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

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VP
Joined: 20 Jul 2017
Posts: 1145
Location: India
Concentration: Entrepreneurship, Marketing
WE: Education (Education)
Mark, walking from point A to B, and Smith walking simultaneously from  [#permalink]

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24 Nov 2019, 09:02
5
00:00

Difficulty:

95% (hard)

Question Stats:

24% (01:30) correct 76% (02:09) wrong based on 37 sessions

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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
Manager
Joined: 24 Sep 2019
Posts: 78
Re: Mark, walking from point A to B, and Smith walking simultaneously from  [#permalink]

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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
VP
Joined: 19 Oct 2018
Posts: 1171
Location: India
Re: Mark, walking from point A to B, and Smith walking simultaneously from  [#permalink]

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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
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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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
Intern
Joined: 24 Nov 2019
Posts: 4
Re: Mark, walking from point A to B, and Smith walking simultaneously from  [#permalink]

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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 ?
Re: Mark, walking from point A to B, and Smith walking simultaneously from   [#permalink] 01 Dec 2019, 01:20
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