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

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

In 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 11-10 = 1 meter

Hence B is the answer
Cheers!
Kudos rain

J

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
But Y had a 10m head start
So total victory margin = 1m

Both run at the same previous speeds !!

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

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

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

already explained .... the ans is B

There is no kudos for being the first?

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



+1 to you

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.

Thought it was some trick question

@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.

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

Given: 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.

Asked: Which of the following is true?

Let Vx & Vy be speeds of X & Y respectively
X and Y run a 100m race, where X beats Y by 10m.
Vx / Vy = 100/90 = 10/9
Vx = 10Vy/9

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.
Time taken by X to complete 110 m = 110/Vx = 110*9/10Vy = 99/Vy
Time taken by Y to complete 100 m = 100/Vy
Distance travelled by Y in 99/Vy = 99/Vy * Vy = 99m
X beats Y be 1 m

IMO B