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

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

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New post 19 Sep 2011, 09:27
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Question Stats:

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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?

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

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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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New post 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.

..._________C___(6)___B_______(12)________A (Finishing Point)

..._____________________C_______(8)______B (Finishing Point)

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
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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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New post 19 Sep 2011, 10:34
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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.

as speeds are constant, we have equation,

x-12/ x-18 = 12/10 > x = 48.

Answer B
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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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New post 20 Sep 2011, 03:12
Can someone better explain this?
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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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New post 20 Sep 2011, 03:42
1
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)

And we know the final lead was 8 meters
So,

6L/(L-12)=8
3L/(L-12)=4
3L=4L-48
L=48m

Ans: "B"
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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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New post 13 Oct 2011, 10:53
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)
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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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New post 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
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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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New post 06 Nov 2015, 22:54
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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
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Three runners A, B and C run a race, with runner A finishing 12m ahead  [#permalink]

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New post 23 Dec 2016, 12:43
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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


Although solution by karishma is more logical and faster approach
i providing one conceptual way for solving...

Let speed of A,B,C be A,B &C respectively
& total distance be D

then for the first case when A finishes,, the time run by all three will be same
D/A=(D-12)/B &&
D/A=(D-18)/C
or (D-12)/B=(D-18)/C----------(i)

again for second case when B finishes,, time taken by B & C will be same
thus D/B=(D-8)/C
C={(D-8)*B}/D-------(ii)

putting value of C from (ii) to (i) and solving for D we get
D=48

Ans B
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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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New post 25 May 2018, 05:58
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



So, when A finished the race, B is 12 m away from finish line and C is 18 m away from finish line. Hence, effectively gap between B and C is 6 m. ----lets call this scenario 1
But when B finished the race, C is 8 m away from B (and hence from the finish line). --- say this is scenario 2
This means, when B covered a distance of 12m to finish the line, it increased the gap between B and C by 2m (from 6 m in first scenario to 8 m in second scenario).
So, every 12 meters distance traveled by B increase the gap between B and C by 2 meters. and as from scenario 2, at the end of the race, there is a gap of 8 meters between B and C.

So, in order to create a gap of 8 meters which is 4 times of 2 meters, B has to run a race which is 4 times of 12 meters = 48 meters
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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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Re: Three runners A, B and C run a race, with runner A finishing 12m ahead   [#permalink] 19 Jul 2019, 15:53
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