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Andrew and Stephen drive on the highway in the same direction at respe

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Andrew and Stephen drive on the highway in the same direction at respe  [#permalink]

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New post 22 Apr 2009, 17:49
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Andrew and Stephen drive on the highway in the same direction at respective rates of 72 and 80 kmh. If Stephen is 4 km behind Andrew, by how much does he have to increase his speed to catch up with Andrew in 20 minutes?

A. 1 kmh
B. 2 kmh
C. 3 kmh
D. 4 kmh
E. 5 kmh

M17-37
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Re: Andrew and Stephen drive on the highway in the same direction at respe  [#permalink]

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New post 19 Oct 2014, 02:58
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bigfernhead wrote:
Andrew and Stephen drive on the highway in the same direction at respective rates of 72 and 80 kmh. If Stephen is 4 km behind Andrew, by how much does he have to increase his speed to catch up with Andrew in 20 minutes?

A. 1 kmh
B. 2 kmh
C. 3 kmh
D. 4 kmh
E. 5 kmh

M17-37


Denote the required acceleration as \(x\). The distance between Andrew and Stephen will be decreasing at \((80 + x - 72)\) kmh. We can compose the equation \(\frac{4}{80 + x - 72} = \frac{20}{60}\) from which \(x = 4\).

Answer: D
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Re: Andrew and Stephen drive on the highway in the same direction at respe  [#permalink]

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New post 25 Jul 2009, 13:42
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I think its much easier to first assume that they are going the same speed. So he has to catch up 4 meters in 1/3 hour. r= d/t = 4/(1/3) = 12 km/hr faster he has to go. He is already going 8km/h faster, so 12 - 8 = 4 km/hr
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Re: Andrew and Stephen drive on the highway in the same direction at respe  [#permalink]

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New post 22 Apr 2009, 18:44
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4
A -- 72 Kmh
B -- 80 Kmh

Relative speed = 8 kmh

distance b/w them = 4 Km

4 /( 80 +x - 72) = 20/60

solve for x = 4 .....

Ans . D
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Re: Andrew and Stephen drive on the highway in the same direction at respe  [#permalink]

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New post 02 May 2009, 07:36
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If x is the distance travelled by Andrew in 20 minutes = 72*20/60 = 24km

Stephen's speed would to travel (24+4)km = 28*60/20 = 84 km/h

So, answer is D - 4km/h
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Re: Andrew and Stephen drive on the highway in the same direction at respe  [#permalink]

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New post 02 May 2009, 15:25
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relative speed=80-72=8
need to catch up 4km

currently Stephen is catching up at a speed of 4km/30mins, or 2km/15mins
he needs to increase the speed, so that he can make 4km in 20 mins instead, or 1km/5mins

x - the relative speed increase we seek
2/15+x=1/5
x=1/15

So, he needs to increase relative speed by 1km per 15mins, or 4km/h
Therefore, the answer is D.
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Re: Andrew and Stephen drive on the highway in the same direction at respe  [#permalink]

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New post 12 Nov 2015, 20:04
4k/8kph=1/2h
4k/(8+x)kph=1/3h
x=4kph faster
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Re: Andrew and Stephen drive on the highway in the same direction at respe  [#permalink]

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New post 03 Feb 2017, 12:47
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bigfernhead wrote:
Andrew and Stephen drive on the highway in the same direction at respective rates of 72 and 80 kmh. If Stephen is 4 km behind Andrew, by how much does he have to increase his speed to catch up with Andrew in 20 minutes?

A. 1 kmh
B. 2 kmh
C. 3 kmh
D. 4 kmh
E. 5 kmh



We can classify this problem as a “catch-up” rate problem, for which we use the formula:

distance of Andrew + 4 = distance of Stephen

We are given that Andrew is driving at a rate of 72 kmh and that Stephen is driving at a rate of 80 kmh. We need to determine how much faster Stephen needs to drive to catch up to Andrew in 20 minutes or 20/60 = 1/3 hour.

If we let r = Stephen’s speed increase, we can say the following:

distance of Stephen = (r + 80)(1/3) = (r + 80)/3

distance of Andrew + 4 = 72 x 1/3 + 4 = 28

We can equate the two distances and determine r.

(r + 80)/3 = 28

r + 80 = 84

r = 4

Answer: D
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Re: Andrew and Stephen drive on the highway in the same direction at respe  [#permalink]

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New post 03 Nov 2017, 11:17
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bigfernhead wrote:
Andrew and Stephen drive on the highway in the same direction at respective rates of 72 and 80 kmh. If Stephen is 4 km behind Andrew, by how much does he have to increase his speed to catch up with Andrew in 20 minutes?

A. 1 kmh
B. 2 kmh
C. 3 kmh
D. 4 kmh
E. 5 kmh

M17-37


given
4/8 = 1/2

to make it 1/3, we need to change 8 to 12
so the difference is 4

thanks
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Re: Andrew and Stephen drive on the highway in the same direction at respe  [#permalink]

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New post 04 Nov 2017, 21:35
Relative speed = 8 kmh

distance b/w them = 4 Km

4 /( 80 +x - 72) = 20/60

Answer D
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Re: Andrew and Stephen drive on the highway in the same direction at respe  [#permalink]

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New post 03 Dec 2018, 03:41
at current speeds,

current time to catch up= 4/8 = 1/2; where 8 is the relative speed and 1/2 means 30 minutes
we want: 4/x = 1/3; where 1/3 means 20 minutes
x=12

s - 72 = 12 --> s=84 --> inc by 4
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Re: Andrew and Stephen drive on the highway in the same direction at respe   [#permalink] 03 Dec 2018, 03:41
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