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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 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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New post 22 Apr 2009, 17:44
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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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New post 02 May 2009, 06:36
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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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New post 02 May 2009, 14:25
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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 25 Jul 2009, 12:42
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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 25 Jul 2009, 12:53
Hey guys, look the solution of benpack03. IT IS VERY FAST.
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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, 13:40
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benpack03 wrote:
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


Yes! That's what I also came up with.
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New post 19 Oct 2014, 01:58
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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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Re: Andrew and Stephen drive on the highway in the same direction at respe [#permalink]

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New post 12 Nov 2015, 19:04
4k/8kph=1/2h
4k/(8+x)kph=1/3h
x=4kph faster
Re: Andrew and Stephen drive on the highway in the same direction at respe   [#permalink] 12 Nov 2015, 19:04
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Andrew and Stephen drive on the highway in the same direction at respe

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