Aptitude - Time and Distance - Discussion

Can we do like this?


(3*40)-(2*40) = D
120-80 = D
Therefore D = 40.

Thanks for your explanation @Akash Kishor.

See take distance =x and rate=y and t= time.
Now overall x/y=t.

We have one condition,
x/y+3 at t-40.

We have second condition,
x/y-y at t+40.

Now equate,
x/y=t and x/y+3=t-40 sub them,
You get x/y- x/y+3=40/60.

Now equate,
x/y=t and x/y-2=t+40 sub them,
You get x/y-2- x/y=40/60.

You will see t getting canceled.

40/60 because convert min into hr.

What if decreasing speed and increasing speed would be same?

When a person is moving with speed x km/hr is reaching office in t time. When he increases speed by 20km/hr, he reaches 40 min early. But when he decreases speed by 20 km/hr, he reaches 40 min late. What is the distance?

Can anyone tell me?

How will you get Y=12?

my calculation is:

3x+2y-4/3y-6 = 3x-2y-6/3y+9,
(3y+9)(3x+2y-4) =( 3y-6)(3x-2y-6),
9xy+6y^2+6y+27x-36 = 9xy-6y^2-6y-18x+36,
12y^2+12y+45x-72 = 0.

please explain me.

Short trick :

3 * (+40) : 2 *(-40)
120: 80 ==> 3 : 2

Then take Avg for the both speed of the Ratio
2(2*3)/1 ==> same direction so 2(XY) / X-Y.

Then we get original speed 12kmhr.
then ,12+3 = 15kmhr
12-2= 10kmhr

Diff for both time 40 - (-40) = 80mins.
The product of speed/ diff of speed *time
15*10/5 * 80/60.
Finally, we get 40 km distance.

Why we have to divide both equations?

Why can't do either addition or subtraction?

Let me know the reason, please.

Let Speed be 's' and Time be 't';
Original Distance= st--------------> (i)
If speed and time changes,
New Speed=s+3
New time= t-2/3 (40/60=2/3)
So again D= (s+3)(t-2/3) = st + 3t - 2/3s - 2
from (i), D=st
=> st= st + 3t - 2/3s - 2
=> 3t - 2/3s = 2--------------> (ii)
Similarly,
D= (s-2)(t+2/3) = st - 2t + 2/3s - 4/3
and D= st
putting value of D,
=> st = st - 2t + 2/3s - 4/3
=> - 2t + 2/3s = 4/3 --------------> (iii)
Now solving (ii) and (iii),
3t - 2/3s = 2
- 2t + 2/3s = 4/3
________________
t = 10/3
Putting value of t in (ii),
=> 3*10/3 - 2/3s = 2,
=> 10-2 = 2/3s,
=> 2/3s = 8,
=> s=12,
So, D=st,
=12*10/3,
= 40 km Ans..

Man goes to his office at a certain speed. If he travels at a speed 4kmph more than his usual speed then he will reach his office early by 12 minutes. If he travels at a speed 8 kmph less than his usual speed he will be late by 36 minutes. Find his original speed.

How would we solve it?

S(s+3)*40/60 = s(s-2)*40/60
----------------------. ------------------
3. 2

S=12 putting in any of side l.h.s or r.h. s we get required answer 40.