Aptitude - Time and Distance - Discussion

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.

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.

@All.

In the above question, only time and speed are varying so distance remains constant.


So, First convert time 40 min=40/60 hr.

Now let's begin.
we know v=d/t. therefore d=vt--> (1)

Now by the first condition.
v+3=d/(t-40/60) .
Therefore d=(v+3)(t-40/60).

By multiplying the brackets.

d=(vt)-(2v/3)+(3t)-2 --> (2).

Now we know distance is constant
Therefore equating equations (1) and (2).

vt = vt-2v/3+3t-2.
Solving this we get -2v/3+3t=2.
now multiply it by 3.
We get -2v+9t=6 ---> (3).

Now By the second condition.

v-2=d/(t+40/60),
therefore d=(v-2)(t+40/60),
therefore d=(vt) +(2v/3)-(2t)-(4/3)--> (4).

Now equating equations (1) and (4)
vt=vt+2v/3-2t-4/3
By solving we get 2v/3-2t=4/3
Now multiply by 3 on both sides
2v-6t=4 ---> (5).

Now adding equations (3) & (5)
-2v+9t=6
+2v-6t=4
We get 3t=10 therfore t=10/3 hr
Now put t=10/3 in equation (5).

2v-6(10/3)=4.
And 2v=4+20.

Therfore
v=12 kmph.

Now finally from equation (1)
d=vt=12*(10/3)
d=40 km..

Hope it helps.

(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.

First we find the speed using this concept.

2s1*s2/s1-s2.
2*3*2/3-2=12km/hr.

Now, if we increase speed by 3km/hr, it will become 12+3=15km/hr and 40min less
same as,if we decrease the speed by 2km/hr then it become 12-2=10km/hr and 40min more
we have s1=15kmhr and s2=10kmhr and the time difference is 80/60.

Let distance be x.
x/10 - x/15 =80/60,
3x-2x/30=4/3,
x=4/3*30=40km/hr.

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

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

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.

Thanks @Sachin Bhagat and @Dinesh.