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by Deepthi Subbu » Mon Nov 29, 2010 9:42 pm
Liam is pulled for over speeding just as he is arriving to work. He explains to the police officer that he cannot afford to be late today , and has arrived at work only 4 minutes before he is to start.The officer explains that if Liam had driven 5mph slower for his whole commute , he would have arrived at work exactly on time . If Liam's commute is 30 miles long , how fast was he actually driving? (Assume that Liam drove at a constant speed for the duration of his commute)

I partly understand the problem but am not sure of the role played by the data 4 mins and the method to solve .

OA 50
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by Night reader » Mon Nov 29, 2010 9:56 pm
Deepthi Subbu wrote:Liam is pulled for over speeding just as he is arriving to work. He explains to the police officer that he cannot afford to be late today , and has arrived at work only 4 minutes before he is to start.The officer explains that if Liam had driven 5mph slower for his whole commute , he would have arrived at work exactly on time . If Liam's commute is 30 miles long , how fast was he actually driving? (Assume that Liam drove at a constant speed for the duration of his commute)

I partly understand the problem but am not sure of the role played by the data 4 mins and the method to solve .

OA 50
Speed=S
Time=T

Set of Equations
Equation I (S-5)*T=30
Equation II S*(T-4/60)=30

T=30/(S-5)
S*30/(S-5)-4S/60=30, S=50

Speed is equal to 50 mph
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by Deepthi Subbu » Tue Nov 30, 2010 12:27 am
Thanks for the solution Night Reader . But can you help me understand what 'and has arrived at work only 4 minutes before he is to start' means , cuz that will help me better understand why T becomes T-4 in equation II
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by Night reader » Tue Nov 30, 2010 1:19 am
Deepthi Subbu wrote:Thanks for the solution Night Reader . But can you help me understand what 'and has arrived at work only 4 minutes before he is to start' means , cuz that will help me better understand why T becomes T-4 in equation II
Rephrasing conditions in the problem:

Liam ... has arrived at work only 4 minutes before he is to start. ... If Liam's commute is 30 miles long...

S*(T-4/60)=30
4 MIN. = 4/60 Hour Format
Liam's actual speed is S, his actual time is T-4/60

Also,

Liam is pulled for over speeding ... The officer explains that if Liam had driven 5mph slower for his whole commute, he would have arrived at work exactly on time. If Liam's commute is 30 miles long...

(S-5)*T=30
Liam's actual speed could be (S-5) mph, then he could safely arrive at work within T (on time, without hurrying)


We need to solve for S (speed). Express T with S ~ S=30/ (T-4/60) and solve the equation.

Find that S=50 mph (answer)
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by GMATGuruNY » Tue Nov 30, 2010 2:56 am
Deepthi Subbu wrote:Liam is pulled for over speeding just as he is arriving to work. He explains to the police officer that he cannot afford to be late today , and has arrived at work only 4 minutes before he is to start.The officer explains that if Liam had driven 5mph slower for his whole commute , he would have arrived at work exactly on time . If Liam's commute is 30 miles long , how fast was he actually driving? (Assume that Liam drove at a constant speed for the duration of his commute)

I partly understand the problem but am not sure of the role played by the data 4 mins and the method to solve .

OA 50
The problem states that Liam arrived 4 minutes early and that, if he had traveled 5mph slower, the trip would have taken 4 more minutes, enabling him to arrive on time.

Since the GMAT would provide us with answer choices, we could plug them in for Liam's actual speed.

Answer choice: actual speed = 50mph
Time at 50mph = d/r = 30/50 = 3/5 hour.
Time at 45mph (5mph slower) = d/r = 30/45 = 2/3 hour.
Difference in time = 2/3 - 3/5 = 1/15 hour = 4 minutes. Success!
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