Explore Exams
Login Signup
Sign in Open App

All questions of LCM and HCF for Mechanical Engineering Exam

Clear all your doubts with EduRev

Sum of 2 numbers is 128 and their HCF is 8. How many numbers of pairs of numbers will satisfy this condition?
  • a)
    2
  • b)
    5
  • c)
    3
  • d)
    6
  • e)
    4
Correct answer is option 'E'. Can you explain this answer?

Preeti Khanna answered
Since HCF = 8 is highest common factor among those numbers
So, let first no = 8x, 2nd number = 8y
So 8x + 8y = 128
x + y = 16
the co-prime numbers which sum up 16 are (1,15), (3,13), (5,11), and (7,9) *co-prime numbers are those which do no have any factor in common. These pairs are taken here because in HCF highest common factor is taken, so the remaining multiples x and y must have no common factors.
Since there are 4 pairs of co-prime numbers here, there will be 4 pairs of numbers satisfying given conditions. These are (8*1, 8*15), (8*3, 8*13) , (8*5, 8*11) , (8*7, 8*9)

Find the least number which when divided by 2, 5, 9 and 12 leaves a remainder 3 but leaves no remainder when same number is divided by 11.
  • a)
    281
  • b)
    357
  • c)
    360
  • d)
    363
  • e)
    224
Correct answer is option 'D'. Can you explain this answer?

Logith S answered
(USE THIS METHOD ONLY IF ANY ONE OF THE OPTIONS REMAINDERS AS 0 and don't do this method if all the options got the remainder as 0).
So the question was by dividing 11 there will be no remainder. I divided the numbers in the option with 11 and I got remainders as
a) remainder is 6
b)remainder is 5
c)remainder is 8
d) remainder is 0
e) remainder is different

so the option d leaves a remainder is 0 the answer is 363.
I solved this problem by using this method.

The product of two numbers is 2028 and their H.C.F. is 13. The number of such pairs is:
  • a)
    1
  • b)
    2
  • c)
    3
  • d)
    5
  • e)
    None of these
Correct answer is option 'B'. Can you explain this answer?

Faizan Khan answered
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.

Find the smallest number of 4 digits which is exactly divisible by 12, 15, 21 and 30.
  • a)
    1580
  • b)
    1420
  • c)
    1260
  • d)
    1056
  • e)
    None of these
Correct answer is option 'C'. Can you explain this answer?

Preeti Khanna answered
Least 4 digit number = 1000
LCM of (12, 15, 21, 30) = 420
On dividing 420 by 1000, leaves remainder = 160
So answer = 1000 + (420 – 160)

HCF and LCM of two numbers are 11 and 385 .If one number lies between 75 and 125 then that number is
  • a)
    123
  • b)
    73
  • c)
    77
  • d)
    154
  • e)
    None of these
Correct answer is option 'C'. Can you explain this answer?

Gaurav Chaurasiya answered
Let the Nos be X and y.
then it is quite clear that,
11x*11y=11*385. (X*Y=HCF*LCM)

•. XY=35 and X and y should be. co-prime
therefore 5*7=x*y would be the required
pair

hence first no is 55 and second no is 77
therefore 77 is the absolute. answer for this question

The HCF of 3 different no is 17, Which of the following cannot be their LCM ?
  • a)
    540
  • b)
    289
  • c)
    340
  • d)
    425
  • e)
    None of these
Correct answer is option 'A'. Can you explain this answer?

Nikita Singh answered
To check whether the given number is it's LCM or not just divide the LCM by HCF .

1) 540 ÷ 17 = 31.76

2) 289 ÷ 17 = 17

3) 340 ÷ 17 = 20

4) 425 ÷ 17 = 25 .


540 is not divisible by 17.


Hence : 540 is not the LCM .

The least multiple of 7, which leaves a remainder of 4, when divided by 6, 9, 15 and 18 is:
  • a)
    68
  • b)
    98
  • c)
    180
  • d)
    364
  • e)
    None of these
Correct answer is option 'D'. Can you explain this answer?

Chirag Nair answered
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.

The L.C.M. of two numbers is 48. The numbers are in the ratio 2 : 3. Then sum of the number is:
  • a)
    30
  • b)
    22
  • c)
    40
  • d)
    60
  • e)
    None of these
Correct answer is option 'C'. Can you explain this answer?

Anaya Patel answered
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.

The product of two numbers is 4107. If the H.C.F. of these numbers is 37, then the greater number is:
  • a)
    124
  • b)
    100
  • c)
    111
  • d)
    175
  • e)
    None of these
Correct answer is option 'C'. Can you explain this answer?

Aarav Sharma answered
The given question is related to finding the greater number out of two numbers whose product is given and the highest common factor (HCF) is also given. Let's solve it step by step.

Given:
Product of two numbers = 4107
HCF of the two numbers = 37

Step 1: Prime Factorization of 4107
To find the prime factorization of 4107, we divide it by prime numbers starting from 2 until we cannot divide any further. The prime factors of 4107 are:

4107 ÷ 3 = 1369
1369 ÷ 37 = 37

So, the prime factorization of 4107 is 3 × 37 × 37.

Step 2: Determining the Numbers
Now that we have the prime factorization, we can determine the two numbers whose product is 4107 and whose HCF is 37. Since the HCF is 37, one of the numbers should be a multiple of 37. Let's assign the factors as follows:

Number 1 = 37 × a
Number 2 = 37 × b

Multiplying the two numbers, we get:
Number 1 × Number 2 = (37 × a) × (37 × b) = 4107

Expanding the equation:
(37 × 37) × (a × b) = 4107

Simplifying:
1369 × (a × b) = 4107

Dividing by 1369 on both sides:
a × b = 3

Step 3: Finding the Greater Number
We need to find the greater number out of the two. To do this, we need to consider the values of a and b. Since a and b are positive integers and their product is 3, the possible values for a and b are:

a = 1, b = 3
a = 3, b = 1

Substituting the values back into the equation, we get the two numbers:

Number 1 = 37 × a = 37 × 1 = 37
Number 2 = 37 × b = 37 × 3 = 111

Comparing the two numbers, we can see that the greater number is 111.

Therefore, the correct answer is option C) 111.

Three boxes of lengths 60m,30m and 45m are to be cut into pieces of equal length. What is the greatest possible length of each pieces ?
  • a)
    15m
  • b)
    18m
  • c)
    5m
  • d)
    3m
  • e)
    None of these
Correct answer is option 'A'. Can you explain this answer?

Sagar Sharma answered
Explanation:
- To find the greatest possible length of each piece, we need to find the greatest common divisor (GCD) of the lengths of the boxes.
- The lengths of the boxes are 60m, 30m, and 45m.
- We can find the GCD of these lengths by finding the GCD of the pairwise combinations of the lengths.
- GCD(60,30) = 30
- GCD(30,45) = 15
- GCD(60,45) = 15
- Therefore, the greatest common divisor of 60, 30, and 45 is 15m.
- So, the greatest possible length of each piece that can be cut from the boxes is 15m.
Therefore, the correct answer is option A) 15m.

Two containers contain 50 and 125 litres of water respectively. Find the maximum capacity of a container which can measure the water in each container an exact number of times(in litres)
  • a)
    25
  • b)
    11
  • c)
    12
  • d)
    15
  • e)
    None of these
Correct answer is option 'A'. Can you explain this answer?

Aarav Sharma answered
Problem: Two containers contain 50 and 125 litres of water respectively. Find the maximum capacity of a container which can measure the water in each container an exact number of times(in litres)

Solution:
We can find the maximum capacity of a container by finding the HCF of the given volumes of water.

Step 1: Prime Factorization
50 = 2 x 5^2
125 = 5^3

Step 2: Identify the common factor(s)
The only common factor is 5.

Step 3: Find the HCF
HCF = 5

Therefore, the maximum capacity of a container is 5 litres, as it can measure 50 litres (10 times) and 125 litres (25 times) an exact number of times.

Answer: (a) 25

A rectangular courtyard 140cm long 525cm wide is to be paved exactly with square tiles, all of the same size. what is the largest size of the tile which could be used for the purpose?
  • a)
    64cm
  • b)
    35cm
  • c)
    21cm
  • d)
    28cm
  • e)
    None of these
Correct answer is option 'B'. Can you explain this answer?

Aarav Sharma answered
Problem: To find the largest size of a square tile that can be used to completely pave a rectangular courtyard.

Given: Length of courtyard = 140cm, width of courtyard = 525cm.

Solution:

To find the largest size of the tile, we need to find the common factor of both the length and width of the courtyard.

Step 1: Find the HCF (Highest Common Factor) of 140 and 525.

We can use the prime factorization method to find the HCF.

140 = 2 × 2 × 5 × 7
525 = 3 × 5 × 5 × 7

Multiplying the common factors, we get:

HCF = 5 × 7 = 35

Step 2: The largest size of the tile that can be used is equal to the HCF.

Therefore, the largest size of the tile that can be used is 35cm.

Hence, option (B) is the correct answer.

Note: It is important to find the HCF of the length and width of the courtyard to ensure that the entire area is covered without leaving any gaps. Using a tile size that is not a factor of both the length and width will result in incomplete paving.

A person has 3 bars whose length are 12,16,24m respectively. He want to cuts the longest possible pieces, all of the same length from each of the 3 bars, what is the length of the each piece, if he is cut without any wastage
  • a)
    12m
  • b)
    20m
  • c)
    6m
  • d)
    4m
  • e)
    None of these
Correct answer is option 'D'. Can you explain this answer?

Aarav Sharma answered
Solution:

To find the length of the longest possible pieces that can be cut from the three bars, we need to find the greatest common divisor (GCD) of the three lengths.

Step 1: Find the factors of the lengths of the bars

The factors of 12 are 1, 2, 3, 4, 6, and 12.

The factors of 16 are 1, 2, 4, 8, and 16.

The factors of 24 are 1, 2, 3, 4, 6, 8, 12, and 24.

Step 2: Find the common factors

The common factors of the three lengths are 1, 2, 4, and 6.

Step 3: Find the largest common factor

The largest common factor is 6. Therefore, the longest possible pieces that can be cut from each of the three bars will be 6 meters long.

Answer: Option D (6m)

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)
    8
  • b)
    11
  • c)
    13
  • d)
    16
  • e)
    None of these
Correct answer is option 'D'. Can you explain this answer?

Pallabi Deshpande answered
LCM of 2, 4, 6, 8 10 and 12 is 120.

So, after each 120 seconds, they would toll together.

Hence, in 30 minutes, they would toll 30*60 seconds / 120 seconds = 15 times

But then the question says they commence tolling together. So, they basically also toll at the "beginning" ("0" second).

So, total tolls together = 15+1 = 16

The least number which should be added to 2497 so that the sum is exactly divisible by 5, 6, 4 and 3 is:
  • a)
    10
  • b)
    14
  • c)
    23
  • d)
    30
  • e)
    None of these
Correct answer is option 'C'. Can you explain this answer?

Rajesh Khatri answered
L.C.M. of 5, 6, 4 and 3 = 60.

On dividing 2497 by 60, the remainder is 37.

L.C.M. of 5, 6, 4 and 3 = 60. Number to be added = (60 - 37) = 23.

The least number, which when divided by 48, 60, 72, 108 and 140 leaves 38, 50, 62, 98 and 130 as remainders respectively, is:
  • a)
    11115
  • b)
    15110
  • c)
    15130
  • d)
    15310
  • e)
    None of these
Correct answer is option 'B'. Can you explain this answer?

Varun Rane answered
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

HCF and LCM of two numbers is 5 and 275 respectively and the sum of these two numbers is 80. Find the sum of the reciprocals of these numbers.
  • a)
    15/242
  • b)
    16/275
  • c)
    32/221
  • d)
    20/268
  • e)
    None of these
Correct answer is option 'B'. Can you explain this answer?

Sagar Sharma answered
Understanding the Problem
To solve for the two numbers given their HCF, LCM, and sum, we can use the relationship between HCF, LCM, and the product of the two numbers.
Key Relationships
- HCF (Highest Common Factor) = 5
- LCM (Least Common Multiple) = 275
- Sum of the two numbers = 80
Finding the Product of the Numbers
Using the relationship:
- HCF × LCM = Product of the two numbers
Therefore, we calculate:
- 5 × 275 = 1375
Let the two numbers be x and y. Hence, we have:
- x + y = 80
- xy = 1375
Forming the Quadratic Equation
From the equations, we can form:
- t^2 - (x+y)t + xy = 0
- t^2 - 80t + 1375 = 0
Calculating the Roots
Using the quadratic formula:
- t = [80 ± sqrt((80)^2 - 4 × 1 × 1375)] / 2
Calculating the discriminant:
- (80)^2 - 4 × 1375 = 6400 - 5500 = 900
Thus, we have:
- t = [80 ± 30] / 2
This gives us:
- t = 55 and t = 25
So, the two numbers are 55 and 25.
Finding the Sum of the Reciprocals
The sum of the reciprocals of the numbers is given by:
- (1/x) + (1/y) = (y + x) / (xy)
Substituting the values:
- (55 + 25) / (55 × 25) = 80 / 1375 = 16 / 275
Conclusion
The sum of the reciprocals of the two numbers is:
- 16/275
Thus, the correct answer is option 'B'.

In a college, all the students can stand in a row, so that each row has 9, 7 and 12 students. Find the least no of students in the school ?
  • a)
    145
  • b)
    265
  • c)
    186
  • d)
    252
  • e)
    None of these
Correct answer is option 'D'. Can you explain this answer?

Aarav Sharma answered
Approach:

- The least number of students in the school will be the LCM of 9, 7, and 12.
- We can find the LCM by prime factorizing the numbers and taking the highest power of each prime factor.
- Once we have the LCM, we can check if it satisfies the given conditions.

Solution:

- Prime factorizing 9: 3^2
- Prime factorizing 7: 7
- Prime factorizing 12: 2^2 * 3
- Taking the highest power of each prime factor: 2^2 * 3^2 * 7 = 252
- So the least number of students in the school is 252.
- We can check if 252 satisfies the given conditions:
- 252/9 = 28 (rows of 9 students)
- 252/7 = 36 (rows of 7 students)
- 252/12 = 21 (rows of 12 students)
- So 252 satisfies the given conditions.
- Therefore, the correct answer is option D) 252.

What is the HCF of 3/5, 7/10, 2/15, 6/21?
  • a)
    5/123
  • b)
    6/121
  • c)
    2/105
  • d)
    1/210
  • e)
    None of these
Correct answer is option 'D'. Can you explain this answer?

Nikita Singh answered
HCF will be HCF of numerators/LCM of denominators
So HCF = HCF of 3,7,2,6)/(LCM of 5,10, 15, 21)
= 1/210

Chapter doubts & questions for LCM and HCF - RRB JE Mock Test Series for Mechanical Engineering (ME) 2026 2025 is part of Mechanical Engineering exam preparation. The chapters have been prepared according to the Mechanical Engineering exam syllabus. The Chapter doubts & questions, notes, tests & MCQs are made for Mechanical Engineering 2025 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests here.

Chapter doubts & questions of LCM and HCF - RRB JE Mock Test Series for Mechanical Engineering (ME) 2026 in English & Hindi are available as part of Mechanical Engineering exam. Download more important topics, notes, lectures and mock test series for Mechanical Engineering Exam by signing up for free.

Top Courses Mechanical Engineering

© EduRev
Education Revolution
Signup to see your scores go up
within 7 days!
Takes less than 10 seconds to signup