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Question
Three friends plan to go for a morning walk. They step off together and their steps measures 48 cm, 52 cm and 56 cm respectively. What is the minimum distance each should walk so that each can cover the same distance in complete steps ten times?
Sum
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Solution
Given, the lengths of the steps are 48 cm, 52 cm, 56 cm, respectively.
Prime factorisation of each length
| 2 | 48 |
| 2 | 24 |
| 2 | 12 |
| 2 | 6 |
| 3 | 3 |
| 1 |
24 × 3
| 2 | 52 |
| 2 | 26 |
| 13 | 13 |
| 1 |
22 × 13
| 2 | 56 |
| 2 | 28 |
| 2 | 14 |
| 7 | 7 |
| 1 |
23 × 7
LCM (48, 52, 56)
= 24 × 3 × 13 × 7
= 48 × 91
= 4368 cm
Since they need to cover the distance ten times, we multiply the LCM by 10.
Minimum distance = 4368 × 10
= 43680 cm
or 436.8 m
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