On a morning walk, three persons step off together and their steps measure 40 cm, 42 cm and 45 cm, respectively.What is the minimum distance each should walk, so that each can cover the same distance in complete steps?

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Given:

Steps of three persons measure 40 cm, 42 cm and 45 cm respectively.

To do:

We have to find the minimum distance each should walk so that he can cover the distance in complete steps.

Solution:

Required distance will be the LCM of the measure of steps of each person.

Calculating the LCM of 40, 42 and 45 using prime factorization method:

Writing the numbers as a product of their prime factors:

Prime factorisation of 40:

• $2\ \times\ 2\ \times\ 2\ \times\ 5 =\ 2^3\ \times\ 5^1$

Prime factorisation of 42:

• $2\ \times\ 3\ \times 7=\ 2^1\ \times\ 3^1\ \times\ 7^1$

Prime factorisation of 45:

• $3\ \times\ 3\ \times\ 5=\ 3^2\ \times\ 5^1$

Multiplying the highest power of each prime number:

• $2^3\ \times\ 3^2\ \times\ 5^1\ \times 7^1=\ 2520$

LCM(40, 42, 45)  $=$  2520

Which means required distance $=$ 2520 cm

$=$ 25 m 20 cm (as, 100 cm  $=$  1 m)

So, the minimum distance each should walk so that each can cover the distance in complete steps is 2520 cm or 25 m 20 cm.

Updated on 10-Oct-2022 13:27:08