Real Numbers

Applications of LCM and HCF

As you have probably been told before, Maths is a subject you understand, not a subject you memorize. So learning the application of a concept helps you fully understand the topic, and also makes it interesting. There are various interesting applications of LCM and HCF. Let us learn more about them.

Suggested Videos

Play
Play
Play
previous arrow
next arrow
previous arrownext arrow
Slider

 

Co-relation between HCF and LCM

So there is an interesting co-relation between H.C.F and L.C.M. of two numbers. The product of the H.C.F. and L.C.M. of any two numbers is always equal to the product of those two numbers. However the same is not applicable to three or more numbers.

Browse more Topics Under Real Numbers

LCM (a,b) * HCF (a,b) = a*b

Example: 

Let the two numbers be 4 and 6.

HCF (4,6) = 2

LCM (4,6) = 12

HCF*LCM = 2*12 = 24

Product of the two numbers = 4*6 = 24

Download NCERT Solutions for Class 10 Maths

Applications of LCM and HCF

Now there are various real-life applications of LCM and HCF. The best way to understand these and grasp the concept of LCM and HCF is to learn via examples. So let us take a look at a few examples which will help you understand LCM and HCF.

Example 1: Find the greatest number that will divide 400, 435 and 541 leaving 9, 10 and 14 as remainders respectively.

Solution: The required number would be HCF of (400-9), (435-10) and (541-14)

So the HCF (391,425,527)

391 = 17 × 23

425 = 5 × 5 × 17

527 = 17 × 31

HCF = 17

Therefore the required number is 17.

Example 2: A, B and C start to jog around a circular stadium. They complete their rounds in 36 seconds, 48 seconds and 42 seconds respectively. After how many seconds will they be together at the starting point?

Solution: The required time is the LCM of all their lap times. This is the earliest when all three will intersect at the same point.

Required time is the LCM (36,48,42)

LCM = 2 × 2 × 3 × 3 × 4 × 7

LCM = 1008

Therefore the required time is 1008 seconds

Example 3: Mr Das has three classes. Each class has 24, 30 and 18 students respectively. Mr Das wants to divide each class into groups so that every group in every class has the same number of students and there are no students left over. What is the maximum number of students he can put into each group?

Solution: We have to find the maximum number of students that can be put into each group. This should give you an indication that here we have to calculate the HCF or GCF.

HFC (24,30,18) = 2 × 3 = 6

Therefore a maximum of 6 students can be put into each group.

Example 4: If the least prime factor of ‘a’ is 3 and the least prime factor of ‘b’ is 7, then find the least prime factor of (a+b)

Solution: Since 3+7 = 10 then the least prime factor of (a+b) has to be 2

Unless a+b is a prime number itself which is greater than 2.

Suppose this is true Then a+b must be an odd number. So one of ‘a’ or ‘b’ must be an even number. Suppose ‘a’ is an even number, then its least prime factor has to be 2, it cannot be 3 or 7. So ‘a’ cannot be an even number. The same logic applies to ‘b’. Hence a+b cannot be a prime number if the least prime factor of ‘a’ and ‘b’ is 3 and 7.

Hence the least prime factor of a+b is 2.

Share with friends

Customize your course in 30 seconds

Which class are you in?
5th
6th
7th
8th
9th
10th
11th
12th
Get ready for all-new Live Classes!
Now learn Live with India's best teachers. Join courses with the best schedule and enjoy fun and interactive classes.
tutor
tutor
Ashhar Firdausi
IIT Roorkee
Biology
tutor
tutor
Dr. Nazma Shaik
VTU
Chemistry
tutor
tutor
Gaurav Tiwari
APJAKTU
Physics
Get Started

Leave a Reply

avatar
  Subscribe  
Notify of

Stuck with a

Question Mark?

Have a doubt at 3 am? Our experts are available 24x7. Connect with a tutor instantly and get your concepts cleared in less than 3 steps.

Download the App

Watch lectures, practise questions and take tests on the go.

Customize your course in 30 seconds

Which class are you in?
No thanks.