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Jonah drove the first half of a 100-mile trip in x hours and the secon

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Jonah drove the first half of a 100-mile trip in x hours and the secon  [#permalink]

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New post 03 Jul 2016, 11:09
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Jonah drove the first half of a 100-mile trip in x hours and the second half in y hours. Which of the following is equal to Jonah’s average speed, in miles per hour, for the entire trip?

A. 50/(x + y)
B. 100/(x + y)
C. 25/x + 25/y
D. 50/x + 50/y
E. 100/x + 100/y

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Re: Jonah drove the first half of a 100-mile trip in x hours and the secon  [#permalink]

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New post 25 Jul 2017, 08:22
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annusngh wrote:
Bunuel wrote:
Jonah drove the first half of a 100-mile trip in x hours and the second half in y hours. Which of the following is equal to Jonah’s average speed, in miles per hour, for the entire trip?

A. 50/(x + y)
B. 100/(x + y)
C. 25/x + 25/y
D. 50/x + 50/y
E. 100/x + 100/y



Hi Bunuel,

I have applied below approach :

{ (50/X) + (50/Y) } / 2

curious to know flaw in my approach.

Thanks,


You cannot add add two different speeds and get an average speed.
Speed is total distance traveled by total time.
say the first half was done in 5 hrs ie. X=5 and second half was done in 25 hrs ie. y=25 then the total time would be 30 hrs.
So avg speed would be 100/30= 3.33 mph.
but according to what you tried it would become
{(50/5)+(50/25)}/2 --->
(10+2)/2=6 (which is incorrect).
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Re: Jonah drove the first half of a 100-mile trip in x hours and the secon  [#permalink]

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New post 03 Jul 2016, 12:05
Bunuel wrote:
Jonah drove the first half of a 100-mile trip in x hours and the second half in y hours. Which of the following is equal to Jonah’s average speed, in miles per hour, for the entire trip?

A. 50/(x + y)
B. 100/(x + y)
C. 25/x + 25/y
D. 50/x + 50/y
E. 100/x + 100/y

Time taken by Jonah to cover first half i.e 50 km of a 100 mile trip = x
time taken by Jonah to cover the second half i.e 50 km of a 100 mile trip = y
Total time taken by Jonah = x+y
Jonah's average speed for the entire trip =total distance /total time
=100/(x+y)
Answer B
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Re: Jonah drove the first half of a 100-mile trip in x hours and the secon  [#permalink]

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New post 04 Mar 2017, 15:03
Bunuel wrote:
Jonah drove the first half of a 100-mile trip in x hours and the second half in y hours. Which of the following is equal to Jonah’s average speed, in miles per hour, for the entire trip?

A. 50/(x + y)
B. 100/(x + y)
C. 25/x + 25/y
D. 50/x + 50/y
E. 100/x + 100/y



Hi Bunuel,

I have applied below approach :

{ (50/X) + (50/Y) } / 2

curious to know flaw in my approach.

Thanks,
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Re: Jonah drove the first half of a 100-mile trip in x hours and the secon  [#permalink]

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New post 09 Jun 2017, 04:14
I see that is the answer, but what is wrong with adding two seperate speed and divide into two?

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Re: Jonah drove the first half of a 100-mile trip in x hours and the secon  [#permalink]

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New post 09 Jun 2017, 04:26
The mistake is that you are assuming the time to be equal for both the legs(thereby dividing it by 2 for final average speed)
Hence, If x=y, the formula you are using is correct.
However, if x>y or x<y, then the average speed will differ from the actual average speed.

Hope that helps!
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Re: Jonah drove the first half of a 100-mile trip in x hours and the secon  [#permalink]

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New post 09 Jun 2017, 05:08
minji391 wrote:
I see that is the answer, but what is wrong with adding two seperate speed and divide into two?

Posted from my mobile device


Adding to pushpitkc explanation:

Hi minji391, annusngh

Since time duration for both the speed is different so you can't simply add and divide it by 2.

Let me explain with some values.

For simplicity consider x = 5hrs and y = 10 hrs.

Speed during 1st half (S1) = 50/5 = 10 miles/hr

Speed during 2nd half (S2) = 50/10 = 5 miles/hr

Average speed = (S1+S2)/2 = (10 + 5)/2 = 7.5, which is wrong.

Actual average speed = 100/(10+5) = 6.66 miles/hr

Please note that time duration is different for each speed. So, you need to apply the weighted average formula.

Weighted average = (5*10 + 10*5)/(5+10) = 6.66 miles/hr.

Read following post for more detail:

weighted-averages

Hope it helps.
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Re: Jonah drove the first half of a 100-mile trip in x hours and the secon  [#permalink]

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New post 13 Jun 2017, 12:09
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Bunuel wrote:
Jonah drove the first half of a 100-mile trip in x hours and the second half in y hours. Which of the following is equal to Jonah’s average speed, in miles per hour, for the entire trip?

A. 50/(x + y)
B. 100/(x + y)
C. 25/x + 25/y
D. 50/x + 50/y
E. 100/x + 100/y


We are given that Jonah drove the first half of a 100-mile trip in x hours and the second half in y hours. We can determine average rate using the following equation:

average rate = total distance/total time

average rate = 100/(x + y)

Answer: B
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Re: Jonah drove the first half of a 100-mile trip in x hours and the secon  [#permalink]

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New post 03 Sep 2018, 08:21
Bunuel wrote:
Jonah drove the first half of a 100-mile trip in x hours and the second half in y hours. Which of the following is equal to Jonah’s average speed, in miles per hour, for the entire trip?

A. 50/(x + y)
B. 100/(x + y)
C. 25/x + 25/y
D. 50/x + 50/y
E. 100/x + 100/y

Must be (C), this is the end result of a formula

Quote:
Average Speed = 2ab/(a + b)

Applicable when one travels at speed a for half the distance and speed b for other half of the distance


Source : https://www.veritasprep.com/blog/2015/0 ... -the-gmat/
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Re: Jonah drove the first half of a 100-mile trip in x hours and the secon &nbs [#permalink] 03 Sep 2018, 08:21
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