Aptitude - Races and Games - Discussion

@Ammu

I think throughout the race a, b&c have respective constant velocities.
That being said this problem can easily be solved using the concept of ratios.

Question is to find the distance btw b&c at the end of the race(when b reaches 100m).
It is given separation btw b & c when b reaches 90m is 18 (ie 90-72).
So, when b reaches 1m mark the Seperation btw them is simply 18÷90=1/5
When b reaches 100m ie end of the race the seperation is (1/5) * 100. =. 20.

I hope it is clear now.

They are asking when B is at 100 m, where is C?

Consider A need x seconds to reach at 100m.

Velocity of A is 100/x.

Given that when A is at 100m, B and C are at 90m and 72m respectively...

So,velocity of B is 90/x (m/sec)
velocity of C is 72/x (m/sec)

Now, For B to reach 100m, 10m again have to run
time required for B to run 10m =10/(90/x) = x/9 sec.
distance covered by C in x/9 sec =(x/9)*(72/x) = 8m.

So when B is at 90+10=100m.
C is at 72+8 = 80m.

B won race by 20m difference with C.

Here, give indicates the distance that participants overtake each other.

So, A beats B by 10m in a 100m race indicates A finishes 100m whereas B finishes 10m behind A in the same time ie.) 90m.

Similiarly, A beats C by 28m indicates C covers 72m in the same time where A finishes 100m.
Now ratio of distance covered in a given time by B and C remains same ie.) 90:72 ie.) 5:4.

When, B covers 100m in a given time, C covers only 80 in the same time. So B beats C by a distance of 20m.

A can give 10 means A can lead B by 10 meter. That's A can give a lead of 10 meter. Got t @Arun.

Melvin its just not distance calculation to simply difference the distance. Here change in velocity of the players maters. For example if you just subtract the distance means B and C are running in same velocity right?

If so how would B can beat C in the race in the beginning? Got t so there is a velocity change in players that's what you are calculating unstintingly.

In a 100 m race, A can give B 10 m, It means A:B=100:90.....equation (1).

In a 100 m race, A can give C 28 m, It means A:C=100:72....equation (2).

In a 100 m race, B can give C=?

{Dividing Equation (1) and Equation (2) we get}.

B:C=90:72 or B:C=5:4 (After Dividing numerator and denominator by 18).

Or B:C=100:80 (After Multiplying numerator and denominator by 20).

In the same race B can give C=20[ANS].

A finished 100m.
At that time,
B is 10m less than A, so B is 100-10=90.
C is 28m less than A, so C is 100-28=72.

The difference b/w in B and C is,
90-72 = 18.

When B is 90m then the difference is 18m.
When B is 1m the difference is 18/90 = 0.2m.
When B is 100m the difference is 0.2 * 100 = 20m.
At that time C will be 80m.
Ans is 20m.

*-------------------*
<---100m------------> A
<-----90-------> B
<-----72---> C

When A completed race(100m) then B ahead 90 and c Ahead 72, and Distance between B & c = 18m.

i.e., When B cover 90 then distance between B & c = 18.
So When B cover 100 then distance between B & c = 18*100/90 =20 m.

@All.

The solution is;

In same time A,B and C travel 100,90,72 meters respectively hence we can write:
100:VA = 90:VB = 72:VC where VA,VB,VC are velocities of A, B, C respectively.
Hence;
VB:VC = 90x:72x that means time taken by B to travel 100m is tb= 100/90x.
In this time C travel ( 100/90x)*72x = 80 m implies B gives C by 20m.

A finished 100m.
At that time.
B is 10m less than A, so B is 90.
C is 28m less than A, so C is 72.
The difference b/w in B and C is.
90-72 = 18.
When B is 90m then the difference is 18m.
When B is 1m the difference is 18/90=0.2m.
When B is 100m the difference is 0.2 * 100=20m.
At that time C will be 80m.
Ans is 20m.

A defeats B by 10m i.e. if A runs 100m then B runs 90m.

A defeats c by 28m i.e. if A runs 100m then C runs 72m.

If B = 90 then C = 72.

If B = 100 then C = ? (cross multiply).

=> C = (72 x 100) / 90 = 80m.

So, if B = 100 C = 80.

ie=> B defeats C by 20 metres.