Mechanical Engineering Exam  >  Mechanical Engineering Tests  >  General Aptitude for GATE  >  Test: HCF and LCM- 1 - Mechanical Engineering MCQ

Test: HCF and LCM- 1 - Mechanical Engineering MCQ


Test Description

10 Questions MCQ Test General Aptitude for GATE - Test: HCF and LCM- 1

Test: HCF and LCM- 1 for Mechanical Engineering 2024 is part of General Aptitude for GATE preparation. The Test: HCF and LCM- 1 questions and answers have been prepared according to the Mechanical Engineering exam syllabus.The Test: HCF and LCM- 1 MCQs are made for Mechanical Engineering 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: HCF and LCM- 1 below.
Solutions of Test: HCF and LCM- 1 questions in English are available as part of our General Aptitude for GATE for Mechanical Engineering & Test: HCF and LCM- 1 solutions in Hindi for General Aptitude for GATE course. Download more important topics, notes, lectures and mock test series for Mechanical Engineering Exam by signing up for free. Attempt Test: HCF and LCM- 1 | 10 questions in 15 minutes | Mock test for Mechanical Engineering preparation | Free important questions MCQ to study General Aptitude for GATE for Mechanical Engineering Exam | Download free PDF with solutions
Test: HCF and LCM- 1 - Question 1

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?

Detailed Solution for Test: HCF and LCM- 1 - Question 1

L.C.M. of 2, 4, 6, 8, 10, 12 is 120.So, the bells will toll together after every 120 seconds, i.e, 2 minutes.In 30 minutes, they will toll together 30/2 + 1 = 16

Test: HCF and LCM- 1 - Question 2

The least multiple of 7, which leaves a remainder of 4, when divided by 6, 9, 15 and 18 is:

Detailed Solution for Test: HCF and LCM- 1 - Question 2

L.C.M. of 6, 9, 15 and 18 is 90.
Let required number be 90k + 4, which is multiple of 7.
Least value of k for which (90k + 4) is divisible by 7 is k = 4.
Required number = (90 x 4) + 4 = 364.

1 Crore+ students have signed up on EduRev. Have you? Download the App
Test: HCF and LCM- 1 - Question 3

The least number, which when divided by 48, 60, 72, 108 and 140 leaves 38, 50, 62, 98 and 130 as remainders respectively, is:

Detailed Solution for Test: HCF and LCM- 1 - Question 3

Here (48 – 38) = 10, (60 – 50) = 10, (72 – 62) = 10, (108 – 98) = 10 & (140 – 130) = 10.
Required number = (L.C.M. of 48, 60, 72, 108, 140) – 10
= 15120 – 10 = 15110

Test: HCF and LCM- 1 - Question 4

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 Test: HCF and LCM- 1 - Question 4

Other number =[11 x 7700]/275 = 308

Test: HCF and LCM- 1 - 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 meet again at the starting point?

Detailed Solution for Test: HCF and LCM- 1 - 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 see i.e., 46 min. 12 sec

Test: HCF and LCM- 1 - Question 6

The L.C.M. of two numbers is 48. The numbers are in the ratio 2 : 3. Then sum of the number is:

Detailed Solution for Test: HCF and LCM- 1 - Question 6

Let the numbers be 2x and 3x.
Then, their L.C.M. = 6x.
So, 6x = 48 or x = 8.
The numbers are 16 and 24.
Hence, required sum = (16 + 24) = 40.

Test: HCF and LCM- 1 - Question 7

The least number, which when divided by 12, 15, 20 and 54 leaves in each case a remainder of 8 is:

Detailed Solution for Test: HCF and LCM- 1 - Question 7

Required number = (L.C.M. of 12, 15, 20, 54) + 8
= 540 + 8
= 548.

Test: HCF and LCM- 1 - Question 8

The product of two numbers is 4107. If the H.C.F. of these numbers is 37, then the greater number is:

Detailed Solution for Test: HCF and LCM- 1 - Question 8

Let the numbers be 37a and 37b. Then, 37a x 37b = 4107
ab = 3.
Now, co-primes with product 3 are (1, 3).
So, the required numbers are (37 x 1, 37 x 3) i.e., (37, 111).
Greater number = 111.

Test: HCF and LCM- 1 - Question 9

The product of two numbers is 2028 and their H.C.F. is 13. The number of such pairs is:

Detailed Solution for Test: HCF and LCM- 1 - Question 9

Let the numbers 13a and 13b.
Then, 13a x 13b = 2028
⇒ ab = 12.
Now, the co-primes with product 12 are (1, 12) and (3, 4).
[Note: Two integers a and b are said to be coprime or relatively prime if they have no common positive factor other than 1 or, equivalently, if their greatest common divisor is 1 ]
So, the required numbers are (13 x 1, 13 x 12) and (13 x 3, 13 x 4).
Clearly, there are 2 such pairs.

Test: HCF and LCM- 1 - Question 10

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 Test: HCF and LCM- 1 - Question 10

L.C.M. of 5, 6, 4 and 3 = 60. On dividing 2497 by 60, the remainder is 37. Number to be added = (60 – 37) = 23

198 videos|165 docs|152 tests
Information about Test: HCF and LCM- 1 Page
In this test you can find the Exam questions for Test: HCF and LCM- 1 solved & explained in the simplest way possible. Besides giving Questions and answers for Test: HCF and LCM- 1, EduRev gives you an ample number of Online tests for practice

Top Courses for Mechanical Engineering

198 videos|165 docs|152 tests
Download as PDF

Top Courses for Mechanical Engineering