Data Sufficiency

Al, Pablo, and Marsha shared the driving on a 1,500-mile trip. Which of the three drove the greatest distance on the trip?

1. Al drove 1 hour longer than Pablo but at an average rate of 5MPH slower than pablo.
2. Marsha drove 9 hours and averaged 50MPH

Thanks

Let speed of Pablo be Sp and Speed of Al be Sa
Let time taken by Pable be Tp and time taken by Al be Ta

Al, Pablo, and Marsha shared the driving on a 1,500-mile trip. Which of the three drove the greatest distance on the trip?

1. Al drove 1 hour longer than Pablo but at an average rate of 5MPH slower than pablo.

Distance travelled by Pablo = Sp * Tp.
Distance travelled by Al = (Sp-5)*(Tp+1)
Distance travelled by Marsha = D (Just another variable)
Sp * Tp + (Sp-5)*(Tp+1) + D = 1500.
Since we don't know the values of these variables, statement I is insufficient to answer the question.

2. Marsha drove 9 hours and averaged 50MPH

Distance travelled by Marsha = D = 50*9 = 450 miles. So the driver who drove th greatest distance will be one between Al and Pablo(and we are not sure who it is).
Statement II is insufficient to answer the question.

From 1 + 2

Sp * Tp + (Sp-5)*(Tp+1) + D = 1500
Sp * Tp + (Sp-5)*(Tp+1) + 450 = 1500
Sp * Tp + (Sp-5)*(Tp+1) = 1050
Sp*Tp + Sp*Tp -5Tp + Sp - 3 = 1050
Since we don't know the values of these variables, statement I+II combined is insufficient to answer the question.

IMO E

Al, Pablo, and Marsha shared the driving on a 1,500-mile trip. Which of the three drove the greatest distance on the trip?

1. Al drove 1 hour longer than Pablo but at an average rate of 5MPH slower than pablo.
2. Marsha drove 9 hours and averaged 50MPH.

Combining the 2 statements:
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.
TRY 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 ≈ 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 ≈ 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, but in the second case Al drives the greatest distance, INSUFFICIENT.

The correct answer is E.