Aptitude - Time and Distance - Discussion

Can somebody please solve this equation ->

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

Hello @Jiya.

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

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

y-2/x - y+3/x = -3/2.

(y-2-y-3)/x = -3/2.

-5/x = -3/2.

x = 10/3.

Please can anyone tell how to write these equation?

x/y - x/y+3 = 40/60.

x/y-2 - x/y = 40/40.

Please give some basic concept.

I don't know when to subtract and when to add.

Please give me idea on this.

Here in equation it is taken minus.

And in previous example it was taken add.

Let distance be the x km and y be the usual speed and t be the time,
=> t = x/y......(1).

Since, speed is inversely prop. to time.
i.e. faster speed -> slower time.

Now, original time(t)- slow time(t1).

=> x/y - x/y+3 = 40/60 => 9x = 2y(y+3)........(2).

And,

fast time(t2) - original time(t)

=> x/y-2 - x/y = 40/60 => 6x = 2y(y-2)........(3)

So dividing 2 by 3 we get y = 12.

Now put y= 12 in (2) we get,

9x = 24(12+3).

=> x = 24*15/9 = 5*8 = 40.

Hence distance = 40.

x/y - x/y+3 = 40/60 => xy+3x-xy = 2/3y(y+3) => 9x = 2y(y+3)......(2).


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


Multiply eq(3)in 2 and solve (2)&(3).

We got y = 3x/10.


Then substitute y = 3x/10 in eq(3).

3x = 3x/10(3x/10 -2).

3x = 3x/10(3x-20/10).

3x = 3x(3x)-60x/100.

300x = 3x(3x)-60x.

3x(3x)-360x = 0.

x(x)-40x = 0.

x(x-40) = 0.

x = 40.

Hi friends please tell me how to solve this equation.

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

How you got y=12 from d above equations I did not understood?

Still am not able to understand, its confusing !

All I have is v+3=s/ (t-2/3).

And v-2=s/ (t+2/3). With these two eqns. How do we solve it?