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.
Divya said:
1 decade ago
Lets try a hit and trial method :
By increasing his speed by 5 km/hr (15-10) , he is saving 2 hours (2pm-12pm) , so when he reaches at 1 pm he increased speed by 2.5 (half of 5) = 12.5, from the options 12 is closest so that must be the answer :).
By increasing his speed by 5 km/hr (15-10) , he is saving 2 hours (2pm-12pm) , so when he reaches at 1 pm he increased speed by 2.5 (half of 5) = 12.5, from the options 12 is closest so that must be the answer :).
Hyndavi said:
1 decade ago
Easy method:
Better take an average speed.
avg speed = 2xy/x+y.
x is 1st speed and y is 2nd speed.
So. here if we c...1pm is between 12pm and 2pm.
Calculating avg speed:
2xy/x+y = (2*10*15)/(10+15).
= 300/25.
= 12km/hr.
Better take an average speed.
avg speed = 2xy/x+y.
x is 1st speed and y is 2nd speed.
So. here if we c...1pm is between 12pm and 2pm.
Calculating avg speed:
2xy/x+y = (2*10*15)/(10+15).
= 300/25.
= 12km/hr.
Rahul Mathur said:
1 decade ago
Take 2 pm as a reference time say--- t.
Let the distance to be covered be--- 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 12kmph.
Let the distance to be covered be--- 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 12kmph.
Vivek said:
2 years ago
Case1: speed = 10kmph.
Ce2: speed = 15kmph.
Given d/10-d/15 = 2,
d=60,
case1 time=60/10, time=6,
case2 time=60/15, time=4,
so, for 12noom time is 4, and for 2 pm is 6
And 1 pm is between them so it will be 5.
And 60/5=12.
Ce2: speed = 15kmph.
Given d/10-d/15 = 2,
d=60,
case1 time=60/10, time=6,
case2 time=60/15, time=4,
so, for 12noom time is 4, and for 2 pm is 6
And 1 pm is between them so it will be 5.
And 60/5=12.
(27)
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.
Pavan sai said:
7 years ago
It's very easy by using relative speed concept there are two things up there.
So, now by using the formula 2xy/x+y.
where x=10 and y=15,
=> 2*10*15/10+15,
=>300/25,
=>12,
So the answer is 12kmph.
So, now by using the formula 2xy/x+y.
where x=10 and y=15,
=> 2*10*15/10+15,
=>300/25,
=>12,
So the answer is 12kmph.
(2)
Aasa sri said:
1 year ago
x/10 - x/15 = 2
1/5(x/2-x/3) = 2
3x-2x/6 = 2 * 5
x/6 = 10
x = 60.
60kmph at 10kmph is 60/10 = 6.
So, 12 noon time is 4 and 2 pm time is 6.
Then 1 pm time is b/w them so 5.
The speed is 60/5 = 12kmph.
1/5(x/2-x/3) = 2
3x-2x/6 = 2 * 5
x/6 = 10
x = 60.
60kmph at 10kmph is 60/10 = 6.
So, 12 noon time is 4 and 2 pm time is 6.
Then 1 pm time is b/w them so 5.
The speed is 60/5 = 12kmph.
(8)
Muhammad ali said:
7 years ago
Why don't you just find the mean value?
With 10kmph reach at 2 p.m.
Witn 15kmph reach at 12noon.
Find mean of both for 1p.m required time.
So, 10+15 = 25/2 = 12.5kmph should be the correct answer.
With 10kmph reach at 2 p.m.
Witn 15kmph reach at 12noon.
Find mean of both for 1p.m required time.
So, 10+15 = 25/2 = 12.5kmph should be the correct answer.
(1)
Anuj said:
3 years ago
10 kmph (2 pm) = t h ----> eq(1)
x kmph (1 pm) = (t-1)h ----> eq(2)
15 kmph (2 pm) = (t-2)h ----> eq(3)
Since, distance is the same for all cases we can equate these equations to get the answer.
x kmph (1 pm) = (t-1)h ----> eq(2)
15 kmph (2 pm) = (t-2)h ----> eq(3)
Since, distance is the same for all cases we can equate these equations to get the answer.
(6)
Erigela vani said:
6 years ago
@Shivasai.
I think here the difference between the two times I. E12noon and 2pm is 2, and also time =d÷s, so we should take distance as "X" and speeds which they are given so x÷10-x÷15=2.
I think here the difference between the two times I. E12noon and 2pm is 2, and also time =d÷s, so we should take distance as "X" and speeds which they are given so x÷10-x÷15=2.
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