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If it took Carlos 1/2 hour to cycle from his house to the [#permalink]
20 Feb 2006, 17:17

00:00

A

B

C

D

E

Difficulty:

95% (hard)

Question Stats:

38% (01:36) correct
62% (01:20) wrong based on 116 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 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.

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.

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.

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.

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).

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 _________________

Data Sufficiency: Was the distance greater? [#permalink]
05 Feb 2007, 20:35

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.

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.

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)

st1-the avg speed at which carlos cycled from his house to the library was greater than 16 feet per second.
st2-the avg speed at which carlos cycled from his house to the library was less than 18 feet per second.

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)

st1-the avg speed at which carlos cycled from his house to the library was greater than 16 feet per second. st2-the 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.

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.

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.

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.

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