Data Sufficiency

Al, Pablo and Marsha shared the driving on a 1500 mile trip. Which of the 3 drove the greatest distance on the trip?
1) Al drove 1 hour longer than Pablo but at an average rate 5 miles per hour slower than Pablo.
2) Marsha drove for 9 hours and averaged 50 miles per hour.

fangtray wrote:Al, Pablo, and Marsha shared the driving on a 1,500 mile trip. Which of the 3 drove the greatest distance on the trip?

1. Al drove 1 hr longer than Pablo at an avg rate of 5 mph slower than pablo
2. Marsha drove 9 hrs and averaged 50 mph.

Statement 1:
No information about Marsha.
INSUFFICIENT.

Statement 2:
Marsha's distance = r*t = 9*50 = 450.
Remaining distance driven by Al and Pablo = 1500-450 = 1050.
It's possible that Al drives 50 miles, while Pablo drives 1000 miles, in which case Pablo drives the greatest distance.
It's possible that Al drives 1000 miles, while Pablo drives 50 miles, in which case Al drives the greatest distance.
INSUFFICIENT.

Statements combined:
Distance driven by Marsha = 50*9 = 450.
Distance traveled by Al and Pablo = 1500 - 450 = 1050.
Let the time for Pablo = t and the time for Al = t+1.
The rate for Pablo is 5mph greater than the rate for Al.
TEST EXTREME CASES.

Case 1: Rate for Pablo = 10mph, rate for Al = 5mph
Since the total distance driven by Pablo and Al is 1050, we get:
10t + 5(t+1) = 1050
15t = 1045
t = 1045/15 ≈ 70, implying that the time for Al = t+1 ≈ 70+1 ≈ 71.
Distance for Pablo ≈ 10*70 ≈ 700, distance for Al ≈ 5*71 ≈ 355.
Pablo drives the greatest distance.

Case 2: Rate for Pablo = 1045mph, rate for Al = 1040mph
Since the total distance driven by Pablo and Al is 1050, we get:
1045t + 1040(t+1) = 1050.
2085t = 10
t = 10/2085 ≈ 1/200, implying that the time for Al = t+1 ≈ 1/200 + 1 ≈ 201/200.
Distance for Pablo ≈ 1045(1/200) ≈ 5, distance for Al ≈ 1040(201/200) ≈ 1040.
Al drives the greatest distance.

Since in the first case Pablo drives the greatest distance, while in the second case Al drives the greatest distance, INSUFFICIENT.

The correct answer is E.