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

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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)
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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

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

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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....

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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)

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

$$Average speed = \frac{100 }{ x + y }$$

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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)

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

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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]