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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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26 Sep 2018, 05:10
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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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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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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 and thus closing the gap together at their relative/combined speed)

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

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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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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.

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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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18 May 2019, 00:13
ScottTargetTestPrep wrote:
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.

Hey ScottTargetTestPrep,

I was hoping if you could share the logic behind this equation d/5 x (t + 1) = d/3 x t..

Since this would mean that the distance travelled by the 2 trains in (t+1) and t hrs respectively would be equal, I'm assuming that reaching this conclusion would require some more calculations or probably a shortcut if you could share
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Joined: 04 Aug 2010
Posts: 475
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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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18 May 2019, 04:14
1
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

Let the distance between P and Q = 5 miles.
Since the first train takes 5 hours (from 6-11am) to travel the 5-mile distance between P and Q, the rate for the first train $$= \frac{d}{t} = \frac{5}{5} = 1$$ mph.:
Traveling at 1 mph from 6-7am, the first train covers 1 mile of the 5-mile distance between P and Q, leaving 4 miles between the two trains.

Time and rate have a RECIPROCAL RELATIONSHIP.
Whereas the first train takes 5 hours to travel between P and Q (from 6am to 11am), the second train takes 3 hours (from 7am to 10am).
Since the TIME RATIO for the two trains $$= \frac{first}{second} = \frac{5}{3}$$, the RATE RATIO for the two trains $$= \frac{first}{second} = \frac{3}{5}$$.

The rate ratio implies the following:
Of every 8 feet traveled when the two trains work together to meet, the first train travels 3 feet, while the second train travels 5 feet.
Thus, the first train travels $$\frac{3}{8}$$ of the remaining 4 miles:
$$\frac{3}{8}*4 = 1.5$$ miles
Since the rate for the first train = 1 mph, the time for the first train to travel 1.5 miles after 7am to meet the second train $$= \frac{d}{r} = \frac{1.5}{1} = 1.5$$ hours.
Thus, the time at which the first train meets the second train = 7am + 1.5 hours = 8:30am.

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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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19 May 2019, 02:10
P______R____________M___________________Q

Let the distance between P and Q be 'd' kms. The times taken by the 1st train(T1) and the 2nd train(T2) to cover 'd' are 5 and 3 hours respectively. By the time T2 started from Station Q, T1 had already been travelling for 1 hr and had thus covered (d/5)kms leaving (4d/5)kms between them. Let R denote the point which T1 had reached when T2 starts from Station Q. So at 7am, T1 and T2 start from R and Q respectively towards each other and meet at Point M covering a distance of (4d/5) kms between them. Since, the ratio of the the speeds of T1 and T2 is 3:5, T2 covers (5/8)ths of RQ(4d/5)=
(4d/5)*(5/8)=(d/2)kms to reach the meeting point M.
Since, T2 covers 'd' kms in 3 hrs, it will reach M 1.5 hrs after it starts, i.e at 8:30 am. Ans: 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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19 May 2019, 05:39
let the distance = lcm(3,5)=15kms
speed of first train,P=3 kmph
speed of second train,Q=5kmph
Distance traveled by train P from 6AM-7AM=3km
Relative distance between two trains at 7AM = 15-3=12
Relative speed= 5+3=8
Time taken to meet= 12/8=1.5hrs
They will meet each other at 8:30AM (7+1hr30min)
Re: A train left a station P at 6 am and reached another station Q at 11 a   [#permalink] 19 May 2019, 05:39
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