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 3 of 10.
Vyshu said:
2 decades 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.
(1)
Raman Sharma said:
8 years ago
Distance = speed * time.
For time ,let say time taken to travel to point A = t.
d = 15*t ------- eqn.1
d = 10(t+2) ------eqn. 2
d = s* (t+1)--------- eqn. 3.
From eqn 1 and eqn 2
15* t = 10(t+2).
after solving, t = 4 means time to travel is 4 hours.
now,
we put value of t = 4 in eqn 3 and eqn. 2.
60 = s(4+1).
s = 12 kmph.
For time ,let say time taken to travel to point A = t.
d = 15*t ------- eqn.1
d = 10(t+2) ------eqn. 2
d = s* (t+1)--------- eqn. 3.
From eqn 1 and eqn 2
15* t = 10(t+2).
after solving, t = 4 means time to travel is 4 hours.
now,
we put value of t = 4 in eqn 3 and eqn. 2.
60 = s(4+1).
s = 12 kmph.
Vinaykumar said:
10 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.
Ashish said:
10 years ago
Let the distance be covered 'd'.
Case I: d/t = 10.
Case II: d/t - 2 = 15.
Solving both equations we have t = 10 and d = 60.
To find d/(t-1) = ?
Replacing values of 'd' and 't' will give speed as 12 kmph.
Case I: d/t = 10.
Case II: d/t - 2 = 15.
Solving both equations we have t = 10 and d = 60.
To find d/(t-1) = ?
Replacing values of 'd' and 't' will give speed as 12 kmph.
Neha said:
9 years ago
We can solve this problem by using avg.speed 2xy/x+y.
2 * (10 + 15)/10 + 15,
=12 km/h.
2 * (10 + 15)/10 + 15,
=12 km/h.
Prawesh Pradhan said:
9 years ago
After finding out the distance.
60/10 - 60/x = 1,
5/x = 1,
x = 5.
60/10 - 60/x = 1,
5/x = 1,
x = 5.
Prawesh Pradhan said:
9 years ago
After finding the distance.
60/10-60/x = 1,
6x-60 = x,
5x = 60,
x = 12.
60/10-60/x = 1,
6x-60 = x,
5x = 60,
x = 12.
Aarusi said:
9 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.
Sagar said:
9 years ago
8 am to 1 pm= 5 hours.
Distance:-60 and time:-5.
Therefore: 60/5 is taken at last step.
i.e: speed:-distance/ time.
Distance:-60 and time:-5.
Therefore: 60/5 is taken at last step.
i.e: speed:-distance/ time.
Ranjith said:
9 years ago
How you are saying starting 8am @Ashish.
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