Carol's commute

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Carol's commute

by alex.gellatly » Mon Jul 02, 2012 9:25 pm
If it took Carols 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 Carols cycled from his house to the library yesterday was greater than 16 feet per second.

2. The average speed at which Carols cycled from his house to the library yesterday was less than 18 feet per second.
Source: — Data Sufficiency |

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by Anurag@Gurome » Mon Jul 02, 2012 10:05 pm
alex.gellatly wrote:If it took Carols 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 Carols cycled from his house to the library yesterday was greater than 16 feet per second.
2. The average speed at which Carols cycled from his house to the library yesterday was less than 18 feet per second.
1/2 hour = 1800 seconds

Statement 1: Speed > 16 fps
Hence, distance covered is greater than 16*1800 feet = (16*1800)/5280 miles = (180/33) miles = (60/11) miles = 5.454545... miles

The distance may be 5.6 miles or 6.5 miles

Not sufficient

Statement 2: Speed < 18 fps
Hence, distance covered is less than 18*1800 feet = (18*1800)/5280 miles = (3*180/88) miles = (3*45/22) miles = (135/22) miles = 6.13... miles

The distance may be 5.6 miles or 6.1 miles

Not sufficient

1 & 2 Together: Speed may be 5.6 miles or 6.1 miles

Not sufficient

The correct answer is E.
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by GMATGuruNY » Tue Jul 03, 2012 2:14 am
If it took Carlos ½ 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 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.
Since the two statements are in terms of FEET PER SECOND, rephrase the question stem in terms of FEET PER SECOND.

1/2 hour = 1800 seconds.
Determine the rate needed to travel 6 miles in 1800 seconds:
(6 miles)/(1800 seconds) * (5280 feet)/(1 mile) = (6*5280)/1800 = 528/30 = 176/10
= 17.6 feet per second.

Thus, to travel MORE than 6 miles in 1800 seconds, the average speed must be GREATER than 17.6 feet per second.
Question rephrased:
Was the average speed greater than 17.6 feet per second?

The two statements combined indicate only that the average speed was between 16 feet per second and 18 feet per second, implying that the average speed could be less than, equal to, or greater than 17.6 feet per second.
INSUFFICIENT.

The correct answer is E.
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