Aptitude - Time and Distance - Discussion

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

Hi friends please tell me how to solve this equation.

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

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.

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.

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.

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.

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.

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 ?