naveenhv wrote: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 16ft per second.
2)The average speed at which Carlos cycles from his house to the library yesterday was less than 18ft per second.
This question takes a lot longer to answer if we get bogged down with calculations. The hard part is to determine just how relaxed we can be with our estimation.
Well, it's important to note that the GMAC is not trying to identify the human calculators out there. They want to identify people with a "feel" for numbers. So, if we recognize and accept the GMAC's mandate, then we can be a little aggressive with our calculations.
Okay, let's take a closer look at the target question: "Is the distance greater than 6 miles?"
Well, we know that if the distance were exactly 6 miles, then Carlos would have been traveling at exactly 12 miles per hour (since he traveled for 1/2 an hour).
So, for the distance to be greater than 6 miles, Carlos must travel faster than 12 mph.
This means that we can rewrite our target question as "Did Carlos travel faster than 12 mph?"
When we glance at the statements, we see that Carlos' speed is given in feet per second, so let's take the target question and rewrite it in terms of feet per second.
Since there are 5280 feet in a mile, we know that there are (12)(5280) feet in 12 miles.
Since there are 3600 seconds in 1 hour, we can rewrite 12miles/hour as (12)(5280)/3600 feet per second.
So, we can rewrite the target question as "Did Carlos travel faster than (12)(5280)/3600 feet per second?"
To simplify (12)(5280)/3600, let's rewrite 5280 as 5300 to get (12)(5300)/3600
Important: Since we replaced 5280 with the slightly larger 5300, we need to keep in mind that our new result will be a little larger than the number we would have calculated if we had kept the original numbers. So, the "real" answer will be a little bit smaller than the results we get when we use 5300
Okay, let's proceed
When we divide top and bottom by 100, we get (12)(53)/36
When we divide top and bottom by 12, we get 53/3
Note: Don't forget that 53/3 is a little bit bigger than the number we would have calculated if we had kept the original numbers. Our "real" result is a little bit less than 53/3, but this won't change the answer to the question.
Aside: If one wants to be super precise, we can see that since we originally, replaced 5280 with 5300, we can now go back and replace 53 with 52.8 to get 52.8/3. Having said that, we don't need to be so precise.
Okay, back to the question.
We know that 51/3 = 17, and we know that 54/3 = 18
So, 53/3 (or 52.8/3) is equal to 17.something.
So, after all of that, we see that 12 mph is equal to 17.something feet per second.
This means we can rewrite the target question one last time: "Did Carlos travel faster than 17.something feet per second?"
At this point we can see that the answer must be
E, for the reasons provided by other posters.
Yes, it looks like I did a lot of work, but if we round 5280 to 5300 and not get bogged down with lengthy calculations, this question can be solved in under 2 minutes.
