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Carlos is cycling...

by yass20015 » Sat Aug 29, 2015 2:09 pm
I got this one from mba.com
If it took Carlos 1/2 h to cycle from his Library yesterday, was the distance that he cycled greater than 6 miles ( 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 less than 18 feet per second

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by GMATGuruNY » Sat Aug 29, 2015 2:16 pm
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.
Note that the units in red CANCEL OUT, leaving the resulting value in terms of 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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by Brent@GMATPrepNow » Sat Aug 29, 2015 2:16 pm
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 less than 18 feet per second.
Target question: Was the distance that he cycled greater than 6 miles?

This question could use some rephrasing. Since the two statements provide speeds in feet per second, let's first see what it means for Carlos to cycle exactly 6 miles in 1/2 an hour.

Distance = 6 miles = (6)(5280) feet
Time = 1/2 hour = 1800 seconds.

Speed = (6)(5280)/1800 = 5280/300 = 176/10
= 17.6 feet per second.

So, in order for Carlos to travel more than 6 miles in 1800 seconds, his average speed must be greater than 17.6 feet per second.

REPHRASED target question: Was Carlo's average speed greater than 17.6 feet per second?

Statement 1: His average speed was greater than 16 feet per second.
So, his speed may have been greater than 17.6 feet per second, or less than 17.6 feet per second.
Since we cannot answer the rephrased target question with certainty, statement 1 is NOT SUFFICIENT

Statement 2: His average speed was less than 18 feet per second.
So, his speed may have been greater than 17.6 feet per second, or less than 17.6 feet per second.
Since we cannot answer the rephrased target question with certainty, statement 2 is NOT SUFFICIENT

Statements 1 and 2 combined:
We now know that his speed is between 16 feet per second and 18 feet per second.
So, once again, his speed may have been greater than 17.6 feet per second, or less than 17.6 feet per second.

Since we still cannot answer the rephrased target question with certainty, the combined statements are NOT SUFFICIENT

Answer = E

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by [email protected] » Sat Aug 29, 2015 3:23 pm
Hi yass20015,

Brent has converted the information into feet/second, which absolutely works. Here's another way to solve this problem though:

We're told that Carlos takes 1/2 hour to cycle to the library. We're asked if the distance was greater than 6 miles. The real question is how fast he was traveling; in essence, was it enough to hit more than 6 miles over the course of 1/2 hour. 6 miles = 6(5280 ft) = 31680 ft., so the question is asking if Carlos traveled more than 31,680 feet.

Fact 1: Carlos' speed was > 16 ft/sec.

Let's convert this...

16 ft/sec x 60 secs x 30 mins = 16 x 1800 = 28,800 feet/half-hour

From this, we know how far Carlos traveled AT THE MINIMUM. But we don't know if he traveled 31,680 feet or not.
Fact 1 is INSUFFICIENT

Fact 2: Carlos' speed was < 18 ft/sec.

Let's convert this...

18 ft/sec x 60 secs x 30 mins = 18 x 1800 = 32,400 feet/half-hour

From this, we know how far Carlos traveled AT THE MAXIMUM. But we don't know if he traveled 31,680 feet or not.
Fact 2 is INSUFFICIENT

Combined, we have the range of distances that Carlos traveled: 28,800 < distance traveled < 32,400. From this, we still don't know if Carlos traveled 31,680 feet or not.
Combined INSUFFICIENT

Final Answer: E

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by Max@Math Revolution » Fri Sep 04, 2015 3:53 am
Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and equations ensures a solution.


If it took Carlos 1/2 h to cycle from his Library yesterday, was the distance that he cycled greater than 6 miles ( 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 less than 18 feet per second
==> transforming the original condition and the question by variable approach method, v:speed rate,
1/2hr=1800sec, v*1800>6miles?
--> v*1800>6*5280ft? --> v>17.6sec?

the important thing about inequality sign in DS questions is that the condition is sufficient if the range of que include the range of con. in v(1/2)=d, we have 2 variables (v,d) and 1 equation (v/2=d) thus we need 1 equation more. Therefore D is likely the answer.


The question is whether 17.6<v?
in case of con 1), 16<v is not sufficient because range of que does not include range of con
in case of con 2), v<18 is not sufficient because range of que does not include range of con
Using both 1) & 2) together, 16<v<18 is not sufficient because range of que does not include range of con
Therefore the answer is E.



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by Matt@VeritasPrep » Fri Sep 04, 2015 6:20 pm
One other way of thinking about this is considering extremes. For instance, take S1. If Carlos traveled faster than 16ft/sec, he could've traveled as fast as we like: say 100,000 ft/sc. So it's certainly POSSIBLE for him to have gone further than 6 miles: thus we need to consider the MINIMUM speed, or 16 ft/sec.

Doing some quick conversions, we find that

16ft / sec =
16*60 ft / min =
16*60*60 ft / hour =
(16*60*60 / 5280) miles hour ≈ 11 miles / hour

So this ISN'T fast enough, and S1 gives us conflicting results. If Carlos traveled at 16 ft/sec, he wouldn't have made it 6 miles, but if he traveled 100,000 ft/sec, he would've.

For S2 we'd use the same approach. He could've traveled 0 ft/sec, in which case he wouldn't have traveled 6 miles, but we need to consider the maximum: 18 ft/sec. When we do, we'll see that this is fast enough to travel more than 6 miles, so S2 also gives conflicting results, etc.
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