On a morning walk,three persons step off toge | vedclass

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(D) To find the minimum distance that each should walk so that each can cover the same distance in complete steps,we need to calculate the Least Commo

Vedclass · Accepted Answer

$2520$

Vedclass · Answer

(D) To find the minimum distance that each should walk so that each can cover the same distance in complete steps,we need to calculate the Least Common Multiple $(LCM)$ of the step measurements: $40\, cm$,$42\, cm$,and $45\, cm$.First,we find the prime factorization of each number:$40 = 2^3 	imes 5$$42 = 2 	imes 3 	imes 7$$45 = 3^2 	imes 5$The $LCM$ is found by taking the highest power of each prime factor present in the numbers:$LCM(40, 42, 45) = 2^3 	imes 3^2 	imes 5^1 	imes 7^1$$LCM = 8 	imes 9 	imes 5 	imes 7$$LCM = 72 	imes 35$$LCM = 2520$Therefore,the minimum distance each should walk is $2520\, cm$.

Part $I$	Part $II$
$1. \text{l.c.m.}(8, 16, 24)$	$a. 36$
$2. \text{g.c.d.}(8, 16, 24)$	$b. 48$
$3. \text{l.c.m.}(6, 12, 18)$	$c. 8$
$4. \text{g.c.d.}(6, 12, 18)$	$d. 12$
	$e. 6$

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? (in $cm$)

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Which of the following groups correctly matches the data of Part $I$ with the data of Part $II$?

Part $I$ Part $II$

$1. \text{l.c.m.}(8, 16, 24)$ $a. 36$

$2. \text{g.c.d.}(8, 16, 24)$ $b. 48$

$3. \text{l.c.m.}(6, 12, 18)$ $c. 8$

$4. \text{g.c.d.}(6, 12, 18)$ $d. 12$

$e. 6$

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