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If it took Carlos 1/2 hour to cycle from his house to [#permalink]
 23 Sep 2010, 14:10
Question Stats:
45% (01:59) correct
54% (01:32) wrong based on 7 sessions
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 ft) (1) The average speed at which Carlos cycles from his house to the library yesterday was greater than 16 feet per second. (2) The average speed at which Carlos cycles from his house to the library yesterday was less than 18 feet per second.
Last edited by Bunuel on 17 Jul 2012, 01:36, edited 3 times in total.
Edited the question and added the OA
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TomB wrote: If it took Carlos 1/2 hour to cycle from his house to library was the distance that he cycle greater than 6 miles(1 mile=5280feet)
1) the avg speed at which Carlos cycled from his house to library was greater than 16 feet per second
2) the avg speed at which Carlos cycled from his house to library was less than 18 feet per second
so we know that 16<x<18. (x = avg speed,x=17) what am i missing? (1) Avg speed > 16fps So distance covered > 16*1800 feet = 28800feet or about 5.45miles So distance > 5.45 miles Not sufficient (2) average speed < 18fps So distance < 18*1800 feet or about 6.14miles So distance < 6.14miles Not sufficient (1)+(2) Distance is between 5.45 and 6.14 miles Not sufficient to say if it is > 6 miles Answer = E
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TomB wrote: If it took Carlos 1/2 hour to cycle from his house to library was the distance that he cycle greater than 6 miles(1 mile=5280feet)
1) the avg speed at which Carlos cycled from his house to library was greater than 16 feet per second
2) the avg speed at which Carlos cycled from his house to library was less than 18 feet per second
so we know that 16<x<18. (x = avg speed,x=17) what am i missing? First of all from 16<x<18 you can not say that x=17. You have the range for x, you can take an average and say that x equals to it. Question is d>6 --> as rt=d (where r is the rate in miles per hour) then question becomes: is rt=d>6 --> or is r*\frac{1}{2}>6, as t=\frac{1}{2} hours --> is r>12 miles/hour? --> 12 \ miles/hour = \frac{12*5280}{60*60} \ feet/second = 17.6 \ feet/sec. Is r>17.6 feet/sec? (1) r>16 feet/sec. Not sufficient. (2) r<18 feet/sec. Not sufficient. (1)+(2) 16<r<18 still not sufficient to say whether r>17.6. Answer: E.
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Bunuel, is there an specific tag or category for this type of question (more / less questions)? (Not only in rate questions). Thank you very much!
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Bunuel wrote: metallicafan wrote: Bunuel, is there an specific tag or category for this type of question (more / less questions)? (Not only in rate questions).
Thank you very much! Not sure, but try min/max questions: search.php?search_id=tag&tag_id=63 and search.php?search_id=tag&tag_id=42Thanks! Kudos for you 
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Folks, I approached this as a YES / NO DS question, since statement i) was NO and ii) indertiminate - I put A.
ie.Statement 1 alone is sufficient.
I though the question was a yes/ no since the question was "Was the distance greater than 6 miles ?"
Whats wrong with my reasoning.
Thank you.
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rjdio wrote: Folks, I approached this as a YES / NO DS question, since statement i) was NO and ii) indertiminate - I put A.
ie.Statement 1 alone is sufficient.
I though the question was a yes/ no since the question was "Was the distance greater than 6 miles ?"
Whats wrong with my reasoning.
Thank you. This is indeed YES/NO DS question, but you don't have a NO answer for (1). If you you convert the question "was the distance that he cycle greater than 6 miles" into the rate you'll get "was rate>17.6 feet/sec?" (Refer to my first post to see how to convert.) (1) r>16 feet/sec. Not sufficient. (2) r<18 feet/sec. Not sufficient. (1)+(2) 16<r<18 still not sufficient to say whether r>17.6. Answer: E. Hope it's clear.
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Thank you Sir.Crystal clear now.
On the test day - one has to be real level headed.
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6 miles = 5280 * 6 = 31680.
1. For average speed greater than 16 feets for second
In 30 minuts distance travelled = 16 * 60 * 30 > 28800 feets
2. For average speed less than 18 feet for second
In 30 minuts distance travelled = 18 * 60 * 30 < 32400 feets
From both we can not say it travelled more than 31680 feets or not.
So answer is (E)
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Re: Data sufficiency [#permalink]
24 Apr 2011, 00:49
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1/2 hour is 30 minutes which is 1800 seconds. (i) If average speed is greater than 16 ft/sec, then he cycled more than 1800*16=28800 feet = 28800/5280=2880/528=720/132=180/44=45/11>4 miles It is not sufficient to say that he cycled more than 6 miles (ii) If average speed is less than 18 ft/sec, then he cycled less than 1800*18=32400 feet = 32400/5280 = 810/132=405/66=135/22=6 3/22 miles>6 miles It is not sufficient to say that he cycled more than 6 miles (i) and (ii) together says that he cycled more than 4 1/11 miles but less than 6 3/22 miles. Again we could not conclude anything about 6 miles. The answer is E
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Re: Data sufficiency [#permalink]
24 Apr 2011, 01:43
agdimple333 wrote: 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)
(a) Average speed at which carlos cycled from his house to the library yesterday was greater than 16 ft / sec
(b) Average speed at which carlos cycled from his house to the library yesterday was less than 18 ft/ sec Distance=Speed*timeQ: D>6\hspace{2} milesI see that Average Speed is given in both statements and time is given in the stem. 6=x*\frac{1}{2}x=12\hspace{2}miles/hrQ: Is the speed more than 12 miles/hr? 1. Speed > 16 ft/sec 16 \hspace{2} \hspace{2} \frac{ft}{sec}=\frac{16*3600}{5280} \hspace{2} \frac{miles}{hr}=\frac{120}{11} \frac{miles}{hr} \approx 11 \hspace{2} \frac{miles}{hr}Speed > 11 miles/h. The speed could be more than 12 miles/hr or less than 12 m/h. Not Sufficient. 2. Speed < 18 ft/sec 18 \hspace{2} \frac{ft}{sec}=\frac{18*3600}{5280} \hspace{2} \frac{miles}{hr}=\frac{135}{11} \hspace{2} \frac{miles}{hr} \approx 12.2 \hspace{2} \frac{miles}{hr}Speed < 12.2 miles/h. The speed could be less than 12 miles/hr or more than 12 m/h. Not Sufficient. Combining both; 11 \hspace{2} \frac{miles}{h}< Speed < 12.2 \hspace{2} \frac{miles}{h}Speed could be 12.1 miles/h or 11.9 miles/h. Not Sufficient. Ans: "E"
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Re: Data sufficiency [#permalink]
24 Apr 2011, 02:51
agdimple333 wrote: 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)
(a) Average speed at which carlos cycled from his house to the library yesterday was greater than 16 ft / sec
(b) Average speed at which carlos cycled from his house to the library yesterday was less than 18 ft/ sec Lots of similar ways to solve this question - and went for the 30 second solve  Firstly saw we're looking at a question dealing with ranges and that the two rates given were not restricted to integer values.... so aware there is a trap of thinking speed is 17/ft/s.. it isnt!Next saw that the only way of solving this is by plugging in the values into "s*t = d" Quick recall or mental calc for an hour's worth of seconds = 60*60 =3600.... (mentally see 6*6 and two zeros) and halved it to 1800. (for some reason my mind goes for half of 36 quicker than 3*6) So we need to do a quick calc to see where 5280*6 falls around 16*1800 and 18*1800. (leave it as feet, no need to waste time in converting back to miles) 5280 * 6 = 31680 <<<< value x we're looking for 1800 * 16 = 28800 1) insuff as prompt only says x is more than this 1800 * 18 = 32400 2) insuff as prompt only says x is less than thistogether we can quickly see that x can be any value between 28800 and 32400 (above and below 31680). insuff. Answer = E
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Last edited by bleemgame on 24 Apr 2011, 08:38, edited 1 time in total.
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Re: Data sufficiency [#permalink]
24 Apr 2011, 08:01
30 * 60 * 16 = 28800/5280 = 5.something, so NO he can be driving at 17.9 ft/s = 1800 * 17.9 = 32220 So Distance can be > 6 miles here (1) is insufficient (2) is insufficient (1) + (2) is insufficient Answer - E
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Re: Data sufficiency [#permalink]
25 Apr 2011, 18:48
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agdimple333 wrote: 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)
(a) Average speed at which carlos cycled from his house to the library yesterday was greater than 16 ft / sec
(b) Average speed at which carlos cycled from his house to the library yesterday was less than 18 ft/ sec One thing that comes to mind when I read the question above is that the statements give the speed in ft/sec while the question stem asks about a speed in miles/hr. It could be easier to convert the speed of the question stem in ft/sec since that involves only one calculation. Was his average speed greater than 6miles/half hour or 12 miles/hr? 12 miles/hour = 12*5280/3600 ft/sec = 17.6 ft/sec Question stem: Was his average speed greater than 17.6 ft/sec? Now it is a direct comparison.
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Re: DS Rate Problem [#permalink]
18 Mar 2012, 21:22
Statement 1: If average speed was more than 16 ft / second = 16/5280 miles /second = 16*3600/5280 = 10.9 miles /hour, then the total distance is more than 10.9*1/2 miles, or more than 5.45 miles. This is insufficient to say whether the distance is more or less than 6 miles. Insufficient. Statement 2: If the average speed was less than 18 ft/second = 18*3600/5280 miles/hour = 12.27 miles/hour, then the total distance was less than 12.27*(1/2) = 6.13 miles. This may or may not be less than 6 miles. Insufficient. Combining both the statements, we know that the distance lies between 5.45 and 6.13 miles, but this is not enough to say whether the distance is more or less than 6 miles. Insufficient. Therefore the answer is (E).
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If it took Carol 1/2 hour to... [#permalink]
27 Nov 2012, 06:03
If it took Carol 1/2 hour to cycle from his house to the library yesterday, was the distance that he cycled
greater than 6 miles? (1mile = 5,280feet)
(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 less than 18 feet
per second
Can somebody explain why it is E??
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Re: If it took Carol 1/2 hour to... [#permalink]
27 Nov 2012, 06:07
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Re: If it took Carol 1/2 hour to...
[#permalink]
27 Nov 2012, 06:07
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