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A driver completed the first 20 miles of a 40-mile trip at an average speed of 50 miles per hour. At what average speed must the driver complete the remaining 20 miles to achieve an average speed of 60 miles per hour for the entire 40-mile trip? (Assume that the driver did not make any stops during the 40-mile trip.)

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A driver completed the first 20 miles of a 40-mile trip at an average speed of 50 miles per hour. At what average speed must the driver complete the remaining 20 miles to achieve an average speed of 60 miles per hour for the entire 40-mile trip? (Assume that the driver did not make any stops during the 40-mile trip.)

\(average \ speed=\frac{total \ distance}{total \ time}=\frac{40}{total \ time}=60\). This implies that for the average time to be 60 miles per hour, the total time must be 40/60 = 2/3 hours.

Now, the first 20 miles were covered in (time) = (distance)/(speed) = 20/50 = 2/5 hours.

Thus, the remaining 20 miles should be covered in 2/3 - 2/5 = 4/15 hours, which means that the remaining 20 miles should be covered at an average speed (distance)/(time) = 20/(4/15) = 75 miles per hour.

Re: A driver completed the first 20 miles of a 40-mile trip at a [#permalink]

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09 Mar 2014, 22:49

1

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Option D. First 20 miles covered at 50 miles/hr. Time taken=D/S=2/5 hrs. Total time=Total D/Avg speed for entire journey.=40/60=2/3 hrs. Time left=2/3-2/5=4/15 hrs. Distance left=20 miles Therefore speed should be D/T=20/(4/15)=75mph

Re: A driver completed the first 20 miles of a 40-mile trip at a [#permalink]

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11 Mar 2014, 09:15

lool wrote:

b2bt wrote:

Average speed concept always confuses me. Anyways there is a formula

\(Avg. Speed = \frac{2(S_1)(S_2)}{S_1+S_2}\)

\(S_1\) being one speed and \(S_2\) being another

By substituting the value you will get 75 miles/hour

Answer D Time Taken - 1:45 Difficulty level - 600

What are (S1) and (S2) ?

Here \(S_1\) = 50 and \(S_2\) = \(x\) Average Speed = 60. Insert these value in formula and you will find x BTW, for this formula to work, the distance covered should be the same. In this case distance = 20 in each case. If you are interested, I can explain you how this formula is derived. Just reply back

A driver completed the first 20 miles of a 40-mile trip at a [#permalink]

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18 Aug 2015, 02:22

Total Distance =40 km Total Time = 20/50 + 20/X (X=average speed for the remaining 20km) Required Average Speed = 60 Formula => Average speed = Total Distance / Total Time --> 40/(20/50 + 20/X)=60; X=75 (D)
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Re: A driver completed the first 20 miles of a 40-mile trip at a [#permalink]

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07 Feb 2017, 12:48

A driver completed the first 20 miles of a 40-mile trip at an average speed of 50 miles per hour. At what average speed must the driver complete the remaining 20 miles to achieve an average speed of 60 miles per hour for the entire 40-mile trip? (Assume that the driver did not make any stops during the 40-mile trip.)

Re: A driver completed the first 20 miles of a 40-mile trip at a [#permalink]

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11 Feb 2017, 01:00

the above problem can be solved with in 50sec. by plugging the answer choices in the avg speed formula as the distance is constant then we use this formula 2xy/x+y where x and y are the avg. speeds. so the only value which satisfies the avg speed equal to 60km/hr. is 75, ex. 2*75*50/125= 60 hence the answer is D thanks if you liked my post please kudos

Re: A driver completed the first 20 miles of a 40-mile trip at a [#permalink]

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17 Sep 2017, 04:31

b2bt wrote:

Average speed concept always confuses me. Anyways there is a formula

\(Avg. Speed = \frac{2(S_1)(S_2)}{S_1+S_2}\)

\(S_1\) being one speed and \(S_2\) being another

By substituting the value you will get 75 miles/hour

Answer D Time Taken - 1:45 Difficulty level - 600

Everyone please not you can only use this approach when the distance is same i.e., If the distance covered is constant then the average speed is Harmonic mean of the values. Else the traditional formula approach must be used.

Re: A driver completed the first 20 miles of a 40-mile trip at a [#permalink]

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09 Oct 2017, 08:57

A table helps solve this problem. This problem involves the fundamental principle of Time x Rate = Distance. Therefore, Time = Distance/Rate. First trip, T = 2/5; Total Trip, T = 2/3 2/5 + 20/s = 2/3 isolate the variable: 20/s = 2/3 - 2/5 20/s = 10/15 - 6/15 20/s = 4/15; looking at the ratio, for every 4 there is 20; thus, 5 5 x 15 = s = 75 Thus, the answer is (D) 75 mph

A driver completed the first 20 miles of a 40-mile trip at an average speed of 50 miles per hour. At what average speed must the driver complete the remaining 20 miles to achieve an average speed of 60 miles per hour for the entire 40-mile trip? (Assume that the driver did not make any stops during the 40-mile trip.)

average rate = (distance 1 + distance 2)/(time 1 + time 2)

where average rate = 60, distance 1 = distance 2 = 20, time 1 = distance 1/rate 1 = 20/50 = 2/5, and time 2 = distance 2/rate 2 = 20/r (where r is the average speed of the remaining 20 miles).

Let’s now determine r:

60 = (20 + 20)/(2/5 + 20/r)

60 = 40/(2r/5r + 100/5r)

60 = 40/[(2r + 100)/5r]

60 = 200r/(2r + 100)

60(2r + 100) = 200r

120r + 6000 = 200r

6000 = 80r

r = 6000/80 = 600/8 = 75

Answer: D
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A driver completed the first 20 miles of a 40-mile trip at an average speed of 50 miles per hour. At what average speed must the driver complete the remaining 20 miles to achieve an average speed of 60 miles per hour for the entire 40-mile trip? (Assume that the driver did not make any stops during the 40-mile trip.)

The total distance is 40 miles, and we want the average speed to be 60 miles per hour. Average speed = (total distance)/(total time) So, we get: 60 = (40 miles)/(total time) Solve equation to get: total time = 2/3 hours So, the TIME for the ENTIRE 40-mile trip needs to be 2/3 hours.

driver completed the first 20 miles of a 40-mile trip at an average speed of 50 miles per hour. How much time was spent on this FIRST PART of the trip? time = distance/speed So, time = 20/50 = 2/5 hours

The ENTIRE trip needs to be 2/3 hours, and the FIRST PART of the trip took 2/5 hours

2/3 hours - 2/5 hours = 10/15 hours - 6/15 hours = 4/15 hours So, the SECOND PART of the trip needs to take 4/15 hours

The SECOND PART of the trip is 20 miles, and the time is 4/15 hours Speed = distance/time So, speed = 20/(4/15) = (20)(15/4) = 75