Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 350,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

Reiko drove from point A to point B at a constant speed, and [#permalink]
31 Aug 2012, 18:02

3

This post received KUDOS

17

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

95% (hard)

Question Stats:

25% (02:28) correct
75% (01:39) wrong based on 713 sessions

Reiko drove from point A to point B at a constant speed, and then returned to A along the same route at a different constant speed. Did Reiko travel from A to B at a speed greater than 40 miles per hour?

(1) Reiko's average speed for the entire round trip, excluding the time spent at point B, was 80 miles per hour.

(2) It took Reiko 20 more minutes to drive from A to B than to make the return trip.

Re: Reiko drove from point A to point B at a constant speed, and [#permalink]
23 Oct 2012, 07:45

15

This post received KUDOS

Expert's post

4

This post was BOOKMARKED

sanjoo wrote:

Bunuell... can u explain this??

Thank u in advance

Reiko drove from point A to point B at a constant speed, and then returned to A along the same route at a different constant speed. Did Reiko travel from A to B at a speed greater than 40 miles per hour?

Say the distance from A to B is \(d\) miles.

(1) Reiko's average speed for the entire round trip, excluding the time spent at point B, was 80 miles per hour -->\(average \ speed=\frac{total \ distance}{total \ time}=\frac{2d}{total \ time}=80\) --> \(total \ time=\frac{d}{40}\). Now, since the time from A to B must be less than the total time (less than \(\frac{d}{40}\)), then Reiko's speed from A to B is \(speed=\frac{distance}{time}=\frac{d}{less \ than \ \frac{d}{40}}=\frac{1}{less \ than \ \frac{1}{40}}>40\) (for example if Reiko's speed from A to B is d/50, so less than d/40, then her speed from A to B is \(\frac{d}{(\frac{d}{50})}=50>40\)). Sufficient.

(2) It took Reiko 20 more minutes to drive from A to B than to make the return trip. Not sufficient.

Re: Speed Time Distance DS question [#permalink]
31 Aug 2012, 23:05

10

This post received KUDOS

3

This post was BOOKMARKED

deepakrobi wrote:

Can somebody please explain the answer? i am having hard time solving it. Reiko drove from point A to point B at a constant speed, and then returned to A along the same route at a different constant speed. Did Reiko travel from A to B at a speed greater than 40 miles per hour?

(1) Reiko's average speed for the entire round trip, excluding the time spent at point B, was 80 miles per hour.

(2) It took Reiko 20 more minutes to drive from A to B than to make the return trip.

Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.

Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.

Both statements TOGETHER are sufficient, but NEITHER one ALONE is sufficient.

EACH statement ALONE is sufficient.

Statements (1) and (2) TOGETHER are NOT sufficient.

Denote by \(D\) the distance between A and B, and by \(S_1\) and \(S_2\) the speeds when traveling from A to B and from B to A, respectively.

(1) The average speed is the total distance divided by the total time, which in our case, translates into the following equation:

\(\frac{2D}{\frac{D}{S_1}+\frac{D}{S_2}}=80\) or \(\, \, \, \, \frac{S_{1}S_{2}}{S_{1}+S_{2}}=40\)

Since \(\frac{S_{2}}{S_{1}+S_{2}}<1,\) it follows that \(40=\frac{S_{1}S_{2}}{S_{1}+S_{2}}<S_1.\)

Sufficient.

(2) Obviously not sufficient.

Answer A _________________

PhD in Applied Mathematics Love GMAT Quant questions and running.

Re: Speed Time Distance DS question [#permalink]
23 Oct 2012, 20:21

8

This post received KUDOS

1

This post was BOOKMARKED

EvaJager wrote:

EvaJager wrote:

deepakrobi wrote:

Can somebody please explain the answer? i am having hard time solving it. Reiko drove from point A to point B at a constant speed, and then returned to A along the same route at a different constant speed. Did Reiko travel from A to B at a speed greater than 40 miles per hour?

(1) Reiko's average speed for the entire round trip, excluding the time spent at point B, was 80 miles per hour.

(2) It took Reiko 20 more minutes to drive from A to B than to make the return trip.

Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.

Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.

Both statements TOGETHER are sufficient, but NEITHER one ALONE is sufficient.

EACH statement ALONE is sufficient.

Statements (1) and (2) TOGETHER are NOT sufficient.

Denote by \(D\) the distance between A and B, and by \(S_1\) and \(S_2\) the speeds when traveling from A to B and from B to A, respectively.

(1) The average speed is the total distance divided by the total time, which in our case, translates into the following equation:

\(\frac{2D}{\frac{D}{S_1}+\frac{D}{S_2}}=80\) or \(\, \, \, \, \frac{S_{1}S_{2}}{S_{1}+S_{2}}=40\)

Since \(\frac{S_{2}}{S_{1}+S_{2}}<1,\) it follows that \(40=\frac{S_{1}S_{2}}{S_{1}+S_{2}}<S_1.\)

Sufficient.

(2) Obviously not sufficient.

Answer A

It seems maybe counter-intuitive, but if you want to travel a given distance at a certain average speed, you cannot travel half of the distance at an average speed not greater than half of that final average speed. Doesn't matter if you travel with the speed of light the other half of the distance, you will not be able to make up for the slow other half. The algebra above proves it.

Does anyone have an intuitive explanation for this?

I understood it this way. If I want my average speed to be, say, x for the entire journey and have covered the half the journey at an average speed less than x/2, then I have already taken more time to cover half the journey than the time needed to cover the entire journey. So, my average speed cannot remain x (has to be less than x).

To say that in numbers, suppose I went on a 200km journey intending to cover at an average speed of 50km/hr (means a time of 4 hours to cover the journey) and I covered the first half i.e. 100km at an average speed less than 25km/hr, which means I have already taken more than 4 hours. So, my total time for the journey cannot be 4 hours. So, my average speed has to be less than 50km/hr.

Re: Speed Time Distance DS question [#permalink]
31 Aug 2012, 22:55

5

This post received KUDOS

1

This post was BOOKMARKED

Statement (1): Let's say, the distance between A and B is 80 miles. It doesn't matter, what distance we assume in statement 1.

The whole trip (160 miles) took 2 hours. If his speed from A to B was NOT greater than 40 miles per hour, he would have needed at least 2 hours just to get from A to B. But we are told that after 2 hours he had already finished the complete trip, so there is no time left to get from B to A. Thus, his speed from A to B must have been greater than 40 miles per hour. --> sufficient

Or, with a little more math: t_1 + t_2 = t Time must be positive, therefore: t_1 < t t_1 = distance / speed_1 = d / s_1 t = 2 * distance / speed = 2 * d / 80 = d / 40 d / s_1 < d / 40 s_1 > 40

Statement (2): Speed = Distance / Time This doesn't tell us anything about the distance between A and B, only about the time. Not sufficient

Can somebody please explain the answer? i am having hard time solving it. Reiko drove from point A to point B at a constant speed, and then returned to A along the same route at a different constant speed. Did Reiko travel from A to B at a speed greater than 40 miles per hour?

(1) Reiko's average speed for the entire round trip, excluding the time spent at point B, was 80 miles per hour.

(2) It took Reiko 20 more minutes to drive from A to B than to make the return trip.

Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.

Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.

Both statements TOGETHER are sufficient, but NEITHER one ALONE is sufficient.

EACH statement ALONE is sufficient.

Statements (1) and (2) TOGETHER are NOT sufficient.

Denote by \(D\) the distance between A and B, and by \(S_1\) and \(S_2\) the speeds when traveling from A to B and from B to A, respectively.

(1) The average speed is the total distance divided by the total time, which in our case, translates into the following equation:

\(\frac{2D}{\frac{D}{S_1}+\frac{D}{S_2}}=80\) or \(\, \, \, \, \frac{S_{1}S_{2}}{S_{1}+S_{2}}=40\)

Since \(\frac{S_{2}}{S_{1}+S_{2}}<1,\) it follows that \(40=\frac{S_{1}S_{2}}{S_{1}+S_{2}}<S_1.\)

Sufficient.

(2) Obviously not sufficient.

Answer A

It seems maybe counter-intuitive, but if you want to travel a given distance at a certain average speed, you cannot travel half of the distance at an average speed not greater than half of that final average speed. Doesn't matter if you travel with the speed of light the other half of the distance, you will not be able to make up for the slow other half. The algebra above proves it.

Does anyone have an intuitive explanation for this? _________________

PhD in Applied Mathematics Love GMAT Quant questions and running.

Re: Reiko drove from point A to point B at a constant speed, and [#permalink]
23 Oct 2012, 21:15

5

This post received KUDOS

Expert's post

2

This post was BOOKMARKED

jordanshl wrote:

I kind of agree with the dissenters that it cannot be A.

If his average round trip was 80mph then it could have taken 30mph and 130mph meaning that the first leg was less than 30mph It could also be 60 and 100 meaning he did it faster than 40

in sufficient

What you missed to note is that avg of 30 mph and 130 mph is 80 mph when he travels at the two speeds for the same TIME. If someone travels at two speeds for the same amount of time, say an hr at one speed and an hr at another speed, then the average speed is (Speed1 + Speed 2)/2

Here, the case is different. The two speeds are for the same distance. He travels at Speed1 from A to B and at Speed2 from B to A (distance in the two cases are same, time taken would be different). So here, the avg is not (Speed1 + Speed 2)/2.

If D is the distance from A to B, Avg Speed = 2D/(D/Speed1 + D/Speed2) = Total distance/Total time

Avg Speed = 2*Speed1*Speed2/(Speed1 + Speed2) (this is the avg when one travels for the same distance)

Now, what happens if one of the speeds is half the avg speed? Say avg speed was 80 mph. Say distance from A to B is 80 miles. Since avg speed for to and fro journey is 80 mph, we need 2 hrs to cover the journey. Now, what happens when the speed while going from A to B is 40 mph? He takes 2 hrs to go from A to B i.e. the entire time allotted for the return trip has gotten used up on the first leg of the journey itself. Of course, it doesn't matter what your speed is in the second leg, you will take more than 2 hrs for the entire journey and hence your avg speed will be less than 80 mph. Hence, in any one leg, your speed cannot be less than half the average.

If you want to see it algebraically,

From above, A = 2*S1*S2/(S1 + S2) If S1 = A/2,

A = 2*(A/2)*S2/(A/2 + S2) S2 = A/2 + S2

There is no value of S2 which will satisfy this equation. _________________

Re: Reiko drove from point A to point B at a constant speed, and [#permalink]
30 Jul 2013, 03:13

3

This post received KUDOS

Expert's post

WholeLottaLove wrote:

Reiko drove from point A to point B at a constant speed, and then returned to A along the same route at a different constant speed. Did Reiko travel from A to B at a speed greater than 40 miles per hour?

(1) Reiko's average speed for the entire round trip, excluding the time spent at point B, was 80 miles per hour. Here is the way I understand it. Please correct me if I am wrong!!

Reiko drives the same distance each way. The average of two speeds is only valid if the two speeds are traveled for the same time. We're not talking about time here we are talking about distance. Lets say he drives one speed from A to B and a much faster speed from B to A. He may have moved faster but because he covered the same distance at a higher speed he did so in less time throwing the average off. When looking at equal distances, the average speed of half of the distance (either from A to B or from B to A) cannot be less than one half of the total average speed (i.e. 80 MPH.) In other words, he must have traveled at least 40MPH for both legs of the trip. SUFFICIENT

I am still having difficultly understanding the algebra for this problem. Could someone explain it to me?

Thanks!

Let the total distance be 2d. Now, the average speed for the first half be \(v_{ab}\) and for the second half be \(v_{ba}\). Thus, \(\frac{2d}{\frac{d}{v_{ab}}+\frac{d}{v_{ba}}} = 80 \to\)

Re: Reiko drove from point A to point B at a constant speed, and [#permalink]
29 Jul 2013, 20:59

2

This post received KUDOS

Expert's post

WholeLottaLove wrote:

Reiko drove from point A to point B at a constant speed, and then returned to A along the same route at a different constant speed. Did Reiko travel from A to B at a speed greater than 40 miles per hour?

(1) Reiko's average speed for the entire round trip, excluding the time spent at point B, was 80 miles per hour. Here is the way I understand it. Please correct me if I am wrong!!

Reiko drives the same distance each way. The average of two speeds is only valid if the two speeds are traveled for the same time. We're not talking about time here we are talking about distance. Lets say he drives one speed from A to B and a much faster speed from B to A. He may have moved faster but because he covered the same distance at a higher speed he did so in less time throwing the average off. When looking at equal distances, the average speed of half of the distance (either from A to B or from B to A) cannot be less than one half of the total average speed (i.e. 80 MPH.) In other words, he must have traveled at least 40MPH for both legs of the trip. SUFFICIENT

I am still having difficultly understanding the algebra for this problem. Could someone explain it to me?

Thanks!

The algebra is only used to explain you the concept. If you understand the concept, you don't need the algebra.

Say, d is the distance from A to B. So Reiko travels 2d (from A to B and then from B to A). His total average speed is 80 mph.

Time taken by Reiko for the entire trip = 2d/80 = d/40

Note that d is the distance of one side of the trip. If the speed of one side is only 40 mph, time taken to go from A to B will be d/40 i.e. the time allotted for the entire trip will get used for one side drive itself (A to B). There will be no time left to go back to A. Hence Reiko's speed for one leg of the journey must be greater than 40. _________________

Re: Reiko drove from point A to point B at a constant speed, and [#permalink]
22 Oct 2012, 05:19

1

This post received KUDOS

I kind of agree with the dissenters that it cannot be A.

If his average round trip was 80mph then it could have taken 30mph and 130mph meaning that the first leg was less than 30mph It could also be 60 and 100 meaning he did it faster than 40

in sufficient _________________

If you find my post helpful, please GIVE ME SOME KUDOS!

Re: Reiko drove from point A to point B at a constant speed, and [#permalink]
23 Oct 2012, 08:43

1

This post received KUDOS

Bunuel wrote:

sanjoo wrote:

Bunuell... can u explain this??

Thank u in advance

Reiko drove from point A to point B at a constant speed, and then returned to A along the same route at a different constant speed. Did Reiko travel from A to B at a speed greater than 40 miles per hour?

Say the distance from A to B is \(d\) miles.

(1) Reiko's average speed for the entire round trip, excluding the time spent at point B, was 80 miles per hour -->\(average \ speed=\frac{total \ distance}{total \ time}=\frac{2d}{total \ time}=80\) --> \(total \ time=\frac{d}{40}\). Now, since the time from A to B must be less than the total time (less than \(\frac{d}{40}\)), then Reiko's speed from A to B is \(speed=\frac{distance}{time}=\frac{d}{less \ than \ \frac{d}{40}}=\frac{1}{less \ than \ \frac{1}{40}}>40\) (for example if Reiko's speed from A to B is d/50, so less than d/40, then her speed from A to B is \(\frac{d}{(\frac{d}{50})}=50>40\)). Sufficient.

(2) It took Reiko 20 more minutes to drive from A to B than to make the return trip. Not sufficient.

Answer: A.

Hope it's clear.

will it not take more than 2 mints to solve?

+1 Bunuel..!! i always luk for ur solution ..:D

Thanks alot..i got it now _________________

Bole So Nehal.. Sat Siri Akal.. Waheguru ji help me to get 700+ score !

Re: Speed Time Distance DS question [#permalink]
24 Oct 2012, 00:38

1

This post received KUDOS

2

This post was BOOKMARKED

chiranjeev12 wrote:

I understood it this way. If I want my average speed to be, say, x for the entire journey and have covered the half the journey at an average speed less than x/2, then I have already taken more time to cover half the journey than the time needed to cover the entire journey. So, my average speed cannot remain x (has to be less than x).

To say that in numbers, suppose I went on a 200km journey intending to cover at an average speed of 50km/hr (means a time of 4 hours to cover the journey) and I covered the first half i.e. 100km at an average speed less than 25km/hr, which means I have already taken more than 4 hours. So, my total time for the journey cannot be 4 hours. So, my average speed has to be less than 50km/hr.

Hope it helps.

CJ

Great explanation!

Which reminds me to try to keep things (algebra inclusive :O) as simple as possible. So, we don't have to go back to the complicated average speed formula. I should have done the following:

If the distance is D, average speed is A, the time needed to cover the distance is T = D/A. Then, the time needed to cover half the distance D/2 with an average speed of A/2 is (D/2)/(A/2) = 2D/2A = T. So, we have already eaten up all our time T on the first half of the journey, nothing left. Therefore, we cannot finish the total D with an average speed A, but with one less than A. _________________

PhD in Applied Mathematics Love GMAT Quant questions and running.

Re: Reiko drove from point A to point B at a constant speed, and [#permalink]
30 Jul 2013, 07:19

1

This post received KUDOS

To do this really quickly, you can just pick any distance, so I choose D=40 miles.

1.) 2*40/(Total Time)=80 -> Total Time=1. Since the Total Time is 1 hour, the time for any leg is less than 1 hour. So it took less than one hour to travel 40 miles from A to B. 40 miles in less than an hour means the rate was greater than 40 miles per hour. Sufficient. 2.) Obviously not sufficient. A

Re: Reiko drove from point A to point B at a constant speed, and [#permalink]
22 Oct 2012, 05:29

jordanshl wrote:

I kind of agree with the dissenters that it cannot be A.

If his average round trip was 80mph then it could have taken 30mph and 130mph meaning that the first leg was less than 30mph It could also be 60 and 100 - NO!!! meaning he did it faster than 40

in sufficient

The average speed is not the average of the speeds. See the formula in the above post:

Re: Reiko drove from point A to point B at a constant speed, and [#permalink]
28 Jun 2013, 02:47

Expert's post

IvyLeague56 wrote:

i did understand the explainations given but doesnt the question tell us to take into account the time spent at Point B ? what about that ?

The question asks: "did Reiko travel from A to B at a speed greater than 40 miles per hour?" Nothing about time spent at point B.

Next, (1) says: "Reiko's average speed for the entire round trip, excluding the time spent at point B, was 80 miles per hour". So, we can ignore the time spent at point B. _________________

Re: Reiko drove from point A to point B at a constant speed, and [#permalink]
29 Jul 2013, 13:36

Reiko drove from point A to point B at a constant speed, and then returned to A along the same route at a different constant speed. Did Reiko travel from A to B at a speed greater than 40 miles per hour?

(1) Reiko's average speed for the entire round trip, excluding the time spent at point B, was 80 miles per hour. Here is the way I understand it. Please correct me if I am wrong!!

Reiko drives the same distance each way. The average of two speeds is only valid if the two speeds are traveled for the same time. We're not talking about time here we are talking about distance. Lets say he drives one speed from A to B and a much faster speed from B to A. He may have moved faster but because he covered the same distance at a higher speed he did so in less time throwing the average off. When looking at equal distances, the average speed of half of the distance (either from A to B or from B to A) cannot be less than one half of the total average speed (i.e. 80 MPH.) In other words, he must have traveled at least 40MPH for both legs of the trip. SUFFICIENT

I am still having difficultly understanding the algebra for this problem. Could someone explain it to me?

Thanks!

gmatclubot

Re: Reiko drove from point A to point B at a constant speed, and
[#permalink]
29 Jul 2013, 13:36

Back to hometown after a short trip to New Delhi for my visa appointment. Whoever tells you that the toughest part gets over once you get an admit is...