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.
Aarushi said:
2 years 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.
(18)
Mouni said:
2 years ago
Can anyone explain this in a simple way?
(5)
Ngawa said:
4 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.
(20)
Gauri said:
5 years ago
Thank you for the explanation @Akhil.
(1)
Aksh said:
6 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)
Prince Priyadarshi said:
6 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)
MOHANAGIRI said:
6 years ago
@Raja.
Clear explanation. Thanks for that.
Clear explanation. Thanks for that.
Raja said:
7 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.
(8)
Siddharth said:
7 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)
Anandu said:
7 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.
(2)
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