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 4 of 10.
Geeta said:
1 decade ago
Simple.
10 = d/t.
d = 10t.
15 = d/t-2.
15 = 10t/t-2.
15t-30 = 10t.
15t-10t = 30.
5t = 30.
t = 6.
S = d/t-1.
S = 10t/t-1 substitute from above.
S = 10*6/6-1.
S = 60/5.
S = 12 km/h.
10 = d/t.
d = 10t.
15 = d/t-2.
15 = 10t/t-2.
15t-30 = 10t.
15t-10t = 30.
5t = 30.
t = 6.
S = d/t-1.
S = 10t/t-1 substitute from above.
S = 10*6/6-1.
S = 60/5.
S = 12 km/h.
Nishit Patel said:
1 year 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.
(73)
Havoc said:
5 years ago
We can also use average speed formula.
Because 12 am to 2 pm time 2 hr.
Thus 12 am to 1 am 1 hr.
So, we can find use avg speed also,
The mid point time of 12 am and 2 pm is 1 pm.
Because 12 am to 2 pm time 2 hr.
Thus 12 am to 1 am 1 hr.
So, we can find use avg speed also,
The mid point time of 12 am and 2 pm is 1 pm.
(3)
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.
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.
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.
Ritik Florian said:
5 years ago
Simply, find the average speed u can get the answer. Since 1 pm comes between 12 noon and 2 pm.
The formula for finding AVG speed is= 2xy/X+y.
i.e. 2*10*15/10+15= 12.
The formula for finding AVG speed is= 2xy/X+y.
i.e. 2*10*15/10+15= 12.
(11)
Aarusi said:
8 years ago
It can be solved easily by applying the average speed formula:
x = 10(given)(speed)
y = 15(given)(speed)
For average speed:
2xy/x + y
2 * 10 * 15/10 + 15 = 12.
x = 10(given)(speed)
y = 15(given)(speed)
For average speed:
2xy/x + y
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.
Tirtha said:
8 years ago
If speed is 10 then he will reach at .2.
If speed 15 then he will reach at 12.
so for 1 pm then we can take avg (10+15)/2=12.5 aprox 12.
Is this correct?
If speed 15 then he will reach at 12.
so for 1 pm then we can take avg (10+15)/2=12.5 aprox 12.
Is this correct?
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