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 |  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 8 of 10.
Xyz said:
2 decades ago
Its 2 hours
12 noon to 2pm = 2 hours
12 noon to 2pm = 2 hours
Nancy said:
1 decade ago
Yes, but why we are considering only 10 km/hr speed?
Why not 15 km/hr?
Why not 15 km/hr?
Arunkumar said:
2 decades ago
How to find that 2. We don't know the starting time. Then how that is calculated.
K.NAVEENA said:
9 years ago
2xy/(x + y) = [(2 * 10 * 15)/(10 + 15)] = 12kmph.
Marco said:
9 years ago
The average speed :
2*10*15/10 + 15.
= 300/25.
= 12 km/h.
2*10*15/10 + 15.
= 300/25.
= 12 km/h.
Kuru said:
1 decade ago
@Amir that was nice, and easy step to solve.
Thanks.
Thanks.
Bharat said:
1 decade ago
x/10-x/15=2. How did 60 comes?
Tarun said:
1 decade ago
Let the traveled distance X.
Then X/10-X/15 = 2.
X = 60 kmph.
Time taken to travel 60 km at 10 kmph is 6 hours.
So the Mr stylish Yankee Robert Peterson a billionaire person started 6 hours before 2 pm. i.e 8 pm.
Speed of Mr equals to 60/(8 pm-1 pm) = 12 kmph.
Then X/10-X/15 = 2.
X = 60 kmph.
Time taken to travel 60 km at 10 kmph is 6 hours.
So the Mr stylish Yankee Robert Peterson a billionaire person started 6 hours before 2 pm. i.e 8 pm.
Speed of Mr equals to 60/(8 pm-1 pm) = 12 kmph.
Param said:
1 decade ago
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.
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.
Harish said:
1 decade ago
At 15 kmph ---> 2 p.m.
_?_kmph ---> 1 p.m.
10 kmph ---> 12 a.m.
So in between 12 am and 2 pm, 1 pm lies in between.
(10+15)/2 = 12.5 which is nearer to 12.
_?_kmph ---> 1 p.m.
10 kmph ---> 12 a.m.
So in between 12 am and 2 pm, 1 pm lies in between.
(10+15)/2 = 12.5 which is nearer to 12.
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