LCM and HCF MCQ Quiz - Objective Question with Answer for LCM and HCF - Download Free PDF

Last updated on May 28, 2025

HCF and LCM i.e. Highest Common Factor and Least Common Multiple are proven to test one’s reasoning & logical skills. They are a part of numerous recruitment processes such as CAT, GATE, Bank and Railway Exams, etc. Testbook brings HCF and LCM MCQs Quiz accompanied by some tips and tricks. These objective questions would chalk out a way to enhance your preparation. Read this article and solve these questions to test your aptitude in this section.

Latest LCM and HCF MCQ Objective Questions

LCM and HCF Question 1:

Find the H.C.F. of p(x) = 2x3 – 3x2 – 2x + 3 and q(x) = 3x2 + 8x + 5.

  1. (x + 1)
  2. (x – 1)
  3. (2x – 3)
  4. (2x + 1)
  5. None of the above

Answer (Detailed Solution Below)

Option 1 : (x + 1)

LCM and HCF Question 1 Detailed Solution

Given:

p(x) = 2x3 – 3x2 – 2x + 3 and q(x) = 3x2 + 8x + 5

Concept:

H.C.F. of two or more equations is the greatest factor that divides each of them exactly.

Calculation:

The factors of p(x) = 2x3 – 3x2 – 2x + 3

⇒ x2 × (2x – 3) – 1 × (2x – 3)

⇒ (x2 – 1) × (2x – 3)

⇒ (x – 1) × (x + 1) × (2x – 3)

And, the factors of q(x) = 3x2 + 8x + 5

⇒ 3x2 + 5x + 3x + 5

⇒ x × (3x + 5) + 1 × (3x + 5)

⇒ (3x + 5) × (x + 1)

∴ The required H.C.F. is (x + 1).

LCM and HCF Question 2:

Least Common Multiple of 6 and 29 is X. Highest Common Factor of 6 and 29 is Y. Then what is the value of (X + 4Y)?

  1. 180
  2. 190
  3. 178
  4. 244
  5. None of the above

Answer (Detailed Solution Below)

Option 3 : 178

LCM and HCF Question 2 Detailed Solution

Given:

Number 1: 6

Number 2: 29

Formula Used:

Least Common Multiple (LCM) of a and b = (a × b) / Highest Common Factor (HCF) of a and b

Highest Common Factor (HCF) of 6 and 29 = 1 (since 6 and 29 are co-prime numbers)

Calculation:

LCM(6, 29) = (6 × 29) / HCF(6, 29)

LCM(6, 29) = (6 × 29) / 1

LCM(6, 29) = 174

HCF(6, 29) = 1

X = LCM(6, 29) = 174

Y = HCF(6, 29) = 1

X + 4Y = 174 + 4 × 1

X + 4Y = 174 + 4

X + 4Y = 178

The value of (X + 4Y) is 178.

LCM and HCF Question 3:

LCM of two numbers is 56 times their HCF, with the sum of their HCF and LCM being 1710. If one of the two numbers is 240, then what is the other number?

  1. 210° 
  2. 171° 
  3. 57° 
  4. More than one of the above
  5. None of the above

Answer (Detailed Solution Below)

Option 1 : 210° 

LCM and HCF Question 3 Detailed Solution

Given:

LCM = 56 × HCF

HCF + LCM = 1710

Formula used:

HCF × LCM = Product of two numbers

Calculation:

Let, HCF = x and LCM = 56x

So, x + 56x = 1710

⇒ 57x = 1710

⇒ x = 1710/57 = 30

Then, HCF = 30 and LCM = 30 × 56 = 1680

One number = 240

Let, another number = a

Therefore, 30 × 1680 = 240 × a

⇒ a = (30 × 1680)/240 = 210

∴ The other number is 210

LCM and HCF Question 4:

The greatest number which divides 68, 140 and 248 leaving the same remainder in each case is

  1. 36
  2. 18
  3. 72
  4. 108

Answer (Detailed Solution Below)

Option 1 : 36

LCM and HCF Question 4 Detailed Solution

Given:

Numbers are 68, 140, and 248.

We need to find the greatest number that divides these numbers leaving the same remainder.

Formula Used:

When a number divides multiple numbers leaving the same remainder, the required number is the HCF (Highest Common Factor) of the differences of these numbers.

Calculation:

Find the differences:

140 - 68 = 72

248 - 140 = 108

248 - 68 = 180

Now, find the HCF of 72, 108, and 180.

Prime factorization:

72 = 23 × 32

108 = 22 × 33

180 = 22 × 32 × 5

Common factors: 22 × 32

HCF = 4 × 9 = 36

The greatest number that divides 68, 140, and 248 leaving the same remainder is 36.

LCM and HCF Question 5:

HCF of 35 and 45 is

  1. 35
  2. 45
  3. 1
  4. 5

Answer (Detailed Solution Below)

Option 4 : 5

LCM and HCF Question 5 Detailed Solution

Given:

HCF of 35 and 45

Formula used:

HCF is calculated by finding the greatest common divisor (GCD) of the given numbers.

Calculation:

Prime factors of 35: 5 × 7

Prime factors of 45: 32 × 5

Common factors: 5

⇒ HCF = 5

∴ The correct answer is option (4).

Top LCM and HCF MCQ Objective Questions

Three piece of timber 143m, 78m and 117m long have to be divided into planks of the same length. What is the greatest possible length of each plank?

  1. 7 m
  2. 11 m
  3. 13 m
  4. 17 m

Answer (Detailed Solution Below)

Option 3 : 13 m

LCM and HCF Question 6 Detailed Solution

Download Solution PDF

Given:

Length of timber1 = 143 m

Length of timber2 = 78 m

Length of timber3 = 117 m

Calculation:

Greatest possible length of each plank = HCF of 143, 78 and 117

143 = 13 × 11

78 = 13 × 2 × 3

117 = 13 × 3 × 3 

HCF is 13

∴ Greatest possible length of each plank is 13 m.

Four bells ring simultaneously at starting and an interval of 6 sec, 12 sec, 15 sec and 20 sec respectively. How many times they ring together in 2 hours?

  1. 120
  2. 60
  3. 121
  4. 112

Answer (Detailed Solution Below)

Option 3 : 121

LCM and HCF Question 7 Detailed Solution

Download Solution PDF

GIVEN:

Four bells ring simultaneously at starting and an interval of 6 sec, 12 sec, 15 sec and 20 sec respectively.

CONCEPT:

LCM: It is a number which is a multiple of two or more numbers.

CALCULATION:

LCM of (6, 12, 15, 20) = 60

All 4 bells ring together again after every 60 seconds

Now,

In 2 Hours, they ring together = [(2 × 60 × 60)/60] times + 1 (at the starting) = 121 times

∴ In 2 hours they ring together for 121 times

Mistake Points

In these type of question we assume that we have started counting the time after first ringing. Due to this when we calculate the LCM it gives us the ringing at 2nd time not the first time. So, we needed to add 1.

Four bells ringing together and ring at an interval of 12 sec, 15 sec, 20 sec, and 30 sec respectively. How many times will they ring together in 8 hours?  

  1. 481
  2. 480
  3. 482
  4. 483

Answer (Detailed Solution Below)

Option 1 : 481

LCM and HCF Question 8 Detailed Solution

Download Solution PDF

Given:

Four bells ringing timing is 12 sec, 15 sec, 20 sec,30 sec 

Calculation:

Four bells ringing timing is 12 sec, 15 sec, 20 sec,30 sec 

Now we have to take LCM of time interval

⇒ LCM of (12, 15, 20, 30) = 60

Total seconds in 8 hours = 8 × 3600 = 28800

Number of times bell rings = 28800/60

⇒ Number of times bell rings = 480

If four bells ring together in starting

⇒ 480 + 1 

∴ The bell ringing 481 times in 8 hours.

Mistake PointsThe bells start tolling together, the first toll also needs to be counted, that is the number of times of tolling since the first time.

The LCM and HCF of 2 numbers are 168 and 6 respectively. If one of the numbers is 24, find the other.

  1. 36
  2. 38
  3. 40
  4. 42

Answer (Detailed Solution Below)

Option 4 : 42

LCM and HCF Question 9 Detailed Solution

Download Solution PDF

We know that,

product of two numbers = L.C.M × H.C.F of those numbers

Let the second number be x.

24 × x = 168 × 6

x = 6 × 7

x = 42

24 mango trees, 56 apple trees and 72 orange trees have to be planted in rows such that each row contains the same number of trees of one variety only. Find the minimum number of rows in which the above mentioned trees may be planted.

  1. 17
  2. 15
  3. 19
  4. 18

Answer (Detailed Solution Below)

Option 3 : 19

LCM and HCF Question 10 Detailed Solution

Download Solution PDF

Given:

24 mango trees, 56 apple trees and 72 orange trees have to be planted in rows such that each row contains the same number of trees of one variety only.

Calculations:

There are 24 mangoes trees, 56 apple trees & 72 Orange trees.

To get the minimum number of rows, we need maximum trees in each row.

In each row, we need the same number of trees

So we need to calculate HCF

HCF of 24, 56 & 72

⇒ 24 = 2³ × 3

⇒ 56 = 2³ × 7

⇒ 72 = 2³ × 3²

HCF = 2³ = 8

Number of minimum rows = (24 + 56 + 72)/8 = 152/8

⇒ 19

∴ The correct choice will be option 3.

The HCF and LCM of two numbers are 24 and 168 and the numbers are in the ratio 1 ∶ 7. Find the greater of the two numbers. 

  1. 168
  2. 144
  3. 108
  4. 72

Answer (Detailed Solution Below)

Option 1 : 168

LCM and HCF Question 11 Detailed Solution

Download Solution PDF

Given:

HCF = 24

LCM = 168

Ratio of numbers = 1 ∶ 7.

Formula:

Product of numbers = LCM × HCF

Calculation:

Let numbers be x and 7x.

x × 7x = 24 × 168

⇒ x2 = 24 × 24

⇒ x = 24

∴ Larger number = 7x = 24 × 7 = 168.

How many multiples of both 3 or 4 are there from 1 to 100 in total?

  1. 55
  2. 50
  3. 58
  4. 33

Answer (Detailed Solution Below)

Option 2 : 50

LCM and HCF Question 12 Detailed Solution

Download Solution PDF

Formula used:

n(A∪B) = n(A) + n(B) - n(A∩B)

Calculation:

On dividing 100 by 3 we get a quotient of 33

The number of multiple of 3, n(A) = 33

On dividing 100 by 4 we get a quotient of 25

The number of multiple of 4, n(B) = 25

LCM of 3 and 4 is 12

On dividing 100 by 12 we get a quotient of 8

The number of multiple of 12, n(A∩B) = 8

The number which is multiple of 3 or 4 = n(A∪B)

Now, n(A∪B) = n(A) + n(B) - n(A∩B)

⇒ 33 + 25 - 8

⇒ 50

∴ The total number multiple of 3 or 4 is 50

Find the sum of the numbers between 550 and 700 such that when they are divided by 12, 16 and 24, leave remainder 5 in each case.

  1. 1980
  2. 1887
  3. 1860
  4. 1867

Answer (Detailed Solution Below)

Option 2 : 1887

LCM and HCF Question 13 Detailed Solution

Download Solution PDF

Given:

The number between 550 and 700 such that when they are divided by 12, 16, and 24, leave the remainder 5 in each case

Concept Used:

LCM is the method to find the Least Common Multiples

Calculations:

⇒ LCM of 12, 16, and 24 = 48

Multiple of 48 bigger than 550 which leaves remainder 5 are

⇒ 1st Number = 48 x 12 + 5 = 581

⇒ 2nd Number = 48 x 13 + 5 = 629

⇒ 3rd Number = 48 x 14 + 5 = 677

⇒ Sum of these numbers are = 581 + 629 + 677 = 1887

⇒ Hence, The sum of the numbers are 1887.

Shortcut Trick Option elimination method:  Subtract the remainder of 5 in every no means in the option 15 we have to subtract because the sum of the three numbers is given.

In this case only 3, no is a possible case

So we have to subtract 15 and then check the divisibility of 16 and 3.

The L.C.M. of \(\frac{2}{4}, \frac{5}{6}, \frac{10}{8}\) is:

  1. \(\frac{1}{5}\)
  2. \(\frac{5}{4}\)
  3. \(\frac{4}{5}\)
  4. \(\frac{5}{2}\)

Answer (Detailed Solution Below)

Option 4 : \(\frac{5}{2}\)

LCM and HCF Question 14 Detailed Solution

Download Solution PDF

Concept used:

LCM of Fraction = LCM of Numerator/HCF of Denominator

Calculation:

​​\(\frac{2}{4}, \frac{5}{6}, \frac{10}{8}\) = \(\frac{1}{2}, \frac{5}{6}, \frac{5}{4}\)

⇒ LCM of (1, 5, 5) = 5

⇒ HCF of (2, 6, 4) = 2

\(\dfrac{LCM\; of\;(1,5,5)}{HCF\;of\;(2,4,6)}\) = 5/2

∴ The correct answer is 5/2.

Mistake Points Please note that LCM means the lowest common multiple. LCM is the lowest number which is completely divisible by all given numbers(2/4, 5/6, 10/8). 

In these types of questions, make sure that you reduce the fractions to their lowest forms before you use their formulae, otherwise, you may get the wrong answer.

If we don't reduce the fractions to their lowest forms then LCM is 5 but the LCM of these 3 numbers is 5/2.

The L.C.M and H.C.F of two numbers is 585 and 13 respectively. Find the difference between the numbers.

  1. 39
  2. 52
  3. 67
  4. 71

Answer (Detailed Solution Below)

Option 2 : 52

LCM and HCF Question 15 Detailed Solution

Download Solution PDF

Given:

H.C.F of numbers = 13

LCM of numbers = 585

Calculation:

 Let the number be 13a and 13b where a and b are co-prime.

L.C.M of 13a and 13b = 13ab

According to question, 13ab = 585

⇒ ab = 45

⇒ ab = 5 × 9

⇒ a = 5 and b = 9 or a = 9 and b = 5

⇒ First number = 13a

⇒ First number = 13 × 5

⇒ First number = 65

⇒ Second number = 13b

⇒ Second number =13 × 9

⇒ Second number = 117

Required difference = 117 - 65 = 52

∴ Required difference = 52

Get Free Access Now
Hot Links: online teen patti real money teen patti 51 bonus teen patti dhani teen patti list teen patti joy 51 bonus