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 2025: A cafeteria offers 5 types of sandwiches. Moreover, for each type of sandwich, a customer can choose one of 4 breads and opt for either small or large sized sandwich. Optionally, the customer may also add up to 2 out of 6 available sauces. The number of different ways in which an

Answer: A · 880. Given 5 sandwiches 4 breads 2 sizes Sauces: choose 0, 1, or 2 from 6 Number of sauce choices: 0 sauces → 1 way 1 sauce → 6 ways 2 sauces → 15 ways Total = 22 ways. Total combinations = 5 × 4 × 2 × 22 = 880. Answer: 880

Q: CAT 2024 Slot 3: The number of distinct integer solutions (x, y) of the equation |x + y| + |x – y| = 2, is

Answer: 8. Since two absolute values are added, for the equation to be true we can have three combinations of $|x + y|, |x - y|$: \[ \{2, 0\}, \{0, 2\}, \{1, 1\} \] Cases and Corresponding Solutions: Case Conditions Solution(s) Case 1 $ x + y = 2 $ and $ x - y = 0 $ $ \{x, y\} = \{1, 1\} $ Case 2 $ x + y = -2 $ and $ x - y = 0 $ $ \{x, y\} = \{-1, -1\} $ Case 3 $ x + y = 0 $ and $ x - y = 2 $ $ \{x, y\} = \{1, -1\} $ Case 4 $ x + y = 0 $ and $ x - y = -2 $ $ \{x, y\} = \{-1, 1\} $ Case 5 $ x + y = 1 $ and $ x - y = 1 $ $ \{x, y\} = \{1, 0\} $ Case 6 $ x + y = 1 $ and \(

Question 1

CAT 2025 Slot 3: The sum of all the digits of the number (10⁵⁰ + 10²⁵ - 123), is

Accepted Answer

Answer: C · 221. Correct Answer: 221 The number 10⁵⁰ is a 1 followed by 50 zeros. The number 10²⁵ is a 1 followed by 25 zeros. On adding , we have 1000000… upto 24zeros 100000…. upto 25 zeros When you subtract 123 from this number, The 25 zeros that follow it are transformed. The first 25 - 3 = 22 zeros all become 9s. The last three digits become 1000 - 123 = 877 So number will be 1000 …upto 24 zeros 999……. upto 22 times 877 Now sum of all digits : 1 (the first digit) + 0 (from the 24 zeros)+ 0 (from the 25 zeros )+ (22 × 9) (from the 22 nines)+ (8 + 7 + 7) (from the last three digits) = 1 + 0 + 0 + 198 + 1 =

Question 2

CAT 2025 Slot 3: For a 4 digit number greater than 1000 , sum of the digits in thousands , hundreds and tens places is 15.And the sum of digits in hundreds , tens and units place is 16.Also the digits in tens place is 6 more than the digit in the unit place. The difference between the largest pos

Accepted Answer

Answer: D · 811. Given conditions for a 4-digit number: a + b + c = 15 —-- I b + c + d = 16 —-- II c = d + 6 —---- III Solving the equations: b = 10 - 2d a = d - 1 Now using values for d: If d = 3: a = 2, b = 4, c = 9, number = 2493 If d = 2: a = 1, b = 6, c = 8, number = 1682 Largest number = 2493, Smallest number = 1682 Difference = 2493 - 1682 = 811 Answer  811

Question 3

CAT 2025 Slot 2: The number of divisors of 2⁶ × 3⁵ × 5³ × 7², which are of the form (3r + 1), where r is a non-negative integer, is

Accepted Answer

Answer: D · 42. N = 2⁶ × 3⁵ × 5³ × 7² We want the number of divisors of N that are of the form (3r + 1), i.e. divisors ≡ 1 (mod 3). Step 1: Kill the 3-factor Any divisor containing 3¹ or higher is divisible by 3 → it cannot be 3r + 1. So exponent of 3 must be 0. We only use: 2ᵃ × 5ᶜ × 7ᵈ a: 0 to 6 → 7 values; c: 0 to 3 → 4 values; d: 0 to 2 → 3 values Step 2: Work mod 3 2 ≡ -1 (mod 3); 5 ≡ -1 (mod 3); 7 ≡ 1 (mod 3) So divisor ≡ (-1)^(a + c) (mod 3). For divisor ≡ 1 (mod 3): (-1)^(a + c) = 1 → a + c must be even. Step 3: Count (a, c) with a + c even a = 0,1,2,3,4,5,6; even: 0,2,4,6 → 4 values; odd: 1,3,5 → 3 v

Question 4

CAT 2025 Slot 2: Suppose a, b, c are three distinct natural numbers, such that 3ac = 8(a + b). Then, the smallest possible value of 3a + 2b + c is _______

Accepted Answer

Answer: 12. Final Answer: 12 Given: 3ac = 8(a + b) ⇒ 8b = 3ac − 8a ⇒ b = (3ac/8) − a Since a, b, c are natural numbers, 3ac must be divisible by 8. Now try small values of a (since we want the minimum value of 3a + 2b + c). Try a = 1 Then 3c must be divisible by 8 ⇒ c = 8 b = (3×1×8)/8 − 1 = 3 − 1 = 2 Value = 3a + 2b + c = 3 + 4 + 8 = 15 Try a = 2 Then 6c divisible by 8 ⇒ c = 4 is the smallest possible b = (3×2×4)/8 − 2 = 3 − 2 = 1 Here a, b, c = 2, 1, 4 are distinct natural numbers Value = 3a + 2b + c = 6 + 2 + 4 = 12 Try a = 3 Then 9c divisible by 8 ⇒ c = 8 b = (3×3×8)/8 − 3 = 9 − 3 = 6 Value = 9 + 12 +

Question 5

CAT 2025 Slot 2: If m and n are integers such that (m + 2n)(2m + n) = 27, then the maximum possible value of 2m − 3n is ___

Accepted Answer

Answer: 17. Given: (m + 2n)(2m + n) = 27 Since m and n are integers, (m + 2n) and (2m + n) must be integer factors of 27. Possible factor pairs (m + 2n, 2m + n) are: (1, 27), (3, 9), (9, 3), (27, 1), (−1, −27), (−3, −9), (−9, −3), (−27, −1) We now solve each pair as a system: m + 2n = A; 2m + n = B From these: Multiply first equation by 2: 2m + 4n = 2A Subtract from second: (2m + n) − (2m + 4n) = B − 2A ⇒ −3n = B − 2A ⇒ n = (2A − B) / 3 Then m = A − 2n m + 2n 2m + n n = (2A − B)/3 m = A − 2n 2m − 3n Valid / Reject 1 27 (2 − 27)/3 = −25/3 — — Reject 3 9 (6 − 9)/3 = −1 3 − 2(−1) = 5 13 Valid 9 3 (18 �� 3)/3

Question 6

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 7

CAT 2025 Slot 1: The number of distinct integers n for which log_(1/4) of (n² − 7n + 11) > 0 is _____.

Accepted Answer

Answer: 0. log_(1/4) of (n² − 7n + 11) > 0 is _____. Using DPAC Strategy: Cycle n² − 7n + 11 < (1/4)^0→ n² − 7n + 11 < 1 Now we need values of n which satisfy 0 < n² − 7n + 11 < 1 Compute E(n) = n² − 7n + 11 for nearby integers: n = 1 → E = 5 (not in (0,1)) n = 2 → E = 1 (equals 1, excluded) n = 3 → E = −1 (negative, excluded) n = 4 → E = −1 (negative, excluded) n = 5 → E = 1 (equals 1, excluded) For n ≤ 0 or n ≥ 6, E ≥ 5 (too large). Therefore no integer n satisfies 0 < E < 1. Final answer: 0. Note: Official OG says answer is 2. Which is challenged by Cetking

Question 8

CAT 2025: The number of distinct integers n for which log (1/4) of (n² − 7n + 11) > 0 is _____.

Accepted Answer

Answer: 0. Using DPAC Strategy: Cycle n² − 7n + 11 < (1/4) 0 → n² − 7n + 11 < 1 Now we need values of n which satisfy 0 < n² − 7n + 11 < 1 Compute E(n) = n² − 7n + 11 for nearby integers: n = 1 → E = 5 (not in (0,1)) n = 2 → E = 1 (equals 1, excluded) n = 3 → E = −1 (negative, excluded) n = 4 → E = −1 (negative, excluded) n = 5 → E = 1 (equals 1, excluded) For n ≤ 0 or n ≥ 6, E ≥ 5 (too large). Therefore no integer n satisfies 0 < E < 1. Final answer: 0. Note: Official OG says answer is 2. Which is challenged by Cetking

Question 9

CAT 2025: A cafeteria offers 5 types of sandwiches. Moreover, for each type of sandwich, a customer can choose one of 4 breads and opt for either small or large sized sandwich. Optionally, the customer may also add up to 2 out of 6 available sauces. The number of different ways in which an

Accepted Answer

Answer: A · 880. Given 5 sandwiches 4 breads 2 sizes Sauces: choose 0, 1, or 2 from 6 Number of sauce choices: 0 sauces → 1 way 1 sauce → 6 ways 2 sauces → 15 ways Total = 22 ways. Total combinations = 5 × 4 × 2 × 22 = 880. Answer: 880

Question 10

CAT 2024 Slot 3: The number of distinct integer solutions (x, y) of the equation |x + y| + |x – y| = 2, is

Accepted Answer

Answer: 8. Since two absolute values are added, for the equation to be true we can have three combinations of $|x + y|, |x - y|$: $$ \{2, 0\}, \{0, 2\}, \{1, 1\} $$ Cases and Corresponding Solutions: Case Conditions Solution(s) Case 1 $ x + y = 2 $ and $ x - y = 0 $ $ \{x, y\} = \{1, 1\} $ Case 2 $ x + y = -2 $ and $ x - y = 0 $ $ \{x, y\} = \{-1, -1\} $ Case 3 $ x + y = 0 $ and $ x - y = 2 $ $ \{x, y\} = \{1, -1\} $ Case 4 $ x + y = 0 $ and $ x - y = -2 $ $ \{x, y\} = \{-1, 1\} $ Case 5 $ x + y = 1 $ and $ x - y = 1 $ $ \{x, y\} = \{1, 0\} $ Case 6 $ x + y = 1 $ and \(

Question 11

CAT 2024 Slot 3: If 10^68 is divided by 13, the remainder is

Accepted Answer

Answer: B · 9. Let’s solve 10^68 ÷ 13 — find the remainder using the cycle method: Step 1: Find powers of 10 modulo 13 10¹ mod 13 = 10 10² mod 13 = (10 × 10) mod 13 = 100 mod 13 100 ÷ 13 = 7 remainder 9 → 10² mod 13 = 9 10³ mod 13 = (10² × 10) mod 13 = (9 × 10) mod 13 = 90 mod 13 90 ÷ 13 = 6 remainder 12 → 10³ mod 13 = 12 10⁴ mod 13 = (12 × 10) mod 13 = 120 mod 13 120 ÷ 13 = 9 remainder 3 → 10⁴ mod 13 = 3 10⁵ mod 13 = (3 × 10) mod 13 = 30 mod 13 30 ÷ 13 = 2 remainder 4 → 10⁵ mod 13 = 4 10⁶ mod 13 = (4 × 10) mod 13 = 40 mod 13 40 ÷ 13 = 3 remainder 1 → 10⁶ mod 13 = 1 ✅ The cycle length is 6: 10^6 mod 13 = 1 S

Question 12

CAT 2024 Slot 3: Find the number of all positive integers up to 500 with non-repeating digits.

Accepted Answer

Answer: 378. Type of Number Digit Positions Choices at Each Step Reasoning Count 1-digit _ 1-9 → 9 choices Any digit 1-9 is valid 9 2-digit Tens digit 1-9 → 9 choices Tens digit can't be 0 81 Units digit 0-9 except tens digit → 9 choices Can't repeat tens digit 3-digit (100-499) Hundreds digit 1-4 → 4 choices First digit ≤ 4 so number ≤ 499 288 Tens digit 0-9 except hundreds digit → 9 choices Must differ from hundreds digit Units digit 0-9 except hundreds & tens digit → 8 choices Must differ from both Final Total: 9 (1-digit) + 81 (2-digit) + 288 (3-digit) = 378

Question 13

CAT 2024 Slot 2: When 3^333 is divided by 11, the remainder is

Accepted Answer

Answer: 5. Modular Arithmetic - Cyclicity Finding the Remainder of $ 3^{333} \div 11 $ Using Cyclicity Step 1: Identify the cycle of $ 3^n \mod 11 $ $$ \begin{align*} 3^1 &\equiv 3 \mod 11 \ 3^2 &\equiv 9 \mod 11 \ 3^3 &\equiv 27 \equiv 5 \mod 11 \ 3^4 &\equiv 15 \equiv 4 \mod 11 \ 3^5 &\equiv 12 \equiv 1 \mod 11 \end{align*} $$ The powers of 3 modulo 11 repeat every 5 terms. Step 2: Reduce the exponent using the cycle Since the cycle length is 5, we compute: $$ 333 \mod 5 = 3 \Rightarrow 3^{333} \equiv 3^3 \mod 11 $$ From our earlier list: $$ 3^3 \equiv 5 \mod 11 $$ ✅ Final Answer: \[ \boxed{5}

Question 14

CAT 2024 Slot 2: A function f maps the set of natural numbers to whole numbers, such that f(xy) = f(x)f(y) + f(x) + f(y) for all x, y and f(p) = 1 for every prime number p. Then, the value of f(160000) is

Accepted Answer

Answer: A · 4095. f(xy) = f(x)f(y) + f(x) + f(y) = (f(x) + 1)(f(y) + 1) – 1 f(160000) = f(400 × 400) = (f(400) + 1)(f(400) + 1) – 1 Now, f(400) = f(20 × 20) = (f(20) + 1)(f(20) + 1) – 1 Now, f(20) = f(4 × 5) = (f(4) + 1)(f(5) + 1) – 1 Now, f(4) = f(2 × 2) = (f(2) + 1)(f(2) + 1) – 1 Since f(p) = 1 where p is a prime number, so Expression Calculation Value f(4) (1 + 1)(1 + 1) - 1 3 f(20) (3 + 1)(1 + 1) - 1 7 f(400) (7 + 1)(7 + 1) - 1 63 f(160000) (63 + 1)(63 + 1) - 1 4095

Question 15

CAT 2024 Slot 1: For any natural number $ n $, let $ a_n $ be the largest integer not exceeding $ \sqrt{n} $. Then the value of $ a_1 + a_2 + \cdots + a_{50} $ is

Accepted Answer

Answer: 217. Floor of Square Roots Sum Solution We are given: For any natural number $ n $, let $ a_n = \left\lfloor \sqrt{n} \right\rfloor $ We need to find: $$ \sum_{n=1}^{50} \left\lfloor \sqrt{n} \right\rfloor $$ Group the terms based on constant value of $ \left\lfloor \sqrt{n} \right\rfloor $: $ \left\lfloor \sqrt{n} \right\rfloor = k $ Range of $ n $ Count of Terms Total Sum Contributed 1 1 to 3 3 $ 1 \times 3 = 3 $ 2 4 to 8 5 $ 2 \times 5 = 10 $ 3 9 to 15 7 $ 3 \times 7 = 21 $ 4 16 to 24 9 $ 4 \times 9 = 36 $ 5 25 to 35 11 $ 5 \times 11 = 55 $ 6 36 to 48 13 \( 6 \times 13 = 78 \

Question 16

CAT 2024 Slot 1: When 10^100 is divided by 7, find the remainder.

Accepted Answer

Answer: D · 4. 10^100 mod 7 - Cyclic Remainder Table Finding the Remainder of $ 10^{100} \div 7 $ Let’s analyze the cyclic pattern of $ 10^n \mod 7 $: Power $ n $ $ 10^n \mod 7 $ 1 $ 3 $ 2 $ 2 $ 3 $ 6 $ 4 $ 4 $ 5 $ 5 $ 6 $ 1 $ Since the cycle length is 6, we reduce: $ 100 \mod 6 = 4 $, so we calculate $ 10^4 \mod 7 $ Therefore, the remainder when $ 10^{100} $ is divided by 7 is $ \boxed{4} $.

Question 17

CAT 2023 Slot 3: The sum of the first two natural numbers, each having 15 factors (including 1 and the number itself), is

Accepted Answer

Answer: 468

Question 18

CAT 2023 Slot 2: Q. For any natural numbers m, n, and k, such that k divides both m + 2n and 3m + 4n, k must be a common divisor of

Accepted Answer

Answer: C · m and 2n. Ans 3

Question 19

CAT 2023 Slot 2: The number of positive integers less than 50, having exactly two distinct factors other than 1 and itself, is:

Accepted Answer

Answer: B · 15. Answer: B. 15 Detailed Solution: Positive integers with exactly two distinct factors other than 1 and itself are of the form $ a \times b $ or $ a^3 $, where $ a $ and $ b $ are prime numbers. Possible combinations include: If one prime is 2, the other can be 3, 5, 7, etc. ∴ Total count = 15.

Question 20

CAT 2023 Slot 2: A container has 40 liters of milk. 4 liters are removed and replaced with 4 liters of water. This process is repeated. The smallest number of times required for milk to be less than water is:

Accepted Answer

Answer: C · 7. Answer: C. 7 Detailed Solution: After each replacement, $ \frac{9}{10} $ of the milk remains. After 7 replacements, quantity of milk = $ 40 \times \left( \frac{9}{10} \right)^7 < 20 $. Thus, 7 replacements are required.

Question 21

CAT 2023 Slot 2: Let both series $ a_1, a_2, a_3, \ldots $ and $ b_1, b_2, b_3, \ldots $ be in arithmetic progression with common differences as prime numbers. If a5=b9, a19=b19 and b2=0, , then a11 equals, find $ a_{19} $:

Accepted Answer

Answer: C · 79. Answer: C. 79 Detailed Solution: Given $ a_5 = 49 $, common differences are 5 and 7 for series $ a_n $ and $ b_n $. $ a_{19} = 49 + 6 \times 5 = 79 $.

Question 22

CAT 2023 Slot 2: Let $ a_n = 13 + 6(n - 1) $ and $ b_n = 15 + 7(n - 1) $. Find the largest three-digit integer common to both sequences:

Accepted Answer

Answer: A · 967. Answer: A. 967 Detailed Solution: First common term = 43; common difference = 42. Largest three-digit term = 967.

Question 23

CAT 2023 Slot 1: A lab experiment measures the number of organisms at 8 am every day. Starting with 2 organisms on the first day, the number of organisms on any day is equal to 3 more than twice the number on the previous day. If the number of organisms on the $ n $-th day exceeds one million,

Accepted Answer

Answer: D · 19. Answer: D. 19 Detailed Solution: Let the number of organisms on day 1 be 2. On day 2, the number of organisms = $ 2 \times 2 + 3 = 7 $. On day 3, the number of organisms = $ 2 \times 7 + 3 = 17 $. On day 4, the number of organisms = $ 2 \times 17 + 3 = 37 $, and so on. The number of organisms on day $ n $ can be represented as: $ 2^{(n-1)} + 3 \times (2^{(n-1)} - 1) $. For the number to exceed one million, we solve for $ n $ such that: $ 5 \times 2^{(n-1)} - 3 \geq 10^6 $. Solving, $ 2^{(n-1)} \geq 200,000 \Rightarrow n - 1 \geq 18 \Rightarrow n = 19 $.

Question 24

CAT 2021 Slot 3: A four-digit number is formed by using only the digits 1, 2 and 3 such that both 2 and 3 appear at least once. The number of all such four-digit numbers is

Accepted Answer

Answer: 50. We will select the 4 digits first and arrange them later. Out of the 4 digits, one of them should be 2 and one of them should be 3. 2, 3, , . So, we just need to select the other two digits… The two digits could be (1,1), (2, 2), (3, 3), (1, 2), (1, 3), or (2, 3). So, the selection of numbers could be… 2, 3, 1, 1 2, 3, 2, 2 2, 3, 3, 3 2, 3, 1, 2 2, 3, 1, 3 2, 3, 2, 3 Each of these selections could be re-arranged in a number of ways. 2, 3, 1, 1 (4!2!4!2! = 12 ways) 2, 3, 2, 2 (4!3!4!3! = 4 ways) 2, 3, 3, 3 (4!3!4!3! = 4 ways) 2, 3, 1, 2 (4!2!4!2! = 12 ways) 2, 3, 1, 3 (4!2!4!2! = 12 ways) 2, 3,

Question 25

CAT 2021 Slot 2: For a 4-digit number, the sum of its digits in the thousands, hundreds and tens places is 14, the sum of its digits in the hundreds, tens and units places is 15, and the tens place digit is 4 more than the units place digit. Then the highest possible 4-digit number satisfying the

Accepted Answer

Answer: 4195. For a 4-digit number we can make 4 placeholders to represent the four digits __ __ __ __ Now, we know that the sum of the thousands, hundreds and the tens place digit is 14, and the sum of the hundreds, tens and the units place digit is 15. So we can conclude that the unit's digit is one unit greater than the thousandth place. To maximize the 4-digit number the value of a should be as large as possible. And the value of the tens digit can be 9 at most. So the maximum value of a is 4. To satisfy the sum of the thousands, hundreds and the tens place digit, the value of the hundreds place should

Question 26

CAT 2021 Slot 1: The natural numbers are divided into groups as (1), (2, 3, 4), (5, 6, 7, 8, 9), ….. and so on. Then, the sum of the numbers in the 15th group is equal to

Accepted Answer

Answer: A · 6119. The natural numbers are divided into groups as (1), (2, 3, 4), (5, 6, 7, 8, 9), ….. and so on. If you observe that the first group ends with 1^2, the second group ends with 2^2and the third group ends with 3^2 Hence, the 15th group ends with 15^2 = 225. The 14th group ends with 14^2 = 196. Therefore, the 15th group contains (197, 198, . . . . . . . . . . ,225) The sum of the numbers in the 15th group =292292(197 + 225) = 6119.

Question 27

CAT 2021 Slot 1: How many three-digit numbers are greater than 100 and increase by 198 when the three digits are arranged in the reverse order?

Accepted Answer

Answer: 70. Let's consider a three-digit number to be ‘abc’. Given that, three-digit numbers increase by 198 when the three digits are arranged in the reverse order. 100c + 10b + a - 100a - 10b - c = 198 99c - 99a = 198 c - a = 2 So, the difference between the hundreds place digit and the units place digit is 2. The possible combinations are: 1 _ 3, 2 _ 4, 3 _ 5, 4 _ 6, 5 _ 7, 6 _ 8, 7 _ 9. We have 10 numbers for each combination. Hence, the total numbers are 70.

Question 28

CAT 2021: The natural numbers are grouped as: (1), (2,3,4), (5,6,7,8,9), ……… Find the sum of all terms in the 15th group.

Accepted Answer

Answer: 6119. Each group nn has 2n−12n - 1 terms. We calculate the starting number (first term) of each group recursively: Group 1: starts at 1 (1 term) Group 2: starts at 2 (3 terms) Group 3: starts at 5 (5 terms) Group 4: starts at 10 (7 terms) Group 5: starts at 17 (9 terms) Group 6: starts at 26 (11 terms) Group 7: starts at 37 (13 terms) Group 8: starts at 50 (15 terms) Group 9: starts at 65 (17 terms) Group 10: starts at 82 (19 terms) Group 11: starts at 101 (21 terms) Group 12: starts at 122 (23 terms) Group 13: starts at 145 (25 terms) Group 14: starts at 170 (27 terms) Group 15: starts at 197 (29 t

Question 29

CAT 2020 Slot 3: How many of the integers 1, 2, … , 120, are divisible by none of 2, 5 and 7?

Accepted Answer

Answer: A · 41. We are aske dto find the number of numbers that are divisible by none of 2, 5 and 7. Instead, let us find the number of numbers that are divisible by atleast one of 2, 5 and 7. Recall that n(A ∪ B ∪ C) = n(A) + n(B) + n(C) - n(A ∩ B) - n(B ∩ C) - n(C ∩ A) + n(A ∩ B ∩ C). Therefore, Number of Multiples of (2 or 5 or 7) = Number of Multiples of (2) + Number of Multiples of (5) + Number of Multiples of (7) - Number of Multiples of (2 and 5) - Number of Multiples of (5 and 7) - Number of Multiples of (7 and 2) + Number of Multiples of (2 and 5 and 7) Number of Multiples of (2 or 5 or 7) = Number o

Question 30

CAT 2020 Slot 3: Let N, x and y be positive integers such that N = x + y, 2 25, then how many distinct values are possible for N?

Accepted Answer

Answer: 6. 2 < x < 10 x can take any of the values from the set {3, 4, 5, 6, 7, 8, 9} 14 < y < 23 y can take any of the values from the set {15, 16, 17, 18, 19, 20, 21, 22} The highest value N (i.e x+y) can take = 9+22 = 31. (at x = 9; y = 22) 30 can be obtained at x = 9; y = 21 29 can be obtained at x = 9; y = 20 28 can be obtained at x = 9; y = 19 27 can be obtained at x = 9; y = 18 26 can be obtained at x = 9; y = 17 25 can be obtained at x = 9; y = 16 But, x+y=25 is not the desired sum, hence the different values of x+y are {31,30,29,28,27,26}. Hence, x+y, and thereby N can take 6 distinct values.

Question 31

CAT 2020 Slot 2: How many 4-digit numbers, each greater than 1000 and each having all four digits distinct, are there with 7 coming before 3?

Accepted Answer

Answer: 315. _ _ _ on the 4 digit 7 has to come before 3 So, ^4C_2 possibilities for 7 coming before 3 into the spaces 7 ___ 3 ____, 2 numbers are already there and 2 remaining spots we have to fill in From 8 digits available (Excluding 7 and 3) So, ^8C_2 ways of choosing After choosing two from 8, that two can be placed in any way For example : 1 and 2 can be arranged as 12 and 21 So, ^4C_2 × ^8C_2 × 2 Now we need to subtract the possibility where 0 comes in the first position 0753, So these are in the form 0__ __ __ In this 3 places 7 before 3 can be placed in ^3C_2 ways and remaining 1 digit Can be chos

Question 32

CAT 2019 Slot 2: How many factors of 2^4 × 3^5 × 10^4 are perfect squares which are greater than 1?

Accepted Answer

Answer: 44. 2^4 x 3^5 x 10^4 2^4 x 3^5 x 2^(4 )x 5^4 2^8 x 3^5 x 5^4 Now, for any number to be a perfect square, it must have an even number of factors. So, if we consider 2^8, only powers of 0, 2, 4, 6, 8 can lead to perfect squares - 5 ways If we consider 3^5, only powers of 0, 2, 4 can lead to perfect squares - 3 ways If we consider 5^4, only powers of 0. 2, 4 can lead to perfect squares - 3 ways So, total number of possibilities = 5 x 3 x 3 ways = 45 ways Since we need to find the number of factors greater than 1, Required number of ways = 45 -1 = 44 ways

Question 33

CAT 2019 Slot 1: The product of two positive numbers is 616. If the ratio of the difference of their cubes to the cube of their difference is 157 : 3, then the sum of the two numbers is

Accepted Answer

Answer: A · 50. Let the two numbers be a and b From the question, we know a x b = 616 and = = = = = - 1 = = = = 3 x 3 x 4 = 36 a - b = 6, ab = 616 616 can be rewritten as 22 x 28, where 28 - 22= 6 using trial and error So, a + b = 28 + 22 = 50

Question 34

CAT 2018 Slot 2: If N and x are positive integers such that N^N = 2^160 and N^2 + 2^N is an integral multiple of 2^x, then the largest possible x is

Accepted Answer

Answer: 10. Given that N and X are positive integers such that N^N = 2^(160) and N^2 + 2^N is an integral multiple of 2^x Now we have to find the largest possible value for x N^N = 2^(160) We can rewrite this 2^(160) as (2^p)^q such that 2^p = N = q pq = 160 We can try by substituting p and q with 2 and 80 respectively (2^2)^(80) = 4^(80) This doesn’t work so we can try by substituting p and q with 4 and 40 respectively (2^4)^(40) = 16^(40) This also doesn’t work so now we can substitute p and q with 5 and 32 respectively Such that (2^5)^(32) = 32^(32) Now we can say that N = 32 N^2 + 2^N ⟹ 32^2 + 2^(32)

Question 35

CAT 2018 Slot 2: If the sum of squares of two numbers is 97, then which one of the following cannot be their product?

Accepted Answer

Answer: A · 64. Given that the sum of squares of two numbers is 97 i.e. a^2 + b^2 = 97 From the given options we have to find which one cannot be their product i.e. ab A. 64 ⟹ 2ab = 128 B. −32 ⟹ 2ab = -64 C. 16 ⟹ 2ab = 32 D. 48 ⟹ 2ab = 96 2ab is found because we know that a^2 + b^2 + 2ab ≥ 0 a^2 + b^2 - 2ab ≥ 0 By this we can know that 97 + 128 works but 97 - 128 doesn’t works so we can understand option A cannot be the product and the rest can be. a^2 + b^2 ≥ |2ab| a^2 + b^2 ≥ 2ab a^2 + b^2 ≥ -2ab ⟹ a2+b2+2 ≥ |ab| So here 2ab should lie between +97 and -97 or ab should be less than 97/2 or greater than −97/2

Question 36

CAT 2018 Slot 2: How many two-digit numbers, with a non-zero digit in the units place, are there which are more than thrice the number formed by interchanging the positions of its digits?

Accepted Answer

Answer: D · 6. Let the two-digit number be xy, which can be expressed as 10x + y Given that the two-digit number is more than thrice the number obtained by interchanging the digits So, 10x + y > 3 × (10y + x) => 10x + y > 30y + 3x => 7x > 29y => x > 297297 × y Approximately, x > 4y Let us fix values for y and check for conditions, For y = 1, x can take the values of 5 , 6 , 7 , 8 , 9 (As, 51 > (3 × 15), 61 > (3 ×16), 71> (3 × 17), 81 > (3 × 18), 91 > (3 × 19)) Similarly, for y = 2, the condition satisfies only for x = 9 (As, 92 > (3 × 29)) The remaining values of y does not satisfy the given conditions. So,

Question 37

CAT 2018 Slot 1: The number of integers x such that 0.25 < 2^x < 200, and 2^x + 2 is perfectly divisible by either 3 or 4, is

Accepted Answer

Answer: 5. Given 0.25 ≤ 2x 0.25 = 1414 =(1212)^2 = 2^(-2) Similarly, 2^6 = 64, 2^7 = 128, 2^8 > 200 Substituting the values and checking for divisibility Number of Integers x = 5

Question 38

CAT 2018 Slot 1: While multiplying three real numbers, Ashok took one of the numbers as 73 instead of 37. As a result, the product went up by 720. Then the minimum possible value of the sum of squares of the other two numbers is

Accepted Answer

Answer: 40. Let the three real numbers be a, b, c. c is taken as 73 instead of 37 So, 73ab = 37ab + 720 36 ab = 720 ab = 720/36 = 20 We know, AM ≥ GM (a2+b2)/2 ≥ ab (a^2 + b^2) ≥ 2ab (a^2 + b^2) ≥ 40 (Or) ab = 20 (a,b) = (√20, √20) (a^2 + b^2) = 20 + 20 = 40

Question 39

CAT 2017 Slot 2: If the product of three consecutive positive integers is 15600 then the sum of the squares of these integers is

Accepted Answer

Answer: D · 1877. Given that the product of three consecutive positive integers is 15600 and we have to find the sum of the squares of the integers As we have two zeroes in 15600 we should have some number ending in zero or in five 156 = 12 × 13 15600 = 12 × 13 × 100 = 12 × 13 × 25 × 4 15600 = 24 × 25 × 26 The product of n (n+1)(n+2) should have a multiple of 5^2 The sum of the squares of these integers 24 , 25 , 26 is 24^2 + 26^2 + 25^2 = 576 + 625 + 676 24^2 + 26^2 + 25^2 = 1877