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 5 of 10.
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.
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.
Bharat said:
1 decade ago
x/10-x/15=2. How did 60 comes?
Kuru said:
1 decade ago
@Amir that was nice, and easy step to solve.
Thanks.
Thanks.
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.
K.NAVEENA said:
9 years ago
2xy/(x + y) = [(2 * 10 * 15)/(10 + 15)] = 12kmph.
Naveen said:
9 years ago
Can anyone explain in easy way? It is hard to understand.
Vinaykumar said:
9 years ago
Simple :.
2 speeds calculate average speed 2xy/(x + y).
So,
2 * 10 * 15/(10 + 15).
12km/hr.
2 speeds calculate average speed 2xy/(x + y).
So,
2 * 10 * 15/(10 + 15).
12km/hr.
Suma said:
9 years ago
If two distance is equal, here two distance is equal.
Then speed = 2s1s2/s1 + s2.
2 * 10 * 15/(10 + 15) = 12.
Then speed = 2s1s2/s1 + s2.
2 * 10 * 15/(10 + 15) = 12.
(1)
Shahul said:
9 years ago
1st case-.
Speed = 10.
Time = t.
Distance = 10 * t -----> (1).
2nd case.
Speed = 15.
Time = t - 2 (he reach at 2pm at t time so take t-2 for 12pm).
D = 15 * (t - 2) -----> (2).
From 1 and 2.
10 * t = 15 * (t - 2).
15t - 10t = 30.
t = 6.
So he need 5hrs to reach at 1pm (take 6hrs to reach 2pm and will take 5hrs to reach at 1).
D = 10 * t = 10 * 6 = 60.
Speed = 60/5 = 12.
Speed = 10.
Time = t.
Distance = 10 * t -----> (1).
2nd case.
Speed = 15.
Time = t - 2 (he reach at 2pm at t time so take t-2 for 12pm).
D = 15 * (t - 2) -----> (2).
From 1 and 2.
10 * t = 15 * (t - 2).
15t - 10t = 30.
t = 6.
So he need 5hrs to reach at 1pm (take 6hrs to reach 2pm and will take 5hrs to reach at 1).
D = 10 * t = 10 * 6 = 60.
Speed = 60/5 = 12.
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