Aptitude - Time and Distance - Discussion

Can somebody please solve this equation ->

X/y-2 + 40/60 = X/y+3 -40/60.

@All.

Certain distance = x.
Some speed = y.

If 3 kmph faster and 40 min less,

(x/y+3)-40/60..........1.

if 2 kmph slower and 40 min more means,

(x/y-2)+40/60..........2.

Now equate these 2 choices.

(x/y+3)-40/60 = (x/y-2)+40/60.

y = 12.

So

x = 40.

I hope you understood this equation.

Please give me your calculation of how did you get y=12 by solving above two equations ?

@Shathi.

Why sivaram method is not correct way?

Speed*time=distance.

Differnce of speed and time= distance.

3*40-2*40=x.

120-80=x.

40=x.

In problem when it says 40 min less then we substract.

But when it says 40 min more then how come we still substract?

Please tell basic concept while deciding wheter to add or substract.

Devide both the equations
2y(y+3) 9x
------- =
y(y-2) 3x

I am not understanding as not able to calculate this lines from the given explanation:
2y(y + 3) = 9x ....(i)
y(y - 2) = 3x ....(ii)

Somebody tell me step by step calculation of how to get above answer.

And also @ Amit :
not understanding this line
How you got Y=12?
and as per your equation:
why didn u consider x/y in the whole sum?
As in the question it is given a certain distance at some speed.
So we should take x/y also.
Your method looks bit simple but I got all this doubts so am asking here .
Please kindly someone explain me explicitly.
thank you :)

3 kmph (faster) - 2 kmph(slower) = 1kmph

1 kmph * 40 minutes = 40.

Can any of you please explain how 2 solve that equations in detail please.

Sivaram:I don't think its a correct way to solve this quesion.