Adam and Bob have exact speeds for walking and running - neither Adam outwalks Bob, nor Bob outruns Adam.
But for a given distance, Adam runs half the time, then walks another half. Bob runs half the distance, then walks another half.
If they were to compete, which distance would Adam win at, and which distances would Bob be the winner at?
35 Comments on Two men running
if adam and bob compete for ‘d’ distances, then adam would win by distance of
( d*(Vr-Vw)^2)/(2*Vr*(Vr+Vw) )
where Vr=running speed
For Adam,
Ta/2*Vr+Ta/2*Vw=d
For Bob,
Tb = d/(2*Vr) +d/(2*Vw)
Distance difference:
(Ta-Tb)*Vw=2*d*Vw/(Vr+Vw)-(d*Vw/2)*(1/Vr+1/Vw)
=2*d*Vw/(Vr+Vw)-(d*Vw/2)*(Vr+Vw)/(Vv*Vw)
=-d*(Vr-Vw)^2/(2*Vr*(Vr+Vw))
So actually Bob will win.
Adam:
Xa = 1/2 * Ta (Vr + Vw)
Bob:
Tb = 1/2 * Xb / Vw + 1/2 * Xb /Vr =>
Xb = Tb * (2*Vw*Vr) /(Vr + Vw)
Adam wins when Xa > Xb & Ta = Tb
=> 1/2 * (Vr + Vw) > (2*Vw*Vr) /(Vr + Vw)
=> (Vr - Vw)^2 > 0
=> if Vr <> Vw, Adam wins!!ÂÂ
if Vr = Vw => Xa = Xb!! Bob never wins :(
Ta/2*Vr+Ta/2*Vw=d
For Bob,
Tb = d/(2*Vr) +d/(2*Vw)
Distance difference:
(Ta-Tb)*Vw=2*d*Vw/(Vr+Vw)-(d*Vw/2)*(1/Vr+1/Vw)
=2*d*Vw/(Vr+Vw)-(d*Vw/2)*(Vr+Vw)/(Vv*Vw)
=-d*(Vr-Vw)^2/(2*Vr*(Vr+Vw))
For any distance Adam wil run longer than Bob hense his average speed is higher. Meaning that Adam always wins.
Let running speed = r, walking speed = w
total distance x = x1 + x2
total time T = t1 + t2
for Adam:
t = T/2
x1 = rT/2
x2 = wT/2
xa = T * (w+r)/2 ——–1
for Bob:
t1 = x/2r
t2 = x/2w
T = t1 + t2
xb = T * 2rw/(w+r) ——–2
divide 1 by 2 ->
xa/xb = (w^2 + 2wr + r^2) / 4wr ;r > w > 0
Hance, xa/xb is always > 1
so Adam always wins!
No Body Wins or Losses. Its Always a Tie regardless of the distance
simpler question is… can Adam cover more than 1/2 of distance in 1/2 of the time… if true, Adam always win… since both of them walk for the rest
We know running is faster than walking… so to cover 1/2 of distance by running and walking… running takes less time… therefore it will take less than 1/2 of the time
Now you let Adam run for 1/2 of the time, it will surpass 1/2 the distance
So Adam always win
With the same velocity at a given distance,
Bob will win at half the distance then at the end of the distance they will have a tie.
Lets say both runs at 10 KMPH and walks at 5 KMPH. And the Distance is 2X.
For Adam
———
Lets say Adam took T1 hour to finish
so (10*T1/2) + (5*T1/2) = 2X
=> 7.5 T1 = 2X
=> X = 7.5/2 T1 = 15/4 T1
For Bob
——-
Lets say Bob took T2 hours
So
X/10 + X/5 = T2
=> T2= 3X/10 = 3/10 * X = 3/10 * 15/4 T1
=> T2 = 9/8 T1
So, T2 will always more than T1. So, Bob will never win.
=>
Sorry, but my previous comment has been messed up a little bit. I try to reenter the confusing parts:
We have to compare these two expressions. x is on both side, we can leave it.
2/(Vr+Vw) and (1/Vr+1/Vw)/2
We need the same expression in the denominator or reckoner of a fraction to compare:
2*(Vr+Vw)/(Vr+Vw)^2 and 2*(Vr+Vw)/(4*Vr*Vw)
Now we need to compare:
(Vr+Vw)^2 and 4*Vr*Vw
Vr^2+2*Vr*Vw+Vw^2 and 4*Vr*Vw
Vr^2+Vw^2 and 2*Vr*Vw
This means if Vr=Vw this will equal and than Ta=Tb
otherwise we can easily prove:
Vr^2+Vw^2 > 2*Vr*Vw |((Vr-Vw)^2>0)
This way the original fraction will be smaller if it has a bigger denominator. So all the time Ta < Tb
If walking speed differs from running speed, Adam always wins.
Adam won some time on first part when he was running ta/tb = d/2Vr/d/2Vw = Vw/Vr
Adan walked on second part so win distance must be (Vw/Vr )* Vw = Vw^2/Vr
always Adams will wins,
for Adams,
Ta*Vr/2+Ta*Vw/2=D;
for Bob
D/(2*Vr)+D/(2*Vw)=Tb;
if we find a relation between Ta & Tb
it will be Ta < Tb.
Hi,
Lets assume the distance to be x kms, and the total time taken by Adam be t and by Bob be t’.
Let walking speed be 10kmps and running be 20kmps (running > walking ) ( we can assume any number as the result will be same coz they both have same walking and running speed)
Now,
Adam :
x = 10 * t/2 + 20 * t/2 ;
x = 15t
t = x / 15
BOB:
t’ = x/2*10 + x/ 2*20 ;
t’ = 3x/40 ;
t’ = x / 13.3333 ;
therefore its clear that watever be the distance t’ > t
hence Adam is always the winner.
Observation:
Both, Adam & Bob start off running at the same speed, hence they are in a tie, until one of them starts walking. Who ever starts walking first looses the challenge. The loosing person will always take longer time to reach the destination, hence figuring out the total time will provide for a winner.
Variables:
R – running speed
W – walking speed
D – distance
Ta – Adam’s total time
Tb – Bob’s total time
Formula:
Distance = Speed * Time
Time = Distance/Speed
Computation:
–ADAM
Total Distance
D = .5*Ta*R + .5*Ta*W
D = .5*Ta(R+W)
–BOB
Total time spent traveling:
Tb = D/2R + D/2W
Solving for Distance (D) yields to:
D = (2RWTb)/(W+R)
Because distance is the same, both equations can be set equal to each other with respect to the time variable as follows:
.5*Ta(R+W) = (2RWTb)/(W+R)
Solving the equation for Tb gives us the following expression:
Tb = Ta * [(R+W)^2]/RW
Focusing on: [(R+W)^2]/RW] we can rewrite it as:
R^2/RW + 2RW/RW + W^2/RW = R/W + 2 + W/R
It is pretty safe to assume that since Running Speed always exceeds Walking Speed R/W will approach 1, and W/R will approach zero, so for simplicity, we’ll conclude that resulting constant X is always greater than 2.
Placing this X value back in the original equation we get:
Ta * X = Tb.
Based on this proof, we can safely make a conclusion that Bob’s time is always higher. Furthermore, please note that in these conditions Ta can never equal to Tb (unless time is zero, but then… well, there’s no race).
Because Ta < Tb, Adam wins, and Bob doesn’t stand a chance. Not only that, Adam claims victory precisely at the point that Bob starts walking, and from the problem statement, we know that it was at the distance D/2.
definitly adam will win d race always coz running takes less time compared to walkin for a same distance n hence ..if v take equal time distance covered in running is always more than
walkin
then compare both of d guys v can c tat adam runs for more distance than bob …so adam wins d race watever b d distance
tats all…
at end both wins because a*Vr+a*Vw=b*Vw+b*vr
and adam wins at start only because a*Vr>b*Vw
but there is no chance of bob winning
a=adam,b=bob,Vr=running speed,Vw=walking speed,a*Vr=adams ruuning speed,b*Vr=bob’s running,so on.
Neither wins until he crosses the finish line. Since the finish line doesn’t move… If they do in fact have the same pace, then whether walking/running or running/walking, they will reach the finish line in a tie.
BTW - I’m assuming the person who typed this question made a typo about Adam running half the TIME and Bob half the DISTANCE. That’s an Apples to Oranges comparison.
But, assuming it’s not a typo: Bob will walk literally half the distance. Adam will run half the TIME. If you spent 30 minutes running, and 30 minutes walking, you will cover more than half the distance RUNNING. Bob will cover exactly HALF the distance running. So given a static length, the more time spent running will win.
Assuming that Adam decide to run initially and bob decide to walk initially.
In this case adam will always be the winner for whole distance and at the finish line bob will cope up with adam and both will tie.
Supposed R for the speed of Run, W for the speed of Walk
Ta is for the time that Adam will use, Tb is for the time that Bob will use.
If they arrive at the same time then have the following
R*0.5*Ta + W*0.5*Ta=R*(W/R+W)*Tb + W*(R/R+W)*Tb
(Speed times time is distance)
Ta*0.5*(R+W) = Tb*(R*W+W*R)/(R+W)
Ta/Tb=2*2R*W/ (R+W) ^2
(R+W) ^2 >=4*R*W
So Bob will arrive at the same time with Adam only if they walk as fast as they run.
Other than that, Adam wins.
let X=x1+x2(distance)
T=t1+t2
r=runnig speed
for adam:
x1=w*(T/2)
x2=r*(T/2)
X=x1+x2
=(r+w)*(T/2)—–>(eqt:1)
let X=x1+x2(distance)
T=t1+t2
r=runnig speed
w=walking speed
for adam:
x1=w*(T/2)
x2=r*(T/2)
X=x1+x2
xa=(r+w)*(T/2)—–>(eqt:1)
for bob:
t1=x1/r
=X/2r
t2=x2/w
=X/2w
T=t1+t2
=(X/2)*(rw/r+w)
=>xb=T*2((r+w)/rw)—–.(eqt2)
from eqt 1 and 2
xb/xa=4/rw
so there comes cases like
if(0<r<1)and (0<w<1)
then bob wins always
if either of runnig speed or walking speed or both are greater than 4
then adam wins always
Let r denote the running speed and w the walking speed.
Adam’s average speed is the arithmetic mean of r and w, i.e. (r+w)/2, whereas Bob’s average speed is the harmonic mean of r and w, i.e. (2*r*w)/(r+w).
The harmonic mean is always less than or equal to the arithmetic mean. The two means are equal when r=w.
So if r!=w, Adam wins. If r=w, it’s a tie. Bob never wins.
Assumptions:
Vr = V Running
Vw = V Walk
if (Vr>Vw) than Adam
if (Vr=Vw) tie
if (Vr<Vw) than Bob
@Mark^^^^
average speed is not (r+w)/2..ie it is not arithmetic mean of running and walking…its rather the total distance covered/total time taken…
and the answer is bob never wins….
Assuming that both walk more slowly than they run, Adam always wins as he runs half the TIME, while Bob runs half the DISTANCE. Since walking is slower, Bob spends more than half his TIME walking. Therefore Bob’s average speed will be closer to his walking speed, while Adam’s speed will be exactly the average of his running and walking speeds.
all it says is that neither outwalk or outrun the other… it doesnt say their speeds dont match, therefor their walking and running speed could be exactly even and they would therefor tie.
It does not say “equal” its says “EXACT”
neither Adam outwalks Bob, nor Bob outruns Adam.
this gives rise to 2 possiblities
case 1 :
may be
Bob outwalks ADAM, or ADAM outruns BOB.
case 2 :
or may be they are equal
so in case 1 adam will be winner
in case 2 both will be equal.
but the catch is ….
the question is
1 the distance at which adam will win ?
at all distances adam will be a winner.
(even if they are equal both will be a winner)
2 the distance at which Bob will win ?
Bob can be a winner(case 2) at any distance
BOB will not be a winner(case1) situations
Adding to Gili,:
> Assumptions:
> Vr = V Running
> Vw = V Walk
>
> if (Vr>Vw) than Adam
> if (Vr=Vw) tie
> if (Vr Tb (i.e. Bob wins), Z > 1
For Ta < Tb (i.e. Adam wins), Z < 1
Z = 1 indicates
4VrVw/(Vr+Vw)^2 = 1
or 4VrVw = Vr^2 + 2VrVw + Vw^2
or Vr^2 - 2VrVr + Vw^2 = 0
or (Vr-Vw)^2 = 0
or Vr-Vw = 0
or Vr = Vw
So, Vr = Vw for a tie
Vr Vw for Adam to win.
Since running is always faster than walking (a fair assumption), Bob never wins.
Its Clear Bob will never win.
Lets Adam travelled Da. Bob travelled Db
Vr -Velocity while running
Vw - Velocity while walking
t= total time of race
We need to find out Da-Db
We have
From Adam
Vr * (t/2) + Vw * (t/2) = Da
t = 2Da / ( Vr + Vw )
From Bob
Let t1 is the time when He ran ,Vr
t2 be the time when he walks,Vw
t1 + t2 = t ,Both take same time
=> Db/(2*Vr) + Db/(2*Vw) = t
=> Db/(2*Vr) + Db/(2*Vw) = 2Da / ( Vr + Vw )
=> Db( Vr + Vw)/(2Vr * Vw) = 2Da / ( Vr + Vw )
=> Db = ( 4 Da * Vr * Vw ) / ( Vr + Vw)^2
We need to calculate, Da-Db
Da - ( 4 Da * Vr * Vw ) / ( Vr + Vw)^2
=> Da*(((Vr+Vw)^2 - 4VrVw))/(Vr+Vw)^2
=> (Da*(Vr-Vw)^2)/(Vr+Vw)^2
Adam won by above distance.
Let, walking speed = Vw
running speed = Vr
Since, Adam runs half the time and walks half the time
therefore, Da(distance travelled by Adam) = (Ta/2)*(Vr+Vw) (Ta = total time taken by Adam)
Similarly for Bob, total time taken =
Tb = (Db/2)*(1/Vr + 1/Vw)
Now considering they run a race, therefore, the distance they travel is equal
i.e. Da = Db
now from above, Ta = 2Da/(Vr+Vw)
Tb = Db(Vr+Vw)/(2*Vr*Vw)
For Adam to win the race,
Ta<Tb
i.e. 2Da/(Vr+Vw) < Db(Vr+Vw)/(2*Vr*Vw)
i.e. 2/(Vr+Vw) Vr*Vw
i.e. ((Vr+Vw)/2) > sqr. root of Vr*Vw
which is always true , since AM > GM
that means Adam always wins
it can result to equality if Vr = Vw
in which case they tie
But practically running is always faster than walking
therefore, Adam always wins
The mathematical equations are definitely good to get the point across, but in an interview I am sure not many people can recite the equations. It is basically common sense. If they are looking for an explanation use the following: If Bob runs half the distance and walks the second half, the second leg will take longer. Since Adam is running half the time, he will always run a further distance than Bob before he starts walking. Since they walk the same speed, Bob will not be able to catch him.
Wouldn’t they just be egual? It says Adam and Bob have exact speeds for running AND walking. So that means they both run just as fast as eachother.
Since Bob and Adam walk and run the same, time or distance does not apply since they both run and walk half of the distance and half of the time. So ultimately they are tied.