Aptitude - Time and Distance - Discussion

Let calculate distance firs say d.

D/10-d/15=2 (he reach 12noon 2hr early).

By calculating we get d = 60km.

Let's speed is S to calculate S difference between travel at first speed rate and S rate is.
60/10-60/S=1 (because he reaches by 1 am, 1hr early).

By calculating this we get.

5S=60.
S = 12kmph.

Simply, find the average speed u can get the answer. Since 1 pm comes between 12 noon and 2 pm.

The formula for finding AVG speed is= 2xy/X+y.
i.e. 2*10*15/10+15= 12.

Let the distance be x.

According to statement 1.
If he travels at 10 kmph and let's imagine the time taken to reach at 2 pm is 't' hrs

i.e, x/10=t ---> equation(1).

According to statement 2.

If he travels at 15 kmph and reaches at 12 noon, then he reaches 2 hrs earlier than the previous time (i.e, t-2 hrs)

then, x/15=t-2 ------> equation(2)

If you get two equations with same two variables, always subtract one equation from the other.

Here, the equations become,
x/10-x/15=t-(t-2) or x/15-x/10=(t-2)-t
either way you will get the same answer i.e, x=60 km.

Just Put this in equation(1) or equation(2) and you will get the original time.
equation(1), 60/10=t, t=6.
equation(2), 60/15=t-2, 4=t-2, t=6.

So if Robert wants to reach the destination at 1 pm, We have to calculate his speed, we got distance and time taken.

If he wants to reach at 1 pm which is 1 hr less than the normal time.
then equation becomes distance/speed=time-1 hr.

i.e, 60/speed=6-1, speed=60/5, speed=12 kmph.
Therefore, the answer is 12 kmph.

We can also use average speed formula.

Because 12 am to 2 pm time 2 hr.
Thus 12 am to 1 am 1 hr.
So, we can find use avg speed also,
The mid point time of 12 am and 2 pm is 1 pm.

1pm is in between 12 and 2 pm.
So, if we find average speed we can also get the answer.
Avg speed = (2*15*10)/(10+15) = 12.

@All.

Use formula like (2 (a.b) /a+b) a is speed 10kmph and b is 15kmph.

Hope you get it.

How 3 came? Please explain.

10 kmph (2 pm) = t h ----> eq(1)
x kmph (1 pm) = (t-1)h ----> eq(2)
15 kmph (2 pm) = (t-2)h ----> eq(3)
Since, distance is the same for all cases we can equate these equations to get the answer.

@Farzana.

Best and neat explanation, Thank you.

Case1: speed = 10kmph.
Ce2: speed = 15kmph.
Given d/10-d/15 = 2,
d=60,
case1 time=60/10, time=6,
case2 time=60/15, time=4,

so, for 12noom time is 4, and for 2 pm is 6
And 1 pm is between them so it will be 5.

And 60/5=12.