The figure below represents a track with identical semi-circular ends

Question

The figure below represents a track with identical semi-circular ends used for a 4-lap relay race involving two 4-person teams (where each team member runs one complete lap around the track). The table below shows the lap times for each runner on Team A and Team B. Assuming that each runner runs at a constant rate, Team A win the race by how many meters?

A. 40 meters

B. \((40 + 10\pi)\) meters

C. \((40 + 20\pi)\) meters

D. \((20 + 10\pi)\) meters

E. \((20 + 20\pi)\) meters

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Official Answer

B

PareshGmat · Accepted Answer

Perimeter = 120 + 120 + 30\pi + 30\pi = 240 + 60\pi Refer the modified diagram below: Untitled.png At 178 seconds, when Team A completed the race, last runner of Team B had 42-35 = 7 seconds to complete the race. We require to find the distance covered in last 7 seconds Distance covered in 42 seconds is 240 + 60\pi , so distance covered in 7 seconds = \frac{7}{42} * (240 + 60\pi) = 40 + 10\pi Answer = B

manpreetsingh86 · Answer

radius of the curved part (semi-circle) = 1/2(60) =30
thus perimeter of the track (120*2) + 2pi(30) = 240 + 60pi

now as per the chart, team A beats team b by 7 sec. obviously, the final lap is run by the last runner of team B which takes 42 seconds to run the complete track.

during these 7 secs, this last runner would have covered 7/42 = 1/6 of the lap.

thus Team A defeated team B by 1/6(240 + 60pi) = 40 + 10 pi

hence B

deya · Answer

Hi,

I am unable to understand the solution of this question. Diameter of the semicircle is not given. However, diameter = 60 is being assumed by all solution providers. Kindly explain it in details. I will be grateful if anyone looks into this.

PareshGmat · Answer

Split the figure in 2 semicircles & 1 rectangle. You get diameter = 60 It has been calculated; not assumed

batyroff · Answer

Why Distance covered in 42 seconds is 240+60π? I think It should be 1/4 of overall distance (240 + 60п)/4

ankitsood5 · Answer

The answer has to be a factor of the perimeter i.e. 240+60pi.
Now 240+60pi = 6(40+10pi). 
Only B satisfies this condition. No other calculation required.

amanvermagmat · Answer

Hi

It is mentioned that its a 4-lap relay race.. which means  every team member would run the entire lap  (which is calculated already by many as 240+60п).

The table shows  lap times for each runner , which means the times depicted are the times taken by each runner to run the complete lap, individually.

hope it helps.

AR15J · Answer

Can anyone please explain why the 4th member alone of team B is considered for running in delayed time?

By the question, I assumed/understood that all people of team B are running at the same time in competition with team A people. Then, why only 4th member race time is considered?

prabsahi · Answer

Please explain why it has to be a factor.

smratikasharma · Answer

Why is the distance to be covered in last 7 seconds found out by using 42 sec (time of team B player) and not by 41sec?

prabsahi · Answer

Team A wins the race (That is after 178th second whatever distance is left will be the,distance left by Team B to cover the distance.So,this lag is from Team B which has to make up.

anant327 · Answer

I did it by a shortcut. So I know that the total distance for one round is (pi + 4) * 60. I just searched in the answer choices the ratio of the constant of pi and the constant in the parenthesis should be 1:4 respectively. Only B matches the answer.

ankitgarg500 · Answer

/185

can anyone plz explain. why the solution cannot be above.

DhanyaAbhirami · Answer

In this example, when we go by average speed method also, we can get a value close to the answer.
Average Speed of Team B = 4(240+60pi)/185
Team A beat team B by 7 sec.
Time team B takes to cover 7 sec:
t =  /185 =  /185
185/28 = 6.6
t ~ (240+60pi)/6.6 ~ (240+60pi)/6 = 40+10pi

Average doesn't tell us about outliers. So we need to use individual speeds when speed is not uniform.
Instead of providing team member's speed in a relay race, this question could be asked by providing the variable speeds of runner A and runner B.

CAUtkarshAgarwal · Answer

I also have the same question

Krunaal · Answer

It's a relay race, runners take turns. All the people are not running together, after a runner completes 1 lap the next runner from the team runs. At any given time, only two runners are running on the track - one from Team A and one from Team B.

Team A and Team B have completed  3*(240+60\pi)  of distance at 137 seconds and 143 seconds, respectively.

Now the last member of Team A took 41 seconds - they completed race at 178 seconds, at this time Team B's last member has just ran for 35 seconds, now if we find how much distance he has covered in this 35 seconds we subtract that from  240+60\pi  and get our answer.

ADisHere · Answer

Heyy
Why is my approach incorrect

Team A and Team b Completed their race in 178 and 185 secs respectively.
So distance covered by A in 178 secs - distance covered by B 178 secs = is what is asked

distance covered by A

Total Distance 4 *( 60pi + 240)

distance covered by B

4 *( 60pi + 240)  
 185 secs 
 
  ? 
 178 secs

Hence cross multiplying we don't get clean integers

4 *( 60pi + 240)*178 / 185

Krunaal · Answer

Total time and distance won't be proportional since each runner has a different constant speed, one lap is  240+60\pi , you can see every runner takes different time to complete a lap. Hence, cross multiplying to find distance covered by B won't work.

To find the distance covered by B, you will have take last runners time and speed into account, when last runner from team A completes  240+60\pi , the last runner from team B has ran for 35 seconds, his speed is  \frac{240+60\pi}{42} . Distance covered by B in 35 seconds =  35*\frac{240+60\pi}{42} . So he has covered 5/6th of  240+60\pi , Team A beat team B by 1/6th of  240+60\pi  =  40+10\pi

Hope it helps.

Depuhan · Answer

This question can be considered as a geometry question right? So is it safe to say that we shouldn't expect this question in Gmat focus?

The figure below represents a track with identical semi-circular ends

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The figure below represents a track with identical semi-circular ends

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