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Three runners A, B and C run a race, with runner A finishing 12m ahead [#permalink]

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19 Sep 2011, 09:27

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Three runners A, B and C run a race, with runner A finishing 12m ahead of runner B and 18m ahead of runner C, while runner B finishes 8m ahead of runner C. Each runner travels entire distance at a constant speed. What was the length of the race?

Re: Three runners A, B and C run a race, with runner A finishing 12m ahead [#permalink]

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19 Sep 2011, 10:34

Let distance of race be x mtrs. Then when A finishes x m , B has run (x- 12)mtrs and C has run x-18 mtrs. so at this point B is 6 m ahead of C. Now to finish race b needs to run another 12 m, so he runs another 12 m. when B finishes race he is 8 m ahead of C. so last 12 m B has run, C has run 10 m.

Re: Three runners A, B and C run a race, with runner A finishing 12m ahead [#permalink]

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20 Sep 2011, 03:34

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jamifahad wrote:

Three runners A,B and C run a race, with runner A finishing 12m ahead of runner B and 18m ahead of runner C, while runner B finishes 8m ahead of runner C. Each runner travels entire distance at a constant speed. What was the length of the race?

A. 36m B. 48m C. 60m D. 72m E. 84m

See if you can logically deduce the answer: This is what happens in the race.

Focus on B and C. B and C had a distance of 6 m between them. When B finished the race by covering another 12 m, he created a gap of 8 m between them i.e. he created another 2 m gap while running 12 m. Since he created a gap of 2 m by running 12 m, he must have created a gap of 8 m by running a total of 12*4 = 48 m _________________

Re: Three runners A, B and C run a race, with runner A finishing 12m ahead [#permalink]

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20 Sep 2011, 03:42

jamifahad wrote:

Three runners A,B and C run a race, with runner A finishing 12m ahead of runner B and 18m ahead of runner C, while runner B finishes 8m ahead of runner C. Each runner travels entire distance at a constant speed. What was the length of the race?

A. 36m B. 48m C. 60m D. 72m E. 84m

|---------------------------L-------------------|L=Length of the track |-----------------------------C------B----------|A |------------------------------------|<--12---->| A finishes 12 m ahead of B |-----------------------------|<------18------->| A finishes 18 m ahead of C |-----------------------------|<-6->|---------->| So, B was 6 meters ahead of C when A finished the race ------------1

|----------------------------------------C------|B |----------------------------------------|<-8-->| B finishes 8 meter ahead of C--------------2

From 1 and 2, we can deduce that B ran a distance of "L-12" meters taking a lead of 6 meters on C.

L-12->6 1 meter-> 6/(L-12) {:This is the per meter lead by B on C}

For the entire race of L meters, the lead would be L meters->L*6/(L-12)

Re: Three runners A, B and C run a race, with runner A finishing 12m ahead [#permalink]

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14 Oct 2011, 07:09

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shankar245 wrote:

Karishma,

can you pls xplain this part

Quote:

Since he created a gap of 2 m by running 12 m, he must have created a gap of 8 m by running a total of 12*4 = 48 m

Im nt able to logically deduce how we got the length of the race?

c ----b -----A(finishes)

Think of the logic this way:

There are 2 people P and Q. They both start running simultaneously at different but uniform speeds. In the time P has covered 2 m, Q has covered only 1 m. What will be the distance between them by the time P covers a total of 4 m? P and Q are running at uniform speeds. Since P created a gap of 1 m in every 2 m he ran, the gap between them must be 2 m now.

Now imagine that P is standing ahead of Q by 6 m. They both start running. What happens by the time P covers 2 m? What is the gap between them? It must be 7 m now since another 1 m gap would have got created.

The logic used here is exactly the same.

B and C had a distance of 6 m between them at one point in time. B covered another 12 m and now the distance between them is 8 m. So basically a gap of 2 m was created by B when he ran 12 m. Overall, at the end of the race, the gap between B and C is 8 m. How much must B have run to create a gap of 8 m ( = 4*2)? He must have run 4*12 = 48 m _________________

Re: Three runners A, B and C run a race, with runner A finishing 12m ahead [#permalink]

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31 Oct 2014, 03:02

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Re: Three runners A, B and C run a race, with runner A finishing 12m ahead [#permalink]

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06 Nov 2015, 22:54

Let d=distance of race. Between A's finish and B's finish, B runs 12 meters while C runs 10 meters. B's rate to C's rate=6/5. When B finishes, he is 8 meters ahead of C. 6/5=d/(d-8) d=48 meters

gmatclubot

Re: Three runners A, B and C run a race, with runner A finishing 12m ahead
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06 Nov 2015, 22:54

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