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Arun Sharma Test: HCF & LCM - CUET Commerce MCQ


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10 Questions MCQ Test General Test Preparation for CUET UG - Arun Sharma Test: HCF & LCM

Arun Sharma Test: HCF & LCM for CUET Commerce 2025 is part of General Test Preparation for CUET UG preparation. The Arun Sharma Test: HCF & LCM questions and answers have been prepared according to the CUET Commerce exam syllabus.The Arun Sharma Test: HCF & LCM MCQs are made for CUET Commerce 2025 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Arun Sharma Test: HCF & LCM below.
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Arun Sharma Test: HCF & LCM - Question 1

Find the lowest common multiple of 24, 36 and 40.

Detailed Solution for Arun Sharma Test: HCF & LCM - Question 1

Arun Sharma Test: HCF & LCM - Question 2

The least number which should be added to 2497 so that the sum is exactly divisible by 5, 6, 4 and 3 is:

Detailed Solution for Arun Sharma Test: HCF & LCM - Question 2

First, we calculate the least common multiple (LCM) of these numbers. The LCM of 5, 6, 4, and 3 is 60. We need to determine the smallest number to add to 2497 such that the result is divisible by 60.

When 2497 is divided by 60, the remainder is 37 (since 2497 divided by 60 equals 41 with a remainder of 37). To achieve divisibility, we must add 23 (60 - 37) to 2497, making the correct answer option C (23).

Arun Sharma Test: HCF & LCM - Question 3

Detailed Solution for Arun Sharma Test: HCF & LCM - Question 3

Arun Sharma Test: HCF & LCM - Question 4

The least number which when divided by 5, 6 , 7 and 8 leaves a remainder 3, but when divided by 9 leaves no remainder, is:

Detailed Solution for Arun Sharma Test: HCF & LCM - Question 4

We need to find the least number that, when divided by 5, 6, 7, and 8, leaves a remainder of 3, and when divided by 9 leaves no remainder (i.e., is divisible by 9).

First, the condition "divided by 5, 6, 7, and 8 leaves a remainder of 3" means the number can be expressed as: Number = LCM of (5, 6, 7, 8) * k + 3, where k is a positive integer.

Calculate the LCM of 5, 6, 7, and 8:

  • Prime factors: 5 = 5, 6 = 2 * 3, 7 = 7, 8 = 2^3
  • LCM = 2^3 * 3 * 5 * 7 = 8 * 3 * 5 * 7 = 840

So, the number can be written as: Number = 840k + 3

Additionally, the number must be divisible by 9 (remainder 0), so: 840k + 3 ≡ 0 (mod 9)

First, find 840 mod 9: 8 + 4 + 0 = 12, 1 + 2 = 3, so 840 ≡ 3 (mod 9)

Now substitute: 3k + 3 ≡ 0 (mod 9) 3k ≡ -3 (mod 9) Since -3 + 9 = 6, this is: 3k ≡ 6 (mod 9)

Divide by 3 (note that 3 and 9 are not coprime, so we check values): k ≡ 2 (mod 3) (because 3 * 2 = 6, and 6 mod 9 = 6)

So, k = 3m + 2 (where m is a non-negative integer).

Substitute k = 3m + 2 into the number: Number = 840 * (3m + 2) + 3 = 2520m + 1680 + 3 = 2520m + 1683

The smallest positive value occurs when m = 0: Number = 1683

Verify:

  • 1683 ÷ 5 = 336 remainder 3 (1680 is divisible by 5, 1683 - 1680 = 3)
  • 1683 ÷ 6 = 280 remainder 3 (1674 is divisible by 6, 1683 - 1674 = 9, but adjust: 1680 ÷ 6 = 280, 1683 - 1680 = 3)
  • 1683 ÷ 7 = 240 remainder 3 (1680 ÷ 7 = 240, 1683 - 1680 = 3)
  • 1683 ÷ 8 = 210 remainder 3 (1680 ÷ 8 = 210, 1683 - 1680 = 3)
  • 1683 ÷ 9 = 187 remainder 0 (exact division)

All conditions are satisfied.

Answer: b) 1683

Arun Sharma Test: HCF & LCM - Question 5

A, B and C start at the same time in the same direction to run around a circular stadium. A completes a round in 252 seconds, B in 308 seconds and c in 198 seconds, all starting at the same point. After what time will they again at the starting point ?

Detailed Solution for Arun Sharma Test: HCF & LCM - Question 5

L.C.M. of 252, 308 and 198 = 2772.

So, A, B and C will again meet at the starting point in 2772 sec. i.e., 46 min. 12 sec.

Arun Sharma Test: HCF & LCM - Question 6

The H.C.F. of two numbers is 11 and their L.C.M. is 7700. If one of the numbers is 275, then the other is:

Detailed Solution for Arun Sharma Test: HCF & LCM - Question 6

Arun Sharma Test: HCF & LCM - Question 7

What will be the least number which when doubled will be exactly divisible by 12, 18, 21 and 30 ?

Detailed Solution for Arun Sharma Test: HCF & LCM - Question 7

Arun Sharma Test: HCF & LCM - Question 8

The ratio of two numbers is 3 : 4 and their H.C.F. is 4. Their L.C.M. is:

Detailed Solution for Arun Sharma Test: HCF & LCM - Question 8

Let the numbers be 3x and 4x. Then, their H.C.F. = x. So, x = 4.

So, the numbers 12 and 16.

L.C.M. of 12 and 16 = 48.

Arun Sharma Test: HCF & LCM - Question 9

The smallest number which when diminished by 7, is divisible 12, 16, 18, 21 and 28 is:

Detailed Solution for Arun Sharma Test: HCF & LCM - Question 9

Required number = (L.C.M. of 12,16, 18, 21, 28) + 7

   = 1008 + 7

   = 1015

Arun Sharma Test: HCF & LCM - Question 10

252 can be expressed as a product of primes as:

Detailed Solution for Arun Sharma Test: HCF & LCM - Question 10

The prime factorization of 252 is calculated as follows:

  • 252 ÷ 2 = 126 (One 2)
  • 126 ÷ 2 = 63 (Two 2s)
  • 63 ÷ 3 = 21 (One 3)
  • 21 ÷ 3 = 7 (Two 3s)
  • 7 is a prime number (One 7)

Thus, 252 = 2 × 2 × 3 × 3 × 7. Therefore, the correct answer is A.

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