Question: A car drives 40 miles on local roads at 20 mph, and 180 miles on the highway at 60 mph, what is the average speed of the entire trip?

1. 36 mph
2. 40 mph
3. 44 mph
4. 52 mph
5. 58 mph

“ A car drives 40 miles on local roads at 20 mph, and 180 miles on the highway” - is a topic of the GMAT Quantitative reasoning section. This question has been taken from the book "GMAT Official Guide Quantitative Review". To solve GMAT Problem Solving questions a student must have good knowledge of qualitative skills. The GMAT Quant section consists of 31 questions in total. The GMAT quant topics in the problem-solving part require calculative mathematical problems that should be solved with proper mathematical knowledge.

Solution and Explanation:

Approach Solution 1:

We are given that a car drives 40 miles at a speed of 20 mph. Also, the car drove 180 miles at a speed of 60 mph.
We know the general relationship between speed, time, and distance.
v(speed) * t(time) = d (distance)
So for the first case,
V1 * t1 = d1
V1 = 20 and d1 = 40
=> 20 * t1 = 40
=> t1 = 40/20
=> t1 = 2s
In the second case,
V2 = 60 and d2 = 180
=> 60 * t2 = 180
=> t2 = 180/60
=> t2 = 3s
Now that the total time taken will be = 3s + 2s = 5s.
let us calculate the total distance traveled,
Total distance = 180 + 40 = 220 miles
Now we know that the average speed of the car will be the total distance traveled / total time taken.
Average speed = (220 / 5) = 44mph
Hence, the correct answer is 44 mph.

Correct Answer: C

Approach Solution 2:

In phase #1 of the trip, the car traveled 40 miles at 20 mph. That time of this phase was:
time = distance/rate = (40 mi)/(20 mph) = 2 hr
In phase #2 of the trip, the car traveled 180 miles at 60 mph. That time of this phase was:
time = distance/rate = (180 mi)/(60 mph) = 3 hr
The total distance of the trip = 40 mi + 180 mi = 220 mi
The total time of the trip = 2 hr + 3 hr = 5 hr
The average speed of the trip is given by
speed = distance/time = (220 mi)/(5 hr) = 44 mph
Hence, the correct answer is 44 mph.

Correct Answer: C

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