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

Question

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.

A. 50 mph
B. 45 mph
C. 48 mph
D. 52 mph
E. 60 mph

OA:  50 miles per hour.

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 =  *  => 450r =    => 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 =   => 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.

akumar5 · Accepted Answer

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

gurpreetsingh · Answer

Check the blue quoted.
Actual speed = r+5 where r = 45
=> r+5 = 50 same answer.

Phoenix9 · Answer

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

thesfactor · Answer

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.)

A. 50 mph
B. 45 mph
C. 48 mph
D. 52 mph
E. 60 mph

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.

MHIKER · Answer

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.

subhashghosh · Answer

@Baten80, the answer is 45, as we have to find the speed on that day.

fluke · Answer

Thus, it should be 50mph, right?

vrk002 · Answer

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

MitDavidDv · Answer

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.

shm0401 · Answer

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

s  = 30
=> s  = 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.

openbox1 · Answer

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.

chun1904 · Answer

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
Christian

shnbanana · Answer

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

Salvetor · Answer

Thanks Akumar for the solution.

sanaexam · Answer

arrived at work only four minutes before he is to start.Can somebody explain what does this statement mean

Bunuel · Answer

Say he starts working at 10:00, then according to the statement he arrived at 9:56.

shashankism · Answer

Let the speed at which Liam drove at office actually be x mph
So, the speed which traffic police suggested = x-5

Distance between his home and office = 30 miles

So, difference of time with these 2 speeds = 4 minutes

So, 30/(x-5) - 30/x = 4/60 
-> 30*5/x(x-5) = 1/15
-> x(x-5) = 30*5*15 
-> x^2 - 5x - 30*5*15 = 0 
-> x^2 - 50x + 45x - 30*5*15 = 0 
-> (x -50)(x+45) = 0 
-> x = 50 (as speed will be +ve)
-> So, he was actually driving at a speed of 50 miles per hour

Answer A

gracie · Answer

let t= L's actual time
30/t=30/(t+1/15)+5
t=3/5 hr
30/(3/5)=50 mph
A

pclawong · Answer

Dear,
I am trying to understand your approach 
but 4 mins is not 1/5 hour
Can you explain on this? Thank you

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