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Intern  Joined: 07 Jul 2015
Posts: 26
Re: 'Distance/Speed/Time' Word Problems Made Easy  [#permalink]

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Thanks for this amazing post!

Now, I am confident with my Speed, Distance Time concepts.

What I want to ask is: Do the practice questions in this post cover the "only types" of questions that could appear in the GMAT? Or has any type of question been left out?
Veritas Prep GMAT Instructor V
Joined: 16 Oct 2010
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Re: 'Distance/Speed/Time' Word Problems Made Easy  [#permalink]

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VICKY69 wrote:
VeritasPrepKarishma
mam i read your weighted average topic in the blog which was ossom..but can u help me with a little confusion which is how do i choose or known in a question that which one is the correct weight that i have to take because if i take wrong weight then i will land up to a wrong answer in topics like time speed distance etc..

Good question.
Check here: http://www.veritasprep.com/blog/2014/12 ... -averages/
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Veritas Prep GMAT Instructor V
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Re: 'Distance/Speed/Time' Word Problems Made Easy  [#permalink]

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kanav06 wrote:
Thanks for this amazing post!

Now, I am confident with my Speed, Distance Time concepts.

What I want to ask is: Do the practice questions in this post cover the "only types" of questions that could appear in the GMAT? Or has any type of question been left out?

There is no exhaustive list of "types of questions" that could appear on GMAT. GMAC keeps coming up with innovative ways of testing the same concepts especially at the higher range of difficulty. But if you understand your concepts well, there is no reason for you to be unable to tackle those questions.
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Joined: 09 Aug 2016
Posts: 62
Re: 'Distance/Speed/Time' Word Problems Made Easy  [#permalink]

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1
sriharimurthy wrote:
Motion in Same Direction (Overtaking): The first thing that should strike you here is that at the time of overtaking, the distances traveled by both will be the same.

Very good work and thanks for your post.

I have a small comment for this statement as is not true for all "Same Direction + Overtaking" cases and I will explain this by using two examples.

Example 1) Assume that we have the classic overtaking question where 2 objects A, B move on the same path with SA > SB and start from the same point. Also assume that B started first and then after x hours A begins to move. At this case the they will meet at some point where their distances (with reference to the starting point) will be equal. Hence we can calculate the duration of the journey for A and B etc...

Example2) Assume exactly the same scenario BUT THE TWIST is that object B has different starting point than A and for the sake of the example lets say that this distance difference is Y (in km). If both objects start moving at the same time they will meet after T time units (i.e. T hrs). The case though is that object A and B they will not have covered the same distance, with reference the starting moving instance, because object B will had covered D km and object A (D+Y) km.

As you can see both question are in the same category but this small twist in example 2 is a game-changer. These small twists are quite confusing and misleading, so perhaps your definitions require some clarification.
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Re: 'Distance/Speed/Time' Word Problems Made Easy  [#permalink]

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VeritasPrepKarishma wrote:
dkj1984 wrote:
Example 2.
An executive drove from home at an average speed of 30 mph to an airport where a helicopter was waiting. The executive boarded the helicopter and flew to the corporate offices at an average speed of 60 mph. The entire distance was 150 miles; the entire trip took three hours. Find the distance from the airport to the corporate offices.

Hi ,

Can someone please let me know what would be the average sped in this case .

Will it be total distance / total time = 150 / 3 or will it be 2ab/a+b = 2(30*60) / 90

Thanks!

Regards,
Divya

I will clear your doubt but first let me give you some unsolicited 'gyan'.
When dealing with formulas, remember two things:
1. Do not learn up formulas without knowing the assumptions made to derive them.
2. Make sure you understand how they are derived and the starting point.

Average speed is always Total Distance/Total Time.

1. The formula 2ab/(a+b) assumes that the distance traveled at speed a is the same as the distance traveled at speed b. Say distance traveled in each case is 1 km.
2. Derivation:
Average Speed = Total Distance/Total Time = (1+1)/(1/a + 1/b) = 2ab/(a+b)
So in case you have three speeds a, b and c, you know how to get the average speed in that case too.

Coming to this question, the formula is not used here because it doesn't say that the distance traveled at the two speeds is the same.
Average Speed = Total distance/Total Time = 150/3 = 50 km/hr

Now there are two ways to handle it:
1. Weighted averages
2. Using algebra

Weighted Averages Method:
This now becomes a weighted average problem since you have two speeds and their average is known. The weights will be the time for which the speeds were maintained.
w1/w2 = (60 - 50)/(50 - 30) = 1:2
So plane travel lasted 2 hrs and car travel lasted 1 hr.
Distance traveled by plane = 2*60 = 120 km

Algebra Method:
Let time for which he traveled by plane is t hrs.
50 = (t*60 + (3-t)*30)/3
150 = 60t - 30t + 90
t = 2 hrs
So plane travel lasted 2 hrs.
Distance traveled by plane = 2*60 = 120 km

For more on weighted averages method (which helps you solve orally), check:
http://www.veritasprep.com/blog/2011/03 ... -averages/

Hello karishma,

Thanks for the solution. However, please explain why the "The weights will be the time for which the speeds were maintained" in the weighted average method. How is the ratio of average speed giving us the ration of time ?
Veritas Prep GMAT Instructor V
Joined: 16 Oct 2010
Posts: 9701
Location: Pune, India
Re: 'Distance/Speed/Time' Word Problems Made Easy  [#permalink]

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1
fmik7894 wrote:
VeritasPrepKarishma wrote:
dkj1984 wrote:
Example 2.
An executive drove from home at an average speed of 30 mph to an airport where a helicopter was waiting. The executive boarded the helicopter and flew to the corporate offices at an average speed of 60 mph. The entire distance was 150 miles; the entire trip took three hours. Find the distance from the airport to the corporate offices.

Hi ,

Can someone please let me know what would be the average sped in this case .

Will it be total distance / total time = 150 / 3 or will it be 2ab/a+b = 2(30*60) / 90

Thanks!

Regards,
Divya

I will clear your doubt but first let me give you some unsolicited 'gyan'.
When dealing with formulas, remember two things:
1. Do not learn up formulas without knowing the assumptions made to derive them.
2. Make sure you understand how they are derived and the starting point.

Average speed is always Total Distance/Total Time.

1. The formula 2ab/(a+b) assumes that the distance traveled at speed a is the same as the distance traveled at speed b. Say distance traveled in each case is 1 km.
2. Derivation:
Average Speed = Total Distance/Total Time = (1+1)/(1/a + 1/b) = 2ab/(a+b)
So in case you have three speeds a, b and c, you know how to get the average speed in that case too.

Coming to this question, the formula is not used here because it doesn't say that the distance traveled at the two speeds is the same.
Average Speed = Total distance/Total Time = 150/3 = 50 km/hr

Now there are two ways to handle it:
1. Weighted averages
2. Using algebra

Weighted Averages Method:
This now becomes a weighted average problem since you have two speeds and their average is known. The weights will be the time for which the speeds were maintained.
w1/w2 = (60 - 50)/(50 - 30) = 1:2
So plane travel lasted 2 hrs and car travel lasted 1 hr.
Distance traveled by plane = 2*60 = 120 km

Algebra Method:
Let time for which he traveled by plane is t hrs.
50 = (t*60 + (3-t)*30)/3
150 = 60t - 30t + 90
t = 2 hrs
So plane travel lasted 2 hrs.
Distance traveled by plane = 2*60 = 120 km

For more on weighted averages method (which helps you solve orally), check:
http://www.veritasprep.com/blog/2011/03 ... -averages/

Hello karishma,

Thanks for the solution. However, please explain why the "The weights will be the time for which the speeds were maintained" in the weighted average method. How is the ratio of average speed giving us the ration of time ?

The weights in case of average speed will ALWAYS be time. Here is a post that tells you why this is so and how to find out the weights in various cases: https://www.veritasprep.com/blog/2014/1 ... -averages/

If distance covered is the same in two cases, ratio of speeds will give the inverse of ratio of time taken. Here is a post on this concept: https://www.veritasprep.com/blog/2011/0 ... os-in-tsd/
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Re: 'Distance/Speed/Time' Word Problems Made Easy  [#permalink]

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Example 5. The presented solution is 21mph. Should this not also be the total average speed?

However the weighted average speed is calculated as 20.57 =((24*3)+(18*4))/(3+4).

You also get 20.57 if you calculate the total average speed with the total distance and the total time =144/7?

Could someone please explain why the table is not consistent in this example?
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Joined: 16 Oct 2010
Posts: 9701
Location: Pune, India
Re: 'Distance/Speed/Time' Word Problems Made Easy  [#permalink]

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1
oli29 wrote:
Example 5. The presented solution is 21mph. Should this not also be the total average speed?

However the weighted average speed is calculated as 20.57 =((24*3)+(18*4))/(3+4).

You also get 20.57 if you calculate the total average speed with the total distance and the total time =144/7?

Could someone please explain why the table is not consistent in this example?

No. The answer 21 is the speed of the boat in calm water. It is not the average speed of the boat for the entire trip.
The speed of the boat upstream was 18 and downstream was 24. For these speeds, 21 would be the weighted average in case the boat had travelled at the two speeds for the same time. But actually, the boat travelled at the two speeds for the same distance. In that case, the average speed is 2ab/(a+b), not (a+b)/2.
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Re: 'Distance/Speed/Time' Word Problems Made Easy  [#permalink]

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Thank you very much @VeritasPrepKarishm
However for me it is still not clear why the total average speed does not equal the boat speed in still water. The current helps you downstream as much as it works against you when you go upstream. This means that when the boat travels the same route without any current it takes the same amount of time 7 hours for 144miles. So in calm water the total average speed is still 20.57mph and not 21mph?
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Re: 'Distance/Speed/Time' Word Problems Made Easy  [#permalink]

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oli29 wrote:
Thank you very much @VeritasPrepKarishm
However for me it is still not clear why the total average speed does not equal the boat speed in still water. The current helps you downstream as much as it works against you when you go upstream. This means that when the boat travels the same route without any current it takes the same amount of time 7 hours for 144miles. So in calm water the total average speed is still 20.57mph and not 21mph?

Ok, think of it in another way:

Speed of boat while going downstream is 24 mph and while going upstream is 18 mph. If the boat travels at each speed for 1 hour, the average speed of the boat is 21 mph, that's correct.
Now think, what if the boat travels for longer time at 18 mph and for shorter time at 24 mph? Will the average speed still be 21 mph for the entire trip?
No, right?
When the boat travels equal distances at the two speeds, this is exactly what happens. It takes longer at slower speed so it travels at the slower speed for longer and at the higher speed for shorter period of time. Hence, as we see, in this case the average speed decreases to 20.57 mph from 21 mph.

Check out this related post: https://www.veritasprep.com/blog/2015/0 ... -the-gmat/
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Re: 'Distance/Speed/Time' Word Problems Made Easy  [#permalink]

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Hi,

I totally understand and agree with the 4th problem's solution. However, I tried solving the 4th problem in a different way, which I think is not wrong, but I am getting an incorrect answer.
I assumed first cyclist distance to be 'x' and 2nd cyclist's to be '45-x'. Then, x/14 = (45-x)/16, and when I solve for 'x' I get 21 hours.

Where am I going wrong in my understanding?

Thanks!
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Re: 'Distance/Speed/Time' Word Problems Made Easy  [#permalink]

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Sprincejindal wrote:
Hi,

I totally understand and agree with the 4th problem's solution. However, I tried solving the 4th problem in a different way, which I think is not wrong, but I am getting an incorrect answer.
I assumed first cyclist distance to be 'x' and 2nd cyclist's to be '45-x'. Then, x/14 = (45-x)/16, and when I solve for 'x' I get 21 hours.

Where am I going wrong in my understanding?

Thanks!

You have assumed that x is the distance traveled by the first cyclist. When you solve, you get x = 21. Since it is the distance, it is 21 miles.
Time taken = 21/14 = 1.5 hrs
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Re: 'Distance/Speed/Time' Word Problems Made Easy  [#permalink]

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VeritasPrepKarishma wrote:
Sprincejindal wrote:
Hi,

I totally understand and agree with the 4th problem's solution. However, I tried solving the 4th problem in a different way, which I think is not wrong, but I am getting an incorrect answer.
I assumed first cyclist distance to be 'x' and 2nd cyclist's to be '45-x'. Then, x/14 = (45-x)/16, and when I solve for 'x' I get 21 hours.

Where am I going wrong in my understanding?

Thanks!

You have assumed that x is the distance traveled by the first cyclist. When you solve, you get x = 21. Since it is the distance, it is 21 miles.
Time taken = 21/14 = 1.5 hrs

WOW! I can't believe I did that mistake. That was really stupid!

Thank you for your prompt response.
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Re: 'Distance/Speed/Time' Word Problems Made Easy  [#permalink]

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