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Two cars, Car1 and Car2 move towards each other from X and Y respectiv

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cleetus
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Two cars, Car1 and Car2 move towards each other from X and Y respectiv [#permalink] New post   09 Nov 2010, 13:07
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Two cars, Car1 and Car2 move towards each other from X and Y respectively with respective speeds of 20 m/s and 15 m/s. After meeting each other Car1 reaches Y in 10 seconds. In how many seconds does Car 2 reach X starting from Y?

A) 15.5 sec
B) 8.4 sec
C) 33.6 sec
D) 31.11 sec
E) 16.8 sec
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D
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Sarang
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Re: Two cars, Car1 and Car2 move towards each other from X and Y respectiv [#permalink] New post   09 Nov 2010, 14:24
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X--------------------------------------|----------------------------Y
Car A(20mps)------------------------->P<---------------Car B(15mps)

Let 2 cars meet each other at point P in t seconds.

Car1 covers distance= 20t. Car2 covers distance=15t. So, total distance XY= 35t.

From P, Car 1 reaches onto Y in 10 secs. So it covers 15t further.
so, 15t/20 = 10
So t=40/3 sec and total distance = (35*40)/3
Hence Car2 will cover total distance in (35*40)/(3*15) = 31.11 sec approx.
:lol:

Answer D
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Re: Two cars, Car1 and Car2 move towards each other from X and Y respectiv [#permalink] New post   09 Nov 2010, 19:37
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cleetus wrote:
Two cars, Car1 and Car2 move towards each other from X and Y respectively with respective speeds of 20 m/s and 15 m/s. After meeting each other Car1 reaches Y in 10 seconds. In how many seconds does Car 2 reach X starting from Y?

A) 15.5 sec
B) 8.4 sec
C) 33.6 sec
D) 31.11 sec
E) 16.8 sec


Slight Variation on the above approach:


____________M_________
X Y
Car 1 Car 2
20 m/s---> <---15 m/s

Speed of Car1 : Speed of Car2 = 4:3 so distance they will cover in same time will be 4:3.
So XM: MY = 4:3
When Car 1 reaches M, it takes 10 secs to reach Y so MY = 20*10 = 200 m
Then XM = 800/3 m
Then XY = 800/3 + 200 = 1400/3 m
Car 2 takes 1400/3*15 = 31.1 sec
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Re: Two cars, Car1 and Car2 move towards each other from X and Y respectiv [#permalink] New post   09 Nov 2010, 14:09
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x ------> z <----- y

\(L_1\) = Distance between x and z
\(L_2\) = Distance between z and y
\(V_1\) = Speed of Car1 = 20m/s
\(V_2\) = Speed of Car2 = 15m/s

z is the point where car1 and car2 meet.

we know that when car 1 and car 2 meet, they would have traveled for the same amount of time t.

so \(\frac{L_1}{V_1} = t\) and \(\frac{L_2}{V_2} = t\)

we can equate the two to see that \(L_1 = \frac{V_1*L_2}{V_2}\)

and we also know that \(\frac{L_2}{V_1} = 10s\)

using these information, we can solve for length of xy. Then just divide that length by speed of car2 to get the time.
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cleetus
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Re: Two cars, Car1 and Car2 move towards each other from X and Y respectiv [#permalink] New post   09 Nov 2010, 15:02
Thanks. that was good way of looking at it.
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Re: Two cars, Car1 and Car2 move towards each other from X and Y respectiv [#permalink] New post   19 Oct 2015, 00:18
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VeritasPrepKarishma wrote:
cleetus wrote:
Two cars, Car1 and Car2 move towards each other from X and Y respectively with respective speeds of 20 m/s and 15 m/s. After meeting each other Car1 reaches Y in 10 seconds. In how many seconds does Car 2 reach X starting from Y?

A) 15.5 sec
B) 8.4 sec
C) 33.6 sec
D) 31.11 sec
E) 16.8 sec


Slight Variation on the above approach:


____________M_________
X Y
Car 1 Car 2
20 m/s---> <---15 m/s

Speed of Car1 : Speed of Car2 = 4:3 so distance they will cover in same time will be 4:3.
So XM: MY = 4:3
When Car 1 reaches M, it takes 10 secs to reach Y so MY = 20*10 = 200 m
Then XM = 800/3 m
Then XY = 800/3 + 200 = 1400/3 m
Car 2 takes 1400/3*15 = 31.1 sec

Responding to a pm:

Quote:
Could you explain how you got 800/3?


Distance XM : distance MY = 4 : 3
We got that distance MY = 200 m. Can we find distance XM then?
3 (MY) on the ratio scale is actually 200.
So 1 will be 200/3 and 4 (XM) will be (200/3)*4 = 800/3

For more on ratio scale, check: https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2011/03 ... of-ratios/
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Re: Two cars, Car1 and Car2 move towards each other from X and Y respectiv [#permalink]
19 Oct 2015, 00:18
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