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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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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]
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]
I see that is the answer, but what is wrong with adding two seperate speed and divide into two?

Posted from my mobile device
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Re: Jonah drove the first half of a 100-mile trip in x hours and the secon [#permalink]
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]
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]
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.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/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 [#permalink]
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Hi All,

We're told that Jonah drove the first half of a 100-mile trip in X hours and the second half in Y hours. We're asked which of the following is equal to Jonah’s average speed, in miles per hour, for the entire trip. This question can be solved in a couple of different ways, including by TESTing VALUES.

Since we're dealing with the trip in two 50-mile pieces, we should TEST values for X and Y that divide evenly into 50. Let's TEST X=5 and Y=10
The first 50 miles were traveled in 5 hours.
The second 50 miles were traveled in 10 hours.
Total Distance = 100 miles and Total Time = 15 hours

Total Dist = (Av. Sp.)(Total Time)
100 miles = (Av. Sp)(15 hours)
100/15 = Av. Sp. = 6 2/3 miles/hour

Based on the design of the answer choices, we don't actually have to do that last extra calculation. Only one of the answers will equal 100/15 when you TEST X=5 and Y=10....

Final Answer:

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


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



Shouldnt we add the speeds separately ? like 50/x for 1st half + 50/y for second ?

why would this be wrong ?
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Re: Jonah drove the first half of a 100-mile trip in x hours and the secon [#permalink]
\(Average speed = \frac{Total distance }{ time 1 + time 2}\)
Time 1 = x
Time 2 = y

\(Average speed = \frac{100 }{ x + y }\)

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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a1997g wrote:
JeffTargetTestPrep 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


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



Shouldnt we add the speeds separately ? like 50/x for 1st half + 50/y for second ?

why would this be wrong ?


Hi a1997g,

There's a specific formula for Average Speed:

Average Speed = (Total Distance)(Total Time)

The reason why you can't just average the speeds in these types of questions is because (in almost all situations) you're dealing with a WEIGHTED Average.

For example, if you drive 60 miles at 60 miles/hour (re: 1 hour) and then another 60 miles at 30 miles/hour (re: 2 hours), then you spent far MORE time traveling 30 miles/hour. In this scenario, total distance = 120 miles and total time = 3 hours..... 120/3 = average speed of 40 miles/hour.

There is only one exception to this: when you travel each speed for the SAME amount of time. For example, if you drive 60 miles at 60 miles/hour (re: 1 hour) and then another 30 miles at 30 miles/hour (re: 1 hour), then total distance = 90 miles and total time = 2 hours..... 90/2 = average speed of 45 miles/hour (and this is the average of 60 and 30).

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


Total distance = 100 miles
Total time taken = x + y hours

Average speed = 100/(x+y) miles/hour

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



Average speed = Total distance/total time

Here, total distance 100 miles and total time is x+y

So, the Average speed = 100/x+y
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Re: Jonah drove the first half of a 100-mile trip in x hours and the secon [#permalink]
average speed = total distance / total time

average speed = 100 / x + y

lets say for example x = 2 y = 3

100 / 5 = 20 mph
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Re: Jonah drove the first half of a 100-mile trip in x hours and the secon [#permalink]
distance = rate * time
rate = distance / time

average speed = total distance / total time

total distance = 100
total time = x + y

100 / x + y

Answer is B.
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Jonah drove the first half of a 100-mile trip in x hours and the secon [#permalink]
Top Contributor
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



\(Average \ speed = \frac{Total \ Distance \ Traveled}{Total \ Time \ Taken}\)

Total Distance=100

Total time = x+y

\(Aaverage \ Speed = \frac{100}{x+y}\)

The answer is \(B\)
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Re: Jonah drove the first half of a 100-mile trip in x hours and the secon [#permalink]
Expert Reply
Keep it simple, folks. Average speed is total distance over total time:

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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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