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A’s speed is 20/17 times that of B. If A and B run a race, what part

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A’s speed is 20/17 times that of B. If A and B run a race, what part  [#permalink]

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New post 08 Apr 2015, 04:07
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A’s speed is 20/17 times that of B. If A and B run a race, what part of the length of the race should A give B as a head start, so that the race ends in a dead heat?

(A) 1/17
(B) 3/17
(C) 1/10
(D) 3/20
(E) 3/10


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Re: A’s speed is 20/17 times that of B. If A and B run a race, what part  [#permalink]

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New post 08 Apr 2015, 09:18
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Bunuel wrote:
A’s speed is 20/17 times that of B. If A and B run a race, what part of the length of the race should A give B as a head start, so that the race ends in a dead heat?

(A) 1/17
(B) 3/17
(C) 1/10
(D) 3/20
(E) 3/10


Kudos for a correct solution.


Picking numbers approach.
From ratio of speed 20/17 we can conclude that when B run 17 miles, A run 20 miles
LCM (17, 20) = 340. Let's imagine that our distance equal to 340 miles

B run that distance by 340/17 = 20 hours
A run this distance by 340/20 = 17 hours

So A should wait 3 hours before start his running (it will be equal to giving head start)
B will run 3 * 17 = 51 miles by this 3 hours

So A should give head start equal to 51 miles
51/340 = 3/20

So answer is D
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A’s speed is 20/17 times that of B. If A and B run a race, what part  [#permalink]

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New post 08 Apr 2015, 05:08
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V = speed of B, 20/17*V = speed of A, X = B's headstart, S = track's length for A, S - X - track lengths for B. Total time they take to complete race is the same so:
\(\frac{17*S}{(20*V)} = \frac{(S - X)}{V}\)
\(X = 3/20*S\) which matches option D
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Re: A’s speed is 20/17 times that of B. If A and B run a race, what part  [#permalink]

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New post 08 Apr 2015, 09:40
Bunuel wrote:
A’s speed is 20/17 times that of B. If A and B run a race, what part of the length of the race should A give B as a head start, so that the race ends in a dead heat?

(A) 1/17
(B) 3/17
(C) 1/10
(D) 3/20
(E) 3/10


Kudos for a correct solution.



A will run 20*K units when B will run 17*K units .
20K - 17K = 3K units

3/20 answer
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Re: A’s speed is 20/17 times that of B. If A and B run a race, what part  [#permalink]

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New post 08 Apr 2015, 10:04
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A and B will be running 20 km.

A can run 20 km/hour and B can run 17 km/hour. If B is allowed to start 3 km ahead, he will have traveled the same distance as A(20km). As such, 3/20 is the part of the length that B has to start at to make it a dead race.

D
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Re: A’s speed is 20/17 times that of B. If A and B run a race, what part  [#permalink]

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New post 08 Apr 2015, 14:53
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Assume speed of B=x, so speed of A=20x/17; time taken by both A & B will be equal;
Speed =distance /time; we can write, Rate=distance/time
Equating both time A & B we get: LA*17/20X=LB/X; LA(length of A); LB(Length of B)
Therefore LB=17LA/20 => we need to calculate LA=LB= LA*(3/20 )

Hence answer is D

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A’s speed is 20/17 times that of B. If A and B run a race, what part  [#permalink]

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New post 09 Apr 2015, 07:31
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given :

Sa/Sb=20/17.

Ergo, Ta/Tb=17/20, Da/Db=20/17

which implies when A covers 100 , B would be at 85. Therefore, B needs to be given a head start of 15. ( so that both finish at 100 at the same time [ dead heat ].

Which is 3/20 of 100.
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Re: A’s speed is 20/17 times that of B. If A and B run a race, what part  [#permalink]

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New post 13 Apr 2015, 05:23
1
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Bunuel wrote:
A’s speed is 20/17 times that of B. If A and B run a race, what part of the length of the race should A give B as a head start, so that the race ends in a dead heat?

(A) 1/17
(B) 3/17
(C) 1/10
(D) 3/20
(E) 3/10


Kudos for a correct solution.


VERITAS PREP OFFICIAL SOLUTION:

We have the ratio of A’s speed and B’s speed. This means, we know how much distance A covers compared with B in the same time.

This is what the beginning of the race will look like:
(Start) A_________B______________________________

If A covers 20 meters, B covers 17 meters in that time. So if the race is 20 meters long, when A reaches the finish line, B would be 3 meters behind him. If we want the race to end in a dead heat, we want B to be at the finish line too at the same time. This means B should get a head start of 3 meters so that he doesn’t need to cover that. In that case, the time required by A (to cover 20 meters) would be the same as the time required by B (to cover 17 meters) to reach the finish line.

So B should get a head start of 3/20th of the race.

Answer (D)
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Re: A’s speed is 20/17 times that of B. If A and B run a race, what part  [#permalink]

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New post 22 May 2015, 07:32
Given: A's speed = (20/17)B

Ignoring distance for now, in order for A and B to finish at the same time, A's speed would need to be reduced by (17/20) so the fractions cancel out and A's speed = B's speed.

Now, s2 (adjusted speed) for A will be 17/20 out of 1 (s1 or original speed), so the ratio for s1/s2 = 1/(17/20) = 20/17

Since (s2/s1) = (t2/t1) = 20/17, we can see that 3/17 (or 1 - 20/17) additional time should be given to B for a head start to allow A and B to finish together.

So we have the additional time (3/17) and the adjusted speed for A (17/20). Using the d = s*t formula, we multiply them together and the 17 cancels out, leaving 3/20 (the answer choice).
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Re: A’s speed is 20/17 times that of B. If A and B run a race, what part  [#permalink]

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New post 19 Aug 2015, 03:52
let b speed = 17
a speed = 20

let the race be of 20 * 17 = 340 m long

time taken by A = 340/20 = 17 sec
time taken by B= 340/17= 20 sec

A defeats B by 3 sec

so B was short of 3*17 = 51 m

so A defeats B by 51m

give a head start of 51 m to B , he needs to cover 340-51=289 m which he covers in 17 seconds and hence we will get a tie ;

So answer is D (51/340 or 3/20)
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Re: A’s speed is 20/17 times that of B. If A and B run a race, what part  [#permalink]

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New post 26 Jul 2017, 02:28
Let's denote the speed of B as 1
Then the speed of A is \(\frac{20}{17}\)

Since the time is the same we can proceed with the following equation:

\(\frac{20}{17}(1-y)=1*1\) (where 1 is the full distance and 1-y is the abridged distance).

\(\frac{20}{17}-\frac{20}{17}y =1\)

\(\frac{20}{17}y=\frac{3}{17}\)

\(y=\frac{3}{20}\)

Answer D
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Re: A’s speed is 20/17 times that of B. If A and B run a race, what part  [#permalink]

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New post 27 Sep 2018, 16:38
Bunuel wrote:
A’s speed is 20/17 times that of B. If A and B run a race, what part of the length of the race should A give B as a head start, so that the race ends in a dead heat?

(A) 1/17
(B) 3/17
(C) 1/10
(D) 3/20
(E) 3/10


We can let the race be 340 ft long and B runs 17 ft/s. So A runs 20 ft/s. So it will take A 17 seconds to run the race. However, in 17 seconds, B only runs 17 x 17 = 289 ft. So A has to give B a 340 - 289 = 51 ft head start, which is 51/340 = 3/20 of the length of the race.

Answer: D
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A’s speed is 20/17 times that of B. If A and B run a race, what part  [#permalink]

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New post 03 Oct 2018, 01:25
Let the total distance = 1

Since A=\(\frac{20}{17}\)B, B=\(\frac{17}{20}\) A.
A & B both run the same total distance, so we can use the product of \(17*20= 340\) of A & B to represent the total distance: 1=\(\frac{340}{340}\)

We know that B covers \(289/340\) of the distance A runs, when running the total distance. Thus, A should give B a head start of \(\frac{340}{340}\) - \(\frac{289}{340}\) = \(\frac{51}{340}\)= \(\frac{3}{20}\).

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