Aptitude - Time and Distance - Discussion

Easy method:

Better take an average speed.

avg speed = 2xy/x+y.

x is 1st speed and y is 2nd speed.

So. here if we c...1pm is between 12pm and 2pm.

Calculating avg speed:

2xy/x+y = (2*10*15)/(10+15).

= 300/25.

= 12km/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.

Hi all, it was not specified that he traveled at the same distance?

Am still confused with the solutions.

Speed1=10km/hr.
Reach at=2 p.m.
Speed2=15km/hr.
Reach at = 12 a.m.

Let dist. be="x"km.
Now,
By formula dist/speed= time.
x/10-x/15=2-12
3x-2x = 60
x = 60km.

Now time taken when travelled with speed 15 km/hr.
15=60/t
t=4 hr.

i.e started at 8am
Then 60/5=x
x=12 kmph Answer.

Nice method @Rahul Mathur.

The only thing is you have calculated value of t=10 wrongly.

It comes out to be 6. Check out.

Take 2 pm as a reference time say--- t.
Let the distance to be covered be--- d.

Case I
d/t=10.

Case II
d/t-2=15.

Solving both equations we have t=10 and d=60.

To find d/t-1=?

Replacing values of d and t will give speed as 12kmph.

We can also solve dis que by this method,
(2*10*15)/(10+15)=average speed in covering same distance
=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

Can any one give me the clear explanation? please.

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.