Author 
Message 
TAGS:

Hide Tags

Intern
Joined: 13 Aug 2012
Posts: 1

Reiko drove from point A to point B at a constant speed, and [#permalink]
Show Tags
31 Aug 2012, 19:02
2
This post received KUDOS
50
This post was BOOKMARKED
Question Stats:
26% (02:27) correct
74% (01:38) wrong based on 1248 sessions
HideShow timer Statistics
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.
Official Answer and Stats are available only to registered users. Register/ Login.



Intern
Joined: 28 Aug 2012
Posts: 46
Location: Austria

Re: Speed Time Distance DS question [#permalink]
Show Tags
31 Aug 2012, 23:55
9
This post received KUDOS
3
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



Director
Joined: 22 Mar 2011
Posts: 612
WE: Science (Education)

Re: Speed Time Distance DS question [#permalink]
Show Tags
01 Sep 2012, 00:05
16
This post received KUDOS
7
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.



Director
Joined: 22 Mar 2011
Posts: 612
WE: Science (Education)

Re: Speed Time Distance DS question [#permalink]
Show Tags
01 Sep 2012, 02:55
7
This post received KUDOS
2
This post was BOOKMARKED
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?
_________________
PhD in Applied Mathematics Love GMAT Quant questions and running.



Senior Manager
Joined: 06 Aug 2011
Posts: 400

Re: Reiko drove from point A to point B at a constant speed, and [#permalink]
Show Tags
21 Oct 2012, 11:30
1
This post received KUDOS
Bunuell... can u explain this?? Thank u in advance
_________________
Bole So Nehal.. Sat Siri Akal.. Waheguru ji help me to get 700+ score !



Intern
Joined: 12 Jun 2012
Posts: 41

Re: Reiko drove from point A to point B at a constant speed, and [#permalink]
Show Tags
22 Oct 2012, 06:19
3
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!



Director
Joined: 22 Mar 2011
Posts: 612
WE: Science (Education)

Re: Reiko drove from point A to point B at a constant speed, and [#permalink]
Show Tags
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: reikodrovefrompointatopointbataconstantspeedand138183.html#p1117785
_________________
PhD in Applied Mathematics Love GMAT Quant questions and running.



Math Expert
Joined: 02 Sep 2009
Posts: 39589

Re: Reiko drove from point A to point B at a constant speed, and [#permalink]
Show Tags
23 Oct 2012, 08:45
28
This post received KUDOS
Expert's post
12
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. Answer: A. Hope it's clear.
_________________
New to the Math Forum? Please read this: All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Senior Manager
Joined: 06 Aug 2011
Posts: 400

Re: Reiko drove from point A to point B at a constant speed, and [#permalink]
Show Tags
23 Oct 2012, 09:43
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 !



Manager
Status: Private GMAT Tutor
Joined: 22 Oct 2012
Posts: 77
Location: India
Concentration: Economics, Finance

Re: Speed Time Distance DS question [#permalink]
Show Tags
23 Oct 2012, 21:21
12
This post received KUDOS
2
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 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? 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
_________________
Website: http://www.GMATwithCJ.com
My articles: My experience with GMAT (Score 780) and My analysis of my ESR Three pillars of a successful GMAT strategy Critical Reasoning and The Life of a GMAT Student The 'Although' Misconception Dear GMAT Aspirant, You need not swim against the tide



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 7438
Location: Pune, India

Re: Reiko drove from point A to point B at a constant speed, and [#permalink]
Show Tags
23 Oct 2012, 22:15
8
This post received KUDOS
Expert's post
6
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.
_________________
Karishma Veritas Prep  GMAT Instructor My Blog
Get started with Veritas Prep GMAT On Demand for $199
Veritas Prep Reviews



Senior Manager
Joined: 06 Aug 2011
Posts: 400

Re: Reiko drove from point A to point B at a constant speed, and [#permalink]
Show Tags
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...
_________________
Bole So Nehal.. Sat Siri Akal.. Waheguru ji help me to get 700+ score !



Director
Joined: 22 Mar 2011
Posts: 612
WE: Science (Education)

Re: Speed Time Distance DS question [#permalink]
Show Tags
24 Oct 2012, 01:38
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.



Math Expert
Joined: 02 Sep 2009
Posts: 39589

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



Intern
Status: RusTinPeace
Joined: 11 Aug 2012
Posts: 35
WE: Sales (Commercial Banking)

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



Math Expert
Joined: 02 Sep 2009
Posts: 39589

Re: Reiko drove from point A to point B at a constant speed, and [#permalink]
Show Tags
28 Jun 2013, 03:47



Senior Manager
Joined: 13 May 2013
Posts: 469

Re: Reiko drove from point A to point B at a constant speed, and [#permalink]
Show Tags
29 Jul 2013, 14:36
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!



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 7438
Location: Pune, India

Re: Reiko drove from point A to point B at a constant speed, and [#permalink]
Show Tags
29 Jul 2013, 21:59
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
Veritas Prep Reviews



Verbal Forum Moderator
Joined: 10 Oct 2012
Posts: 629

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
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!
_________________
All that is equal and notDeep Dive Inequality
Hit and Trial for Integral Solutions



Intern
Joined: 30 May 2012
Posts: 21
Concentration: Finance, Strategy
GPA: 3.39

Re: Reiko drove from point A to point B at a constant speed, and [#permalink]
Show Tags
30 Jul 2013, 08:19
2
This post received KUDOS
1
This post was BOOKMARKED
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]
30 Jul 2013, 08:19



Go to page
1 2
Next
[ 36 posts ]




