Aptitude - Time and Distance - Discussion

Good thanks for explaining.

@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 am not understanding this. Please help me to get it.

(a(a-2)/2) * 40 = (a(a+3)/3) *40.
a=12kmph this is original speed.

Ddistance = (12(12-10)/2)*40/60.
Distance = 40km.

Divide both the equations.

2y(y+3) 9x.
------- =
y(y-2) 3x.

Solution:

2(y+3)÷y(y-2) = 9x÷ 3x.
2(y+3)÷(y-2) = 3,
2(y+3) = 3(y-2),
2y+6= 3y-6.
3y-2y = 6+6.
Y =12.

I have easier solution to this:

Distance is same so first consider speed x and time t:

As per the question,
Increasing speed by 3 i.e. (x+3) will decrease time by 40 min i.e (t-40/60).
Therefore,
Distance = (x+3)(t-40/60).

Now Decreasing speed by 2 i.e. (x-2) will decrease time by 40 min i.e (t+40/60).
Therefore,

Distance = (x+3)(t-40/60).

Since Distance would be same:

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

solving this, we get

-2x/3 + 3t -2 = 2x/3 + -2t - 4/3.

3t -2x/3 =2 -->(2)
2x/3 + -2t = 4/3 --->(3)

Solving two-equation, we get t=10/3 and x=12. Putting this value in eq. 1, we get,

Distance = 40.

Can you please explain how to slove this equation and get 2y(y+3)=9x?

Thanks @Sachin Bhagat and @Dinesh.

3times more -> Dist= 3 * 40= 120.
2tines less -> Dist= -2* 40= -80.
So, 120 - 80 = 40kmph.

Distance is same.
Therefore d = d.
3(x-40) = 2(x+40),
3x+120 = 2x+80,
3x-2x = 120-80,
X = 40km.