted Answer

Answer: A · (−∞, −3) ∪ (3, ∞). Use X maro G Strategy put x = -5 and x = -10 Option Interval Test x = −5 Test x = −10 Status A (−∞, −3) ∪ (3, ∞) Included, LHS = 16 > 0 Included, LHS = 89 > 0 ✅ Possible B (−9, −3) ∪ (3, ∞) Included, LHS = 16 > 0 Not included ❌ Eliminated C (−∞, −9) ∪ (3, ∞) Not included Included, LHS = 89 > 0 ❌ Eliminated D (−∞, −9) ∪ (9, ∞) Not included Included, LHS = 89 > 0 ❌ Eliminated

Question 3

CAT 2025 Slot 1: Let 3 ≤ x ≤ 6 and [x²] = ([x])², where [x] is the greatest integer not exceeding x. If S represents all feasible values of x, then a possible subset of S is

Accepted Answer

Answer: D · (3, √10] ∪ [5, √26) ∪ {6}. Using X maro For n = 3 → x must be before 4 → [3, √10) For n = 4 → x must be before 5 → [4, √17) For n = 5 → x must be before 6 → [5, √26) For n = 6: only x = 6 (since x ≤ 6) Only Option 4 matches: (3, √10] ∪ [5, √26) ∪ {6} Note: ideally Set should include 4 as well.

Question 4

CAT 2025 Slot 1: The number of distinct pairs of integers (x, y) satisfying the inequalities x > y ≥ 3 and x + y < 14 is ____.

Accepted Answer

Answer: 16. We fix y and find all possible x values that satisfy both conditions: x > y x + y < 14 y Allowed x values Count 3 4, 5, 6, 7, 8, 9, 10 7 4 5, 6, 7, 8, 9 5 5 6, 7, 8 3 6 7 1 Total 16

Question 5

CAT 2024 Slot 1: In the XY-plane, the area, in sq. units, of the region defined by the inequalities $ y \geq x + 4 $ and $ -4 \leq x^2 + y^2 + 4(x - y) \leq 0 $ is:

Accepted Answer

Answer: B · $ 2\pi $. Area of Region Defined by Inequalities Finding the Area Defined by Given Inequalities Given inequalities: $ y \geq x + 4 $ $ -4 \leq x^2 + y^2 + 4(x - y) \leq 0 $ Step 1: Simplify the second inequality $$ x^2 + y^2 + 4x - 4y = (x + 2)^2 - 4 + (y - 2)^2 - 4 = (x + 2)^2 + (y - 2)^2 - 8 $$ So the inequality becomes: $$ -4 \leq (x + 2)^2 + (y - 2)^2 - 8 \leq 0 \Rightarrow 4 \leq (x + 2)^2 + (y - 2)^2 \leq 8 $$ This represents an annular region (ring) centered at $ (-2, 2) $, with: Inner radius: $ \sqrt{4} = 2 $ Outer radius: $ \sqrt{8} = 2\sqrt{2} $ Step 2: Apply the line constraint \( y

Question 6

CAT 2021 Slot 3: The number of distinct pairs of integers (m, n) satisfying |1+mn| < |m+n| < 5 is

Accepted Answer

Answer: 12. |1 + mn| 1, or |n| > 1 and m = 0 Now for the case when m = 0 and |n| > 1 |m + n| 1 |m + n| < 5 |0 + m| < 5 So m can be ±±2, ±±3, ±±4 Again we have 6 cases. Hence the answer is 12.

Question 7

CAT 2021 Slot 2: For all possible integers n satisfying 2.25 ≤ 2 + 2^n + 2 ≤ 202, the number of integer values of 3 + 3^n + 1 is

Accepted Answer

Answer: 7. 2.25 ≤ 2 + 2^n + 2 ≤ 202 So, 2 + 2 ^n+2 has to be either lesser than or equal to 202. 2^7 = 128 and so will satisfy the inequality, but 2^8 wont. Hence, the value of n can be 5 at most. Further, subtracting 2 from the first two parts of the inequality, 0.25 ≤ 2^n+2 Or, 2^-2 ≤ 2^n+2 Hence, n ≥ -4 So from the inequality we can say that the integral value of n can be [-4, 5]. For 3 + 3^n+1 to have integral values, n but be greater than or equal to -1 and go up till 5. Hence the number of values of n can be -1, 0, 1, 2, 3, 4, 5. 7 values

Question 8

CAT 2021 Slot 1: The number of integers n that satisfy the inequalities |n - 60| < |n - 100| < |n - 20| is

Accepted Answer

Answer: D · 19. |n - 60| < |n - 100| < |n - 20| To solve this question, we need to know the basic concept that ‘if x and y are two points on the number line, then the distance between these two points can be represented by |x - y| or |y - x|.’ Let's take the first part of the inequality |n - 60| < |n - 100| This inequality holds good for the values of n below 80. So this is the turning point. Hence ‘n’ should be less than 80. Now Let's check for the later part of the inequality. |n - 100| < |n - 20| This inequality holds good for the values of n above 60. So, 60 is the turning point here. Hence ‘n’ should be

Question 9

CAT 2020 Slot 2: If x and y are non-negative integers such that x + 9 = z, y + 1 = z and x + y < z + 5, then the maximum possible value of 2x + y equals

Accepted Answer

Answer: 23. x + x + 8 < x + 9 + 5 2x + 8 < x + 14 x < 6 Maximum possible of x = 5, then y = 13 2x + y = 2(5) + 13 = 23

Question 10

CAT 2018 Slot 2: If a and b are integers such that 2x^2 - ax + 2 ＞ 0 and x^2 - bx + 8 ≥ 0 for all real numbers x, then the largest possible value of 2a - 6b is

Accepted Answer

Answer: 36. Given that a and b are integers such that 2x^2 − ax + 2 > 0 and x^2 − bx + 8 ≥ 0 for all real numbers x We have to find the largest possible value of 2a – 6b Using the completion of squares method 2x^2 − ax + 2 > 0 can be written as x^2 - a/2x + 1 > 0 We know that (x – k )^2 = x^2 – 2kx + k^2 Here we have a/2 = 2k or k = a/4 This a^2/16is added on both sides ⟹ x^2 - a^2/16 + a^2/16 > a^2/16 - 1 ⟹ (x - a/4)^2 > a^2/16 - 1 [ this a^2/16 - 1 always holds good if it is negative or zero] a^2/16 - 1 < 0 a^2 < 16 a ∈(−4 , 4) Since we have to find the largest value as possible and it cannot be 4 so a

Question 11

CAT 2017 Slot 1: For how many integers n, will the inequality (n – 5) (n – 10) – 3(n – 2) ≤ 0 be satisfied

Accepted Answer

Answer: 11. We have to find for how many integers n, will the inequality (n – 5)(n – 10) – 3(n – 2) ≤ 0 be satisfied (n – 5)(n – 10) – 3(n – 2) ≤ 0 n^2 – 15n + 50 – 3n + 6 ≤ 0 n^2 – 18n + 56 ≤ 0 (n – 4)(n – 14) ≤ 0 n = (4 , 5 , 6 , 7 , 8 , 9 , 10 , 11 , 12 , 13 , 14) So n has to be between the two positive numbers 4 and 14 which is equal to 11.