EXERCISE 3.7 (POST 5)

GOOD MORNING BOYS:

Today, let us discuss Questions 10 and 11 of the Exercise 3.7.

Copy it in your Maths copy.

 Do the Revision work sheet questions in the same copy.

10) Find the LCM of the following numbers: Observe a common property in the obtained LCMs. Is the LCM the product of two numbers in each case?

Yes, LCM is the product of two numbers in each case

CONCEPT:
If the given two numbers are co-primes, then no need to do prime factorization.Simply, multiply both the given numbers and you get the LCM.

Note: Two given numbers are co-prime, if they are divided only by 1. This means The common factor of a co-prime numbers is always 1.

a) 9 and 4 (COMMON FACTOR IS 1, THUS THEY ARE COPRIME)
Therefore LCM of 9 and 4 = 9 x 4 = 36
==============================================================

b) 12 and 5
LCM = 12 x 5 = 60
==============================================================

c) 6 and 5
LCM = 6 x 5 = 30

==============================================================

d) 15 and 4
LCM = 15 x 4 = 60

==============================================================

11) Find the LCM of the following numbers in which one number is the factor of the other:

CONCEPT:

When a number is a factor of the other given number, then their LCM will be the larger number.

a)     5|5, 20
         2|1, 4
          2|1, 2
            |1, 1

Therefore: LCM = 5 x 2 x 2 = 20
[Between 5 and 20, 20 is the larger number. Concept is fulfilled]

==============================================================

b)            2|6, 18
                3|3, 9
                 3|1, 3
                  |1, 1

Therefore LCM = 2 x 3 x 3 =18
======================================================================
c)             2|12, 48
                  2|6, 24
                    2|3, 12
                      3|1, 4
                        2|1, 2   
                          |1, 1

Therefore LCM = 2 x 2 x 2 x 3 x 2 = 48
==============================================================d)          3|9, 45 
               3|3, 15
                5|1, 5
                   |1, 1

Therefore LCM = 3 x 3 x 5 = 45
==============================================================

Revision Work Sheet:

1) Find the HCF of 513, 1134 and 1215: 
[Hint: Take 3 as your divisor]

2) Find the least number which is exactly divisible by 12, 15 and 20:
 [Hint: least number means LCM]

3) Find the largest number of 4 digits divisible by 12, 15 and 18
 [Hint: Find LCM, divide the largest 4 digit number by the LCM, subtract the remainder from the largest number]

4) Product of two numbers = 4107
    HCF of both = 37
    Find the greater number.
[Hint: Here, LCM is the greatest number. Concept: Product of HCF and LCM = Product of the two numbers]

5) The traffic lights at three different road crossings change after efery40 seconds, 72 seconds and 108 seconds respectively.   If they all change simultaneously at 5:20 hours, then find the time at which they will change simultaneously.
[Hint:Find the LCM. After that much time the traffic light will change again]

6) A rectangular courtyard 455 cm width and 525 cm length is paved exactly with square tiles of same size.  Find the largest size of the tile used for this purpose?
[Hint: Find the HCF]

7) John, Steve and Jerome start at same time, same point and in same direction to run around a circular ground.  John completes a round in 250 seconds, Steve in 300 seconds and Jerome in 150 seconds.  Find after what time will they meet again at the starting point? Give the answer in minutes.
[Hint: Find the LCM]

Hope you can do the above with the hints provided. If you can't, let me know.
That's all for today
Good Morning & Thank you Boys
Take Care
Boys and Motivation in School

Comments

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