Aptitude - Time and Distance - Discussion

For simplifying method, follow easy steps which are lengthy but clear understanding.

(x/y)-(x/y+3)=40/60 ---> (1)
(x/y-2)-(x/y)=40/60 ---> (2)

Let solve eq. (1)
(x/y)-(x/y+3) = 40/60,
(x(y+3)-y(x)) / y(y+3) = 2/3,
((xy+3x-xy) / (y^2+3y)) = 2/3,
3(3x)=2(y^2+3y),
9x=2y(y+3) ---------> (A).

Now let solve eqn. (2)
(x/y-2)-(x/y) = 40/60,
((xy-x(y-2)) /(y(y-2))) = 2/3,
((xy-xy+2x) / (y^2-2y)) = 2/3,
3(2x) = 2(y^2-2y)
6x = 2y(y-2)
now divide by 2 we get
3x = y(y-2) --------------> (B)
now we get two equations as per given in answer are
9x = 2y(y+3) -------------->(A).
3x = y(y-2) -------------->(B).

Now we divide eqn (A) by (B) we get;
(2y(y+3)) / y(y-2) = 9x/3x,
2(y+3) / (y-2) = 3,
y+3 / y-2 = 3/2,
2(y+3) = 3(y-2),
y = 12.

After putting y=12 in equation (A) we get x = 40.

d=(S1S2/S1-S2)*t.
(S(S+3) / 3)*(40/60)=(S(S-2)/2)*(40/60).
2S+6= 3S-6.
S=12Kmph.
d=( (12*15)/3)*40/60.
d=40km.

Distance= s* t.

x= actual speed,
x +3= 40 min (faster),
x-2=40 min (slower),
d= 3 * 40= 120,
d=2*40=80,
120 -80= 40.

It's an actual answer.

@JAHNAVI.

Initially, distance is the same so to find time and speed, we take speed x and time and now implement the distance formula.

First, speed is greater than 3 we take x+3 and if the speed is increased time is decreased so t-40/60 we convert 40 min to an hour divided by 60. Second x-2 and t+40/60 solve these equations you get the why and then x.

I didn't understand this solution of the above mentioned question.Could you please give a little better explanation.

Thanks @Amit Kr. Soni.

The solution is;

2y(y + 3) = 9x --> (i)
y(y - 2) = 3x --> (ii)

So, 2y(y + 3) = 3*3x = 2y(y + 3) = 3* y(y - 2).

After Solving y=12 put y value any above 1 and 2 equation, and get x=40.

(40*3)-(40*2),
120-80 = 40.

Thanks @Sachin Bhagat.

d = st--->1.
d = (s+3)(t-2/3)------>2.
d = (s-2)(t+2/3)------>3.
Solving 1and 2 we get 2s/3 = 3t-2----->4.
2s/3 = 2t-4/3--------->5.
By solving 4and 5 we get t = 10/3 and s = 12 so d = st = 40km.