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In the time taken by Morton to take 4 steps, Lindsey takes 5 steps. If [#permalink]

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12 Jan 2018, 12:21

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40% (01:33) correct 60% (01:39) wrong based on 30 sessions

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In the time taken by Morton to take 4 steps, Lindsey takes 5 steps. If Morton's three steps are equal to Lindsey's four steps and Lindsey's speed is 7.5 miles/hour, what is Morton's(approximate) speed?

Re: In the time taken by Morton to take 4 steps, Lindsey takes 5 steps. If [#permalink]

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12 Jan 2018, 14:06

Let Morton's step distance be M Let Lindsey's step distance be L.

Given: 3M = 4L M = (4/3)*L

in a given time, while morton takes 4 steps, lindsey takes 5 steps

in given time x: Lindsey distance = 5L, Morton distance = 4M = 4 * (4/3) * L = (16/3) * L so basically, the ratio of distance of the two: 16/3 : 5 => 16 : 15

so if in one hour, lindsey covers 7.5 miles, then in one hour morton will cover (i.e speed of morton) (16/15) = x / 7.5 x = 8 m/hr (C)

Please post OA and OE, to check if i had made any mistake.

In the time taken by Morton to take 4 steps, Lindsey takes 5 steps. If Morton's three steps are equal to Lindsey's four steps and Lindsey's speed is 7.5 miles/hour, what is Morton's(approximate) speed?

Let's get Morton's steps to same base... LCM of 3 and 4 is 12 Morton's 3 steps = L's 4... So M's 3*4 or 12 =L's 4*4=16 steps... But when M takes 4, L takes 5, So when M takes 4*3, L takes 5*3 That is when K covers 15 steps, M covers 16 equivalent steps

Speed M: L =16:15... If 7.5 is 15, 16 will be 16*7.5/15=8
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