Andrew and Stephen are driving on a highway in the same direction at

Question

Andrew and Stephen are driving on a highway in the same direction at speeds of 72 km/h and 80 km/h, respectively. If Stephen is 4 km behind Andrew, by how much must he increase his speed to catch up with Andrew in 20 minutes?

A. 1 km/h
B. 2 km/h
C. 3 km/h
D. 4 km/h
E. 5 km/h

M17-37

Bunuel · Accepted Answer

Andrew and Stephen are driving on a highway in the same direction at speeds of 72 km/h and 80 km/h, respectively. If Stephen is 4 km behind Andrew, by how much must he increase his speed to catch up with Andrew in 20 minutes?

A. 1 km/h
B. 2 km/h
C. 3 km/h
D. 4 km/h
E. 5 km/h

Let's assume Stephen needs to increase his speed by  x  km/h, which means he would drive at  80 + x  km/h. The relative speed between Stephen and Andrew would then be  80 + x - 72  km/h. To close the 4 km gap, the time required would be  \frac{distance}{speed} = \frac{4}{80 + x - 72}  hours. Since this time should be 1/3 of an hour (or 20 minutes), we have the equation  \frac{4}{80 + x - 72} = \frac{1}{3} . Solving for  x , we find  x = 4 .

Answer: D.

BTheBestU · Answer

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

tkarthi4u · Answer

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

peraspera · Answer

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.

benpack03 · Answer

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

ScottTargetTestPrep · Answer

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

KarishmaB · Answer

This is simply a relative speed problem. 4 km needs to be covered within 20 mins i.e. 1/3rd of an hour. So relative speed required = \frac{4}{1/3} = 12 km/hr Andrew's speed is 72 km/hr so Stephen speed should be 12 km/hr more than this i.e. Stephen's speed should be 84 km/hr. Stephen's current speed is 80 km/hr so he should increase his speed by 4 km/hr Answer (D) Check these videos for relative speed: https://youtu.be/EDsHa_CULmg https://youtu.be/wrYxeZ2WsEM

kanikaa9 · Answer

We know meeting time is delta distance/delta speed

In this case delta distance = 4 KM
delta speed we have to find
Meeting time is 20 mins or 1/3 hr (we are given)

Solving we get delta speed as 12, so the difference between slower and faster should be 12 kmph, so faster should be 72+12 = 84 kmph, Hence Stephen should increase his speed by 4 KMPH

rohitz98 · Answer

Current differnce in speed is 8 km/hr 
For catching up 4 km it takes 30 min

For catching up in 20 min the speed should be 4/(1/3) = 12 km/hr
So the speed should be increased by 4 km/hr  (Ans: Option D)

Dereno · Answer

Relative speed = 80-72 = 8 kmph. ( same direction).

Distance between them is 4 km.

To cover 4 km at speed 8 kmph takes 30 mins.

It’s given the time should be 20 mins. Time is reduced by 1/3.

We know time is inversely proportional to speed. Time reduced by 1/3, will result in speed increasing by 1/2.

Relative speed increase by half = 8+4 = 12 kmph.

Stephen’s new speed - 72 = 12 kmph.

Hence, Stephen’s new speed = 72+12 = 84 kmph.

Speed of Stephen old = 80 kmph —————> new Stephen’s speed = 84 kmph.

Hence increase is  4 kmph. Option D

Andrew and Stephen are driving on a highway in the same direction at

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