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Barbara walked up and down a 25 meter long

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by Vincen » Thu Feb 08, 2018 8:36 am
Barbara walked up and down a 25 meter long corridor. When she walked one way, she walked at 4 meters per second, and when she walked the other way, her speed was 2 meters per second. What was her average speed in meters per second?

$$A.\ \ \ 1\frac{1}{3} $$
$$B.\ \ \ 1\frac{2}{3}\ \ .$$
$$C.\ \ \ 2\frac{1}{3}\ \ .$$
$$D.\ \ \ 2\frac{2}{3}\ \ .$$
$$E.\ \ \ 3$$

The OA is the option D.

Experts, I don't understand how to solve this PS question.

May you give me some help here, please?
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by [email protected] » Thu Feb 08, 2018 9:59 am
Hi Vincen,

We're told that Barbara walked up and down a 25 meter long corridor. When she walked one way, she walked at 4 meters per second, and when she walked the other way, her speed was 2 meters per second. We're asked for her average speed in meters per second for the entire trip. This question is a standard "Average Speed" question and can be solved with a bit of math.

Walking the 25 meters at 4 meters/second:
Distance = (Rate)(Time)
25m = (4m/sec)(T)
25/4 seconds = Time

Walking the 25 meters at 2 meters/second:
Distance = (Rate)(Time)
25m = (2m/sec)(T)
25/2 = 12.5 seconds = Time

Total Distance = 25m + 25m = 50 meters
Total Time = 25/4 + 25/2 = 25/4 + 50/4 = 75/4 seconds

Total Distance = (Av. Speed)(Total Time)
50m = (Av. Speed)(75/4 seconds)
(50)(4/75) = Av. Speed
200/75 =
8/3 =
2 1/3 meters/second = Average Speed

Final Answer: D

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barbara

by GMATGuruNY » Thu Feb 08, 2018 10:33 am
Vincen wrote:Barbara walked up and down a 25 meter long corridor. When she walked one way, she walked at 4 meters per second, and when she walked the other way, her speed was 2 meters per second. What was her average speed in meters per second?

$$A.\ \ \ 1\frac{1}{3} $$
$$B.\ \ \ 1\frac{2}{3}\ \ .$$
$$C.\ \ \ 2\frac{1}{3}\ \ .$$
$$D.\ \ \ 2\frac{2}{3}\ \ .$$
$$E.\ \ \ 3$$
Since the distance traveled at each speed is THE SAME, we can determine the average speed by plugging in ANY VALUE for the distance.
Let the distance in each direction = 4 meters.
When Barbara travels the 4-meter corridor at a rate of 4 meters per second, her time = d/r = 4/4 = 1 second.
When Barbara travels the 4-meter corridor at a rate of 2 meters per second, her time = d/r = 4/2 = 2 seconds.
Average speed = (total distance)/(total time) = (4+4)/(1+2) = 8/3 meters per second.

The correct answer is D.
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hi

by Jeff@TargetTestPrep » Mon Feb 12, 2018 4:48 pm
Vincen wrote:Barbara walked up and down a 25 meter long corridor. When she walked one way, she walked at 4 meters per second, and when she walked the other way, her speed was 2 meters per second. What was her average speed in meters per second?

$$A.\ \ \ 1\frac{1}{3} $$
$$B.\ \ \ 1\frac{2}{3}\ \ .$$
$$C.\ \ \ 2\frac{1}{3}\ \ .$$
$$D.\ \ \ 2\frac{2}{3}\ \ .$$
$$E.\ \ \ 3$$
We can use the average rate formula:

average rate = total distance/total time

average = 50/(25/4 + 25/2)

average = 50/(25/4 + 50/4)

average = 50/(75/4)

average = 200/75 = 8/3 = 2 2/3

Answer: D

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