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A train left a station P at 6 am and reached another station Q at 11 a

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A train left a station P at 6 am and reached another station Q at 11 a  [#permalink]

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New post 26 Sep 2018, 05:10
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Question Stats:

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A train left a station P at 6 am and reached another station Q at 11 am. Another train left station Q at 7 am and reached P at 10 am. At what time did the two trains pass one another?

(A) 7:50 am
(B) 8:13 am
(C) 8:30 am
(D) 8:42 am
(E) 9:03 am

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A train left a station P at 6 am and reached another station Q at 11 a  [#permalink]

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New post 26 Sep 2018, 14:13
Bunuel wrote:
A train left a station P at 6 am and reached another station Q at 11 am. Another train left station Q at 7 am and reached P at 10 am. At what time did the two trains pass one another?

(A) 7:50 am
(B) 8:13 am
(C) 8:30 am
(D) 8:42 am
(E) 9:03 am



For the 1st train total time = 11 - 6 = 5 hrs

For the second train , total time = 10 -7 = 3 hrs.

***** speed and distance are not given . Assign variables for distance.

let the distance be x km.

Average speed of 1st train = x/5 kmh

Average speed of 2nd train = x / 3 kmh

***the point the trains meet is equal to total distance.

let they meet y hrs after 6 am.

y*(x/5) + y*(x-1/3) = y...............x - 1 . second train starts its journey 1 hr later. So this train covered less distance.

x/5 + x-1/3 = 1

3x + 5x - 5 = 15

8x = 20

x = 20/8

x = 5/2

x = 2 hrs 30 mins.

So 8.30 is our time both train meet. (6 +2.30) = 8.30 am.

The best answer is C.

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A train left a station P at 6 am and reached another station Q at 11 a  [#permalink]

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New post 26 Sep 2018, 18:13
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Bunuel wrote:
A train left a station P at 6 am and reached another station Q at 11 am. Another train left station Q at 7 am and reached P at 10 am. At what time did the two trains pass one another?

(A) 7:50 am
(B) 8:13 am
(C) 8:30 am
(D) 8:42 am
(E) 9:03 am

Assign a distance. Then use the "gap" and relative speed approach

Distance?
Call the trains B and C.
They travel the same distance.

B takes 5 hours (from 6 to 11 a.m.)
C takes 3 hours (from 7 to 10 a.m.)

Let distance between P and Q = 15 miles (LCM of 5 and 3)

Use each train's time to find rates.

(1) Each train's rate?
\(r*t=D\)
B, rate: \((r*5hrs)=15mi\)
B, rate: \(r=\frac{15mi}{5hrs}=3\) mph

C, rate: \((r*3hrs)=15mi\)
C, rate: \(r=\frac{15mi}{3hrs}=5\) mph

(2) Distance "gap" between B and C?
B is at station P
C is at station Q
Initial distance = 15 miles

But B travels alone for 1 hour
-- B covers \((3mph*1hr)=3\) miles
-- B shortens the distance between them from 15 to 12 miles

That 12 miles is a "gap." When B and C at relative speed cover the distance, they close that gap and pass one another.

(3) Relative rate/speed?
Opposite directions (towards or away): ADD rates
Relative rate/speed: \((3+5)=8\) mph

(2) Time required for them to pass? \(R*T=D\), so
\(T=\frac{D}{R}\)

\(T=\frac{12mi}{8mph}=\frac{3}{2}=1.5\) hours

(4) Clock time at which they pass one another?

Calculate clock time from the start time of the second train (then both trains are moving)

Train C left at 7:00 a.m.
1.5 hours later is 8:30 a.m.

Answer C
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Re: A train left a station P at 6 am and reached another station Q at 11 a  [#permalink]

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New post 02 Oct 2018, 19:13
Bunuel wrote:
A train left a station P at 6 am and reached another station Q at 11 am. Another train left station Q at 7 am and reached P at 10 am. At what time did the two trains pass one another?

(A) 7:50 am
(B) 8:13 am
(C) 8:30 am
(D) 8:42 am
(E) 9:03 am


We can let t = the number of hours the second train traveled before passing the first train and d = the distance between the two stations. We see that the rate of the first train is d/5 and that of the second train is d/3, and we can create the equation:

d/5 x (t + 1) = d/3 x t

Dividing by d we have:

1/5 x (t + 1) = t/3

Multiplying by 15 we have:

3(t + 1) = 5t

3t + 3 = 5t

3 = 2t

3/2 = 1.5 = t

So the trains passed each other at 8:30am.

Alternate Solution:

We can let t = the time of the second train and d = the distance between the two stations. We note that the rate of the first train is d/5 and the rate of the second train is d/3. At 7:00am, the first train had been traveling for one hour; thus, it traveled a distance of d/5 x 1 = d/5. Therefore, at 7:00am, the distance between the two trains is d - d/5 = 4d/5.

Since the two trains are traveling towards each other, the distance between the two trains is decreasing by d/5 + d/3 = 8d/15 miles each hour. Thus, a distance of 4d/5 will be covered in (4d/5)/(8d/15) = 3/2 = 1.5 hours. Thus, the two trains will meet at 7:00am + 1.5 hours = 8:30am.

Answer: C
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Re: A train left a station P at 6 am and reached another station Q at 11 a &nbs [#permalink] 02 Oct 2018, 19:13
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