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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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Question Stats:

87% (01:30) correct 13% (03:13) wrong based on 69 sessions

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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
[Reveal] Spoiler: OA

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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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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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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

Kudos [?]: 14 [0], given: 111

Train X leaves New York at 1 A.M. and travels east at a speed of x mil   [#permalink] 04 Nov 2017, 21:54
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