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DS- Average rate (m09q15)

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DS- Average rate (m09q15) [#permalink] New post 29 Nov 2007, 13:29
In the morning, John drove to his mother's house in the village at an average speed of 60 kmh. When he was going back to town in the evening, he drove more cautiously and his speed was lower. If John went the same distance in the evening as in the morning, what was John's average speed for the entire trip?

1. In the evening, John drove at a constant speed of 40 kmh.
2. John's morning drive lasted 2 hours.

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A

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I got the answer to this, I just want to see what you guys calculate the answer to be. I come up with a number too large so I wanna see where my calculations are wrong. Thx.
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 [#permalink] New post 29 Nov 2007, 13:44
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A.

1. Vm=60, Ve=40

Va=2L/t=2L/(tm+te)=2L/(L/Vm+L/Ve)=2VmVe/(Vm+Ve)=2*40*60/100=48kmh
SUFF.


2) John's morning drive lasted 2 hrs - information about evening trip absents.
INSUFF.
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 [#permalink] New post 29 Nov 2007, 14:47
walker wrote:
A.

1. Vm=60, Ve=40

Va=2L/t=2L/(tm+te)=2L/(L/Vm+L/Ve)=2VmVe/(Vm+Ve)=2*40*60/100=48kmh
SUFF.


2) John's morning drive lasted 2 hrs - information about evening trip absents.
INSUFF.


Ok ya I got 48 as well. My notes were so sloppy that I it looked like the time was 48hours... so i was like what the heck is goin on??? owell still got the answer.

Thx walk.
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 [#permalink] New post 15 Dec 2007, 18:03
didnt actually solve for the value, but got A

We can express the overall equation for avg speed just in terms of one variable, which is the total distance. Youll see that this variable cancels out anyways, and once you have the speed for the 2nd leg of the trip, youre good.
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Re: DS- Average rate (m09q15) [#permalink] New post 08 Apr 2010, 17:53
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distance = speed x time
(1) 60x = 40y................................(1) x & y are corresponding times.
we need (60x + 40y/(x+y)..........(2)
Suff.

(2)60x2 = 120km. Nothing about evening time or speed.
Insuff
A is it.
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Re: DS- Average rate (m09q15) [#permalink] New post 09 Apr 2010, 00:12
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We just need the speed at which john travelled in morning and evening to find the his average speed
(2*60*40)/60+40 = 48km/hr
Sufficient

Stmt 2 is not sufficient - no information regarding evening travel

IMO A
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Re: DS- Average rate (m09q15) [#permalink] New post 11 Apr 2010, 21:01
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Just Started my prep.

Was thinking when the average speeds needs to be calculated, from basic logic you need the average speed in the morning and average speed in the evening OR if you dont have that information you need to calculate.

1) provides me that information ( I dont need to calculate anything, calculation seems more like a trap for this questions to waste time).
2) distance in this case has no relevance since the problem was solved in step 1, step 2 only needs to be analyzed for checking if this option contributes to statement 1 or independently resolves the question asked.

My 2 cents.

By the way the calculation seems interesting :

why ((2)*60*40)/60+40 = 48km/hr
wouldn't the average be 60+40/2 we already have from questions and statement 1 the average speed.
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Re: DS- Average rate (m09q15) [#permalink] New post 12 Apr 2011, 06:12
(1)
x = distance
Average Speed = 2x/(x/60 + x/40)

Sufficient

(2)

The speed during return trip is not given, so insufficient.

Answer - A
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Re: DS- Average rate (m09q15) [#permalink] New post 12 Apr 2011, 09:30
Answer i got is A

1- Total distance/total time - distance can be assumed and time can be determined using rate - suff
2- Need to know the rate - it cannot be assumed - insuff
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Re: DS- Average rate (m09q15) [#permalink] New post 12 Apr 2011, 19:39
1. Sufficient

Avg speed = 2x/((x/40)+(x/60))
2. Not sufficient. Evening trip info missing.

Answer is A.
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Re: DS- Average rate (m09q15) [#permalink] New post 12 Apr 2011, 20:13
GMATBLACKBELT wrote:
In the morning, John drove to his mother's house in the village at an average speed of 60 kmh. When he was going back to town in the evening, he drove more cautiously and his speed was lower. If John went the same distance in the evening as in the morning, what was John's average speed for the entire trip?

1. In the evening, John drove at a constant speed of 40 kmh.
2. John's morning drive lasted 2 hours.

[Reveal] Spoiler: OA
A

Source: GMAT Club Tests - hardest GMAT questions

I got the answer to this, I just want to see what you guys calculate the answer to be. I come up with a number too large so I wanna see where my calculations are wrong. Thx.


As helpful as it is to see everyone repeat the same equation - I was wondering if someone could specifically walk me through how they arrived at ((2)*60*40)/60+40 = 48km/hr.

I understand that the end result is correct and how since we are using average speeds its irrespective of distance or time. I just am a little confused on the logic behind the equation.

Is the logic something like 2x (because it represents both legs of the journey) x 60km/h (morning speed) x 40km/h (evening speed) / 60+40.

Let me know. Thanks.
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Re: DS- Average rate (m09q15) [#permalink] New post 12 Apr 2011, 22:36
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u0422811 wrote:
GMATBLACKBELT wrote:
In the morning, John drove to his mother's house in the village at an average speed of 60 kmh. When he was going back to town in the evening, he drove more cautiously and his speed was lower. If John went the same distance in the evening as in the morning, what was John's average speed for the entire trip?

1. In the evening, John drove at a constant speed of 40 kmh.
2. John's morning drive lasted 2 hours.

[Reveal] Spoiler: OA
A

Source: GMAT Club Tests - hardest GMAT questions

I got the answer to this, I just want to see what you guys calculate the answer to be. I come up with a number too large so I wanna see where my calculations are wrong. Thx.


As helpful as it is to see everyone repeat the same equation - I was wondering if someone could specifically walk me through how they arrived at ((2)*60*40)/60+40 = 48km/hr.

I understand that the end result is correct and how since we are using average speeds its irrespective of distance or time. I just am a little confused on the logic behind the equation.

Is the logic something like 2x (because it represents both legs of the journey) x 60km/h (morning speed) x 40km/h (evening speed) / 60+40.

Let me know. Thanks.


Average \hspace{2} speed=\frac{Total \hspace{2} Distance}{Total \hspace{2} Time}

Average speed for the entire trip needs to be found.

Let the distance of the drive one way be "D"
Distance for the entire trip = D+D=2D

Time taken for the entire trip = Time taken to reach John's Mom's+Time taken for the return journey

Time = \frac{Distance}{Speed}

Time taken to reach Mom's: \frac{D}{60}
Time taken for the return trip: \frac{D}{40} ---- Using St1

Average \hspace{2} speed=\frac{2D}{\frac{D}{60}+\frac{D}{40}}
Average \hspace{2} speed=\frac{2}{\frac{1}{60}+\frac{1}{40}}
Average \hspace{2} speed=\frac{2*60*40}{60+40}

2.
Using 2, we would know the exact distance "D" as

D= speed*time = 60*2=120km.
But, as we saw earlier, the distance gets canceled from numerator and denominator and won't help us determine the average speed for the entire trip. We need to know how long did it take for both trips.
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Re: DS- Average rate (m09q15) [#permalink] New post 13 Apr 2011, 08:41
I think 1 is sufficient to answer :)
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Re: DS- Average rate (m09q15) [#permalink] New post 16 Apr 2011, 13:59
Easy one standard equation 2xy/(x+y) will give you the answer!!
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Re: DS- Average rate (m09q15) [#permalink] New post 17 Apr 2011, 12:15
A, formula 2S1S2/(S1 + S2) S1, S2 being the speeds in each direction
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Re: DS- Average rate (m09q15) [#permalink] New post 18 Apr 2011, 05:20
Ans A

A. 2*60*40/100 = 48
B. Dist = 60*2 = 120 km; No info on evening's speed. Insufficient
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Re: DS- Average rate (m09q15) [#permalink] New post 17 Apr 2013, 04:12
Expert's post
In the morning, John drove to his mother's house in the village at an average speed of 60 kilometers per hour. When he was going back to town in the evening, he drove more cautiously and his speed was lower. If John went the same distance in the evening as in the morning, what was John's average speed for the entire trip?

The average speed equals to \frac{total \ distance}{total \ time}.

Say the distance between the town and the village is d kilometers, and the speed from the village to town is x kilometers per hour, then \frac{total \ distance}{total \ time}=\frac{d+d}{\frac{d}{60}+\frac{d}{x}} --> d can be reduced, and we get speed=\frac{2}{\frac{1}{60}+\frac{1}{x}}. So, as you can see we only need to find the average speed from the village to town.

(1) In the evening, John drove at a constant speed of 40 kilometers per hour. Sufficient.

(2) John's morning drive lasted 2 hours. We know nothing about his evening drive. Not sufficient.

Answer: A.
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Re: DS- Average rate (m09q15) [#permalink] New post 17 Apr 2013, 04:20
No need to solve anything, that's the point of these questions: all you need to know is if (1) or (2) or neither are sufficient. Its obvious just by first pass that (1) is enough. Given average speeds for entire trip you can get the average speed. Length of trip is not relevant.

Answer is A
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Re: DS- Average rate (m09q15) [#permalink] New post 17 Apr 2013, 05:20
1) As the onward and return journey covered same distance and we know speed for both onward and return journey, we can calculate average speed. Sufficient.

2) There is no information on speed or time taken in the return journey. Not sufficient.

Answer is A.
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Re: DS- Average rate (m09q15) [#permalink] New post 18 Apr 2013, 00:00
In the morning, John drove to his mother's house in the village at an average speed of 60 kmh. When he was going back to town in the evening, he drove more cautiously and his speed was lower. If John went the same distance in the evening as in the morning, what was John's average speed for the entire trip?

1. In the evening, John drove at a constant speed of 40 kmh.
2. John's morning drive lasted 2 hours.


Let distance be x
Case 1:
u1 = 60, t1 = x/60
u2 = 40, t2 = x/40

Hence we calculate avg speed => total dist/total time and get an integer value

SUFFICIENT


Case 2
u2 is unknowd so we cannot get the exact avg value.
INSUFFICIENT


Hence A
Re: DS- Average rate (m09q15)   [#permalink] 18 Apr 2013, 00:00
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