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Liam is pulled over for speeding just as he is arriving at work.He 10 Dec 2018, 07:13
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quite simply, just plug in -
distance = 30 miles
speed = 50 mph
time = 30/50 = 3/5 hrs = 36 mins

distance = 30 miles
speed = 45 miles
time = 30/45 = 2/3 hrs = 40 mins.

40 - 36 = 4 mins

this is exactly what we need.
so 50 is the answer.
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Liam is pulled over for speeding just as he is arriving at work.He 28 Jan 2020, 11:24
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thesfactor
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.
Show more

When you convert 4 mins to hour, you will get 1/15. But you have mentioned 1/12.Hope this helps.

Posted from my mobile device
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Liam is pulled over for speeding just as he is arriving at work.He 26 Apr 2021, 20:52
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Phoenix9
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

OA:
Show Spoiler
50 miles per hour.

Solution:
Show 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.
Show more
--------------------------------------------------

Can someone help me understand this: "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"

before he is to start - meaning?
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Liam is pulled over for speeding just as he is arriving at work.He 27 Apr 2021, 00:58
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summerbummer
Phoenix9
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

OA:
Show Spoiler
50 miles per hour.

Solution:
Show 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.
--------------------------------------------------

Can someone help me understand this: "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"

before he is to start - meaning?
Show more

He is pulled over for speeding. So the officer may have asked him for the reason for speeding. He said that he was speeding because he cannot be late for work today (so say if he starts work at 8:00 AM, he had to reach by 8 only, not 8:01). And that, he has arrived just 4 mins before he is to start (so he has arrived at 7:56 AM) even after speeding.
The officer explains that had he driven at 5 mph slower, he would have arrived exactly at 8:00 AM (and still would not be late).

30 miles commute ..... Speed S ............ Reach at 7:56 AM
30 miles commute ..... Speed S - 5....... Reach at 8:00 AM

30/(S - 5) - 30/S = 4/60 = 1/15

We have 15 in denominator so let's look for S with factors 3 and 5. Options (A) and (B) look promising though (A) looks better.
Option (A): Put S = 50
30/45 - 30/50 = 2/3 - 3/5 = 1/15
Correct.

Think why other options didn't look as good.
B. 45 mph
30/40 would give 4 in denominator with simplified numerators. We need 15 in denominator which has no 4.

C. 48 mph
43, a prime in denominator.

D. 52 mph
47, a prime in denominator

E. 60 mph
55 in denominator which has 11 as factor. But 15 has no 11.
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Liam is pulled over for speeding just as he is arriving at work.He 27 Apr 2021, 09:55
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VeritasKarishma
summerbummer
Phoenix9
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

OA:
Show Spoiler
50 miles per hour.

Solution:
Show 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.
--------------------------------------------------

Can someone help me understand this: "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"

before he is to start - meaning?

He is pulled over for speeding. So the officer may have asked him for the reason for speeding. He said that he was speeding because he cannot be late for work today (so say if he starts work at 8:00 AM, he had to reach by 8 only, not 8:01). And that, he has arrived just 4 mins before he is to start (so he has arrived at 7:56 AM) even after speeding.
The officer explains that had he driven at 5 mph slower, he would have arrived exactly at 8:00 AM (and still would not be late).

30 miles commute ..... Speed S ............ Reach at 7:56 AM
30 miles commute ..... Speed S - 5....... Reach at 8:00 AM

30/(S - 5) - 30/S = 4/60 = 1/15

We have 15 in denominator so let's look for S with factors 3 and 5. Options (A) and (B) look promising though (A) looks better.
Option (A): Put S = 50
30/45 - 30/50 = 2/3 - 3/5 = 1/15
Correct.

Think why other options didn't look as good.
B. 45 mph
30/40 would give 4 in denominator with simplified numerators. We need 15 in denominator which has no 4.

C. 48 mph
43, a prime in denominator.

D. 52 mph
47, a prime in denominator

E. 60 mph
55 in denominator which has 11 as factor. But 15 has no 11.
Show more


Hi VeritasKarishma,

Thanks for your explanation, it works wonder as always.

Please note that I used some kind of strategy for solving these type of questions (S * T = D) after going through some videos/blogs etc. of a few experts on YouTube. (The reason is because of the fractions, I simply hate them but I have to deal with them anyways).
Please let me know about any flaw in this strategy or if it will work for all scenarios in such questions?

Let's say the speed is S miles per hour and Time is T minutes for the first case (over-speeding). Note the usage of minutes because we have been given 4 minutes as the time difference.

So, ATQ
S * T = 30 * 60 (We need to multiply by 60 because we have taken time in minutes).
So, S * T = 1800

Algebraically, the second equation ATQ will be -> (S - 5) (T + 4) = 1800
S-slow = S-5 kmph
and T-slow = T+4
Now, we have to factorize 1800 into such two numbers, such that when we increase one of those factors (speed) by 5, the other factor (time) is decreased exactly by 4 and also the product remains constant i.e. 1800.

Let's say (by some hit & trial and looking at the answer choices), I took S = 45 and T = 40, because 45 * 40 (= 1800)
Now, if this (45 factor) is the faster speed (over-speeding), I'm going to decrease the speed for other case. If I decrease (Speed = 45) by 5, it leads to New Speed, S-slow as 40 but the corresponding increase in time will be 5 minutes (because moving from T-fast at 40 minutes to T-slow at 45 minutes). Actually, Speed and Time have interchanged their values.
But this doesn't satisfy our case, increase in time is exactly 4 minutes and not 5 minutes.

Now, If I increase (Speed = 45) by 5, it leads to New Speed as 50 and the decrease in time (T = 4 minutes) because 50 * 36 = 1800.
This is the only combination that works and hence shows that our answers will be S-slow = 45 and T-slow = 40. and S-fast = 50 and T-fast = 36. So, our final answer would be S-fast = 50 (Option Choice:A)

The only trouble for me is that it worked as we are given the options and I would have tested smartly within 2 minutes. If 45 and 50 wouldn't have worked in such cases, I would have chosen 60 as the next value (kind of plugging and chugging).

But does it work well for all such hard questions (700+ Level questions) so that I can solve them in approx. 2-2.5 minutes?
Do, I need to take extra care / precaution, in the method I'm trying to make-up / follow?
TIA.

Thanks and Regards,
Ravish.
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Liam is pulled over for speeding just as he is arriving at work.He 28 Apr 2021, 01:49
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ravish844

Let's say the speed is S miles per hour and Time is T minutes for the first case (over-speeding). Note the usage of minutes because we have been given 4 minutes as the time difference.

So, ATQ
S * T = 30 * 60 (We need to multiply by 60 because we have taken time in minutes).
So, S * T = 1800

Algebraically, the second equation ATQ will be -> (S - 5) (T + 4) = 1800
S-slow = S-5 kmph
and T-slow = T+4
Now, we have to factorize 1800 into such two numbers, such that when we increase one of those factors (speed) by 5, the other factor (time) is decreased exactly by 4 and also the product remains constant i.e. 1800.

Let's say (by some hit & trial and looking at the answer choices), I took S = 45 and T = 40, because 45 * 40 (= 1800)
Now, if this (45 factor) is the faster speed (over-speeding), I'm going to decrease the speed for other case. If I decrease (Speed = 45) by 5, it leads to New Speed, S-slow as 40 but the corresponding increase in time will be 5 minutes (because moving from T-fast at 40 minutes to T-slow at 45 minutes). Actually, Speed and Time have interchanged their values.
But this doesn't satisfy our case, increase in time is exactly 4 minutes and not 5 minutes.

Now, If I increase (Speed = 45) by 5, it leads to New Speed as 50 and the decrease in time (T = 4 minutes) because 50 * 36 = 1800.
This is the only combination that works and hence shows that our answers will be S-slow = 45 and T-slow = 40. and S-fast = 50 and T-fast = 36. So, our final answer would be S-fast = 50 (Option Choice:A)

The only trouble for me is that it worked as we are given the options and I would have tested smartly within 2 minutes. If 45 and 50 wouldn't have worked in such cases, I would have chosen 60 as the next value (kind of plugging and chugging).

But does it work well for all such hard questions (700+ Level questions) so that I can solve them in approx. 2-2.5 minutes?
Do, I need to take extra care / precaution, in the method I'm trying to make-up / follow?
TIA.

Thanks and Regards,
Ravish.
Show more

Nothing wrong with your method - you made two equations and then tried the options.
But, I would "almost never" take two variables in any GMAT question. My comfort zone is 0 variables but sometimes I do take 1.
Plugging in and trying would be more time consuming with 2 variables.
Questions with the kind of equation I made above usually have helpful options. Otherwise solving the quadratic can be a bit painful and that is usually not GMAT's game.
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Liam is pulled over for speeding just as he is arriving at work.He 17 Feb 2022, 00:58
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KarishmaB Vinit800HBS please could you help me understand where have I gone wrong with the below approach?

In Liam's case:
Rate = R = ? (To find)
Time = T - 4
D = 30

In the hypothetical case of the Police:
Rate = R - 5
Time = T
D = 30

Since D = R x T we can say:

R (T - 4) = (R - 5) T
RT - 4R = RT - 5T

So we get 4R = 5T.

Now we know that D = R x T Or 30 = R (T - 4)

So T - 4 = \(\frac{30}{R }\)
T = \(\frac{30}{R}\) + 4

4R = 5 { \(\frac{30}{R}\) + 4 }

4\(R^2\) = 150 + 20R OR 2\(R^2\) = 75 + 10R
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Liam is pulled over for speeding just as he is arriving at work.He 17 Feb 2022, 01:46
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Hoozan
KarishmaB Vinit800HBS please could you help me understand where have I gone wrong with the below approach?

In Liam's case:
Rate = R = ? (To find)
Time = T - 4
D = 30

In the hypothetical case of the Police:
Rate = R - 5
Time = T
D = 30

Since D = R x T we can say:

R (T - 4) = (R - 5) T
RT - 4R = RT - 5T

So we get 4R = 5T.

Now we know that D = R x T Or 30 = R (T - 4)

So T - 4 = \(\frac{30}{R }\)
T = \(\frac{30}{R}\) + 4

4R = 5 { \(\frac{30}{R}\) + 4 }

4\(R^2\) = 150 + 20R OR 2\(R^2\) = 75 + 10R
Show more

Hi Hoozan,

One fundamental flaw that i see is that the value for time that you have taken is in terms of minutes (T-4).

Considering that the distance is given in terms of miles and the answer that we have to find would be in terms of "miles per hour", there is bound to be some issue creeping in if you don't take appropriate units for all the variables. I would suggest you to convert T-4 to T - (4/60).


More importantly, If from the question, you have figured out that "Distance in both the scenarios is going to be the same", and the difference in speeds/rates will result in the difference in time, why don't you get rid of the Time variable completely and express everything in terms of "Rate". Try this.

\(\frac{ Distance }{ Rate as per officer } - \frac{ Distance }{ Rate as per Liam } = Difference in Time in Hours\)
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Liam is pulled over for speeding just as he is arriving at work.He 17 Apr 2024, 22:33
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shm0401
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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­But it says speed has been oversped . how can you take just S in first equation because it arrived early 
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Liam is pulled over for speeding just as he is arriving at work.He 12 May 2024, 11:09
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will such questions actually feature on the GMAT, can not figure doing this in less than 2 mins. OR is this just good practice
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Liam is pulled over for speeding just as he is arriving at work.He 28 Jul 2025, 11:17
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