Hi All,

I have a question regarding a problem from Manhattan Strategy Guide: Word Translations (3).
Chapter 2. Rates. Page 37.

Problem:
Liam is pulled over for speeding just as he is arriving at work.He explains to the police officer that he could not afford to be late today, and has arrived at work only four 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.)

Solution:
Of the many ways to solve this problem, two are as follows:

(Method 1)
Assume the actual speed of Liam to be r. Distance travelled is 30 miles. So, time taken is 30/r. In the hypothetical case, speed of Liam is (r-5). Distance remains the same. So, time taken is (30/r)+(1/15) because 4 minutes is 1/15th hour. So, translating these values into equations, the hypothetical scenario becomes:

distance = speed * time
30 = [(30/r)+(1/15)]*[r-5] => 30 = [(450+r)/15r]*[r-5] => 450r = [450+r][r-5] => r^2 -5r-2250 = 0 => (r-50)(r+45) = 0 => r = 50.

(Method 2)
This is the method used in the Manhattan guide. Speed in the actual case is considered to be (r+5). Time taken is therefore 30/(r+5). Speed in the hypothetical case is considered to be r. Time taken is 30/r. Because we know time taken in the hypothetical scenario is 4 minutes more, 30/r = [(30/(r+5))+(1/15)] => 30/r = [((450+r+5)/(15r+75)] => 30(15r+75) = r(455+r) => r^2 +5r-2250 = 0 => (r+50)(r-45) = 0 => r=45.

Can anyone please explain to me why both these methods DON'T yield the same answer? Isn't the first method more appropriate because the hypothetical scenario is the one in which we should assume the speed to be 5mph less than the actual and time taken is 4 minutes more than the actual?

Thanks.

My bad. Was dumb enough to ignore the (r+5) part. Thanks though

Let required speed=s
30/s-30/(s-5)=1/5 (4 minutes= 4/60=1/5 hour)
s^2-5s+2250=0
s=50 ans.
_________________

I am student of everyone-baten
Collections:-
PSof OG solved by GC members: http://gmatclub.com/forum/collection-ps-with-solution-from-gmatclub-110005.html
DS of OG solved by GC members: http://gmatclub.com/forum/collection-ds-with-solution-from-gmatclub-110004.html
100 GMAT PREP Quantitative collection http://gmatclub.com/forum/gmat-prep-problem-collections-114358.html
Collections of work/rate problems with solutions http://gmatclub.com/forum/collections-of-work-rate-problem-with-solutions-118919.html
Mixture problems in a file with best solutions: http://gmatclub.com/forum/mixture-problems-with-best-and-easy-solutions-all-together-124644.html

Thus, it should be 50mph, right?
_________________

~fluke

Find out what's new at GMAT Club - latest features and updates

Let t be the number of hours he would need to reach office on time.

when he is driving with over speed, he reached office 4 min earlier! so the equation for this is s(t - 4/60) = 30

where s is the speed and 30 is the distance.

if he decreases his speed by 5mph then he would have reached his office on time: (s-5)t = 30

if you solve above equations, you will arrive at t = 2/3 hr and s = 50mph

therefore answer is A

s(t - 4/60) = 30 ---- (1)
(s-5)t = 30 --- (2)
Therefore t = 30 /(s-5)
substitute t in (2)

s[(30/(s-5)) - (1/15)] = 30
=> s[ (450 - s + 5) / (15(s-5)) ] = 30
=> 450s - s^2 + 5s = 450s - 2250
=> s^2 - 5s - 2250 = 0;
=> s^2 - 50s + 45s - 2250 = 0
=> (s - 50) ( s + 45) = 0
=> s = 50; s = -45
Speed cant be negative
therefore s = 50.