The circumference of the fore and hind wheels of a carriage are $6 \frac{3}{14}$ m and $8 \frac{1}{18}$ m respectively. At any given moment a chalk mark is put on the point of contact of each wheel with the ground. Find the distance travelled by the carriage so that both the chalk marks are again on the ground at the same time

218 m

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217.5 m

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218.25 m

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217 m

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Solution

The correct option is B $217.5 m$
Required distance travelled = $L C M o f (6 \frac{3}{14} m, 8 \frac{1}{181} m)$
$(\frac{87 = 3 \times 29}{145 = 5 \times 29}) = L C M o f (\frac{87}{14}, \frac{145}{18})$
$(\frac{14 = 2 \times 7}{18 = 2 \times 9}) = \frac{L C M o f (87, 145)}{H C F o f (14, 18)}$
$= \frac{3 \times 5 \times 29}{2} = \frac{435}{2} = 217.5 m$

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The correct option is B 217.5mRequired distance travelled = LCMof(6314m,81181m)(87=3×29145=5×29)=LCMof(8714,14518)(14=2×718=2×9)=LCMof(87,145)HCFof(14,18)=3×5×292=4352=217.5m