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 9 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.
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.
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.
Ammad Arshad said:
1 decade ago
10km/h = d/x; -- (1);
15km /h = d/(x-2);--- (2);
By comparing both equation 1 and 2 we get total time.
15(x-2) = 10x;
15x -30 = 10x;
5x = 30;
x = 6h;
Put the value of x in equation 1 we got distance.
d = 60km.
As from the state we know that.
y = d/(x-1); -- (3);
Put the value of d and x in equation 3.
y = 60/(6-1).
y = 60/5.
y = 12 kmph Answer.
15km /h = d/(x-2);--- (2);
By comparing both equation 1 and 2 we get total time.
15(x-2) = 10x;
15x -30 = 10x;
5x = 30;
x = 6h;
Put the value of x in equation 1 we got distance.
d = 60km.
As from the state we know that.
y = d/(x-1); -- (3);
Put the value of d and x in equation 3.
y = 60/(6-1).
y = 60/5.
y = 12 kmph Answer.
Shahul said:
10 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.
Karthika said:
1 decade ago
Same distance but with different speeds.
The difference in time is 2 hours.
Let the distance be x.
X/10-x/15 = 2.
=> x = 60 kms.
Calculate time in first case i.e. 60/10 = 6 hours.
=> he took 6 hours to reach the destination.
The question is at 1 p.m => speed to reach destination in 5 hours.
We know that speed is inversely proportional to time.
s1/s2 = t2/t1.
10/x = 5/6.
=> x = 12 kmph.
The difference in time is 2 hours.
Let the distance be x.
X/10-x/15 = 2.
=> x = 60 kms.
Calculate time in first case i.e. 60/10 = 6 hours.
=> he took 6 hours to reach the destination.
The question is at 1 p.m => speed to reach destination in 5 hours.
We know that speed is inversely proportional to time.
s1/s2 = t2/t1.
10/x = 5/6.
=> x = 12 kmph.
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.
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