Aptitude - Races and Games - Discussion

Awesome explanation. Thanks @Shivakrishna.

Let total distance is "y" mtr.
As per the question A gives B a start of 80 mtr.
So we can say that A runs "y" mtr while B covers (y-80) mtr.
Now B's speed is X m/sec (say)
And A's speed is 5X/3 m/sec
Now as they both have to reach the finish point at the same time then we can say that time taken by both the players to reach the same spot is equal.
Time taken by A = y/(5X/3) secs --> (i)
and B = (y-80)/X secs --> (ii)
Now eqating the equation i and ii we get ,
y/5X/3= (y-80)/X
3y= 5y-400
2y= 400
y= 200 m (ans.)

If A runs 1 times as fast as B then how can be the ratio is 5:3?

It should be 2:1.

A runs 1 2/3 times faster than B so it can be written as (1*3+2)/3 : 1 => 5/3 :1.On taking LCM we have 5 : 3

In the question, its is given that A runs 1 2/3 times as fast as B.

They why the speed of A : B = 5 : 3 is taken in solution.

Speed of A:B = 5:3.

Time A = Time B.

Distance A/Speed A = Distance B/Speed B.

Distance A/Distance B = 5/3.

A/A+80 = 5/3.

3A = 5A+400.

2A = 400 (Neglecting negative sign as its distance).

A = 200 m.

Given a is 1(2/3) as fast as b.

Ratio of their speeds a:b = 5:3.

Speed is directly proportional to distance traveled.

d1:d2 = 5:3.

Assume a travels 5 m in t time then b travels 3 m in the same time.

That's it and continue as explained in the answer.

Hello @Shivakrishna A and B to reach at same time i.e., 4 has not mentioned in the problem see the problem clearly write.

The difference between A and B is 80m.

Let x be the total distance to cover in race.

In the problem he given A and B to reach at same time. i.e.4.

Time taken by A = time taken by B.

x/5 = (x-80)/3.

3x = 5x - 400.

2x = 400.

x = 200.

x is B speed then A speed is 5x/3 let pole is at distance y.

Then y-80/x=t for b y/5x/3=t for a equate both you get y as 200,

Why to equate? because both should come at same time.