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.
Anshu said:
1 decade ago
By,
speed 1 (10 kmph) he would reach at point A at 2 P.M.
speed 2 (15 kmph) he would reach at point A at 12 noon.
we know that speed = distance time ^ -1
let the distanc be x
then
x/10 - x/15 = 2
3x - 2x /30 = 2
x/30 = 2
x = 60
the distance is 60 km
@the speed of 10 kmph he would require 10 hours to reach point A.
therefore he left his point of starting 6 hours ago.
time was 14:00 hours - 6 hours = 8 hours. or 8 A.M.
total between 6 a.m. and
08:00 hours - 13:00 hours = 05:00 hours
time = 05:00 hours
distance = 60 km [ we have already found out]
speed = ?
speed = distance time^-1
60 / 5 - 12
therefore, the speed = 12 kmph.
speed 1 (10 kmph) he would reach at point A at 2 P.M.
speed 2 (15 kmph) he would reach at point A at 12 noon.
we know that speed = distance time ^ -1
let the distanc be x
then
x/10 - x/15 = 2
3x - 2x /30 = 2
x/30 = 2
x = 60
the distance is 60 km
@the speed of 10 kmph he would require 10 hours to reach point A.
therefore he left his point of starting 6 hours ago.
time was 14:00 hours - 6 hours = 8 hours. or 8 A.M.
total between 6 a.m. and
08:00 hours - 13:00 hours = 05:00 hours
time = 05:00 hours
distance = 60 km [ we have already found out]
speed = ?
speed = distance time^-1
60 / 5 - 12
therefore, the speed = 12 kmph.
Mano said:
1 decade ago
I agree to vishu comment. How that step came ?
Vishu said:
1 decade ago
x/10-x/15=2
=>3x-2x=60
I didn't get the proper method.
How it comes anyone could tell me briefly it will be helpful for me.
=>3x-2x=60
I didn't get the proper method.
How it comes anyone could tell me briefly it will be helpful for me.
Venkat said:
1 decade ago
Thank you Raja.
Raja said:
1 decade ago
Simple one
2pm-12=2 hour
Similarly 15km-10km=5km
5km/2hour=2.5km/h
initial 10+2.5=12.5 nearer to 12
2pm-12=2 hour
Similarly 15km-10km=5km
5km/2hour=2.5km/h
initial 10+2.5=12.5 nearer to 12
Srinu said:
1 decade ago
Hi vyshu
2 is time between 12noon and 1pm why subtracting
12noon - 2 pm = 2 it is time
x/15-x/10=2 x=60km
12noon time =60/15=4hrs(before) mean 8am
find speed 1pm----------8am to 1pm time is 5 hrs ,distance 60km,speed =? speed=60/5= 12kmph
2 is time between 12noon and 1pm why subtracting
12noon - 2 pm = 2 it is time
x/15-x/10=2 x=60km
12noon time =60/15=4hrs(before) mean 8am
find speed 1pm----------8am to 1pm time is 5 hrs ,distance 60km,speed =? speed=60/5= 12kmph
Pardhu said:
2 decades ago
Thank you aamir. Is this applicable for all these type methods. ?
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)
Aamir said:
2 decades ago
We can also solve through avearge speed its easy .x=10,y=15
2xy/x+y
2*10*15/10+15=
300/25=12
2xy/x+y
2*10*15/10+15=
300/25=12
(2)
Santhosh said:
2 decades ago
How to solve the equation first?
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