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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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18 Sep 2010, 22:05
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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 The solution will be posted on Monday. Please post explanation along with your answer. Happy Weekend Questions Archive : happyweekendquestions100374.html
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Re: Happy weekend question 18 Sept 2010
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18 Sep 2010, 22:23
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 Answer: B I think I deserve kudos
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Re: Happy weekend question 18 Sept 2010
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26 Sep 2010, 06:24
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 Answer is B
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Re: Happy weekend question 18 Sept 2010
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18 Sep 2010, 22:47
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 The solution will be posted on Monday. Please post explanation along with your answer. Happy Weekend Questions Archive : happyweekendquestions100374.htmlLets 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 But Y had a 10m head start So total victory margin = 1m
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Re: Happy weekend question 18 Sept 2010
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28 Sep 2010, 02:55
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
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20 Jan 2011, 06:15
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
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26 Nov 2013, 05:49
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 The solution will be posted on Monday. Please post explanation along with your answer. Happy Weekend Questions Archive : happyweekendquestions100374.htmlIn 100 meters X beats Y by 10 meters In 110 meters X will beat Y by 11 meters. Since Y had an advantage of 10 meters. Then difference is 1110 = 1 meter Hence B is the answer Cheers! Kudos rain J



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Re: Happy weekend question 18 Sept 2010
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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
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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 Answer is B
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Re: Happy weekend question 18 Sept 2010
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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
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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
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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
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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 The solution will be posted on Monday. Please post explanation along with your answer. Happy Weekend Questions Archive : happyweekendquestions100374.htmlLets 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 But Y had a 10m head start So total victory margin = 1mMany 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
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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
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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
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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
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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
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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
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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. Answer B. Thanks, GyM
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Re: X and Y run a 100m race,where X beats Y by 10m. To do a favo
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Re: X and Y run a 100m race,where X beats Y by 10m. To do a favo
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