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If it took Carlos 1/2 hour to cycle from his house to [#permalink]
23 Sep 2010, 13:10

4

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00:00

A

B

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D

E

Difficulty:

95% (hard)

Question Stats:

52% (02:08) correct
48% (01:09) wrong based on 759 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.

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

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 cannot say that x=17. You have the range for x, you cannot 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 \(r>17.6\).

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 \(r>17.6\).

Re: Data sufficiency [#permalink]
23 Apr 2011, 23:49

1

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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 _________________

If my post is useful for you not be ashamed to KUDO me! Let kudo each other!

Re: Data sufficiency [#permalink]
24 Apr 2011, 00:43

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

\(Distance=Speed*time\)

Q: \(D>6\hspace{2} miles\) I see that Average Speed is given in both statements and time is given in the stem.

Re: Data sufficiency [#permalink]
24 Apr 2011, 01:51

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 this

together we can quickly see that x can be any value between 28800 and 32400 (above and below 31680). insuff. Answer = E _________________

Re: Data sufficiency [#permalink]
25 Apr 2011, 17:48

5

This post received KUDOS

Expert's post

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. _________________

Re: DS Rate Problem [#permalink]
18 Mar 2012, 20:22

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.

Re: If it took Carlos 1/2 hour to cycle from his house to [#permalink]
29 Jul 2013, 14:06

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)

The statements appear to utilize feet, so let's convert miles to feet.

5280*6 = 31,680ft.

Was the distance greater than 31,680 feet?

(1) The average speed at which Carlos cycles from his house to the library yesterday was greater than 16 feet per second.

1hour = 60 minutes 1/2hour = 30 minutes 1 minute = 60 seconds 30minutes*60seconds = 1800 seconds. 1800seconds*16feet/second = 28800 feet. In 30 minutes Carlos travels minimally, 28800 feet. However, we don't know if he travels more than 28800 feet in that 30 minute timespan as the question states simply that he travels greater than 16feet/second. INSUFFICIENT

(2) The average speed at which Carlos cycles from his house to the library yesterday was less than 18 feet per second. 1hour = 60 minutes 1/2hour = 30 minutes 1 minute = 60 seconds 30minutes*60seconds = 1800 seconds. 1800seconds*18feet/second = 32400 feet.

Here, Carlos travels less than 32400 feet. This means he could travel greater than 6 miles (31680 feet) or he may travel less. INSUFFICIENT

1+2) He travels more than 16feet/second but less than 18feet/second. There is a possibility he could travel more or less than the 31680 feet the question asks. INSUFFICIENT

Re: If it took Carlos 1/2 hour to cycle from his house to [#permalink]
08 Aug 2014, 15:21

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