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Aptitude - Time and Distance - Discussion

Discussion Forum : Time and Distance - General Questions (Q.No. 15)
Time and Distance
15.
A man covered a certain distance at some speed. Had he moved 3 kmph faster, he would have taken 40 minutes less. If he had moved 2 kmph slower, he would have taken 40 minutes more. The distance (in km) is:
35
36 2
3
37 1
2
40
Answer: Option
Explanation:

Let distance = x km and usual rate = y kmph.

Then, x - x = 40     2y(y + 3) = 9x ....(i)
y y + 3 60

And, x - x = 40     y(y - 2) = 3x ....(ii)
y -2 y 60

On dividing (i) by (ii), we get: x = 40.

Discussion:
118 comments Page 8 of 12.
Sivaram said:   1 decade ago
There is simple method
3*40=120
2*40=80

Then subtract two we get 40 this simple
Uroosa said:   8 years ago
@Sivaram.

How can you multiply km/hr with minutes without changing it to hours?
Puneet negi said:   8 years ago
How can we get the answer in km when the time is given in minutes? @Shivraman.
Jiya said:   1 decade ago
Can somebody please solve this equation ->

X/y-2 + 40/60 = X/y+3 -40/60.
Atchu said:   9 years ago
@Sivaram.

Is it the correct way and suitable to all this type of questions?
Prasanna said:   1 decade ago
Can any of you please explain how 2 solve that equations in detail please.
Gursewak singj said:   9 years ago
It is tough to find the solution. Can anybody solve it in a simple way?
Singh said:   9 years ago
Simplest formulae (S + 3)/(s - 2) = 3t2/2t1,
Solving we get s = 12km/h.
Mani said:   8 years ago
Can we do like this?


(3*40)-(2*40) = D
120-80 = D
Therefore D = 40.
Sukanya said:   5 years ago
Can you please explain how to slove this equation and get 2y(y+3)=9x?
(9)
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