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On a particular trip, Jane drove x miles at 30 miles per hour and y
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On a particular trip, Jane drove x miles at 30 miles per hour and y miles at 40 miles per hour. Was her average speed for the trip greater than 35 miles per hour? (1) y > x + 60 (2) y/x > 4/3
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Originally posted by iamdp on 10 Jun 2015, 05:19.
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On a particular trip, Jane drove x miles at 30 miles per hour and y
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10 Jun 2015, 06:38
dpo28 wrote: On a particular trip, Jane drove x miles at 30 miles per hour and y miles at 40 miles per hour. Was her average speed for the trip greater than 35 miles per hour?
(1) y > x + 60 (2) y/x > 4/3 On a particular trip, Jane drove x miles at 30 miles per hour and y miles at 40 miles per hour. Was her average speed for the trip greater than 35 miles per hour?\(average \ speed = \frac{total \ distance}{total \ time}=\frac{x+y}{(\frac{x}{30}+\frac{y}{40})}=\frac{120(x+y)}{4x+3y}\) Thus the question asks whether \(\frac{120(x+y)}{4x+3y}>35\). After simplifying the question becomes: is \(\frac{y}{x}>\frac{4}{3}\). (1) y > x + 60. If x = 1 and y = 100, then y/x > 4/3. If x = 1000 and y = 1200, then y/x < 4/3. Not sufficient. (2) y/x > 4/3. Directly gives an answer to our question. Sufficient. Answer: B. Hope it's clear. Alternatively, notice that 35 miles per hour is halfway between 30 and 40 miles per hour. Now, for the average speed to be exactly 35 miles per hour, the amount of time spent to cover x miles (x/30 hours) must equal to the amount of time spent to cover y miles (y/40 hours). The average speed to be greater than 35 miles per hour, the amount of time spent to cover x miles at the slower speed (x/30 hours) must be less than the amount of time spent to cover y miles at the faster speed(y/40 hours): x/30 < y/40. Then continue as above.
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On a particular trip, Jane drove x miles at 30 miles per hour and y
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10 Jun 2015, 06:37
dpo28 wrote: On a particular trip, Jane drove x miles at 30 miles per hour and y miles at 40 miles per hour. Was her average speed for the trip greater than 35 miles per hour?
(1) y > x + 60 (2) y/x > 4/3 35 miles this is average speed if both distance will take equal time. For example x = 300 miles on the speed 30 mph  10 hours and y = 400 on the speed 40 mph  10 hours So for have average speed more than 35 miles Jane should spend more time on distance y than on distance x time = distance / speed \(\frac{y}{40}>\frac{x}{30}\) > \(\frac{y}{x}>\frac{4}{3}\) 1) this statement insufficient because we can't find ratio of distances x and y from given information 2) This statement gives us direct answer on the question. Sufficient Answer is B
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On a particular trip, Jane drove x miles at 30 miles per hour and y
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10 Jun 2015, 23:03
Hi Bunueli found another explanation for this question it says that avg speed ie 4 is one away from 3 and 4 away from 8. Therefore time spent travelling at 3 m/hr is 4/5 of total. I know this explanation is in compliance with the weighted average formula. Can you plz elaborate on this method how it is 4/5 and not 1/5
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On a particular trip, Jane drove x miles at 30 miles per hour and y
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dpo28 wrote: Hi Bunueli found another explanation for this question it says that avg speed ie 4 is one away from 3 and 4 away from 8. Therefore time spent travelling at 3 m/hr is 4/5 of total. I know this explanation is in compliance with the weighted average formula. Can you plz elaborate on this method how it is 4/5 and not 1/5 Step 1: The distance between 3 and 8 is 5 Step 2: Realize that Weighted Average of 3 and 8 is 4 Step 3: Distance of 3 from Average 4 is 1 AND Distance of 8 from Average 4 is 4 Step 4: S1 / S2 = T2 / T1 by property because Speed S is Inversely proportional to Time T Step 5: i.e. Speed 3 will take 4/5 of Total time and 8 will take 1/5 of Total time I hope it clears!!!
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Originally posted by GMATinsight on 11 Jun 2015, 02:51.
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Re: On a particular trip, Jane drove x miles at 30 miles per hour and y
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11 Jun 2015, 03:09
GMATinsight wrote: Step 1: The distance between 3 and 8 is 5
Step 2: Realize that Weighted Average of 3 and 8 is 4
Step 3: Distance of 3 from Average 4 is 1 AND Distance of 8 from Average 4 is 5
Step 4: S1 / S2 = T2 / T1 by property because Speed S is Inversely proportional to Time T
Step 5: i.e. Speed 3 will take 4/5 of Total time and 8 will take 1/5 of Total time
I hope it clears!!!
Hello GMATinsightI think there is typo in 3 step: "Step 3: Distance of 3 from Average 4 is 1 AND Distance of 8 from Average 4 is 4"
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Re: On a particular trip, Jane drove x miles at 30 miles per hour and y
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11 Jun 2015, 05:09
Seriously, how can you simplify so fast to this; Quote: Thus the question asks whether 120(x+y)/(4x+3y)>35. After simplifying the question becomes: is y/x>4/3.



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Re: On a particular trip, Jane drove x miles at 30 miles per hour and y
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11 Jun 2015, 05:14
tjerkrintjema wrote: Seriously, how can you simplify so fast to this; Quote: Thus the question asks whether 120(x+y)/(4x+3y)>35. After simplifying the question becomes: is y/x>4/3. 120(x+y)/(4x+3y)>35 Since x and y are Positive Numbers hence you can cross multiply 120x + 120 y > 35*(4x+3y) i.e. 120x + 120 y > 140x + 105y i.e. 15y > 20x i.e. 3y > 4x i.e. (y/x) > (4/3) I hope it clears your doubt!
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Re: On a particular trip, Jane drove x miles at 30 miles per hour and y
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11 Jun 2015, 05:29
GMATinsight wrote: tjerkrintjema wrote: Seriously, how can you simplify so fast to this; Quote: Thus the question asks whether 120(x+y)/(4x+3y)>35. After simplifying the question becomes: is y/x>4/3. 120(x+y)/(4x+3y)>35 Since x and y are Positive Numbers hence you can cross multiply 120x + 120 y > 35*(4x+3y) i.e. 120x + 120 y > 140x + 105y i.e. 15y > 20x i.e. 3y > 4x i.e. (y/x) > (4/3) I hope it clears your doubt! Wow I was going at it the whole wrong way. Thanks I guess a better strategy is always to get rid of the fraction first.



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Re: On a particular trip, Jane drove x miles at 30 miles per hour and y
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09 Sep 2015, 22:13
Bunuelyou have mentioned "If x = 100 and y = 200, then y/x < 4/3." In this case y/x = 2 and it is not less than 4/3. we might need something like x = 240 and y = 320 to prove option 1 is not sufficient. in this case y/x = 4/3 and thus not less than 4/3 Please share your thoughts.



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Re: On a particular trip, Jane drove x miles at 30 miles per hour and y
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14 Sep 2015, 02:35
Bunuel wrote: dpo28 wrote: On a particular trip, Jane drove x miles at 30 miles per hour and y miles at 40 miles per hour. Was her average speed for the trip greater than 35 miles per hour?
(1) y > x + 60 (2) y/x > 4/3 On a particular trip, Jane drove x miles at 30 miles per hour and y miles at 40 miles per hour. Was her average speed for the trip greater than 35 miles per hour?\(average \ speed = \frac{total \ distance}{total \ time}=\frac{x+y}{(\frac{x}{30}+\frac{y}{40})}=\frac{120(x+y)}{4x+3y}\) Thus the question asks whether \(\frac{120(x+y)}{4x+3y}>35\). After simplifying the question becomes: is \(\frac{y}{x}>\frac{4}{3}\). (1) y > x + 60. If x = 1 and y = 100, then y/x > 4/3. If x = 1000 and y = 1200, then y/x < 4/3. Not sufficient. (2) y/x > 4/3. Directly gives an answer to our question. Sufficient. Answer: B. Hope it's clear. Alternatively, notice that 35 miles per hour is halfway between 30 and 40 miles per hour. Now, for the average speed to be exactly 35 miles per hour, the amount of time spent to cover x miles (x/30 hours) must equal to the amount of time spent to cover y miles (y/40 hours). The average speed to be greater than 35 miles per hour, the amount of time spent to cover x miles at the slower speed (x/30 hours) must be less than the amount of time spent to cover y miles at the faster speed(y/40 hours):x/30 < y/40. Then continue as above. Could anyone please explain the logic behind the highlighted part. I m not able to get how more than average speed correlates to time taken to cover specific distances, x and y.



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Re: On a particular trip, Jane drove x miles at 30 miles per hour and y
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14 Sep 2015, 16:29
Harley1980 wrote: dpo28 wrote: On a particular trip, Jane drove x miles at 30 miles per hour and y miles at 40 miles per hour. Was her average speed for the trip greater than 35 miles per hour?
(1) y > x + 60 (2) y/x > 4/3 35 miles this is average speed if both distance will take equal time. For example x = 300 miles on the speed 30 mph  10 hours and y = 400 on the speed 40 mph  10 hours So for have average speed more than 35 miles Jane should spend more time on distance y than on distance x time = distance / speed \(\frac{y}{40}>\frac{x}{30}\) > \(\frac{y}{x}>\frac{4}{3}\) 1) this statement insufficient because we can't find ratio of distances x and y from given information 2) This statement gives us direct answer on the question. Sufficient Answer is B Harley1980Could you please elaborate on how does one say that " average speed more than 35 miles Jane should spend more time on distance y than on distance x"



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earnit wrote: Harley1980Could you please elaborate on how does one say that " average speed more than 35 miles Jane should spend more time on distance y than on distance x" Hello earnitThis is weighted average question We can imagine scales: on one side  on another side speed 40 mph  30mph if both side have equal weight (in our case weight is time) when speed will be in middle: speed 40 mph 30mph 10 hours 10 hours Average = 35 mph =============================== if we shift our weight to left: more time with speed 40 mph then average will grow speed 40 mph30mph 20 hours0 hours Average = 40 mph =============================== if we shift our weight to right: more time with speed 30 mph then average will decline speed 40 mph 30mph 0 hours  20 hours Average = 30 mph So if we want average to be more than 35 mph we should put more weight on 40 mph scale (distance y in this question) Try to read Karishma's articles on this topic. She like this topic and write really good posts about it. At first weighted average was very difficult for me but when I read her articles and solve like 50 tasks this topic became very intuitive: http://www.veritasprep.com/blog/2014/12 ... averages/http://www.veritasprep.com/blog/2011/04 ... gebrutes/http://www.veritasprep.com/blog/2011/03 ... averages/http://www.veritasprep.com/blog/2014/09 ... difficult/http://www.veritasprep.com/blog/2013/12 ... question/There is much more articles I just post first that I find. You can use search by words "weighted average"
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Re: On a particular trip, Jane drove x miles at 30 miles per hour and y
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14 Sep 2015, 22:54
Harley1980 wrote: earnit wrote: Harley1980Could you please elaborate on how does one say that " average speed more than 35 miles Jane should spend more time on distance y than on distance x" Hello earnitThis is weighted average question We can imagine scales: on one side  on another side speed 40 mph  30mph if both side have equal weight (in our case weight is time) when speed will be in middle: speed 40 mph 30mph 10 hours 10 hours Average = 35 mph =============================== if we shift our weight to left: more time with speed 40 mph then average will grow speed 40 mph30mph 20 hours0 hours Average = 40 mph =============================== if we shift our weight to right: more time with speed 30 mph then average will decline speed 40 mph 30mph 0 hours  20 hours Average = 30 mph So if we want average to be more than 35 mph we should put more weight on 40 mph scale (distance y in this question) Try to read Karishma's articles on this topic. She like this topic and write really good posts about it. At first weighted average was very difficult for me but when I read her articles and solve like 50 tasks this topic became very intuitive: http://www.veritasprep.com/blog/2014/12 ... averages/http://www.veritasprep.com/blog/2011/04 ... gebrutes/http://www.veritasprep.com/blog/2011/03 ... averages/http://www.veritasprep.com/blog/2014/09 ... difficult/http://www.veritasprep.com/blog/2013/12 ... question/There is much more articles I just post first that I find. You can use search by words "weighted average" Harley1980 +1 explanation. Terrific stuff. I am aware of this concept but just needed that missing link that connects weighted average with this qs. (\(c1w1+c2w2)/ (w1+w2)\) = \(Cavg\) If w1>w2 then Cavg tilts towards c1 and vice versa. Thank you so much for clearing the clutter and making me see the plain logic behind.



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16 Mar 2017, 15:02
Bunuel, If y=41miles and x=40 miles the average speed will be 34,7 (less than 35!) 41 miles = ~1 hour 40 miles = ~1 hour and 20 minutes total distance: 81 miles/ 2hours and 20 minutes = 37,7 miles. Am I doing something wrong? Bunuel wrote: dpo28 wrote: On a particular trip, Jane drove x miles at 30 miles per hour and y miles at 40 miles per hour. Was her average speed for the trip greater than 35 miles per hour?
(1) y > x + 60 (2) y/x > 4/3 On a particular trip, Jane drove x miles at 30 miles per hour and y miles at 40 miles per hour. Was her average speed for the trip greater than 35 miles per hour?\(average \ speed = \frac{total \ distance}{total \ time}=\frac{x+y}{(\frac{x}{30}+\frac{y}{40})}=\frac{120(x+y)}{4x+3y}\) Thus the question asks whether \(\frac{120(x+y)}{4x+3y}>35\). After simplifying the question becomes: is \(\frac{y}{x}>\frac{4}{3}\). (1) y > x + 60. If x = 1 and y = 100, then y/x > 4/3. If x = 1000 and y = 1200, then y/x < 4/3. Not sufficient. (2) y/x > 4/3. Directly gives an answer to our question. Sufficient. Answer: B. Hope it's clear. Alternatively, notice that 35 miles per hour is halfway between 30 and 40 miles per hour. Now, for the average speed to be exactly 35 miles per hour, the amount of time spent to cover x miles (x/30 hours) must equal to the amount of time spent to cover y miles (y/40 hours). The average speed to be greater than 35 miles per hour, the amount of time spent to cover x miles at the slower speed (x/30 hours) must be less than the amount of time spent to cover y miles at the faster speed(y/40 hours): x/30 < y/40. Then continue as above.



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Re: On a particular trip, Jane drove x miles at 30 miles per hour and y
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latinamerica2016 wrote: Bunuel, If y=41miles and x=40 miles the average speed will be 34,7 (less than 35!) 41 miles = ~1 hour 40 miles = ~1 hour and 20 minutes total distance: 81 miles/ 2hours and 20 minutes = 37,7 miles. Am I doing something wrong? Bunuel wrote: dpo28 wrote: On a particular trip, Jane drove x miles at 30 miles per hour and y miles at 40 miles per hour. Was her average speed for the trip greater than 35 miles per hour?
(1) y > x + 60 (2) y/x > 4/3 On a particular trip, Jane drove x miles at 30 miles per hour and y miles at 40 miles per hour. Was her average speed for the trip greater than 35 miles per hour?\(average \ speed = \frac{total \ distance}{total \ time}=\frac{x+y}{(\frac{x}{30}+\frac{y}{40})}=\frac{120(x+y)}{4x+3y}\) Thus the question asks whether \(\frac{120(x+y)}{4x+3y}>35\). After simplifying the question becomes: is \(\frac{y}{x}>\frac{4}{3}\). (1) y > x + 60. If x = 1 and y = 100, then y/x > 4/3. If x = 1000 and y = 1200, then y/x < 4/3. Not sufficient. (2) y/x > 4/3. Directly gives an answer to our question. Sufficient. Answer: B. Hope it's clear. Alternatively, notice that 35 miles per hour is halfway between 30 and 40 miles per hour. Now, for the average speed to be exactly 35 miles per hour, the amount of time spent to cover x miles (x/30 hours) must equal to the amount of time spent to cover y miles (y/40 hours). The average speed to be greater than 35 miles per hour, the amount of time spent to cover x miles at the slower speed (x/30 hours) must be less than the amount of time spent to cover y miles at the faster speed(y/40 hours): x/30 < y/40. Then continue as above. Are you talking about the second statement? If yes, then you cannot use y = 41 miles and x = 40 miles because these values do not satisfy the condition y/x > 4/3.
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Re: On a particular trip, Jane drove x miles at 30 miles per hour and y
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23 Sep 2018, 11:59
Explanation: For average speed problems, you always need to remember that the average speed is total distance divided by total time, not a simple average of the speeds (unless the times are the same). Or in weighted average terms: it is the time that affects the weighting of the two speeds when you average them together.
To answer this question, you need to determine at which speed more time was spent. If you can prove more time was spent at one speed, then you can answer the question or in other words, this question is really: is time for y greater than the time for x.
Algebraically this can be written as: is y/40>x/30 or is y/x>4/3
From this you realize clearly that statement (2) is sufficient. For statement (1) you cannot prove this, because you do not know how big x and y are. If x and y were really large distances, then the average might not be greater than 35, and if x and y were really small distances then it would be.
Answer is (B).



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On a particular trip, Jane drove x miles at 30 miles per hour and y
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27 Oct 2018, 01:53
Hi chetan2uWhile analyzing statement 1, is there any algebraic method of solving these two equation y/x>3/4 and y>x+60, instead of picking numbers or using weighted average concept ?
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