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A runs 25% faster than B and is able to allow B a lead of 7 meters to

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A runs 25% faster than B and is able to allow B a lead of 7 meters to  [#permalink]

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New post 14 Mar 2016, 07:23
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Re: A runs 25% faster than B and is able to allow B a lead of 7 meters to  [#permalink]

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New post 14 Mar 2016, 21:48
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Bunuel wrote:
A runs 25% faster than B and is able to allow B a lead of 7 meters to end a race in dead heat. What is the length of the race?

A. 10 meters
B. 15 meters
C. 25 meters
D. 35 meters
E. 45 meters

The ratio their speeds is 5:4; therefore the ratio of distance covered by them will also be 5:4
The difference in the ratio of distance is 1, that corresponds to the lead of 7 meters.
Therefore, length of race =5*7=35 meters
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Re: A runs 25% faster than B and is able to allow B a lead of 7 meters to  [#permalink]

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New post 14 Mar 2016, 20:08
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let d=length of race
r=B's rate
because times of A and B are equal,
then d/(5r/4)=(d-7)/r
d=35 meters
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Re: A runs 25% faster than B and is able to allow B a lead of 7 meters to  [#permalink]

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New post 30 Sep 2018, 12:25
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Another approach would be to let speed of B be 4 and speed of A = 4 * 1.25 = 5

Now B starts at 7 then move 4s

7,11,15,19,23,27,31,35

A starts at 0,5,10,15,20,25,30,35

Answer choice D.

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Re: A runs 25% faster than B and is able to allow B a lead of 7 meters to  [#permalink]

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New post 30 Sep 2018, 16:09
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1

Solution



Given:
• A runs 25% faster than B
• A allow B a lead of 7 meters to end a race in dead heat

To find:
• What is the length of the race?

Approach and Working out:
• Let speed of B= S then A’s speed= 5/4 S
Now, if we assume race track to be of length x then A ran x metres in the same time as B took to cover x-7.
• x/ 5S /4= x-7/ S
• 4X/5S= x-7/S
• 5x- 35=4x
• x= 35
Answer: D
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Re: A runs 25% faster than B and is able to allow B a lead of 7 meters to  [#permalink]

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New post 30 Sep 2018, 20:43
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Bunuel wrote:
A runs 25% faster than B and is able to allow B a lead of 7 meters to end a race in dead heat. What is the length of the race?

A. 10 meters
B. 15 meters
C. 25 meters
D. 35 meters
E. 45 meters


Let the length of the race be \(d\), B runs \((d-7)\) meters and A runs \(d\) meters, in \(s\) mph and \(1.25s\) mph respectively.

\(time = \frac{distance}{speed}\)

Time taken by A = \(\frac{d}{1.25s}\)

Time taken by B = \(\frac{d-7}{s}\)

Since the race was a dead heat \(\frac{d}{1.25s} = \frac{d-7}{s}\)

\(ds = 1.25ds - 8.75s\)

\(d = 1.25d - 8.75\)

\(d = \frac{8.75}{0.25} = 35\).
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Re: A runs 25% faster than B and is able to allow B a lead of 7 meters to  [#permalink]

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New post 01 Oct 2018, 06:08
Bunuel wrote:
A runs 25% faster than B and is able to allow B a lead of 7 meters to end a race in dead heat. What is the length of the race?

A. 10 meters
B. 15 meters
C. 25 meters
D. 35 meters
E. 45 meters


If B's usual rate of running the race is 4x, A's rate will be 5x

Since the distances covered by both of them are equal provided A covers
7m less, we can equate the distances. So, \(5x = 4x + 7\) -> \(x = 7\)

Therefore, the total length of the race is \(5x\) = 35 meters(Option D)
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Re: A runs 25% faster than B and is able to allow B a lead of 7 meters to  [#permalink]

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New post 01 Oct 2018, 13:21
Bunuel wrote:
A runs 25% faster than B and is able to allow B a lead of 7 meters to end a race in dead heat. What is the length of the race?

A. 10 meters
B. 15 meters
C. 25 meters
D. 35 meters
E. 45 meters



\(\frac{x}{125} = \frac{x-7}{100}\)

\((x-7)125 = 100x\)

\(125x-100x=875\)

\(25x=875\)

\(x = 35\)
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Re: A runs 25% faster than B and is able to allow B a lead of 7 meters to  [#permalink]

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New post 03 Oct 2018, 17:38
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Bunuel wrote:
A runs 25% faster than B and is able to allow B a lead of 7 meters to end a race in dead heat. What is the length of the race?

A. 10 meters
B. 15 meters
C. 25 meters
D. 35 meters
E. 45 meters


Let the length of the race be represented by d and let the rate of runner B be represented by r. Since A runs 25% faster than B, the rate of A is 1.25r.

Since the times are the same, we can create the equation:

time for runner A = time for runner B

d/1.25r = (d - 7)/r

rd = 1.25r(d - 7)

d = 1.25d - 8.75

8.75 = 0.25d

35 = d

Answer: D
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Re: A runs 25% faster than B and is able to allow B a lead of 7 meters to  [#permalink]

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New post 12 Oct 2018, 07:05
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Alternative Approach.

Using the Gap formula, we Suppose that B is 7 meters ahead of A when starting the race. It means that
Meeting time = Gap/(Va-Vb) = 7/(1.25-1) = 28 meters.

Now the problem says that A is ahead of B at the moment of ending the race by 7 meters. So the race distance is = 28 + 7 = 35

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Re: A runs 25% faster than B and is able to allow B a lead of 7 meters to  [#permalink]

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New post 29 Jun 2019, 21:05
Bunuel wrote:
A runs 25% faster than B and is able to allow B a lead of 7 meters to end a race in dead heat. What is the length of the race?

A. 10 meters
B. 15 meters
C. 25 meters
D. 35 meters
E. 45 meters


A can cover d in t time. B can cover d-7 in t time.
So here time t is constant for A and B, let d be the distance for track-

d/125 = (d-7)/100

d= 35
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Re: A runs 25% faster than B and is able to allow B a lead of 7 meters to   [#permalink] 29 Jun 2019, 21:05
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