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A driver completed the first 20 miles of a 40-mile trip at a

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

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09 Mar 2014, 15:05
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The Official Guide For GMAT® Quantitative Review, 2ND Edition

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.)

(A) 65 mph
(B) 68 mph
(C) 70 mph
(D) 75 mph
(E) 80 mph

Problem Solving
Question: 142
Category: Algebra Applied problems
Page: 80
Difficulty: 650

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

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09 Mar 2014, 15:05
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SOLUTION

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.)

(A) 65 mph
(B) 68 mph
(C) 70 mph
(D) 75 mph
(E) 80 mph

$$average \ speed=\frac{total \ distance}{total \ time}=\frac{40}{total \ time}=60$$. 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.

Answer: 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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09 Mar 2014, 19:32
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Answer = (D) 75 mph

Time = Distance / Speed

Setting up equation as shown in fig
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Re: A driver completed the first 20 miles of a 40-mile trip at a  [#permalink]

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09 Mar 2014, 18:40
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Speed = Distance/ Time

Time elapsed for 20Miles = (20/50) *60 = 24 Min
Remaining distance = 20 Miles

Remaining time:
@ 60 Miles/ hr, 40 Miles would take - 40 Minutes

So remaining time = 40-24 = 16 Mins

Speed required to cover 20 miles in 16 mins = (20/16)*60 = 75 Miles/hr
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Re: A driver completed the first 20 miles of a 40-mile trip at a  [#permalink]

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09 Mar 2014, 22:15
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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
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Re: A driver completed the first 20 miles of a 40-mile trip at a  [#permalink]

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09 Mar 2014, 23:49
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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
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Re: A driver completed the first 20 miles of a 40-mile trip at a  [#permalink]

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11 Mar 2014, 08:12
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) ?
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Re: A driver completed the first 20 miles of a 40-mile trip at a  [#permalink]

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11 Mar 2014, 10:15
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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
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A driver completed the first 20 miles of a 40-mile trip at a  [#permalink]

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18 Aug 2015, 03:22
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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A driver completed the first 20 miles of a 40-mile trip at a  [#permalink]

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22 Nov 2015, 16:49
My way was the same as everyone else's until the end. Still got the right answer though.

(Total Distance 40 / Total time ) = 60
So 40/60th of an hour means 40 minutes.

20/50 = 24 minutes

40-24=16

24:16 is 3:2 so 2x=50 so 3x=75
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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, 13:48
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.)

(A) 65 mph
(B) 68 mph
(C) 70 mph
(D) 75 mph
(E) 80 mph

r=speed needed to average 60 mph for trip
total distance=40 miles
total time=2/5+20/r
40/(2/5+20r)=60
r=75 mph
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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11 Feb 2017, 02:00
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
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Re: A driver completed the first 20 miles of a 40-mile trip at a  [#permalink]

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17 Sep 2017, 05:31
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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.
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Re: A driver completed the first 20 miles of a 40-mile trip at a  [#permalink]

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09 Oct 2017, 09:57
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

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

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12 Oct 2017, 18:02
Bunuel wrote:
The Official Guide For GMAT® Quantitative Review, 2ND Edition

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.)

(A) 65 mph
(B) 68 mph
(C) 70 mph
(D) 75 mph
(E) 80 mph

We can use the following formula:

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

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10 Dec 2017, 10:52
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Bunuel wrote:
The Official Guide For GMAT® Quantitative Review, 2ND Edition

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.)

(A) 65 mph
(B) 68 mph
(C) 70 mph
(D) 75 mph
(E) 80 mph

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

Answer: D

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

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15 Dec 2017, 09:47
$$rate * time = distance$$

$$50 * \frac{2}{5} = 20$$

$$x * \frac{20}{x} =20$$

$$total distance=20 +20$$

$$total time = 2/5+20/x$$

$$\frac{(20+20)}{(2/5+20/x)}=60$$

solving for x gives 75
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Re: A driver completed the first 20 miles of a 40-mile trip at a  [#permalink]

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12 Apr 2018, 11:33
VeritasPrepKarishma
Hi. I'm sorry for bothering you again. I just wanted to know how the scale method can be used for this specific problem? I'd be grateful.
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A driver completed the first 20 miles of a 40-mile trip at a  [#permalink]

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12 Apr 2018, 11:41
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VeritasPrepKarishma

I figured out. Sorry!

It's like:

t1/t2=(x-60)/(60-50)

Where t1= 2/5 & t2= 20/x

The equation looks something like this after solving a bit
2x/5*20=(x-60)/10
x=5(x-60)
X=5x-300
4x=300
x= 75

Option 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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12 Apr 2018, 21:18
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asfandabid wrote:
VeritasPrepKarishma

I figured out. Sorry!

It's like:

t1/t2=(x-60)/(60-50)

Where t1= 2/5 & t2= 20/x

The equation looks something like this after solving a bit
2x/5*20=(x-60)/10
x=5(x-60)
X=5x-300
4x=300
x= 75

Option D

Though you have applied the scale method correctly, I would like to point out that the best method to use here is the formula 2ab/(a+b)
When one drives at speed a for half the distance and at speed b for the other half of the distance, the average speed is given by 2ab/(a+b)

60 = 2*50*b/(50 + b)
b = 3000/40 = 75

Check this post for these formulas: https://www.veritasprep.com/blog/2015/0 ... -the-gmat/
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