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 3 of 10.
Abishek jesuraj said:
7 years ago
Well I would make it simple !!
first, let us find the time (in hrs)
let t be the time
distance/speed = time, distance we don't know so take it as x.
here there are two references given with respect to time.
he once reached at 2pm which can be taken as t .
he again reached at 12pm so that can be taken as t - 2.
So now there are two times , (t) and (t-2).
substitute this in formula (above distance\speed = time).
x/10 - x/15 = (t) - (t-2) (difference of two times, two distances etc),
now, (15x - 10x) / 150 = 2,
5x = 300,
x=60km.
Now we have found the distance travelled
Again substitute this in formula
speed = d/t.
1st case
10= 60 / t = 6hrs.
2nd case
15 = 60 / t = 4hrs
so if in the frst case he has travelled for 6 hours and reached at 2pm. so he would have started at 8am.
Now the main problem, at what speed he must travel so that he can reach at 1pm?
he starts at 8 am and he should reach at 1 pm so his travel must be 5 hrs.
Now substitute 5hrs in the formula speed =d/t.
speed = 60/5,
speed = 12 kmph.
first, let us find the time (in hrs)
let t be the time
distance/speed = time, distance we don't know so take it as x.
here there are two references given with respect to time.
he once reached at 2pm which can be taken as t .
he again reached at 12pm so that can be taken as t - 2.
So now there are two times , (t) and (t-2).
substitute this in formula (above distance\speed = time).
x/10 - x/15 = (t) - (t-2) (difference of two times, two distances etc),
now, (15x - 10x) / 150 = 2,
5x = 300,
x=60km.
Now we have found the distance travelled
Again substitute this in formula
speed = d/t.
1st case
10= 60 / t = 6hrs.
2nd case
15 = 60 / t = 4hrs
so if in the frst case he has travelled for 6 hours and reached at 2pm. so he would have started at 8am.
Now the main problem, at what speed he must travel so that he can reach at 1pm?
he starts at 8 am and he should reach at 1 pm so his travel must be 5 hrs.
Now substitute 5hrs in the formula speed =d/t.
speed = 60/5,
speed = 12 kmph.
Naveen said:
9 years ago
Can anyone explain in easy way? It is hard to understand.
Shahul said:
9 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.
Ashish said:
9 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:
8 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:
8 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:
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.
Sagar said:
8 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:
8 years ago
How you are saying starting 8am @Ashish.
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