# HCF and LCM questions

HCF and LCM questions are very much important for a clear understanding of their concepts. In this section, we will see different questions based on HCF and LCM.

Before that, you must know the methods and concepts of HCF and LCM. Go through the given link for basic concepts of them.
HCF and LCM – concepts, definitions and methods.

Now let us have a look at the questions.

#### Questions Based on finding the HCF and LCM

Question 1: Find the LCM and HCF of 135 and 225.
Solution: Here, we will find the LCM and HCF by prime factorization method.
135 = 3 × 3 × 3 × 5 =32 × 5
225 = 3 × 3× 5 × 5 = 32×52
Therefore,

LCM = 32×52 =675
and HCF = 32× 5 = 45

Question 2: The greatest number that exactly divides 105, 1001, and 2436 is?
Solution: We will find the HCF by division method.

Let us see the HCF of 105 and 1001. The HCF of 105 and 1001 is 7.
Now let us find the HCF of 7 and 2436. The HCF of 7 and 2436 is 7.
So, the HCF of 105, 1001 and 2436 is 7.

Question 3: The least number which when divided by 4, 6, 8, 12 and 16 leaves a remainder of 2 in each case is?
Solution: First we will find the LCM of 4, 6, 8, 12 and 16.

LCM of 4, 6, 8, 12 and 16 = 48

Now, as the number when divided by 4, 6, 8, 12 and 16 leaves remainder 2 in each case.

So, the required number is = 48 +2 = 50

Question 4: The least number which when divided by 4,5 and 6 leaves remainder 1,2, and 3 respectively is given by?
Solution: Here, If we subtract the remainder from the numbers-
4-1 =3
5-2 =3
and 6-3 =3

We can see that the remainder in each case is less than the divisor by 3.

Hence, when 3 is added to the required number, it will be exactly divisible by 4,5 and 6.

So, the LCM of 4,5 and 6 = 60
And the required number = 60 – 3  = 57.

##### Question: 5 Find the least number of five digit exactly divisible by 12, 15 and 18.Solution:

We know,
Least number of five digit = 10000
Here LCM of 12,15 and 18 = 180

On dividing 10000 by 180 the remainder is 100.
So, the required least number of five digit = 10000 -100 + 180 = 10080

Question 6: The greatest number which will divide 410, 751, and 1030 to leave remainder 7 in each case will be?
Solution: As, 7 is the remainder in each case.
So, required number must divide-
410-7 = 403
751 – 7 = 744
and 1030 – 7 = 1023

So the HCF of 403, 744 and 1023 = 31.
Thus, the required number is 31.

Question 7: Find the greatest possible length which can be used to measure exactly the lengths 4 m 95 cm, 9 m, and 16 m 65 cm.
Solution: Convert the following in cm.

We have to measure 495 cm, 900 cm, and 1665 cm.
Hence, we have to find the HCF of 495, 900 and 1665.

495 = 32 × 5 × 11
900 =22 × 32 × 52
and 1665 = 32 × 5 × 37
HCF = 32 × 5 = 45

So, the HCF of 495, 900, and 1665 is 45.

Question 8: Three bells rings at intervals of 36 seconds, 40 seconds and 48 seconds respectively. They start ringing together after every ——–?
Solution: LCM of 36, 40 and 48 seconds = 720 seconds = 720/60 = 12 minutes

Thus, they will ring after every 12 minutes.

#### Questions Based on the relationship between HCF and LCM

Question 9: The HCF and LCM of two numbers are 12 and 336 respectively. If one number is 84, then the other number is?
Solution: Given, HCF = 12 and LCM =336

We know, ⇒ 84 × second number = 12 × 336
⇒ Second number = (12 × 336) / 84
Second number = 48

Question 10: The product of co-primes is 117. Find their LCM?
Solution: We know the HCF of co-primes is 1.
Given, Product of numbers = 117. 117 = 1 x LCM
LCM = 117

Question 11: The LCM of two numbers is 495 and their HCF is 5. If sum of the numbers is 100, then find their difference.
Solution: Let the two numbers be x and ( 100 – x).

Given HCF = 5 and LCM =495
We know ⇒ x × ( 100 – x ) = 5 × 495
100x – x2 = 2475
x2 – 100x -2475 = 0
(x-45) (x -55 ) = 0
⇒   x= 45 and 55.

So the numbers are 45 and 55. The difference between them is 10.

#### Questions based on ratios

Question 12: The HCF and LCM of two numbers are 21 and 84 respectively. If the ratio of the two numbers is 1:4. Then find the numbers.
Solution: Given, HCF = 21 and LCM = 84
So, We know if the numbers are given in ratio x:y. Then

1st number = HCF × x = 21 × 1 = 21

2nd number = HCF × y = 21 × 4 = 84

Question 13: If the ratio of two numbers is 2:3 and their LCM is 54. Then the sum of the two numbers is?
Solution:  Given ratio of numbers = 2:3
and LCM = 54
We know if the numbers are given in ratio x:y. Then

LCM = HCF × x ×y
54 = HCF × 2 × 3
HCF = 9

Now,
First number = HCF × 2 = 9 × 2 = 18
Second number = HCF × 3 = 9 × 3 = 27

So, the sum of the numbers is 45.

I hope this questions will be helpful. You can also download these questions along with the theory of HCF and LCM in the pdf form. Click on the link below to download the pdf.