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# Reiko drove from point A to point B at a constant speed, and

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Reiko drove from point A to point B at a constant speed, and [#permalink]

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31 Aug 2012, 19:02
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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.
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Re: Reiko drove from point A to point B at a constant speed, and [#permalink]

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23 Oct 2012, 08:45
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sanjoo wrote:
Bunuell... can u explain this??

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.

Hope it's clear.
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Re: Speed Time Distance DS question [#permalink]

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01 Sep 2012, 00:05
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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.

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Re: Speed Time Distance DS question [#permalink]

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23 Oct 2012, 21:21
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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.

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.

Hope it helps.

CJ
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Re: Speed Time Distance DS question [#permalink]

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31 Aug 2012, 23:55
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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
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Re: Reiko drove from point A to point B at a constant speed, and [#permalink]

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23 Oct 2012, 22:15
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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.
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Director Joined: 22 Mar 2011 Posts: 612 WE: Science (Education) Followers: 90 Kudos [?]: 760 [6] , given: 43 Re: Speed Time Distance DS question [#permalink] ### Show Tags 01 Sep 2012, 02:55 6 This post received KUDOS 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? _________________ PhD in Applied Mathematics Love GMAT Quant questions and running. Verbal Forum Moderator Joined: 10 Oct 2012 Posts: 630 Followers: 72 Kudos [?]: 920 [5] , given: 136 Re: Reiko drove from point A to point B at a constant speed, and [#permalink] ### Show Tags 30 Jul 2013, 04:13 5 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, $$\frac{2d}{\frac{d}{v_{ab}}+\frac{d}{v_{ba}}} = 80 \to$$ $$\frac{1}{\frac{1}{v_{ab}}+\frac{1}{v_{ba}}} = 40 \to$$ $$\frac{1}{40}= {\frac{1}{v_{ab}}}+{\frac{1}{v_{ba}}}$$ We know that both $$v_{ab}$$ and $$v_{ba}$$ are positive.Thus, $$\frac{1}{40}- {\frac{1}{v_{ab}}}={\frac{1}{v_{ba}}}$$ and $$\frac{1}{40}- {\frac{1}{v_{ab}}}>0 \to$$ $$v_{ab}>40$$. Sufficient. PS:Noticed Karishma's post just now. No need for algebra now! _________________ Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 6473 Location: Pune, India Followers: 1756 Kudos [?]: 10468 [2] , given: 205 Re: Reiko drove from point A to point B at a constant speed, and [#permalink] ### Show Tags 29 Jul 2013, 21:59 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. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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Re: Reiko drove from point A to point B at a constant speed, and [#permalink]

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21 Oct 2012, 11:30
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Bunuell... can u explain this??

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Re: Reiko drove from point A to point B at a constant speed, and [#permalink]

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22 Oct 2012, 06:19
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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
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Re: Reiko drove from point A to point B at a constant speed, and [#permalink]

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23 Oct 2012, 09:43
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Bunuel wrote:
sanjoo wrote:
Bunuell... can u explain this??

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.

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
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Re: Speed Time Distance DS question [#permalink]

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24 Oct 2012, 01:38
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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.
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Re: Reiko drove from point A to point B at a constant speed, and [#permalink]

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30 Jul 2013, 08:19
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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.
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Re: Reiko drove from point A to point B at a constant speed, and [#permalink]

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18 Dec 2015, 21:16
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nishantsharma87 wrote:

EvaJager : I did get (S1*S2)/(S1+S2)=40. I am unable to entirely digest other explanations but may have understood what you have mentioned. Not sure if I have, so writing it here to get your comments and to also ask you another query.

First query:
S1/(S1 + S2) should be lesser than 1 because the numerator S1 is lesser than the denominator S1+S2 (Because both S1 and S2 have to be positive). Thus, S1*(a quantity less than 1) = 40. This means S1 has to be greater than 40.
This way, S2 should ALSO be greater than 40 right?

Second query:
Another doubt I have which may sounds silly but is bugging me a bit after reading other users' explanations that say that even though the average of both speeds is 80 mph, neither of the speeds can be lesser than 40mph. Why is that? Can one of the speeds not be 130mph and the other 30 mph (both averaging 80 mph), where the latter is less than 40 mph ?

Whoever replies, I would highly appreciate your detailed response.

Thanks,

Nishant

I have discussed this concept in detail here: http://www.veritasprep.com/blog/2015/07 ... -the-gmat/

In short, this should help you: If I want my speed to be 100 mph over the entire return journey of 100 miles each side, I should take a total of 2 hours - 1 hour for each side. What if in the first leg of the journey itself, I take more than 2 hours (drive at a speed of less than 50 mph)? Can any speed I pick on the return leg make up for the lost time? Can my average speed every be 100 mph now?
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Re: Reiko drove from point A to point B at a constant speed, and [#permalink]

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22 Oct 2012, 06:29
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:

reiko-drove-from-point-a-to-point-b-at-a-constant-speed-and-138183.html#p1117785
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Re: Reiko drove from point A to point B at a constant speed, and [#permalink]

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23 Oct 2012, 23:41
athirasateesh wrote:
sanjoo can u explain this for me

Hey..u can see bunuel and karishma explanation..its clear..

A is sufficient..1/less than 40>40..as u put any value in denominator less than than 1/40 u l c the rate will be more than 40..

so we get the ans from A..yes speed was more than 40 from A to B...
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Re: Reiko drove from point A to point B at a constant speed, and [#permalink]

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27 Jun 2013, 01:02
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Re: Reiko drove from point A to point B at a constant speed, and [#permalink]

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28 Jun 2013, 03:12
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 ?
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Re: Reiko drove from point A to point B at a constant speed, and [#permalink]

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28 Jun 2013, 03:47
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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.
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11 A squirrel climbs a straight wire from point A to point C. If B is the 6 30 Mar 2015, 05:19
9 Rachel drove the 120 miles from A to B at a constant speed. What was 11 23 Feb 2015, 03:22
6 B is a point equidistant from points A and C. How far is B 4 05 Aug 2013, 07:06
2 If ab different from 0 and points (-a,b) and (-b,a) are in 3 05 Apr 2010, 19:02
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