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

Question

Three runners A,B and C run a race with A finishing 20 m ahead of B and 34 m ahead of C, while B finishes 21 m ahead of C. Each runner travels the entire distance at a constant speed. What was the length of the race.

A) 40 m
B) 50 m
C) 60 m
D) 80 m
E) 90 m

A) 40 m
B) 50 m
C) 60 m
D) 80 m
E) 90 m

KarishmaB · Accepted Answer

(I am assuming it is 34 m.) I have said this before and I will repeat it again - In races questions, make a diagram. Ques.jpg 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. Ques1.jpg 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.

fluke · Answer

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"

DeeptiM · Answer

I am sorry but i dont seem to get this part

144144 · Answer

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

GyanOne · Answer

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.

Bunuel · Answer

Similar question to practice: three-runners-a-b-and-c-run-a-race-with-runner-a-finishing-12m-ahead-120830.html

mathivanan · Answer

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

gracie · Answer

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

keatross · Answer

Hi Bunuel,

Can you let me know if this method is viable?

I reverse-plugged answer   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

TheNightKing · Answer

Bunuel  can you please edit the question?

Thank you!

Bunuel · Answer

________________________
Done.

effatara · Answer

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

subaru2019 · Answer

can u explain this??

Posted from my mobile device

subaru2019 · Answer

Posted from my mobile device

GMATGuruNY · Answer

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.

C

Fdambro294 · Answer

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-

Fdambro294 · Answer

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)
__________________
(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

luisdicampo · Answer

Deconstructing the Question  
Let the length of the race be  D .
The runners travel at constant speeds, so the ratio of distances covered by any two runners is constant.

Step 1: Analyze the situation when A finishes  
When A covers distance  D :
* B is 20m behind, so B covers  D - 20 .
* C is 34m behind, so C covers  D - 34 .

Ratio of speeds of B and C ( S_B : S_C ):
 \frac{S_B}{S_C} = \frac{D - 20}{D - 34}  ... (Equation 1)

Step 2: Analyze the situation when B finishes  
When B covers distance  D :
* C is 21m behind, so C covers  D - 21 .

Ratio of speeds of B and C ( S_B : S_C ):
 \frac{S_B}{S_C} = \frac{D}{D - 21}  ... (Equation 2)

Step 3: Solve for D  
Equate the two ratios:
 \frac{D - 20}{D - 34} = \frac{D}{D - 21}

Cross-multiply:
 (D - 20)(D - 21) = D(D - 34) 
 D^2 - 21D - 20D + 420 = D^2 - 34D 
 D^2 - 41D + 420 = D^2 - 34D

Subtract  D^2  from both sides:
 -41D + 420 = -34D 
 420 = 41D - 34D 
 420 = 7D 
 D = 60

The length of the race is 60 meters.

Answer:  C

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

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