Aptitude - Time and Distance - Discussion

Discussion Forum : Time and Distance - General Questions (Q.No. 12)
12.
Robert is travelling on his cycle and has calculated to reach point A at 2 P.M. if he travels at 10 kmph, he will reach there at 12 noon if he travels at 15 kmph. At what speed must he travel to reach A at 1 P.M.?
8 kmph
11 kmph
12 kmph
14 kmph
Answer: Option
Explanation:

Let the distance travelled by x km.

Then, x - x = 2
10 15

3x - 2x = 60

x = 60 km.

Time taken to travel 60 km at 10 km/hr = 60 hrs = 6 hrs.
10

So, Robert started 6 hours before 2 P.M. i.e., at 8 A.M.

Required speed = 60 kmph. = 12 kmph.
5

Discussion:
93 comments Page 2 of 10.

Raja said:   1 decade ago
Simple one

2pm-12=2 hour

Similarly 15km-10km=5km

5km/2hour=2.5km/h

initial 10+2.5=12.5 nearer to 12

Venkat said:   1 decade ago
Thank you Raja.

Vishu said:   1 decade ago
x/10-x/15=2
=>3x-2x=60

I didn't get the proper method.

How it comes anyone could tell me briefly it will be helpful for me.

Mano said:   1 decade ago
I agree to vishu comment. How that step came ?

Anshu said:   1 decade ago
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.

TEJU said:   1 decade ago
Can any one give me the clear explanation? please.

Ankit said:   1 decade ago
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

Saba Azmat said:   1 decade ago
We can also solve dis que by this method,
(2*10*15)/(10+15)=average speed in covering same distance
=12 kmph

Rahul Mathur said:   1 decade ago
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.

Aarti said:   1 decade ago
Nice method @Rahul Mathur.

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

It comes out to be 6. Check out.


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