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Three runners A, B and C run a race, with runner A finishing 12m ahead
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19 Sep 2011, 09:27
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62% (02:32) correct 38% (02:34) wrong based on 163 sessions
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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
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20 Sep 2011, 03:34
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
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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 x18 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,
x12/ x18 = 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
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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
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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 LL=Length of the track CBA <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 CB <8> B finishes 8 meter ahead of C2 From 1 and 2, we can deduce that B ran a distance of "L12" meters taking a lead of 6 meters on C. L12>6 1 meter> 6/(L12) {: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/(L12) And we know the final lead was 8 meters So, 6L/(L12)=8 3L/(L12)=4 3L=4L48 L=48m Ans: "B"
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Re: Three runners A, B and C run a race, with runner A finishing 12m ahead
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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
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14 Oct 2011, 07:09
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
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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/(d8) d=48 meters



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Three runners A, B and C run a race, with runner A finishing 12m ahead
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23 Dec 2016, 12:43
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=(D12)/B && D/A=(D18)/C or (D12)/B=(D18)/C(i) again for second case when B finishes,, time taken by B & C will be same thus D/B=(D8)/C C={(D8)*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
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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
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19 Jul 2019, 15:53
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Re: Three runners A, B and C run a race, with runner A finishing 12m ahead
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