Aptitude - Time and Distance - Discussion

3 * 40 = 120,
2 * 40 = 80,
120-80 = 40.
Does this process will come? Please tell me.

Hi,
3*40 =120
2*40 = 80
120-80 = 40.

In this one quantity is in hr and second one is in minutes how you cancel the different units.
km/hr*min not relevant.

LET'S SOLVE IN VERY EASY MANNER.

Easiest formula :- DISTANCE = (s1*s2)/(speed difference) * time difference
So, now
D= {s*(s+3)/3} * 40/60 = {s*(s-2)/2} * 40/60.
=> (s+3)/3 = (s-2)/2 [ s and 40/60 will cancel out each other at both side],
=> 2s+6 = 3s-6,
=> s = 12 km/hr.
And now,
D= (12*15)/3 * 40/60 = 40 km Ans. [ according to formula].

3/x+3/2/x-2= 40/40.
On solving we get x = 12.

Please, can anyone solve this equation in details?

On dividing (i) by (ii), we get: x = 40.

@Sivaram.

How can you multiply km/hr with minutes without changing it to hours?

@Akash.

I can't understand why have you divided by e3 and 2 on both side of equations. Can you explain please?

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.

How can we get the answer in km when the time is given in minutes? @Shivraman.

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..