Three runners A,B and C run a race with A finishing 20 m ahead of B an

(I am assuming it is 34 m.)
I have said this before and I will repeat it again - In races questions, make a diagram.


This is what the race looks like when A reaches the end point. Notice that B has already created a distance of 14 m between himself and C. When B reaches the end point, ie. after another 20 m, distance between him and C is 21 m i.e. another 7 m.

So in every 20 m B is creating a distance of 7 m between himself and C. To create a distance of 21 m between himself and C, he must have run a total of 60m. The race must be of 60 m.
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Just use the formula we just saw :D

\(\frac{L(N-M)}{L-M}=21\)

\(\frac{L(34-20)}{L-20}=21\)

\(\frac{14L}{L-20}=21\)

\(L=60\)

Or:

B runs (L-20)m -> Attains a lead of 34-20=14 over C
1 meter-> 14/(L-20) m lead
L meters-> 14L/(L-20) m lead

We know,
14L/(L-20)=21
L=60

Ans: "C"

I am sorry but i dont seem to get this part

I did it a bit different, more complicated... but this is what I would have done in a real test I guess.

B/C= 20/(34-21) = 20/13
A/B = X/(x-20)
A/C= X/(X-34)

Solving for A and B we get
x^2-60x=0
x(x-60)=0 we know x is not zero so x =60

144144,

Your method is correct too, and I understand when you say that is what you would have done in a real test. However, I would urge you to focus on how you can equate the values you get for B/C through the three equations, leading you to the answer. This might save you some time on the test on similar questions.

Similar question to practice: three-runners-a-b-and-c-run-a-race-with-runner-a-finishing-12m-ahead-120830.html
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let d be the distance covered by C; B covers d+14 and A covers d+34
when B reached the finishing point, C had covered another 13 m so that B beat C by 21m
the ratio of the speeds of B and C is 20/13
Therefore, (d+14)13 = 20d
solving d=26
Hence, length of the race is 26+34=60

For runners B and C the ratios of distance attained at start of same time period to distance attained at end of period are equal.
Let L=length of race.
L-20/L=L-34/L-21
7L=420
L=60 meters

Hi Bunuel,

Can you let me know if this method is viable?

I reverse-plugged answer [C] and assumed runner A finished the race in 1 min at a rate of 60m/min. This implied runner B finished the race finished in 1.5min at 40m/min. And finally, runner C would have travelled 39m in 1.5min. This satisfies each stipulation in the question.

Let me know what you think.

Thanks

Bunuel can you please edit the question?

Thank you!

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Done.
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My solution is based on the fact that the ratio of the speeds of two moving objects is equal to the ratio of the distances covered by each in any given time.

Let 'x' meters be the length of the race-track and V1, V2 and V3 be the speeds of A, B and C respectively.
V1/V2=x/x-20....(i)
V1/V3=x/x-34....(ii)

Dividing (ii) by (i):
V2/V3=(x-20)/(x-34)....(iii)

Since B completes the race (i.e. runs 'x' meters) when C is still 21 meters behind:
V2/V3=x/(x-21)....(iv)

Equating (iii) and (iv):
(x-20)/(x-34)=x/(x-21)....> x^2-41x+420=x^2-21x....> 7x=420...> x=60 meters

can u explain this??



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Since A finishes 20 meters ahead of B, the distance traveled by B in B's final leg = 20 meters.
Since the remaining distance for C decreases from 34 meters to 21 meters, the distance traveled by C during B's final leg = 34-21 = 13 meters.
Thus, the distance ratio for B and C = \(\frac{B's-distance}{C's-distance}\) = \(\frac{20}{13}\) = \(\frac{60}{39}\).
In the green fraction, B's distance - C's distance = 60-39 = 21 meters.
Implication:
For B to finish 21 meters ahead of C, B must travel a TOTAL DISTANCE OF 60 METERS, while C travels only 39 meters.


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1st) A Finishes

B is -20 miles from Finishing Line
C is -34 miles from Finishing Line

B has a +14 mile LEAD > over C


2nd) B Finishes the Last 20 miles of his race

C is -21 miles from Finishing Line

B NOW has a +21 mile LEAD > over C


Concept: in the 20 miles that B Ran ----------- B was able to Increase his LEAD over C by +7 miles

in the entire X Distance Race that B Ran --------- B was able to finish with a LEAD over C of +21 miles


Set up the Proportion and Solve for X Distance = Length of Race

20/7 = X/21

X = (20 * 21) / 7 = 20 * 3 = 60 km

The Length of the Race = 60 km

Answer -C-

2nd Method, using More of a 'Distance Covered in Same Time' Concept:


Let the Length of the Track = X


1st) A Finishes the Race

in the Same Time that it takes A to finish the Race:

Distance A Covers / Distance B Covers = X / (X - 20) = A / B

Distance A Covers / Distance C Covers = X / (X - 34) = A / C

2nd)DIVIDE these 2 Equations:

(A/B) = X / (X - 20)
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(A/C) = X / (X - 34)


TRICK - Do NOT Multiply Large Numbers through. We know the A.C. must be a Whole Number so Keep the Products in "Factor Form" and then Cancel Later

After Dividing the 2 Equations, you end up with:

C / B = (X - 34) / (X - 20) (equation 1)


2nd) B finishes his Last 20 miles and C is -21 miles behind

Distance B can Cover in the SAME TRAVEL TIME as C = 20 Miles
Distance C can Cover in the SAME TRAVEL TIME as B = 13 Miles

C / B = 13 / 20 (equation 2)


Set equation 1 = equation 2

(X - 34) / (X - 20) = 13 / 20

20X - 20 * 34 = 13X - 20 * 13

7X = 20*34 - 20*13 ------ take 20 as Common Factor

7X = 20 * (34 - 13)

7X = 20 * (21) ------Divide Both Sides of Equation by 7

X = 20 * (3) = 60

60 km = Total Length of Track

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