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

Last updated on Jun 7, 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:

Find the HCF of  and 

Answer (Detailed Solution Below)

Option 2 :

LCM and HCF Question 3 Detailed Solution

Given:

Numbers: , ,

Formula Used:

HCF of fractions = HCF of numerators / LCM of denominators

Calculation:

Convert mixed fractions to improper fractions:

 = (4 × 5 + 1)/5 = 21/5

 = (5 × 7 + 2)/7 = 37/7

 = (7 × 9 + 4)/9 = 67/9

Extract numerators and denominators:

Numerators: 21, 37, 67

Denominators: 5, 7, 9

Find HCF of numerators:

HCF(21, 37, 67) = 1 (since all are co-prime)

Find LCM of denominators:

LCM(5, 7, 9):

Prime factorization:

5 = 51

7 = 71

9 = 32

LCM = 51 × 71 × 32 = 315

HCF = HCF of numerators / LCM of denominators

HCF = 1 / 315

The HCF of , , and is 1/315 .

LCM and HCF Question 4:

Three lightships flash simultaneously at 8:00 a.m. The first lightship flashes every 24 seconds, the second lightship flashes after every 60 seconds, and the third lightship every 132 seconds. At what time will the three lightships next flash together?

  1. 8 am 18 minutes
  2. 8 am 22 minutes
  3. 8 am 10 minutes 
  4. 8 am 25 minutes

Answer (Detailed Solution Below)

Option 2 : 8 am 22 minutes

LCM and HCF Question 4 Detailed Solution

Given:

The first lightship flashes every 24 seconds.

The second lightship flashes every 60 seconds.

The third lightship flashes every 132 seconds.

All three flash together at 8:00 a.m.

Formula Used:

To find the next time all three lightships flash together, calculate the Least Common Multiple (LCM) of the three time periods.

Next Time = Starting Time + (LCM of time periods in seconds).

Calculation:

Find LCM of 24, 60, and 132:

Prime factorization:

24 = 23 × 3

60 = 22 × 3 × 5

132 = 22 × 3 × 11

LCM = Highest powers of all prime factors = 23 × 3 × 5 × 11

⇒ LCM = 8 × 3 × 5 × 11

⇒ LCM = 1320 seconds

Convert 1320 seconds to minutes:

1320 ÷ 60 = 22 minutes

Next flash time = 8:00 a.m. + 22 minutes

⇒ 8:22 a.m.

The three lightships will next flash together at 8:22 a.m.

LCM and HCF Question 5:

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 5 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

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

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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

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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

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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

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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

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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

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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.

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 12 Detailed Solution

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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.

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 13 Detailed Solution

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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

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 14 Detailed Solution

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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

Answer (Detailed Solution Below)

Option 4 :

LCM and HCF Question 15 Detailed Solution

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Concept used:

LCM of Fraction = LCM of Numerator/HCF of Denominator

Calculation:

​​ = 

⇒ LCM of (1, 5, 5) = 5

⇒ HCF of (2, 6, 4) = 2

 = 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.

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