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Reiko drove from point A to point B at a constant speed, and [#permalink]
 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: Speed Time Distance DS question [#permalink]
31 Aug 2012, 23:55
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: Speed Time Distance DS question [#permalink]
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}}=40Since \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
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Re: Speed Time Distance DS question [#permalink]
01 Sep 2012, 02:55
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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}}=40Since \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?
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Re: Reiko drove from point A to point B at a constant speed, and [#permalink]
21 Oct 2012, 11:30
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Bunuell... can u explain this??
Thank u in advance
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Re: Reiko drove from point A to point B at a constant speed, and [#permalink]
22 Oct 2012, 06:19
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]
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]
23 Oct 2012, 08:45
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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.
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Re: Reiko drove from point A to point B at a constant speed, and [#permalink]
23 Oct 2012, 09:43
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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
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Re: Speed Time Distance DS question [#permalink]
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}}=40Since \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. Hope it helps. CJ
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Re: Reiko drove from point A to point B at a constant speed, and [#permalink]
23 Oct 2012, 22:15
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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Re: Reiko drove from point A to point B at a constant speed, and [#permalink]
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: Speed Time Distance DS question [#permalink]
24 Oct 2012, 01:38
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: Speed Time Distance DS question
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24 Oct 2012, 01:38
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