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Train X leaves New York at 1 A.M. and travels east at a speed of x mil

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Train X leaves New York at 1 A.M. and travels east at a speed of x mil  [#permalink]

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New post 31 Oct 2017, 00:03
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Train X leaves New York at 1 A.M. and travels east at a speed of x miles per hour. If train Z leaves New York at 2 A.M. and travels east, at what rate of speed will train Z have to travel in order to catch train X at exactly 5:30 A.M.?

(A) 5x/6
(B) 9x/8
(C) 6x/5
(D) 9x/7
(E) 3x/2

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Re: Train X leaves New York at 1 A.M. and travels east at a speed of x mil  [#permalink]

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New post 01 Nov 2017, 17:07
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Bunuel wrote:
Train X leaves New York at 1 A.M. and travels east at a speed of x miles per hour. If train Z leaves New York at 2 A.M. and travels east, at what rate of speed will train Z have to travel in order to catch train X at exactly 5:30 A.M.?

(A) 5x/6
(B) 9x/8
(C) 6x/5
(D) 9x/7
(E) 3x/2


Since we have a catch-up problem, we can use the formula: Distance 1 = Distance 2.

The rate of train X is x, and the time is 4.5 hours. We can let the rate of train Z = r, and the time is 3.5 hours. Thus:

4.5x = 3.5r

45x = 35r

9x = 7r

9x/7 = r

Answer: D
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Re: Train X leaves New York at 1 A.M. and travels east at a speed of x mil  [#permalink]

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New post 31 Oct 2017, 01:28
I would go for (D).

Train X travels with a speed of x=d/4.5 (1.00am to 5.30am), Train Z travels with a speed of v=d/3.5 (2.00am to 5.30am).

Now if d=4.5 {miles}, Train X travels with a speed of x=1. Train Z travels with a speed of v=4.5/3.5=9/7.

So, for x=1, 9x/7=9/7, hence option (D).
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Train X leaves New York at 1 A.M. and travels east at a speed of x mil  [#permalink]

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New post 03 Nov 2017, 08:37
Bunuel wrote:
Train X leaves New York at 1 A.M. and travels east at a speed of x miles per hour. If train Z leaves New York at 2 A.M. and travels east, at what rate of speed will train Z have to travel in order to catch train X at exactly 5:30 A.M.?

(A) 5x/6
(B) 9x/8
(C) 6x/5
(D) 9x/7
(E) 3x/2

This question can be approached with the "close the gap in a chase" method, and by choosing numbers.

Train A, when traveling alone, creates the distance between A and B. That's the gap.

When Train B starts moving, the chase is on.
Time = 3.5 hours
B chases A, closes the gap, only while both move.

B closes the gap at a relative speed, r, of (B's rate - A's rate). What is r? rt = D, so...

The gap will close at D/t = r.

To find r, pick an easy "gap" distance: 70 mi (because t = 3.5)

A travels for 1hr, A's rate = 70 mph = x

The distance/gap of 70 mi is closed by relative speed of (B-A)

Relative speed? D/t = r
70/3.5 = 20
The difference between A's speed and B's speed is 20 mph.

B is chasing; B travels at 20 mph faster than A

B's speed: (70 + 20) = 90 mph

With x = 70, check choices until the answer is 90.

(A) 5x/6 = 350/6 = 5_. No

(B) 9x/8 = 6300/8 = 7_. No

(C) 6x/5 = 4200/5 = 8_. No

(D) 9x/7 = 6300/7= 90. MATCH

(E) 3x/2 = 210/2 = 105. No

Answer D
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Train X leaves New York at 1 A.M. and travels east at a speed of x mil  [#permalink]

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New post 04 Nov 2017, 21:54
let the speed of train X=10m/hr
It has traveled from 1 am till 5.30am, total of 4.5 hrs
Distance=10*4.5=45 miles
This distance has to be covered by train z in 3.5 hrs(it travels from 2 am to 5.30am)
let the speed of z=s
45=sx3.5=450/35=90/7
Now in answer choice plugin value of speed of train X=10
the correct answer is D=90/7
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Re: Train X leaves New York at 1 A.M. and travels east at a speed of x mil  [#permalink]

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New post 08 Jun 2018, 06:44
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Bunuel wrote:
Train X leaves New York at 1 A.M. and travels east at a speed of x miles per hour. If train Z leaves New York at 2 A.M. and travels east, at what rate of speed will train Z have to travel in order to catch train X at exactly 5:30 A.M.?

(A) 5x/6
(B) 9x/8
(C) 6x/5
(D) 9x/7
(E) 3x/2


Train X leaves at 1 A.M. and Train Z leaves New York at 2 A.M.
When they meet at 5:30 A.M., Train X's travel time will be 4.5 hours, and Train Z's travel time will be 3.5 hours
Also recognize that, when the trains meet, they both will have traveled the same distance.

We know that Train X travels at x mph, and we want to find Train Z's speed.
So, let z = Train Z's speed (in mph)

Now let's create an equation we can work with!

Let's start with a word equation:
Distance traveled by Train X = Distance traveled by Train Z
distance = (time)(speed)

So, we get: (4.5 hours)(x mph) = (3.5 hours)(y mph)
Simplify: 4.5x = 3.5y
Solve for y to get: = 4.5x/3.5
Check answer choices....not there.
So, take 4.5x/3.5, and multiply top and bottom by 2 to get the EQUIVALENT fraction: 9x/7
Check answer choices....answer: D

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Re: Train X leaves New York at 1 A.M. and travels east at a speed of x mil &nbs [#permalink] 08 Jun 2018, 06:44
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