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Re: Reiko drove from point A to point B at a constant speed, and
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30 Jul 2013, 07:32
I realize that I don't need algebra if I understand the concept (I wish all GMAT problems were like that! ) but I feel like knowing the algebra plus the underlying concepts strengthens my understanding of the material as a whole. Thanks for the response! VeritasPrepKarishma wrote: 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.



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Re: Reiko drove from point A to point B at a constant speed, and
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15 Dec 2013, 23:13
how about this, dist atobtoa=240 miles 120 miles each atob and btoa, given a) avg speed = 80 mph => 3 hrs to complete the entire trip (so 1.5 hrs each side atob and btoa considering 1.5 hrs only to get avgerage to 1.5 as avg speed given is 80 mph). now if we consider speed as 41 mph(least possible integer value), reiko will need almost 3(175.61 min to be precise) hrs to complete(task was 3 hrs for to and fro trip) atob thus speed has to be > 40 mph b) NS.



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Re: Reiko drove from point A to point B at a constant speed, and
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31 Dec 2013, 04:31
deepakrobi 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.
(2) It took Reiko 20 more minutes to drive from A to B than to make the return trip. OK, here's the deal Let's say that each leg was 120 miles So the question is asking if he took less than 3 hours on his way to B (120/40<3?) Statement 1 240/8 = 3 So if Reiko took 3 hours on the whole trip then he must have taken less than 3 hours on the way to B Therefore, this statement is sufficient Statement 2 We really don't get much out of here, we need more information Insufficient Hence answer is A Hope it helps! Cheers J



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Re: Speed Time Distance DS question
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02 Jun 2014, 18:48
Can the last part be the following? \(\frac{2D}{\frac{D}{S_1}+\frac{D}{S_2}}=80\) > \(\frac{1}{\frac{1}{S_1}+\frac{1}{S_2}}=40\) So we get \(\frac{1}{S_1}+\frac{1}{S_2}= \frac{1}{40}\) and of course \(\frac{1}{S_1}< \frac{1}{40}\) So s1>40 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 counterintuitive, 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?[/quote]



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Reiko drove from point A to point B at a constant speed, and
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13 Nov 2015, 06:34
I have been struggling through the concept, really. Here is my simplified solution for ST1 for the guys that are as math gifted as me. I tried to go away from too many formulas and terms that were exploitted too heavily in the solutions above in the thread and created even more abstraction. So you know the general formula, right: Average speed =\(\frac{Total Distance}{Total time}\)(a) Therefore 80 = \(\frac{D+D}{Total time}\) which is simplified to 40 = \(\frac{D}{Total time}\). From this we pull out Total time = \(\frac{D}{40}\) (b) And now is simple reasoning: Total time should be greater than the time spent on the first half of the route (A to B), right? (Unless Reiko teleported in no time from B back to A). Hence ("Speed 1" stands for the searched speed from A to B) \(\frac{D}{40}\)>\(\frac{D}{Speed 1}\)  so when does this inequality hold true? Right, when the denominator in the second fraction is larger than the one in the first. From this we derive that Speed 1 > 40
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Re: Reiko drove from point A to point B at a constant speed, and
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18 Nov 2015, 00:49
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 counterintuitive, 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? 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



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Reiko drove from point A to point B at a constant speed, and
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18 Dec 2015, 09:38
deepakrobi 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.
(2) It took Reiko 20 more minutes to drive from A to B than to make the return trip. Couldn't solve this one due to time limit during the MGMAT CAT. It's just too tough to present this one as a DS question.... We can pick any value for distance: let's say 80 km one way > 160 km whole trip Total Distance = 160 Average Speed = 80 \(Total Time = \frac{Total Distance}{Average Speed} = 2 Hours\) = Time 1 (from A to B) + Time 2 (from B to A) > So Time 1 must be < 2 Hours (Total time) Speed 1= \(\frac{80}{< 2 Hours}> 40 km/h\)
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Re: Reiko drove from point A to point B at a constant speed, and
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18 Dec 2015, 20:16
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 ... thegmat/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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Reiko drove from point A to point B at a constant speed, and
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26 Dec 2016, 07:14
EvaJager wrote: 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
Great explanation but I think the part that is not so intuitive is the following: Since \(\frac{S_{2}}{S_{1}+S_{2}}<1,\) it follows that \(40=\frac{S_{1}S_{2}}{S_{1}+S_{2}}<S_1.\) As you dont explain why <1 ..... Fair enough for some people the relationship between PART TO WHOLE has to be <1 but some ppl they cannot see this clearly. Also the \(40=\frac{S_{1}S_{2}}{S_{1}+S_{2}}<S_1.\) is quite convoluted. So if you can please elaborate your solution. In any case I think that since we end up with the following equation \(40=\frac{S_{1}S_{2}}{S_{1}+S_{2}}\) we can rearrange such that \(S_11=\frac{40S_{2}}{S_{2}40}\)IS quite obvious that: 1) S2 CANNOT BE 0 otherwise it would never made the second leg and also forces S1 to be zero which doesn't make sense 2) S2 CANNOT BE smaller than 40 because then we have NEGATIVE S1 and strictly speaking speed is a positive quantity 3) S2 CANNOT BE 40 otherwise S_1 is not defined because of the 0 in the denominator of the fraction ( 40S_2 / 0 is not allowed ) Hence since S2 > 40 then S1 > 40



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Reiko drove from point A to point B at a constant speed, and
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16 Feb 2018, 01:18
deepakrobi 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.
(2) It took Reiko 20 more minutes to drive from A to B than to make the return trip. Got wrong as couldn't drill down to the core. However, now I solved like this. Stmt 1: Let speed from A to B is ' a' and speed from B to A is ' b' Avg speed for the entire journey is 2ab/(a+b) = 80 => ab/(a+b) = 40 => ab = 40a + 40b => a = 40a/b + 40. Since, a and b positive(Speed), 40a/b will be +ve. Thus, a > 40 => Sufficient. Stmt 2: Don't know the time or speed. Hence, Not Sufficient. Thanks.



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Re: Reiko drove from point A to point B at a constant speed, and
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08 Aug 2018, 09:16
sanjoo wrote: 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... But why should A to B must be greater than 40 km/hr and not B to A ? we cannot deduce from option A that A to B avg speed was greater than 40 and B to A less than 40.




Re: Reiko drove from point A to point B at a constant speed, and &nbs
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