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:
95 comments Page 7 of 10.
Mohit said:
1 decade ago
Let time at 12noon be x hours.
Time at 2pm will be x+2.
Speed = distance*time.
So, 15*x = 10*(x+2).
x = 20/5 = 4km.
Distance = 15*4 = 60km.
Speed at 1pm = 60/5 = 12kmph.
Time at 2pm will be x+2.
Speed = distance*time.
So, 15*x = 10*(x+2).
x = 20/5 = 4km.
Distance = 15*4 = 60km.
Speed at 1pm = 60/5 = 12kmph.
Shashank said:
1 decade ago
We can use net speed formula = 2xy/x+y.
Because the distance travelled is same,
So => 2*10*15/(10+15) = 12.
Because the distance travelled is same,
So => 2*10*15/(10+15) = 12.
KISHORE said:
1 decade ago
@Shilpa. We have to go with subtraction in case the objects moving in same direction and go with addition in case the objects moving in opposite direction.
Shilpa said:
1 decade ago
Why we are subtracting here i.e., x/10-x/15=2;?
And in the next problem why we are adding. Can anyone explain when should add and when should subtract.
And in the next problem why we are adding. Can anyone explain when should add and when should subtract.
Sree said:
1 decade ago
For example :
A man travels 20 km at a speed of 15 km/hr and returns back at a speed of 10 km/hr. What is the average speed of the man in his journey ?
Sol:
Speed = 2*15*10/15+10.
= 300/25.
=12 km/hr.
I Think now you came to know where we can use the average formula.
A man travels 20 km at a speed of 15 km/hr and returns back at a speed of 10 km/hr. What is the average speed of the man in his journey ?
Sol:
Speed = 2*15*10/15+10.
= 300/25.
=12 km/hr.
I Think now you came to know where we can use the average formula.
Sree said:
1 decade ago
Average formula can be used when a person travels two equal distances at different speeds.
Pinky said:
1 decade ago
When came to know that on where we have to use this avg formula ?
Please anyone answer.
Please anyone answer.
Sidhu said:
1 decade ago
It can be solve by using 2xy/x+y.
Leelakrishna said:
1 decade ago
Since the distance traveled to be same equate d = v*t.
i.e. 10*(14-t) = 15*(12-t).
From above eq find t i.e. t = 8.
Substitute in 10*(14-8) = 60km.
So velocity = 60/5 =12kmph.
i.e. 10*(14-t) = 15*(12-t).
From above eq find t i.e. t = 8.
Substitute in 10*(14-8) = 60km.
So velocity = 60/5 =12kmph.
Ajit said:
1 decade ago
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
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
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers