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Intern
Joined: 22 Jan 2006
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If it took Carlos 1/2 hour to cycle from his house to the [#permalink]
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20 Feb 2006, 18:17
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Question Stats:
37% (01:31) correct
63% (01:17) wrong based on 166 sessions
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If it took Carlos 1/2 hour to cycle from his house to the library yesterday, was the distance that he cycled greater than 6 miles? ( Note: 1 mile=5280 feet. 1) The average speed at which Carlos cycled from his house to the library yesterday was greater than 16 feet per second. 2) The average speed at which Carlos cycled from his house to the library yesterday was greater than 18 feet per second. OPEN DISCUSSION OF THIS QUESTION IS HERE: ifittookcarlos12hourtocyclefromhishouseto129287.html
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Intern
Joined: 18 Feb 2006
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A or B alone should be enough.
30 min * (60 sec / 1 min) * (x ft / 1 sec) * (1 mile / 5,280 ft)
= (30 * 60 * x) / 5,280
= 1800x/5280
= 15x/44
Plug 'n' chug x = 16 or 18, and you can get an answer.



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Joined: 14 Dec 2005
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If it took Carlos 1/2 hour to cycle from his house to the library yesterday, was the distance that he cycled greater than 6 miles? (Note: 1 mile=5280 feet.
1) The average speed at which Carlos cycled from his house to the library yesterday was greater than 16 feet per second.
let's assume it is 16 feet.
D = speed * time..
D = 16 feet per second * 30 minutes
= 16 feet/second * 1800 seconds
= 18000 + 10800 feet = 28800/5280 < 6 miles..
D = 18 feet then answer is different
therefore A is not sufficient.
2) The average speed at which Carlos cycled from his house to the library yesterday was greater than 18 feet per second.
sufficient.
my answer is B.



Intern
Joined: 22 Jan 2006
Posts: 18

Neither of you folks have the correct answer. I will let y'all know whenever someone gets right. Since GMAT Prep software does not offer explanations, I thought this might be a good try. I failed the question and would like to know how they got to their answer.



Manager
Joined: 13 Dec 2005
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me too ... i also got B ....looks like either we all are missing something or the gmat answer is wrong ....
well gmatprep answer cannot be wrong ?? then what are we missing here



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Joined: 07 Jul 2004
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1) Speed = greater than 16 feet per second. If the speed was 16 feet per second, he would have cycled 16 * 1800 = 28800 feet = 5.45 miles. So he could have cycled more, or less than 6 miles. Insufficient.
2) If speed = 18 feet per second, he would have cycled 18*1800 = 6.1 miles. So he cycled at more than 18 feet per second, he would definitely have gone more than 6 miles. Sufficient.
Ans B
Note: I don't want to read too deeply into the meaning of the term 'average speed'. In most circumstances, this is how I would work the question. However, it could also mean Carlos did not travel at a constant speed but average out over the 1/2hours, this is his speed he cycled at.



Intern
Joined: 07 Feb 2006
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Here's another way to look at it: [#permalink]
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20 Feb 2006, 20:20
Suppose Carlos only pedals fast enough to travel 6 miles in 1/2 hour (or 12 mph). Then, converting units and multiplying:
12mph * (5280 feet / mi) * (1 hr/3600 sec) = (5280/300) ft/sec = 17.6 ft/sec
Therefore, to go less than 17.6 ft/sec would mean traveling less than 6 miles (ruling out the first answer, 16 ft/sec), and faster than 17.6 ft/sec would mean exceeding 6 miles (therefore ruling in the second answer, 18 ft/sec).



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You made a typo here in the second statmenet. On the GMATPREP 2nd statement is
2) The average speed at which Carlos cycled from his house to the library yesterday was less than 18 feet per second.
If original then this is the solution:
E
Time = Distance/Speed i.e 1800 = D/S i.e D = S * 1800/5280 = 45S/132 where S is in feet/second, D is in miles and T is in seconds
St1: if S is greater than 16 feet/second. i.e D> 45*16/132 i.e D> 180/33: INSUFF
St2: D< 45*18/132 i.e D<135/22: INSUFF
Combined: 180/33 < D < 135/22: INSUFF
In your version this is the solution:
B
Time = Distance/Speed i.e 1800 = D/S i.e D = S * 1800/5280 = 45S/132 where S is in feet/second, D is in miles and T is in seconds
St1: if S is greater than 16 feet/second. i.e D> 45*16/132 i.e D> 180/33: INSUFF
St2: D> 45*18/132 i.e D>135/22: SUFF
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Intern
Joined: 22 Jan 2006
Posts: 18

You are right Dahiya. Sorry guys for the typo. The second statement should read as follows:
2) The average speed at which Carlos cycled from his house to the library yesterday was less than 18 feet per second.
The official answer is E.
Thanks y'all for the contribution!



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Re: DS: GMATPrep [#permalink]
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24 Jun 2006, 10:58
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OasisNYK wrote: If it took Carlos 1/2 hour to cycle from his house to the library yesterday, was the distance he cycled greater than 6 miles? (Note: 1 mile = 5280 sq feet)
1) The avg speed at which he travelled was greater than 16 feet per second 2) the avg speed at which he travelled was less than 18 feet per second
1. If average speed was 16 feet/s, in 1/2 hr, Carlos Moya travelled
16 x 60 x 30 = 28,800 feet = 5.5 miles approx. Since speed was greater than 16feet/s, this is not sufficient
2. If average speed was 18 feet/s, in 1/2 hr, Carlos Santana travelled
18 x 60 x 30 = 32,400 feet = 6.1 miles approx. Since speed was less than 16feet/s, this is not sufficient
Combining both, speed between 5.5 mi/hr and 6.1 mi/hr
So not sufficient, E
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Data Sufficiency: Was the distance greater? [#permalink]
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05 Feb 2007, 21:35
How do you solve?
If it took Carlos 1/2 hour to cycle from his house to the library yesterday, was the distance that he cycled greater than 6 miles? (Note: 1 mile = 5,280 feet)
(1) The average speed at which Carlos cycled from his house to the library yesterday was greater than 16 ft. per second.
(2) The average speed at which Carlos cycled from his house to the library yesterday was less than 18 feet per second.



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I would solve it with good old fashion elbow grease.
60 seconds in a minute. 30 minutes in a 1/2 hour.
5280 x 6 = 31,680 or the target number
16 x 60 x 30 = 28800 is the minimum distance (Not Sufficient)
18 x 60 x 30 = 32,400 is the maximum distance (Not Sufficient)
Together they equal 28,800< Disstance < 32,400
Therefore E.
I think this is right, but if anyone has a shorter way that would be great



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DS: Carlos' distance that he cycled [#permalink]
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04 Apr 2007, 08:50
If it took Carlos 1/2 hr to cycle from his house to the library yesterday, was the distance that he cycled greater than 6 miles?
(1 mile=5280 feet)
1. The average speed Carlos cycled from his house to the library was greater than 16 feet per second.
2.The average speed Carlos cycled from his house to the library was smaller than 18 feet per second.



Manager
Joined: 11 Mar 2007
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i would say D. Both A or B is sufficient.
This is a YES/NO question.
Based on the avg speed given in A or B we can calculate the MAX & MIN speeds. Based on just calculated speed & time (as given in problem) we can calculate the max/min distance & thus say YES/NO to the question.
Any thoughts ?? BTW whats the correct answer ??



Current Student
Joined: 31 Aug 2007
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DSrates [#permalink]
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26 Sep 2007, 16:57
if it took carlos 1/2 hour to cycle from his house to the library yesterday, was the distance that he cycled greater than 6 miles? (1mi=5280ft)
st1the avg speed at which carlos cycled from his house to the library was greater than 16 feet per second.
st2the avg speed at which carlos cycled from his house to the library was less than 18 feet per second.



Director
Joined: 18 Jul 2006
Posts: 524

6 miles in 1/2 hr
=> 12 miles in 1 hr
So question is, did he drive at more than 12 m/h or 17.6 feet/sec.
From 1. speed could be 17, 18, 19...Not suff
From 2. speed could be 17.9, 17, 16, 15.... Not suff
From 1 & 2: speed between 16 and 18...NOT suff again.
E.



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Re: DSrates [#permalink]
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26 Sep 2007, 19:08
young_gun wrote: if it took carlos 1/2 hour to cycle from his house to the library yesterday, was the distance that he cycled greater than 6 miles? (1mi=5280ft)
st1the avg speed at which carlos cycled from his house to the library was greater than 16 feet per second. st2the avg speed at which carlos cycled from his house to the library was less than 18 feet per second.
U gotta rephrase this question. We need to find Carlo's would be rate.
so R=6/1/2 > did Carlos travel 12 m/h or greater???
S1: his avg speed was 16ft per sec.
We need to take our rephrased question a step further. 12m/h *5280 = 63360ft/hr now divide by 60 twice (once for minutes and then once again for seconds) So 63360/3600 = 17.6 ft/sec.
B/c his avg speed was greater 16ft per sec, it could be 17.6 or just slightly greater than 16ft per sec... INsuff.
S2: Less than 18ft per sec. Could be 1ft/per sec could be 17.6 Insuff.
1 and 2:
Let R equal Carlo's rate so 16<R<18 Could be anywhere between these numbers. INsuff.
Ans E.



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St1:
speed > 16 ft per second
Speed could be 20 ft/second, then the distance = 17 * 1800 = 30600 ft = 5.7miles
Speed could be 100 ft/second, then the distance = 100 * 1800 = 180000 = 34 miles
Insufficient.
St2:
speed < 18 ft per second
Speed could be 2 ft/second, then the distance = 2* 1800 = 3600 = 0.68 miles
Speed could be 100 ft/second, then the distance = 34 miles
Insufficient.
St1&St2:
16 < speed < 18
Speed could be 17 ft/second, then the distance = 17 * 1800 = 5.7 miles
Speed could be 17.9 ft/secdon, the the distance = 32220 ft = 6.1 miles
Insufficient.
Ans E



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5280*6 = 31680 feet in 6 miles
16*60*30 = 16*1800 = 28800
so he went over 28,800 feet
18*60*30 = 18*1800 = 32400
so he went less than 32,400 feet
taken together we know he went between 28,800 and 32,400 feet. since 31,680 falls between these two numbers we don't know if he went more than 6 miles or not.
Answer E



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Another (maybe easier) way to do it would be to take...
5280/60 = 88
88/30 = 2 and 28/30 or 2 and 14/15 with is very close to 3, but not quite. we'll call it 2.95
so that's how many feet per second Carlos must walk to go 1 mile in 30 minutes
2.95*6 = (6*3)(6*.05) = 17.70
which gives us 17.7 feet per second for 30 minutes is 6 miles.
it's not exact (answer is 17.6, but it's close enough to get the right answer
17.7 is greater than 16 and less than 18. which means neither of two two statements rules anything out.
Answer E.







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