# Aptitude - Races and Games - Discussion

Discussion Forum : Races and Games - General Questions (Q.No. 12)

12.

A runs 1 times as fast as B. If A gives B a start of 80 m, how far must the winning post be so that A and B might reach it at the same time?

Answer: Option

Explanation:

Ratio of the speeds of A and B = | 5 | : 1 = 5 : 3. |

3 |

Thus, in race of 5 m, A gains 2 m over B.

2 m are gained by A in a race of 5 m.

80 m will be gained by A in race of | 5 | x 80 | m | = 200 m. | |

2 |

Winning post is 200 m away from the starting point.

Discussion:

15 comments Page 1 of 2.
Md. Iqbal Hasan said:
3 years ago

Awesome explanation. Thanks @Shivakrishna.

Mainak Mukherjee said:
6 years ago

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.)

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.)

Salman said:
6 years ago

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

It should be 2:1.

It should be 2:1.

Anu said:
7 years ago

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

Navneet said:
8 years ago

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.

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

Pranky said:
8 years ago

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.

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.

(1)

Anonmyous said:
8 years ago

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.

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.

Anjali said:
1 decade ago

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

Shivakrishna said:
1 decade ago

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.

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.

Vinay sv said:
1 decade ago

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.

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.

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