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# In a race of 4800 meters run on a circular track of 400 meters length

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Math Expert
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In a race of 4800 meters run on a circular track of 400 meters length  [#permalink]

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06 Jan 2020, 07:07
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45% (medium)

Question Stats:

43% (01:45) correct 57% (02:12) wrong based on 21 sessions

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In a race of 4800 meters run on a circular track of 400 meters length, the ratio of the speed of the two athletes is 3 : 5. If they run in the same direction, how many times do they meet in the entire race?

A. 4
B. 5
C. 6
D. 7
E. 8

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Re: In a race of 4800 meters run on a circular track of 400 meters length  [#permalink]

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07 Jan 2020, 02:54
Bunuel wrote:
In a race of 4800 meters run on a circular track of 400 meters length, the ratio of the speed of the two athletes is 3 : 5. If they run in the same direction, how many times do they meet in the entire race?

A. 4
B. 5
C. 6
D. 7
E. 8

Explanation:

Let the slower athlete speed be 3m/min, the faster = 5m/min

The faster athlete overtakes the slower for first time after 400/2 = 200min i.e. when it has covered 1000 meters. After every 200 minutes Faster athlete will overtake slower athlete

Now the total time taken by Faster athlete to cover entire race = 4800/5 = 960 mins

Hence total no of times he will overtake = 960/200 = 4.8 i.e 4 times

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Re: In a race of 4800 meters run on a circular track of 400 meters length  [#permalink]

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07 Jan 2020, 14:42
These questions are a little confusing because i'm not sure what points are supposed to be considered significant.
Does the very start count as a data point? What about the end? Since one guy finishes first, does he "meet" with the other guy after every lap thereafter?

The answer is either 4 (the number of times the faster runner laps the slower runner during the race)
1000M:600M
2000M:1200M
3000M:1800M
4000M:2400M
Or every 2.5 laps:1.5 laps

5 times if you count just the start as significant. 6 if you count just the final end as significant.
I believe they meet 10 times if you count every lap after the first runner finishes as significant, and that's not an answer choice.
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Re: In a race of 4800 meters run on a circular track of 400 meters length  [#permalink]

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07 Jan 2020, 22:07
Bunuel wrote:
In a race of 4800 meters run on a circular track of 400 meters length, the ratio of the speed of the two athletes is 3 : 5. If they run in the same direction, how many times do they meet in the entire race?

A. 4
B. 5
C. 6
D. 7
E. 8

Let's start by calculating the time through relative speed formula : Relative speed = Relative distance / time
So, here the relative speed is (2) {the difference between ratio of the speed of the two athletes}
Thus, 2=400/T
T=400/2 = 200 mins

Now, we calculate how much distance will the faster athlete cover in 200 mins => 200 * 5 = 1000 mtrs.

since, it will take 1000 mtrs for the faster athlete to cross the other one in a 4800 mtr race. It becomes quite clear that there will be total of 4 points each after the faster athelete run 1000 mtrs.

The correct ans is option A.
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Re: In a race of 4800 meters run on a circular track of 400 meters length  [#permalink]

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08 Jan 2020, 01:37
$$Formula: S = \frac{D }{ T}$$

$$\frac{S1}{S2}=\frac{3}{5}$$ --> S1 = 3k and S2 = 5k (k >0)

D2 = D1 + n x 400 (n>0) Since S2 > S1, there is no way that they can meet within the 1st lap.

When the 2 will meet, the time will be the same --> T1 = T2 --> $$\frac{D1 }{ S1} = \frac{D2 }{ S2}$$ or D1 x S2 = D2 x S1

D1 x 5k = (D1 + 400n) x 3k
5 x D1 = 3 x D1 + 1200n
2 x D1 = 1200n

D1=600n
D2=1000n

for n=5, D2=5000 > 4800. So 0 < n < 5 --> They meet 4 times

Hope it helps!
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Re: In a race of 4800 meters run on a circular track of 400 meters length  [#permalink]

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09 Jan 2020, 11:15
Bunuel wrote:
In a race of 4800 meters run on a circular track of 400 meters length, the ratio of the speed of the two athletes is 3 : 5. If they run in the same direction, how many times do they meet in the entire race?

A. 4
B. 5
C. 6
D. 7
E. 8

We can let the speed of the faster person = 10 and that of the slower person = 6. Notice that the faster person will finish the race in 4800/10 = 480 seconds.

The faster person must overtake the slower person when he runs exactly 1 lap, 2 laps, etc. more than the slower person. Let t be the time it takes when the faster person overtake the slower person We can create the equations:

10t = 6t + 400 (for 1 lap), 10t = 6t + 800 (for 2 laps), etc.

Notice that none of the values of t can be more than 480 seconds, the time when the faster person finishes his race. Solving the first equation, we have:

4t = 400

t = 100

That is, it takes 100 seconds for the faster person to overtake the slower person by 1 lap.

Solving the second equation, we have:

4t = 800

t = 200

That is, it takes 200 seconds for the faster person to overtake the slower person by 2 laps.

We can see that the values of t are increasing by 100, so it will take 300 seconds for the faster person to overtake the slower person by 3 laps, 400 seconds by 4 laps. Since the next one will be 500 seconds and recall that t can’t be more than 480, then there are only 4 times when the faster person overtakes the slower person.

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Re: In a race of 4800 meters run on a circular track of 400 meters length  [#permalink]

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10 Jan 2020, 03:03
Bunuel wrote:
In a race of 4800 meters run on a circular track of 400 meters length, the ratio of the speed of the two athletes is 3 : 5. If they run in the same direction, how many times do they meet in the entire race?

A. 4
B. 5
C. 6
D. 7
E. 8

TWIN QUESTION: https://gmatclub.com/forum/in-a-race-of ... 14091.html
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Re: In a race of 4800 meters run on a circular track of 400 meters length   [#permalink] 10 Jan 2020, 03:03
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