A certain road trip was completed in two parts...

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A certain road trip was completed in two parts interrupted by an hour-long lunch break. During the first portion of the trip, the average speed was 55 miles per hour and the average speed during the second part of the trip was 63 miles per hour. If the trip was 480 miles long and took a total of 8 hours (excluding the time stopped for lunch) then which of the following is the amount of time spent driving before the lunch break?

A. 2
B. 3
C. 4
D. 5
E. 6

The OA is B.

I don't have clear this PS question. I appreciate if any expert explains it to me. Thank you so much.

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by Brent@GMATPrepNow » Mon Mar 05, 2018 9:49 am
swerve wrote:A certain road trip was completed in two parts interrupted by an hour-long lunch break. During the first portion of the trip, the average speed was 55 miles per hour and the average speed during the second part of the trip was 63 miles per hour. If the trip was 480 miles long and took a total of 8 hours (excluding the time stopped for lunch) then which of the following is the amount of time spent driving before the lunch break?

A. 2
B. 3
C. 4
D. 5
E. 6
Let's start with a word equation

(distance traveled during FIRST part of trip) + (distance traveled during SECOND part of trip) = 480 miles

Let t = time spend on FIRST part of trip (in hours)
So, 8-t = time spend on SECOND part of trip (since the ENTIRE driving time was 8 hour)

Distance = (speed)(time)

So, we can write: (55)(t ) + (63)(8 - t) = 480
Expand: 55t + (504 - 63t) = 480
Simplify: -8t + 504 = 480
Subtract 504 from both sides: -8t = -24
Solve: t = 3

Answer: B

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by [email protected] » Mon Mar 05, 2018 8:40 pm
Hi swerve,

We're told that a certain road trip was completed in two parts interrupted by an hour-long lunch break. During the first portion of the trip, the average speed was 55 miles per hour and the average speed during the second part of the trip was 63 miles per hour. We're also told the trip was 480 miles long and took a total of 8 hours (excluding the time stopped for lunch). We're asked for the number of hours spent driving BEFORE the lunch break. This question can be solved in a number of different ways; here's how you can TEST THE ANSWERS.

To start, since the trip was 480 miles and took 8 hours of drive time, the average speed for the trip was 480/8 = 60 miles/hour.

With the two given speeds (55 mph and 63 mph) - and since 63 is closer to 60 than 55 is to 60 - we will need to speed MORE time at 63 mph than we do at 55 mph to end up with an average speed of 60 mph. Thus, the correct answer has to be LESS than 4 hours. Eliminate Answers C, D and E.

Let's TEST Answer B: 3 hours
IF... we spent 3 hours at 55 miles per hour, then we travel (3)(55) = 165 miles
we then spend 5 hours at 63 miles per hour, so we travel (5)(63) = 315 miles
165 miles + 315 miles = 480 miles
This is an exact match for what we were told, so this MUST be the answer.

Final Answer: C

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by Jeff@TargetTestPrep » Tue Mar 06, 2018 10:51 am
swerve wrote:A certain road trip was completed in two parts interrupted by an hour-long lunch break. During the first portion of the trip, the average speed was 55 miles per hour and the average speed during the second part of the trip was 63 miles per hour. If the trip was 480 miles long and took a total of 8 hours (excluding the time stopped for lunch) then which of the following is the amount of time spent driving before the lunch break?

A. 2
B. 3
C. 4
D. 5
E. 6
We are given that the average speed before lunch was 55 mph. If we let t equal the time spent traveling before lunch, the distance covered is 55t.

Since the trip takes 8 hours, excluding the lunch break, the second part of the trip takes (8 - t) hours, and the distance covered in this part is 63(8 - t).

Since the sum of the distances in the first and second parts must equal the total length of the trip, which is 480 miles, we can create the following equation and determine t:

55t + 63(8 - t) = 480

55t + 504 - 63t = 480

24 = 8t

t = 3

Answer: B

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