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Speed and Distance

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Speed and Distance

by winnerhere » Sat Sep 05, 2009 1:23 am
During a marathon an athlete's speed reduces half by the end of every hour.However at the end of every three hours,he is given energy drinks to help him replenish his salts.After this he can run at his peak speed.At his peak speed he can complate the race in 700 minutes but the cycle of the reduction in speed,and the subsequent increase continues.When does he actually complete the race if he starts off at his peak speed.

1.1200 minutes

2.1300 minutes

3.1400 minutes

4.1180 minutes

5.1160 minutes
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by chetanojha » Sat Sep 05, 2009 10:23 am
IMO: E i.e. 1160.

Definetly not ideal for GMAT sample below. Hope somebody will translate into formula too.

Let's say athlete will run 4200 mtrs in 700 minutes. This give rate of 6. This is from where you will start counting as shown in the picture.
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Answer: 1200

by iamaneophyte » Sat Sep 05, 2009 11:07 am
I will surely post a short cut for this , if my answer is correct.

Is it 1200 ?
neophyte
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by PussInBoots » Sat Sep 05, 2009 11:41 am
I got 1,200 as well
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by bharathh » Sat Sep 05, 2009 12:54 pm
1200

Here's my method

I assumed that the runner at peak speed covers 1 unit distance per minute at peak speed.

So in 700 mins he covers 700 units of distance at peak speed.

With the given conditions, he will cover 60 units of distance in the first hour, 30 in the second hour and 15 in the third hour. = 105 units in 3 hrs. Then he gets back to top speed ...

So basically he will complete 105 units of speed every 3 hrs.

He will complete 700 units of distance in ~ (700/105)*3 ~= 7*3 = 21 hrs

21 hrs = 1260 minutes. Closest answer is 1200.
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chetanojha

by winnerhere » Sun Sep 06, 2009 5:18 am
I have followed the same method.But I get the answer 1200

Ive considered the distance as 4200m.So the distance travelled in fir three hours,second three hours etc is 6,3,1.5

So the average speed for every three hours is 10.5/3 = 3.5

Now the (total distance/average speed)=4200/3.5 = 1200

Whats the logical flaw in my method.

OA is 1160

In other logic I took the average distance travelled in three hours is (360+180+90)/3=210 per hour

so 4200/210 = 20 hours = 1200 minutes.

I dont know where i flawed ibn my logic
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by bharathh » Sun Sep 06, 2009 5:53 am
What is the source and the explanation given?
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I made a rough calculation last time, he is my solution

by iamaneophyte » Sun Sep 06, 2009 11:01 am
700 mins -> 11.67 hr.

Lets suppose that the total distance is 11.67 km and there by he travel 1km in 1 hr provided that he is travelling at peak speed.

From the question cycle repeates for 3 hrs and there by he can travel a distance of 1+0.5+0.25 for the three hours

Therefore 35/3 -7/4(x) where 7/4 is the distance for 3 hrs

A multiple of 6 will result in a postive distance which leaves 7/6 km. So he is again at the peak speed and there by covers 1km.

Rest of 1/6 can be coverd with 1/2 of the speed so it amounts to 20 min

Therfore total time (1080+60+20)=1160. Hope this helps for you winnerhere.


(
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by gmatv09 » Sun Sep 06, 2009 2:59 pm
Used the following technique:
Let us say the tot distance = 7000 mts
Peak speed = 7000/700 = 10 mts/min

t=0 speed(s)= 10 mts/min
therefore in first hr dist covered = 600 mts
second hr = 300 mts
third hr = 150 mts

so every 3 hrs the dist = 1050 mts

1050 * 6 = 6300 mts (dist covered in 6 laps of 3 hrs)
Tot time for 6300 mts = 6 * 3 = 18 hrs = 1080 mins
Remaining 700 mts, 80 mins

Tot time = 1160 mins
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by blr_gmat_prep » Thu Sep 10, 2009 2:38 am
The race should be completed in 700 mts if runner operates at full speed.

1st hour = 60 mins (at full Speed)
2nd hour = 30 mins (at full speed since speed is halved)
3rd hour = 15 minutes (since speed is further halved)

The cycle contiues

So in 180 mins he completes (60+30+15) mins of race.
=> time taken to complete 700 mins of race = 700*180/105 = 1200.
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