# Aptitude - Races and Games - Discussion

Discussion Forum : Races and Games - General Questions (Q.No. 1)

1.

In a 100 m race, A can give B 10 m and C 28 m. In the same race B can give C:

Answer: Option

Explanation:

A : B = 100 : 90.

A : C = 100 : 72.

B : C = | B | x | A | = | 90 | x | 100 | = | 90 | . |

A | C | 100 | 72 | 72 |

When B runs 90 m, C runs 72 m.

When B runs 100 m, C runs | 72 | x 100 | m | = 80 m. | |

90 |

B can give C 20 m.

Discussion:

43 comments Page 1 of 5.
Ngawa said:
3 years ago

Difference between B and C

28 - 10 =18.

Now,

18 /90 * 100= 20.

Thus give 20m in 100 game.

28 - 10 =18.

Now,

18 /90 * 100= 20.

Thus give 20m in 100 game.

(11)

Aarushi said:
6 months ago

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.

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.

(6)

Raja said:
6 years ago

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.

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.

(6)

Mouni said:
1 year ago

Can anyone explain this in a simple way?

(3)

Aksh said:
5 years ago

A gives B 10m means A beats B by 10m, A wins the race, by completing 100m, ie B will be at 10.

(2)

Siddharth said:
6 years ago

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.

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.

(2)

Gauri said:
4 years ago

Thank you for the explanation @Akhil.

(1)

Prince Priyadarshi said:
5 years ago

@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.

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.

(1)

Anandu said:
6 years ago

@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.

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.

(1)

Aarti said:
8 years ago

How to solve this problems in easiest way?

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