Speed, distance,time

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RiyaR: Senior | Next Rank: 100 Posts; Posts: 98; Joined: Sat May 31, 2014 8:29 pm; Thanked: 1 times; Followed by:2 members

Speed, distance,time

by RiyaR » Wed Aug 27, 2014 9:01 am

Two men drove from Hartford to Boston(100miles) leaving at the same time. Mr. Clifford averaged 30miles an hour for the first 20 miles and 40 miles an hour for the rest of the journey. Mr. Sherman drove at a steady pace. They arrived at the same time. What was Mr.Sherman's average speed?
A)35.2
B)24.2
C)25
D)37.2
E)35

This was my understanding of the problem: Since they both reach at the same time we can assume that the time taken is the same. We also know that the distance is constant. So in order to find Mr.Clifford's average speed for the whole journey we can use the formula:(m+n)pq/mq+np. Can you tell me where I am going wrong in this sum? The answer seems to be D, but Im not able to arrive at it....

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Source: Beat The GMAT — Problem Solving | Wed Aug 27, 2014 9:01 am

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by [email protected] » Wed Aug 27, 2014 9:09 am

Hi RiyaR,

This question gives us lots of data to work with, so solving it is really just about doing the math correctly (and in a simple way; complexity is unnecessary).

Let's start with Mr. Clifford:
-He's going to travel 100 miles.
-He travels 30mph for the first 20 miles

D = R x T
20 = 30 (T)
2/3 hr. = T

-He travels 40mph for the rest of the journey

D = R x T
80 = 40 (T)
2 hrs. = T

We now know the total time was 2/3 + 2 hours = 8/3 hours

Now we deal with Mr. Sherman:
-He's going to travel 100 miles at a CONSTANT SPEED. Using the 8/3 hours for total time, we have..

D = R x T
100 = R (8/3)
3(100)/8 = R
37.5 = R

Final Answer: [spoiler]D - I assume those ".2"s are supposed to be ".5"s[/spoiler]

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Rich

Contact Rich at [email protected]

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GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pm; Location: New York, NY; Thanked: 13060 times; Followed by:1906 members; GMAT Score:790

by GMATGuruNY » Wed Aug 27, 2014 9:24 am

RiyaR wrote:Two men drove from Hartford to Boston (100miles) leaving at the same time. Mr. Clifford averaged 30 miles an hour for the first 20 miles and 40 miles an hour for the rest of the journey. Mr. Sherman drove at a steady pace. They arrived at the same time. What was Mr.Sherman's average speed?

The distance traveled at 40mph (80 miles) is 4 times the distance traveled at 30mph (20 miles).
The total distance can be ANY VALUE, as long as the distance traveled at 40mph is 4 times the distance traveled at 30mph.

Let the distance traveled at 30mph = 30 miles and the distance traveled at 40mph = 120 miles, for a total distance of 150 miles.
Time to travel 30 miles at a speed of 30mph = d/r = 30/30 = 1 hour.
Time to travel 120 miles at a speed of 40mph = d/r = 120/40 = 3 hours.
Average speed for the entire 150 miles = d/(total time) = 150/(1+3) = 150/4 = 75/2 = 37.5 mph.

Last edited by GMATGuruNY on Wed Aug 27, 2014 10:33 am, edited 1 time in total.

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