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Mark, walking from point A to B, and Smith walking simultaneously from

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VP
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Mark, walking from point A to B, and Smith walking simultaneously from  [#permalink]

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New post 24 Nov 2019, 09:02
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Question Stats:

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Mark, walking from point A to B, and Smith walking simultaneously from B to A, met at a point and reached the opposite ends of their destinations in 10 and 40 minutes respectively after the time of meeting. What is the ratio of speeds of Mark and Smith ?

A. 5 : 1
B. 4 : 1
C. 3 : 1
D. 2 : 1
E. None of these
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Re: Mark, walking from point A to B, and Smith walking simultaneously from  [#permalink]

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New post 24 Nov 2019, 14:00
Since the 2 people met at a common point, the ratio is a simple comparison of how long after the meeting that each person made it to their final destination.
40:10= 4:1
Ans=B
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Re: Mark, walking from point A to B, and Smith walking simultaneously from  [#permalink]

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New post 24 Nov 2019, 14:12
Speed of Mark= \(v_M\)

Speed of Smith= \(v_S\)

Time taken by Mark and Smith to meet= t

\(\frac{v_M*t}{v_S}\) : \(\frac{v_S*t}{v_M}\) = 40:10

\((\frac{v_M}{v_S})^2\)= 4/1

\((\frac{v_M}{v_S})\)= 2:1


Dillesh4096 wrote:
Mark, walking from point A to B, and Smith walking simultaneously from B to A, met at a point and reached the opposite ends of their destinations in 10 and 40 minutes respectively after the time of meeting. What is the ratio of speeds of Mark and Smith ?

A. 5 : 1
B. 4 : 1
C. 3 : 1
D. 2 : 1
E. None of these
VP
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Joined: 20 Jul 2017
Posts: 1145
Location: India
Concentration: Entrepreneurship, Marketing
WE: Education (Education)
Re: Mark, walking from point A to B, and Smith walking simultaneously from  [#permalink]

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New post 27 Nov 2019, 07:47
1
Dillesh4096 wrote:
Mark, walking from point A to B, and Smith walking simultaneously from B to A, met at a point and reached the opposite ends of their destinations in 10 and 40 minutes respectively after the time of meeting. What is the ratio of speeds of Mark and Smith ?

A. 5 : 1
B. 4 : 1
C. 3 : 1
D. 2 : 1
E. None of these


Let Mark & Smith walking towards each meet at point C and time taken to meet = t
Let the speeds of Mark & Smith be \(S_m\) & \(S_s\) respectively.
and let the distance AC = x & BC = y

--> \(x = S_mt\) & \(y = S_st\)
--> \(\frac{x}{y} = \frac{S_mt}{S_st} = \frac{S_m}{S_s}\) ....... (1)

Given, the distance covered by Smith from C to A in 40 min and that of by Mark from C to B in 10 min
--> \(x = S_s*40\) & \(y = S_m*10\)
--> \(\frac{x}{y} = \frac{S_s40}{S_m10} = \frac{4S_s}{S_m}\) ....... (2)

From (1) & (2),
\(\frac{S_m}{S_s} = \frac{4S_s}{S_m}\)
--> \(\frac{S_m^2}{S_s^2} = 4\)
--> \(\frac{S_m}{S_s} = \sqrt{4} = 2\)
--> \(S_m : S_s = 2 : 1\)

Option D
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Re: Mark, walking from point A to B, and Smith walking simultaneously from  [#permalink]

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New post 01 Dec 2019, 01:20
nick1816 wrote:
Speed of Mark= \(v_M\)

Speed of Smith= \(v_S\)

Time taken by Mark and Smith to meet= t

\(\frac{v_M*t}{v_S}\) : \(\frac{v_S*t}{v_M}\) = 40:10

\((\frac{v_M}{v_S})^2\)= 4/1

\((\frac{v_M}{v_S})\)= 2:1


Dillesh4096 wrote:
Mark, walking from point A to B, and Smith walking simultaneously from B to A, met at a point and reached the opposite ends of their destinations in 10 and 40 minutes respectively after the time of meeting. What is the ratio of speeds of Mark and Smith ?

A. 5 : 1
B. 4 : 1
C. 3 : 1
D. 2 : 1
E. None of these



How did u come at this part \(\frac{v_M*t}{v_S}\) : \(\frac{v_S*t}{v_M}\) = 40:10 ?
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Re: Mark, walking from point A to B, and Smith walking simultaneously from   [#permalink] 01 Dec 2019, 01:20
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