Aptitude - Time and Distance - Discussion

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.

Well I would make it simple !!
first, let us find the time (in hrs)
let t be the time

distance/speed = time, distance we don't know so take it as x.

here there are two references given with respect to time.
he once reached at 2pm which can be taken as t .
he again reached at 12pm so that can be taken as t - 2.

So now there are two times , (t) and (t-2).
substitute this in formula (above distance\speed = time).
x/10 - x/15 = (t) - (t-2) (difference of two times, two distances etc),
now, (15x - 10x) / 150 = 2,
5x = 300,
x=60km.

Now we have found the distance travelled

Again substitute this in formula
speed = d/t.

1st case
10= 60 / t = 6hrs.

2nd case
15 = 60 / t = 4hrs
so if in the frst case he has travelled for 6 hours and reached at 2pm. so he would have started at 8am.

Now the main problem, at what speed he must travel so that he can reach at 1pm?

he starts at 8 am and he should reach at 1 pm so his travel must be 5 hrs.

Now substitute 5hrs in the formula speed =d/t.

speed = 60/5,
speed = 12 kmph.

Hi,

Suppose Robert has started traveling on his cycle = x hr
Suppose distance to be covered is = y km

case 1:-

speed = 10 kmph
time to reach point A = 2 pm or 14 (no am or pm : just standard time)

hence
distance = speed * time
y = 10 (14 - x) -- eq 1


case 2 :-

speed = 15 kmph
time to reach point A = 12 noon or 12 (no am or pm : just standard time)

hence
distance = speed * time
y = 15 (12 - x) -- eq 2

since distance traveled is the same hence equating eq1 and eq2.

10(14 - x) = 15(12 -x)

hence solving

2(14 - x) = 3(12 -x)

28 - 2x = 36 -3x

x = 8

hence Robert started at 8 hrs i.e 8 AM.

Now equating the value of x in eq1 or eq2 to get the value of distance travelled.

y = 10(14 -8)

hence y = 60 and hence distance = 60 km.

Now,

time to reach at A : - 1 pm or 13 hrs
speed = z kmph (??)
distance = 60 km

60 = z(13 -8)
60 = 5z

z = 60/5

hence z = 12 or speed = 12 kmph.

Hope this helps.

Thanks

By,
speed 1 (10 kmph) he would reach at point A at 2 P.M.
speed 2 (15 kmph) he would reach at point A at 12 noon.
we know that speed = distance time ^ -1
let the distanc be x
then
x/10 - x/15 = 2
3x - 2x /30 = 2
x/30 = 2
x = 60
the distance is 60 km
@the speed of 10 kmph he would require 10 hours to reach point A.
therefore he left his point of starting 6 hours ago.
time was 14:00 hours - 6 hours = 8 hours. or 8 A.M.
total between 6 a.m. and
08:00 hours - 13:00 hours = 05:00 hours
time = 05:00 hours
distance = 60 km [ we have already found out]
speed = ?
speed = distance time^-1
60 / 5 - 12
therefore, the speed = 12 kmph.

We need to calculate the distance, then we can calculate the time and finally speed .

Lets solve this,

Let the distance travelled by x km.

Time = Distance/Speed.

x10-15 = 2[because, 2 pm - 12 noon = 2 hours] 3x-2x=60 =>x=60.

Time = Distance*Speed,

Time@10km/hr = 60/10 = 6 hours

So 2 P.M. - 6 = 8 A.M

i.e. Robert starts at 8 A.M.


He have to reach at 1 P.M.
i.e,diff between 8 A.M to 1P.M in 5 hours

So, Speed = distance/time => 60/5 = 12 km/hr

It travelled the same distance. Let total time taken in first case is t.

Therefore,

Distance = 10*t..........{1st case}.

= 15*(t-2) .....{2nd case}.

= x*(t-1).......{3rd case} let x be the speed.

Now,equating 1st case and 2nd case as both are the same distance.

10*t=15(t-2).

=> 10t=15t-30.

=> 5t=30.

=> t=6.

So distance = 10*6=60.

Therefore 60=x*(6-1)......{3rd case}.

=>60=5x.

=>x=12 kmph.

Consider starting time as x.

We know that time = (distance/speed),
When he reach at 2 P.M i.e.,14.00 (railway time)
(14-x) = (distance/ 10)

When he reach at 12 P.M
(12-x) = (distance/15)

As we know that., distance is the same.
So, from above equations we get x=8. so he started cycling at 8 A.M.

So from this, we can calculate distance = 60 km.
To reach at 1 P.M, the time taken is 5 hrs.
Speed = (60/5) = 12 kmph.

Distance is same in both the cases:

So :: let's the time taken to reach at 2 pm be t.

And then time taken to reach at 12 pm will be t-2(2-12 = 2 hours).

As distance is same:

t*10 = t-2*15.

Giving t = 6 hours (to reach at 2, time taken is 6 hours).

Total distance = 6*10 = 60 KM.

Now to reach at 1 PM speed = d/t.

s = 60/t-1 = 60/6-1 = 12.

Why 6-1? Because (2 pm - 1 pm = 1 hour and we calculated t as 6).

Hope this helps.

Same distance but with different speeds.

The difference in time is 2 hours.

Let the distance be x.

X/10-x/15 = 2.

=> x = 60 kms.

Calculate time in first case i.e. 60/10 = 6 hours.

=> he took 6 hours to reach the destination.

The question is at 1 p.m => speed to reach destination in 5 hours.

We know that speed is inversely proportional to time.

s1/s2 = t2/t1.

10/x = 5/6.

=> x = 12 kmph.

Let the distance has been travelled be x km.
At the speed of 10 kmph, it will reach 2pm.(equation 1.)
At the speed of 15, it will reach at 12 noon.(equation 2)
And,

At the speed of ?, it will reach at 1pm. (equation 3)
Let's balance the equation.1 and 2,
x/10-x/15=2 hours,
i.e. x=60km.

Now again balance the equation 2 and 3,
60/?-60/15=1 hours,
?=12,

Thus, 12 kmph is the answer.