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.?
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 | ![]() |
= 6 hrs. |
10 |
So, Robert started 6 hours before 2 P.M. i.e., at 8 A.M.
![]() |
![]() |
60 | ![]() |
= 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
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.
=>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.
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
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
(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.
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.
The only thing is you have calculated value of t=10 wrongly.
It comes out to be 6. Check out.
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