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 10 of 10.
Aasa sri said:
11 months ago
x/10 - x/15 = 2
1/5(x/2-x/3) = 2
3x-2x/6 = 2 * 5
x/6 = 10
x = 60.
60kmph at 10kmph is 60/10 = 6.
So, 12 noon time is 4 and 2 pm time is 6.
Then 1 pm time is b/w them so 5.
The speed is 60/5 = 12kmph.
1/5(x/2-x/3) = 2
3x-2x/6 = 2 * 5
x/6 = 10
x = 60.
60kmph at 10kmph is 60/10 = 6.
So, 12 noon time is 4 and 2 pm time is 6.
Then 1 pm time is b/w them so 5.
The speed is 60/5 = 12kmph.
(7)
Rabin said:
10 months ago
Thanks for the explanation @Farzana.
(4)
Nishit Patel said:
9 months ago
The question is what's the average speed of Robert.
Avg Speed = 2xy/x+y.
= 2 × 10 × 15/10 + 15.
= 300/25.
= 12 km/h.
Avg Speed = 2xy/x+y.
= 2 × 10 × 15/10 + 15.
= 300/25.
= 12 km/h.
(49)
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