BrentGMATPrepNow wrote:
AbdurRakib wrote:
Al and Ben are drivers for SD Trucking Company. One snowy day, Ben left SD at 8:00 a.m. heading east and Al left SD at 11:00 a.m. heading west. At a particular time later that day, the dispatcher retrieved data from SD’s vehicle tracking system. The data showed that, up to that time, Al had averaged 40 miles per hour and Ben had averaged 20 miles per hour. It also showed that Al and Ben had driven a combined total of 240 miles. At what time did the dispatcher retrieve data from the vehicle tracking system?
A. 1:00 p.m.
B. 2:00 p.m.
C. 3:00 p.m.
D. 5:00 p.m.
E. 6:00 p.m.
Let's start with a "word equation"
(
Ben's travel distance) + (
Al's travel distance) = 240 miles
Let
t = Al's travel time (in hours)
So,
t + 3= Ben's travel time (since Ben spent 3 more hours driving)
Distance = (rate)(time)
So, our word equation becomes...
(
20)(
t + 3) + (
40)(
t) = 240 miles
Expand: 20t + 60 + 40t = 240
Simplify: 60t + 60 = 240
Subtract 60 from both sides: 60t = 180
Solve: t = 3
So, AL traveled for
3 hours (which means BEN traveled for
6 hours)
So, the dispatcher retrieved data at 2pm (since 8am +
6 hours = 2pm)
Answer: B
Cheers,
Brent
BrentGMATPrepNowThank you for this helpful explanation. I would be so appreciative if you can correct me as to where I am making an error in setting up my rt=d chart below to solve this problem.
R * T = D
-----------------------------------------------
Ben: (20 mi/hr) (T +3) = D
Al: (40 mi/hr) T = D
------------------------------------------------
I wasn't sure what to set D equal to, but I know that the combined distance was 240 but just left each equation equal to D above. But, I am not sure where I went wrong based on what else I have.
=(60 mi/hr)*(2T+3)=240
--> t=1/2 is my answer based on how I set up my chart.
Thank you for your time and help in advance.