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.
Sree said:
1 decade ago
For example :
A man travels 20 km at a speed of 15 km/hr and returns back at a speed of 10 km/hr. What is the average speed of the man in his journey ?
Sol:
Speed = 2*15*10/15+10.
= 300/25.
=12 km/hr.
I Think now you came to know where we can use the average formula.
A man travels 20 km at a speed of 15 km/hr and returns back at a speed of 10 km/hr. What is the average speed of the man in his journey ?
Sol:
Speed = 2*15*10/15+10.
= 300/25.
=12 km/hr.
I Think now you came to know where we can use the average formula.
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.
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.
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.
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.
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.
Nancy said:
1 decade ago
Yes, but why we are considering only 10 km/hr speed?
Why not 15 km/hr?
Why not 15 km/hr?
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.
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.
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.
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