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# A train traveled from station X to station Y at a constant speed of 88

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Math Expert
Joined: 02 Sep 2009
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A train traveled from station X to station Y at a constant speed of 88  [#permalink]

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23 Jun 2016, 03:01
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Difficulty:

55% (hard)

Question Stats:

67% (02:30) correct 33% (02:24) wrong based on 321 sessions

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A train traveled from station X to station Y at a constant speed of 88 feet per second. Is the distance that the train traveled from station X to Station Y greater than 40 miles? (1 mile = 5,280 feet)

(1) It took less than 45 minutes for the train to travel from station X to Station Y.
(2) It took more than 42 minutes for the train to travel from station X to Station Y

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Re: A train traveled from station X to station Y at a constant speed of 88  [#permalink]

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23 Jun 2016, 03:40
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3
Bunuel wrote:
A train traveled from station X to station Y at a constant speed of 88 feet per second. Is the distance that the train traveled from station X to Station Y greater than 40 miles? (1 mile = 5,280 feet)

(1) It took less than 45 minutes for the train to travel from station X to Station Y.
(2) It took more than 42 minutes for the train to travel from station X to Station Y

Speed of train = 88feet/sec = (88x60)feet/min
=> (88x60x)/5280] mile/ min
=> 1mile/ min

Statement 1) It took less than 45min. Train could have reached after 39 min (travelling 39 miles) or after 41mins (travelling 41miles) .. Insufficient
Statement 2) It took more than 42min. The train travelled more than 42miles .. Sufficient

Choice B) alone is sufficient to ans the question.

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A train traveled from station X to station Y at a constant speed of 88  [#permalink]

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Updated on: 05 Nov 2018, 05:41
1
1
Bunuel wrote:
A train traveled from station X to station Y at a constant speed of 88 feet per second. Is the distance that the train traveled from station X to Station Y greater than 40 miles? (1 mile = 5,280 feet)

(1) It took less than 45 minutes for the train to travel from station X to Station Y.
(2) It took more than 42 minutes for the train to travel from station X to Station Y

$$speed of train = \frac{88feet}{sec}= \frac{60.0012 miles}{hour} = \frac{1mile}{minute}$$

(1) It took less than 45 minutes for the train to travel from station X to Station Y.
INSUFFICIENT:- If the Train took less than 45 minutes then it has covered less than 45 mile. But we cannot determine precisely how many miles. Less than 45 can be anything from 1 to 44. Train may reach in 30 minutes or 38 minutes or 44 minutes hence distance can be 30 miles, 38 miles or 44 miles. Anything less than 45 miles.
(2) It took more than 42 minutes for the train to travel from station X to Station Y
SUFFICIENT:- We can confidently answer "YES, The distance is more than 40 Miles"

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Originally posted by LogicGuru1 on 23 Jun 2016, 03:43.
Last edited by LogicGuru1 on 05 Nov 2018, 05:41, edited 1 time in total.
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Re: A train traveled from station X to station Y at a constant speed of 88  [#permalink]

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23 Jun 2016, 12:03
Bunuel wrote:
A train traveled from station X to station Y at a constant speed of 88 feet per second. Is the distance that the train traveled from station X to Station Y greater than 40 miles? (1 mile = 5,280 feet)

(1) It took less than 45 minutes for the train to travel from station X to Station Y.
(2) It took more than 42 minutes for the train to travel from station X to Station Y

Converting the speed to mile/hr:-

88*60*60/5280= 60 miles/hr

(1) It took less than 45 minutes for the train to travel from station X to Station Y.

distance = speed*time

since time <45 minutes, distance will be <60*45/60, i.e. <45.

But it may or may not be <40 miles.

(2) It took more than 42 minutes for the train to travel from station X to Station Y

distance> 60*42/60 i.e. >42.

Sufficient.

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Re: A train traveled from station X to station Y at a constant speed of 88  [#permalink]

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25 Mar 2017, 15:53
Quick question, how do you guys convert the mileage of the train so fast?

I wrote it out and the conversion alone took me forever. I'm obviously missing something...

88ft/1sec x 60sec/1min x 60min/1hr x 1mile/5280ft = 60

Any help or shortcuts? Thanks!
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Re: A train traveled from station X to station Y at a constant speed of 88  [#permalink]

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20 Jul 2017, 04:47
Mavs1205 wrote:
Quick question, how do you guys convert the mileage of the train so fast?

I wrote it out and the conversion alone took me forever. I'm obviously missing something...

88ft/1sec x 60sec/1min x 60min/1hr x 1mile/5280ft = 60

Any help or shortcuts? Thanks!

just divide 5280 by 88, and you will get number of seconds the train needs to cover 1 mile. that's it...

hope this helps
cheers ...
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Re: A train traveled from station X to station Y at a constant speed of 88  [#permalink]

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20 Jul 2017, 06:28
Bunuel wrote:
A train traveled from station X to station Y at a constant speed of 88 feet per second. Is the distance that the train traveled from station X to Station Y greater than 40 miles? (1 mile = 5,280 feet)

(1) It took less than 45 minutes for the train to travel from station X to Station Y.
(2) It took more than 42 minutes for the train to travel from station X to Station Y

R = 88 feet per second => 60 miles per hour.
D > 40 ?

1) t < 45 minutes
t = 44 minutes.
D > 40 miles (Yes)
t = 20 minutes
D < 40 miles (No)
Insufficient.

2) t > 42 minutes
t = 42.1 minutes
D = $$\frac{42.1}{60}$$ * 60 = 42.1 miles
Sufficient.

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Re: A train traveled from station X to station Y at a constant speed of 88  [#permalink]

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30 Jul 2017, 17:45
1
1
Bunuel wrote:
A train traveled from station X to station Y at a constant speed of 88 feet per second. Is the distance that the train traveled from station X to Station Y greater than 40 miles? (1 mile = 5,280 feet)

(1) It took less than 45 minutes for the train to travel from station X to Station Y.
(2) It took more than 42 minutes for the train to travel from station X to Station Y

We can let d = the distance between station X and station Y. We need to determine whether d > 40 miles.
Since the distance is given in miles but the speed, or rate, is given in feet per second, we need to convert the distance from miles to feet. Since 1 mile = 5,280 feet, 40 miles = 40 x 5,280 = 211,200 feet.

Since rate x time = distance and we have a rate of 88 ft/s, we can change the question to:

Is 88t > 211,200 feet?

Is t > 2,400 seconds?

Since 2,400 seconds = 2,400/60 = 40 minutes, the question becomes:

Is t > 40 minutes?

Statement One Alone:

It took less than 45 minutes for the train to travel from station X to Station Y.

Since less than 45 minutes could mean 42 minutes or 38 minutes, statement one does not tell us whether t > 40 minutes. Statement one alone is not sufficient to answer the question.

Statement Two Alone:

It took more than 42 minutes for the train to travel from station X to Station Y.

Statement two tells us that t is greater than 40 minutes.

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Re: A train traveled from station X to station Y at a constant speed of 88  [#permalink]

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04 Nov 2018, 11:02
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Re: A train traveled from station X to station Y at a constant speed of 88   [#permalink] 04 Nov 2018, 11:02
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