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 3 of 10.
Harsha said:
1 decade ago
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.
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.
Tbabe said:
1 decade ago
Hi all, it was not specified that he traveled at the same distance?
Am still confused with the solutions.
Am still confused with the solutions.
Abhi said:
1 decade ago
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.
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.
Hyndavi said:
1 decade ago
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.
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.
Divya said:
1 decade ago
Lets try a hit and trial method :
By increasing his speed by 5 km/hr (15-10) , he is saving 2 hours (2pm-12pm) , so when he reaches at 1 pm he increased speed by 2.5 (half of 5) = 12.5, from the options 12 is closest so that must be the answer :).
By increasing his speed by 5 km/hr (15-10) , he is saving 2 hours (2pm-12pm) , so when he reaches at 1 pm he increased speed by 2.5 (half of 5) = 12.5, from the options 12 is closest so that must be the answer :).
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
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.
Sidhu said:
1 decade ago
It can be solve by using 2xy/x+y.
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.
Sree said:
1 decade ago
Average formula can be used when a person travels two equal distances at different speeds.
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