Numbers in MBA CET

Quant Numbers 10 mins Mahayagya TestDownload
Numbers TopicSlot 1Slot 2Slot 3Slot 4Slot 5Slot 6Total
HCF & LCM14212515
Sum of Numbers13412415
Remainder Theorem111216
Divisibility Rules1225
Unit Digit11114
Grand Total599541345
MBA CET 2026: HCF, LCM & Number Systems | Interactive Quiz

MBA CET 2026 Numbers

MBA CET 2026 Quant Actual Questions
Q1. The number 63_42_0 is a 7-digit number divisible by both 3 and 5. If the first missing digit is greater than the tens digit, then which of the following options gives the correct missing digits respectively?
  • A) 3,1
  • B) 6,4
  • C) 9,7
  • D) 9,6
  • E) 5,9
Correct Option: D Rationale: Divisibility by 5 is already satisfied (ends in 0). For divisibility by 3, digit sum = 15 + 9 + 6 = 30, which is divisible by 3, and 9 > 6. Hence the missing digits are 9 and 6.
Q2. Three positive integers are such that the difference between the largest and the smallest number is 180. The HCF of the three numbers is the greatest possible, and the smallest number is 90. If the middle number differs from the smallest number by 60, what is the HCF of the three numbers?
  • A) 14
  • B) 21
  • C) 28
  • D) 30
  • E) 35
Correct Option: D Rationale: The three numbers are 90, 150 and 270. HCF = HCF(90, 150, 270) = 30.
Q3. Three bells ring at intervals of 30 seconds, 45 seconds, and 60 seconds. After how much time will they ring together again?
  • A) 90 seconds
  • B) 180 seconds
  • C) 120 seconds
  • D) 240 seconds
  • E) 300 seconds
Correct Option: B Rationale: They ring together after LCM(30, 45, 60) = 180 seconds.
Q4. How many 3-digit numbers are there between 100 and 300, having the first and the last digit as 2?
  • A) 9
  • B) 10
  • C) 11
  • D) 12
  • E) 13
Correct Option: B Rationale: The numbers are 202, 212, 222, … , 292. The middle digit can be 0 to 9, giving 10 numbers.
Q5. The HCF of 2923 and 3239 is:
  • A) 37
  • B) 47
  • C) 73
  • D) 79
  • E) 22
Correct Option: D Rationale: 3239 − 2923 = 316 and 2923 = 79 × 37, while 316 = 79 × 4. Hence HCF = 79.
Q6. The HCF of 2472, 1284 and a third number n is 12. If their LCM is 2³ × 3² × 5 × 103 × 107, then which of the following could be the value of n?
  • A) 2² × 3² × 7
  • B) 2² × 3³ × 103
  • C) 2² × 3² × 5
  • D) 3³ × 103
  • E) 2² × 3²
Correct Option: C Rationale: 2472 = 2³ × 3 × 103 and 1284 = 2² × 3 × 107. To obtain the required LCM, n must introduce 3² and 5 without introducing any new prime, so n = 2² × 3² × 5.
Q7. The smallest perfect square number that is exactly divisible by 2, 3, 4, 6, 9, 18, 30 and 60 is k², where k is a natural number. What is the sum of all prime factors of k?
  • A) 8
  • B) 9
  • C) 10
  • D) 11
  • E) 12
Correct Option: C Rationale: LCM of the given numbers = 2² × 3² × 5 = 180. The smallest perfect-square multiple is 2² × 3² × 5² = 900 = 30². Prime factors of k (which is 30) = 2, 3 and 5 → sum = 10.
Q8. Find the remainder when 73 + 75 + 78 + 57 + 197 is divided by 34.
  • A) 2
  • B) 4
  • C) 5
  • D) 7
  • E) 9
Correct Option: B Rationale: 73 + 75 + 78 + 57 + 197 = 480. Since 480 = 34 × 14 + 4, the remainder is 4.
Q9. If n is any odd number greater than 1, then n(n − 1):
  • A) is divisible by 15 always
  • B) is divisible by 24 always
  • C) is divisible by 48 always
  • D) is divisible by 100 always
  • E) None of these
Correct Option: E (None of these) Rationale: The statement is not true for all odd n. For example, n = 3 gives n(n − 1) = 6, which is not divisible by 15, 24, 48, 96 or 100.
Q10. If the seven-digit number 84392X6 is divisible by 9, find the single-digit value of X.
  • A) 3
  • B) 4
  • C) 5
  • D) 6
  • E) 7
Correct Option: B Rationale: Sum of digits = 8 + 4 + 3 + 9 + 2 + X + 6 = 32 + X. For divisibility by 9, 32 + X = 36 → X = 4.
Q11. What least number must be added to 1056, so that the sum is completely divisible by 23?
  • A) 2
  • B) 4
  • C) 3
  • D) 5
  • E) 1
Correct Option: A Rationale: 1056 ÷ 23 leaves remainder 21. Therefore, 23 − 21 = 2 must be added.
Q12. Find the difference between the place value and the face value of the digit 7 in the number 4782.
  • A) 700
  • B) 7
  • C) 693
  • D) 782
  • E) 707
Correct Option: C Rationale: Place value of 7 = 700 and face value = 7. Difference = 700 − 7 = 693.
Q13. What is the units digit of the product 15 × 22 × 37 × 44?
  • A) 2
  • B) 5
  • C) 0
  • D) 4
  • E) 8
Correct Option: C Rationale: The product contains 15 × 22, whose units digits give 5 × 2 = 10. Therefore, the entire product ends in 0.
Q14. Six bells commence tolling together and toll at intervals of 2, 4, 6, 8, 10 and 12 seconds respectively. In 30 minutes, how many times do they toll together?
  • A) 15
  • B) 16
  • C) 30
  • D) 31
  • E) 12
Correct Option: B Rationale: LCM(2, 4, 6, 8, 10, 12) = 120 seconds. In 30 minutes = 1800 seconds, they meet 1800/120 + 1 = 16 times, including the starting time.
Q15. Three numbers are in the ratio 3 : 4 : 5, and their LCM is 2400. Their HCF is:
  • A) 40
  • B) 80
  • C) 120
  • D) 200
  • E) 15
Correct Option: A Rationale: Let the numbers be 3x, 4x and 5x. Their LCM = 60x = 2400 → x = 40. Hence their HCF = 40.
© 2026 MBA CET Preparation Portal by Cetking – Targeted Practice for MBA Success.
MBA CET 2026: HCF, LCM & Number Systems | Interactive Quiz

MBA CET 2026 Numbers

Q1. The number 63_42_0 is a 7-digit number divisible by both 3 and 5. If the first missing digit is greater than the tens digit, then which of the following options gives the correct missing digits respectively?
  • A) 3,1
  • B) 6,4
  • C) 9,7
  • D) 9,6
  • E) 5,9
Correct Option: D Rationale: Divisibility by 5 is already satisfied (ends in 0). For divisibility by 3, digit sum = 15 + 9 + 6 = 30, which is divisible by 3, and 9 > 6. Hence the missing digits are 9 and 6.
Q2. Three positive integers are such that the difference between the largest and the smallest number is 180. The HCF of the three numbers is the greatest possible, and the smallest number is 90. If the middle number differs from the smallest number by 60, what is the HCF of the three numbers?
  • A) 14
  • B) 21
  • C) 28
  • D) 30
  • E) 35
Correct Option: D Rationale: The three numbers are 90, 150 and 270. HCF = HCF(90, 150, 270) = 30.
Q3. Three bells ring at intervals of 30 seconds, 45 seconds, and 60 seconds. After how much time will they ring together again?
  • A) 90 seconds
  • B) 180 seconds
  • C) 120 seconds
  • D) 240 seconds
  • E) 300 seconds
Correct Option: B Rationale: They ring together after LCM(30, 45, 60) = 180 seconds.
Q4. How many 3-digit numbers are there between 100 and 300, having the first and the last digit as 2?
  • A) 9
  • B) 10
  • C) 11
  • D) 12
  • E) 13
Correct Option: B Rationale: The numbers are 202, 212, 222, … , 292. The middle digit can be 0 to 9, giving 10 numbers.
Q5. The HCF of 2923 and 3239 is:
  • A) 37
  • B) 47
  • C) 73
  • D) 79
  • E) 22
Correct Option: D Rationale: 3239 − 2923 = 316 and 2923 = 79 × 37, while 316 = 79 × 4. Hence HCF = 79.
Q6. The HCF of 2472, 1284 and a third number n is 12. If their LCM is 2³ × 3² × 5 × 103 × 107, then which of the following could be the value of n?
  • A) 2² × 3² × 7
  • B) 2² × 3³ × 103
  • C) 2² × 3² × 5
  • D) 3³ × 103
  • E) 2² × 3²
Correct Option: C Rationale: 2472 = 2³ × 3 × 103 and 1284 = 2² × 3 × 107. To obtain the required LCM, n must introduce 3² and 5 without introducing any new prime, so n = 2² × 3² × 5.
Q7. The smallest perfect square number that is exactly divisible by 2, 3, 4, 6, 9, 18, 30 and 60 is k², where k is a natural number. What is the sum of all prime factors of k?
  • A) 8
  • B) 9
  • C) 10
  • D) 11
  • E) 12
Correct Option: C Rationale: LCM of the given numbers = 2² × 3² × 5 = 180. The smallest perfect-square multiple is 2² × 3² × 5² = 900 = 30². Prime factors of k (which is 30) = 2, 3 and 5 → sum = 10.
Q8. Find the remainder when 73 + 75 + 78 + 57 + 197 is divided by 34.
  • A) 2
  • B) 4
  • C) 5
  • D) 7
  • E) 9
Correct Option: B Rationale: 73 + 75 + 78 + 57 + 197 = 480. Since 480 = 34 × 14 + 4, the remainder is 4.
Q9. If n is any odd number greater than 1, then n(n − 1):
  • A) is divisible by 15 always
  • B) is divisible by 24 always
  • C) is divisible by 48 always
  • D) is divisible by 100 always
  • E) None of these
Correct Option: E (None of these) Rationale: The statement is not true for all odd n. For example, n = 3 gives n(n − 1) = 6, which is not divisible by 15, 24, 48, 96 or 100.
Q10. If the seven-digit number 84392X6 is divisible by 9, find the single-digit value of X.
  • A) 3
  • B) 4
  • C) 5
  • D) 6
  • E) 7
Correct Option: B Rationale: Sum of digits = 8 + 4 + 3 + 9 + 2 + X + 6 = 32 + X. For divisibility by 9, 32 + X = 36 → X = 4.
Q11. What least number must be added to 1056, so that the sum is completely divisible by 23?
  • A) 2
  • B) 4
  • C) 3
  • D) 5
  • E) 1
Correct Option: A Rationale: 1056 ÷ 23 leaves remainder 21. Therefore, 23 − 21 = 2 must be added.
Q12. Find the difference between the place value and the face value of the digit 7 in the number 4782.
  • A) 700
  • B) 7
  • C) 693
  • D) 782
  • E) 707
Correct Option: C Rationale: Place value of 7 = 700 and face value = 7. Difference = 700 − 7 = 693.
Q13. What is the units digit of the product 15 × 22 × 37 × 44?
  • A) 2
  • B) 5
  • C) 0
  • D) 4
  • E) 8
Correct Option: C Rationale: The product contains 15 × 22, whose units digits give 5 × 2 = 10. Therefore, the entire product ends in 0.
Q14. Six bells commence tolling together and toll at intervals of 2, 4, 6, 8, 10 and 12 seconds respectively. In 30 minutes, how many times do they toll together?
  • A) 15
  • B) 16
  • C) 30
  • D) 31
  • E) 12
Correct Option: B Rationale: LCM(2, 4, 6, 8, 10, 12) = 120 seconds. In 30 minutes = 1800 seconds, they meet 1800/120 + 1 = 16 times, including the starting time.
Q15. Three numbers are in the ratio 3 : 4 : 5, and their LCM is 2400. Their HCF is:
  • A) 40
  • B) 80
  • C) 120
  • D) 200
  • E) 15
Correct Option: A Rationale: Let the numbers be 3x, 4x and 5x. Their LCM = 60x = 2400 → x = 40. Hence their HCF = 40.
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