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# X and Y run a 100m race,where X beats Y by 10m. To do a favo

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X and Y run a 100m race,where X beats Y by 10m. To do a favo  [#permalink]

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18 Sep 2010, 22:05
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67% (02:22) correct 33% (01:58) wrong based on 247 sessions

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X and Y run a 100m race,where X beats Y by 10m. To do a favor to Y, X starts 10m behind the starting line in a second 100m race. Both run at the same previous speed. Which of the following is true?

A) Y Beats X by 10m
B) X Beats Y by 1m
C) X and Y both reach the finishing line simultaneously
D) Y Beats X by 1m
E) Y Beats X by 11m

Happy Weekend Questions Archive : happy-weekend-questions-100374.html

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Re: Happy weekend question- 18 Sept 2010  [#permalink]

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18 Sep 2010, 22:23
8
B
Let's suppose that the speed of X is 10m/s and that the speed of Y is 9m/s.
So, in the first race, the results are the following:
X crosses the finish line in 10s. In other words, X has run 100m in 10 s.
In 10s, Y has run 90 meters.
Therefore, there is the difference of 10 meters mentioned in the question.

Now, let's review the second race.
In 11 seconds, X can complete the 100 meters race. We have to remember that X is 10 meters behind the starting line.
But in 11 seconds, Y has covered 99 meters. Also, it is important to notice that in 11 seconds Y cannot arrive to the finish line.

So, 100 - 99 = 1 meter

I think I deserve kudos
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##### General Discussion
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Re: Happy weekend question- 18 Sept 2010  [#permalink]

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26 Sep 2010, 06:24
2
1
Here's the solution.

Distance traveled is proportional to speed if time taken is same.

$$\frac{D_{X}}{D_{Y}} = \frac{V_{X}}{V_{Y}}$$

$$\frac{100}{90}= \frac{D_{X}}{D_{Y}} = \frac{10}{9}$$

In the second case

$$\frac{D_{X}}{D_{Y}} =\frac{V_{X}}{V_{Y}} = \frac{10}{9}$$

Since $$D_{X} > D_{Y}$$ => X beats Y by some distance => X must have traveled 110 m

$$\frac{10}{9} =\frac{D_{X}}{D_{Y}}=\frac{110}{D_{Y}}$$

$$D_{Y} = 99$$, Hence X beats Y by 1 m

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Re: Happy weekend question- 18 Sept 2010  [#permalink]

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18 Sep 2010, 22:47
1
gurpreetsingh wrote:
Q3. X and Y run a 100m race,where X beats Y by 10m. To do a favor to Y, X starts 10m behind the starting line in a second 100m race. Both run at the same speed. Which of the following is true?

A) Y Beats X by 10m
B) X Beats Y by 1m
C) X and Y both reach the finishing line simultaneously
D) Y Beats X by 1m
E) Y Beats X by 11m

Happy Weekend Questions Archive : happy-weekend-questions-100374.html

Lets say X takes t seconds to finish the race. This means his speed = 100/t and the difference in the speed of X and Y is 10/t

The second time around, the speeds are constant, the time to complete the race is Min{110/(100/t) [if X wins], 100/(90/t) [if Y wins]} = Min{t*1.1,t*1.1111} = t*1.1. This also tells us that X wins the race, as it takes X less time to reach the finish from 110m than it takes Y to reach from 100m

Finally the margin. Difference in speeds = 10/t. Time = 1.1t. Hence Margin = 1.1t * 10/t = 11m
So total victory margin = 1m

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Re: Happy weekend question- 18 Sept 2010  [#permalink]

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28 Sep 2010, 02:55
1
hemanthp wrote:
gurpreetsingh wrote:
Q3. X and Y run a 100m race,where X beats Y by 10m. To do a favor to Y, X starts 10m behind the starting line in a second 100m race. Both run at the same speed. Which of the following is true?

A) Y Beats X by 10m
B) X Beats Y by 1m
C) X and Y both reach the finishing line simultaneously
D) Y Beats X by 1m
E) Y Beats X by 11m

Gurupreet, in your question it says "Both run at the same speed". How can X win if he starts 10m behind Y and run at the same speed as Y?
did you print that by mistake? if not, then A is the answer.

Both run at the same previous speeds !!
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Re: Happy weekend question- 18 Sept 2010  [#permalink]

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20 Jan 2011, 06:15
1
2

When X runs 100m, Y runs 90m.
When X runs 110m (10+100), Y runs $$= \frac{90*110}{100} = 99m$$

So X beat Y by 1m
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Re: Happy weekend question- 18 Sept 2010  [#permalink]

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26 Nov 2013, 05:49
1
1
gurpreetsingh wrote:
Q3. X and Y run a 100m race,where X beats Y by 10m. To do a favor to Y, X starts 10m behind the starting line in a second 100m race. Both run at the same previous speed. Which of the following is true?

A) Y Beats X by 10m
B) X Beats Y by 1m
C) X and Y both reach the finishing line simultaneously
D) Y Beats X by 1m
E) Y Beats X by 11m

Happy Weekend Questions Archive : happy-weekend-questions-100374.html

In 100 meters X beats Y by 10 meters
In 110 meters X will beat Y by 11 meters.
Then difference is 11-10 = 1 meter

Cheers!
Kudos rain

J
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Re: Happy weekend question- 18 Sept 2010  [#permalink]

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18 Sep 2010, 23:32
My attempt:

Given that X beats Y by 10m. Hence at some time instant, X has covered 100m and Y has covered only 90m.

Time = Distance/Speed. Let speed of X be x1 and Y be y1.

Now at the same time instant => 100/x1 = 90/y1

x1= (10/9)y1.

Second attempt. X attempts to cover 110 meters and allows Y to compete for 100 meters.

Again we need to see at the exact time instant when either X or Y has completed their respective distance.

Again at some time instant, t => 110/x1

Substituting x1 => 110/((10/9)*y1) = 99/y1. We see that X with his speed of x1 has only covered 99 meters whereas Y with a speed of y1 has covered his ground - 100m. Hence Y has won with one meter distance.

Answer: D (Y beats X by 1 meter).
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Re: Happy weekend question- 18 Sept 2010  [#permalink]

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19 Sep 2010, 07:06
ezhilkumarank wrote:
My attempt:

Given that X beats Y by 10m. Hence at some time instant, X has covered 100m and Y has covered only 90m.

Time = Distance/Speed. Let speed of X be x1 and Y be y1.

Now at the same time instant => 100/x1 = 90/y1

x1= (10/9)y1.

Second attempt. X attempts to cover 110 meters and allows Y to compete for 100 meters.

Again we need to see at the exact time instant when either X or Y has completed their respective distance.

Again at some time instant, t => 110/x1

Substituting x1 => 110/((10/9)*y1) = 99/y1. We see that X with his speed of x1 has only covered 99 meters whereas Y with a speed of y1 has covered his ground - 100m. Hence Y has won with one meter distance.

Answer: D (Y beats X by 1 meter).

You have gone wrong in the second last step
110/x1 = 99/y1 means that in the time X covers 110m, Y covers only 99
Hence X wins the race even when starting 10m behind and by the time he finishes Y has only reached the 99 mark which means X wins by 1m

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Re: Happy weekend question- 18 Sept 2010  [#permalink]

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19 Sep 2010, 15:14
In Same time, X covers 100 m and Y covers 90 m. So when they start second time, both will meet at 90 m point. Then X will overtake. Based on answer choices, X beats Y in only option A and option B. Option A is ruled as Y difference between two will be less than 10.

So B
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Re: Happy weekend question- 18 Sept 2010  [#permalink]

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26 Sep 2010, 20:43
already explained .... the ans is B
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Re: Happy weekend question- 18 Sept 2010  [#permalink]

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27 Sep 2010, 16:54
There is no kudos for being the first?
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Re: Happy weekend question- 18 Sept 2010  [#permalink]

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27 Sep 2010, 19:06
shrouded1 wrote:
gurpreetsingh wrote:
Q3. X and Y run a 100m race,where X beats Y by 10m. To do a favor to Y, X starts 10m behind the starting line in a second 100m race. Both run at the same speed. Which of the following is true?

A) Y Beats X by 10m
B) X Beats Y by 1m
C) X and Y both reach the finishing line simultaneously
D) Y Beats X by 1m
E) Y Beats X by 11m

Happy Weekend Questions Archive : happy-weekend-questions-100374.html

Lets say X takes t seconds to finish the race. This means his speed = 100/t and the difference in the speed of X and Y is 10/t

The second time around, the speeds are constant, the time to complete the race is Min{110/(100/t) [if X wins], 100/(90/t) [if Y wins]} = Min{t*1.1,t*1.1111} = t*1.1. This also tells us that X wins the race, as it takes X less time to reach the finish from 110m than it takes Y to reach from 100m

Finally the margin. Difference in speeds = 10/t. Time = 1.1t. Hence Margin = 1.1t * 10/t = 11m
So total victory margin = 1m

Many thanks for correcting me. +1 to you - shrouded1.

metallicafan wrote:
There is no kudos for being the first?

+1 to you
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Re: Happy weekend question- 18 Sept 2010  [#permalink]

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28 Sep 2010, 02:17
gurpreetsingh wrote:
Q3. X and Y run a 100m race,where X beats Y by 10m. To do a favor to Y, X starts 10m behind the starting line in a second 100m race. Both run at the same speed. Which of the following is true?

A) Y Beats X by 10m
B) X Beats Y by 1m
C) X and Y both reach the finishing line simultaneously
D) Y Beats X by 1m
E) Y Beats X by 11m

Gurupreet, in your question it says "Both run at the same speed". How can X win if he starts 10m behind Y and run at the same speed as Y?
did you print that by mistake? if not, then A is the answer.
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Re: Happy weekend question- 18 Sept 2010  [#permalink]

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28 Sep 2010, 03:19
Thought it was some trick question
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Re: Happy weekend question- 18 Sept 2010  [#permalink]

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10 Oct 2010, 04:21
@gurpreet
How come time taken is same???
in "Distance traveled is proportional to speed if time taken is same"
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Re: Happy weekend question- 18 Sept 2010  [#permalink]

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10 Oct 2010, 04:26
utin wrote:
@gurpreet
How come time taken is same???
in "Distance traveled is proportional to speed if time taken is same"

In a race, both of them will travel for same time.

Suppose you are faster than me. You beat me by 1 m means when you reached the line I was 1 m behind.
total time consumed by us is same.
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X and Y run a 100m race,where X beats Y by 10m. To do a favo  [#permalink]

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01 Feb 2016, 14:26
let rate of X=10 m/s=.1 seconds/meter
rate of Y=9 m/s
110/10=11 seconds=time of X in race 2
100/9=11.1 seconds=time of Y in race 2
X beats Y by .1 seconds=1 meter
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Re: X and Y run a 100m race,where X beats Y by 10m. To do a favo  [#permalink]

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24 Jun 2018, 00:19
Let x be speed of X & y be the speed of Y, such that x > y

When X covers 100m, Y has ran 90m, hence time take by both to cover their respective distances is same

100/x = 90/y or x/y = 10/9.........................(i)

Now X starts the race 10 m behind Y, hence when X covers 100 m, Y will have covered 90m & they both will be at same point.

From this point, to the finish line they both have to travel 10m, since X is faster he will reach the finish line first in 10/x time.

During this time the suppose D is distance covered by Y.

Hence we have 10/x = D/y, Substituting (i) we get D = 9m

Hence X beats Y by 1m.

Thanks,
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Re: X and Y run a 100m race,where X beats Y by 10m. To do a favo  [#permalink]

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Re: X and Y run a 100m race,where X beats Y by 10m. To do a favo   [#permalink] 08 Aug 2019, 07:13