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?
2520 cm (i.e., 25.2 m)
Step 1: Understand the problem.
Each person takes steps of different lengths: 40 cm, 42 cm, and 45 cm. We want a distance that is a common multiple of all three step lengths. The minimum such distance will be the Least Common Multiple (LCM) of 40, 42, and 45.
Step 2: Prime factorisation of each number.
Factorise 40:
\(40 = 2^3 \times 5\)
Factorise 42:
\(42 = 2 \times 3 \times 7\)
Factorise 45:
\(45 = 3^2 \times 5\)
Step 3: Take the highest power of each prime factor.
From 40: we need \(2^3\) and \(5\).
From 42: we need \(3\) and \(7\).
From 45: we need \(3^2\) (this is larger than the single \(3\) from 42).
Step 4: Multiply them together.
LCM = \(2^3 \times 3^2 \times 5 \times 7\)
= \(8 \times 9 \times 5 \times 7\)
= \(2520\)
Step 5: Interpret the result.
The minimum distance that each can walk in complete steps is 2520 cm.
Convert into metres:
\(2520\,\text{cm} = 25.2\,\text{m}\).
Final Answer: The minimum distance is 2520 cm or 25.2 m.