Problem Solving

some simpler method

hey_thr67 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.

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.