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Train A and B, 455 miles apart, are traveling toward each ot

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Train A and B, 455 miles apart, are traveling toward each ot  [#permalink]

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New post 07 Oct 2013, 05:26
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Train A and B, 455 miles apart, are traveling toward each other at constant rates and in the same time zone. If train A left at 4 pm traveling at a speed of 60 miles per hour, and train B left at 5:45 pm and traveling at 45 miles per hour, then at what time would they pass each other?

A. 7:20 pm
B. 8:55 pm
C. 9:05 pm
D. 9:20 pm
E. 9:25 pm

Any faster approaches?

train A travels for an hour and 45 mins so 7/4. distance travelled = 7/4 * 60 = 105.

since they track is 455 miles and Train A has travelled 105 miles. Now 350 miles left.

Since when they meet distance would be equal so 60t + 45t = 350

t = 10/3 so its 3 hours and 20 mins + 5 : 45 = 9:05 pm

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Re: Train A and B, 455 miles apart, are traveling toward each ot  [#permalink]

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New post 07 Oct 2013, 06:40
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fozzzy wrote:
Train A and B, 455 miles apart, are traveling toward each other at constant rates and in the same time zone. If train A left at 4 pm traveling at a speed of 60 miles per hour, and train B left at 5:45 pm and traveling at 45 miles per hour, then at what time would they pass each other?

A. 7:20 pm
B. 8:55 pm
C. 9:05 pm
D. 9:20 pm
E. 9:25 pm

Any faster approaches?

train A travels for an hour and 45 mins so 7/4. distance travelled = 7/4 * 60 = 105.

since they track is 455 miles and Train A has travelled 105 miles. Now 350 miles left.

Since when they meet distance would be equal so 60t + 45t = 350

t = 10/3 so its 3 hours and 20 mins + 5 : 45 = 9:05 pm


It's pretty much it.

In 1 hour and 45 minutes (7/4 hours) A traveled 7/4*60 = 105 miles.

Distance to cover 455 - 105 = 350 miles.

Combined rate (60 + 45) = 105 miles per hour, which mean that they will cover 350 miles in 350/105 = 10/3 hours (3 hours and 20 minutes).

5:45 pm + 3 hours and 20 minutes = 9:05 pm.

Answer: C.
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Re: Train A and B, 455 miles apart, are traveling toward each ot  [#permalink]

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New post 26 Oct 2013, 07:48
We have: 455 = 60*1.75 + x*105. So we get x = (455-105)/105= 10/3 = 3h 20mins.
So the the answer should be 5:45 + 3:20=9:05 (C)
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Re: Train A and B, 455 miles apart, are traveling toward each ot  [#permalink]

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New post 31 Dec 2013, 12:09
fozzzy wrote:
Train A and B, 455 miles apart, are traveling toward each other at constant rates and in the same time zone. If train A left at 4 pm traveling at a speed of 60 miles per hour, and train B left at 5:45 pm and traveling at 45 miles per hour, then at what time would they pass each other?

A. 7:20 pm
B. 8:55 pm
C. 9:05 pm
D. 9:20 pm
E. 9:25 pm

Any faster approaches?

train A travels for an hour and 45 mins so 7/4. distance travelled = 7/4 * 60 = 105.

since they track is 455 miles and Train A has travelled 105 miles. Now 350 miles left.

Since when they meet distance would be equal so 60t + 45t = 350

t = 10/3 so its 3 hours and 20 mins + 5 : 45 = 9:05 pm


Nice problem, here we go

First, since A has a headstart then in that 1hr45min or 1.75hrs he travels 105 miles

Then remaining distance to be traveled will be 455 - 105 = 350 miles

Now, using relative rates (105)(t) = 350

This gives 10/3 hours

Now 5.45pm + 10/3 hours gives us 9.05pm

Hence answer is C

Hope it helps!
Cheers!
J :)
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Re: Train A and B, 455 miles apart, are traveling toward each ot  [#permalink]

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New post 08 Jan 2014, 09:04
Why is this different from the way that you are solving it?

D1= D2 then 60T = 45( T + 1.45h)

Once we know the time we could add it up to the starting time to know when is the meeting time i think.

Thank you very much in advance,
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Re: Train A and B, 455 miles apart, are traveling toward each ot  [#permalink]

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New post 10 Jan 2014, 03:57
guimooow wrote:
Why is this different from the way that you are solving it?

D1= D2 then 60T = 45( T + 1.45h)

Once we know the time we could add it up to the starting time to know when is the meeting time i think.

Thank you very much in advance,


What is T in your equation? Which distances are you equating?

If you want to solve for T it should be: 60T + 45(T-7/4) = 455, where T is the total time for A. Though it's much better to solve with step by step approach explained here: train-a-and-b-455-miles-apart-are-traveling-toward-each-ot-161174.html#p1275213
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Resources:
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: Train A and B, 455 miles apart, are traveling toward each ot  [#permalink]

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New post 02 Mar 2016, 19:22
fozzzy wrote:
Train A and B, 455 miles apart, are traveling toward each other at constant rates and in the same time zone. If train A left at 4 pm traveling at a speed of 60 miles per hour, and train B left at 5:45 pm and traveling at 45 miles per hour, then at what time would they pass each other?

A. 7:20 pm
B. 8:55 pm
C. 9:05 pm
D. 9:20 pm
E. 9:25 pm



i tried to visualize it...
so train A leaves at 4PM. at 5, 60 of the distance would be covered. distance between trains = 395.
in another 45 mins, the train would travel another 45 miles. distance between them = 350.
train B starts...now the distance between them would shrink by 60+45 or 105 miles per hour.
350/105 = 10/3.
so additional 3 hours and 20 minutes after train B starts:
9:05PM
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Re: Train A and B, 455 miles apart, are traveling toward each ot  [#permalink]

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Re: Train A and B, 455 miles apart, are traveling toward each ot &nbs [#permalink] 28 Jul 2018, 03:10
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