Aptitude - Races and Games - Discussion

Please anyone explain this in simple way.

Does anyone know how can we solve this question by having the difference in distance values only( i.e, using 10 and 28)?

Please tell me.

I am not getting it.

At last how the answers came as 20? please explain.

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.

Thank you for your explanation @Akhil.

Given A 100, B 10, C 28.

A gives B means 100 - 10 = 90 = B.
A gives C means 100 - 28 = 72 = C.
A : B = 100 : 90.
A : C = 100 : 72.

B : C = 90 : 72 -------> 90/72 = B/C.


Consider given one as 100.
B gives C is

B : C = 100 : 72/90.
C/B = 72/90 * 100 = 80m.

We have to compare between B & C.

If B goes 90.

C goes 72.

To be 100 we have to multiply by 100 and divide by 90.

90 * 100/90 ------> 72 * 100/90.

100 ------> 80.

So, B can give C = 100 - 80 = 20.

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.

Thank you all for helping me.

Thanks, @Abhishek Kushwaha.