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Liam is pulled over for speeding just as he is arriving at work.He

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Liam is pulled over for speeding just as he is arriving at work.He [#permalink] New post 21 Jun 2010, 10:13
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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.)

OA:
[Reveal] Spoiler:
50 miles per hour.


Solution:

[Reveal] Spoiler:
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.
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Re: Liam is pulled over for speeding just as he is arriving at work.He [#permalink] New post 21 Jun 2010, 10:35
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Phoenix9 wrote:
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.


Check the blue quoted.
Actual speed = r+5 where r = 45
=> r+5 = 50 same answer.
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Re: Liam is pulled over for speeding just as he is arriving at work.He [#permalink] New post 21 Jun 2010, 10:45
My bad. Was dumb enough to ignore the (r+5) part. Thanks though ;-)
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Re: Liam is pulled over for speeding just as he is arriving at work.He [#permalink] New post 01 Apr 2011, 09:22
Liam is pulled over for speeding just as he is arriving at work. He explains that he could not afford to be late today, and has arrived at work only 5 minutes before he is to start. The officer explains that if he had driven 5mph slower for his whole commute, he would have arrived on time. If his commute is 30 miles, how fast was he actually driving?

What is wrong with the way I'm trying to solve this problem?

Actual:
Speed = s
time = t - 1/12

Hypothetical:
Speed = s-5
time = t

Since distances are equal, equate the two. I cannot seem to get the correct answer...Please help.
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Re: Liam is pulled over for speeding just as he is arriving at work.He [#permalink] New post 01 Apr 2011, 11:30
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.
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Re: Liam is pulled over for speeding just as he is arriving at work.He [#permalink] New post 01 Apr 2011, 18:31
@Baten80, the answer is 45, as we have to find the speed on that day.
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Re: Liam is pulled over for speeding just as he is arriving at work.He [#permalink] New post 01 Apr 2011, 21:59
subhashghosh wrote:
@Baten80, the answer is 45, as we have to find the speed on that day.


Thus, it should be 50mph, right?
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Re: Liam is pulled over for speeding just as he is arriving at work.He [#permalink] New post 28 Jun 2011, 10:26
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 5 mph 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.)
A. 50 mph
B. 45 mph
C. 48 mph
D. 52 mph
E. 60 mph
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Re: Liam is pulled over for speeding just as he is arriving at work.He [#permalink] New post 28 Jun 2011, 12:11
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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
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Re: Liam is pulled over for speeding just as he is arriving at work.He [#permalink] New post 28 Jun 2011, 21:13
Shalom vrk002,
Shalom! I like your method the best because of its simplicity. However, could you post the step whereby you solve both equations? The only way I see that I could get the answer by using your method is plugging in all the possible answers until I see the one that is the solution.
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Re: Liam is pulled over for speeding just as he is arriving at work.He [#permalink] New post 29 Jun 2011, 01:50
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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.
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Re: Liam is pulled over for speeding just as he is arriving at work.He [#permalink] New post 03 Aug 2011, 07:00
plugging the numbers would be much faster, sometimes, especially if you can't intuitively guess that 2250 = 45*50

Just solve to 30/s + 1/15 = 30(s-5) and plug in the options. Its much much faster. In this case you have to be careful for whether u use s-5 or s+5. The latter can't be used cause you're searching for the faster speed.
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Re: Liam is pulled over for speeding just as he is arriving at work.He [#permalink] New post 16 Nov 2014, 21:49
Phoenix9 wrote:
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.



Hi all,

I'm reviewing rates & work and even though I feel I pretty much got it figured out, questions like this one let me doubt myself.

My question for this one is: Phoenix9 introduced 2 approaches to solve the question that use the information given differently, in approach 1 the fact that Liam should go 5mph slower is marked as (r-5) in the hypothetical case. In approach 2 it's (r+5) in the actual case. No questions until here.

I, however, tried like this:

Real case: R: r+5 ; T: (30/r - 1/15); Hypothetical case: R: R ; t: (30/r)

The equation than comes to two negative values for r, which is unsolvable.

My question is: what's the mistake in my approach?



Thanks
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Liam is pulled over for speeding just as he is arriving at work.He [#permalink] New post 17 Nov 2014, 03:08
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case 1
rate = x
time =30/x
distance = 30

case 2
rate = x-5
time = 30/(x-5)
distance = 30


given that, 30/(x-5) - 30/x =4/60
150/x(x-5)=1/15
x^2-5x-2250=0
x=50,-45

so x=50mph
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Re: Liam is pulled over for speeding just as he is arriving at work.He [#permalink] New post 24 Jan 2015, 11:56
Phoenix9 wrote:
My bad. Was dumb enough to ignore the (r+5) part. Thanks though ;-)

Can someone explain to me why he would be 4 mins faster and break down the equations? I'm not getting why 30/(x-5) - 30/x = 4/60
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Re: Liam is pulled over for speeding just as he is arriving at work.He [#permalink] New post 08 Feb 2015, 12:15
Thanks Akumar for the solution.
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Re: Liam is pulled over for speeding just as he is arriving at work.He   [#permalink] 08 Feb 2015, 12:15
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