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 | ![]() |
= 6 hrs. |
10 |
So, Robert started 6 hours before 2 P.M. i.e., at 8 A.M.
![]() |
![]() |
60 | ![]() |
= 12 kmph. |
5 |
Discussion:
93 comments Page 1 of 10.
Prathamesh Pendal said:
4 years ago
1pm is in between 12 and 2 pm.
So, if we find average speed we can also get the answer.
Avg speed = (2*15*10)/(10+15) = 12.
So, if we find average speed we can also get the answer.
Avg speed = (2*15*10)/(10+15) = 12.
(56)
Nishit Patel said:
9 months ago
The question is what's the average speed of Robert.
Avg Speed = 2xy/x+y.
= 2 × 10 × 15/10 + 15.
= 300/25.
= 12 km/h.
Avg Speed = 2xy/x+y.
= 2 × 10 × 15/10 + 15.
= 300/25.
= 12 km/h.
(48)
Farzana said:
5 years ago
Let the distance be x.
According to statement 1.
If he travels at 10 kmph and let's imagine the time taken to reach at 2 pm is 't' hrs
i.e, x/10=t ---> equation(1).
According to statement 2.
If he travels at 15 kmph and reaches at 12 noon, then he reaches 2 hrs earlier than the previous time (i.e, t-2 hrs)
then, x/15=t-2 ------> equation(2)
If you get two equations with same two variables, always subtract one equation from the other.
Here, the equations become,
x/10-x/15=t-(t-2) or x/15-x/10=(t-2)-t
either way you will get the same answer i.e, x=60 km.
Just Put this in equation(1) or equation(2) and you will get the original time.
equation(1), 60/10=t, t=6.
equation(2), 60/15=t-2, 4=t-2, t=6.
So if Robert wants to reach the destination at 1 pm, We have to calculate his speed, we got distance and time taken.
If he wants to reach at 1 pm which is 1 hr less than the normal time.
then equation becomes distance/speed=time-1 hr.
i.e, 60/speed=6-1, speed=60/5, speed=12 kmph.
Therefore, the answer is 12 kmph.
According to statement 1.
If he travels at 10 kmph and let's imagine the time taken to reach at 2 pm is 't' hrs
i.e, x/10=t ---> equation(1).
According to statement 2.
If he travels at 15 kmph and reaches at 12 noon, then he reaches 2 hrs earlier than the previous time (i.e, t-2 hrs)
then, x/15=t-2 ------> equation(2)
If you get two equations with same two variables, always subtract one equation from the other.
Here, the equations become,
x/10-x/15=t-(t-2) or x/15-x/10=(t-2)-t
either way you will get the same answer i.e, x=60 km.
Just Put this in equation(1) or equation(2) and you will get the original time.
equation(1), 60/10=t, t=6.
equation(2), 60/15=t-2, 4=t-2, t=6.
So if Robert wants to reach the destination at 1 pm, We have to calculate his speed, we got distance and time taken.
If he wants to reach at 1 pm which is 1 hr less than the normal time.
then equation becomes distance/speed=time-1 hr.
i.e, 60/speed=6-1, speed=60/5, speed=12 kmph.
Therefore, the answer is 12 kmph.
(39)
Akash said:
4 years ago
@All.
Use formula like (2 (a.b) /a+b) a is speed 10kmph and b is 15kmph.
Hope you get it.
Use formula like (2 (a.b) /a+b) a is speed 10kmph and b is 15kmph.
Hope you get it.
(30)
Vivek said:
1 year 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)
Ritik Florian said:
5 years ago
Simply, find the average speed u can get the answer. Since 1 pm comes between 12 noon and 2 pm.
The formula for finding AVG speed is= 2xy/X+y.
i.e. 2*10*15/10+15= 12.
The formula for finding AVG speed is= 2xy/X+y.
i.e. 2*10*15/10+15= 12.
(11)
Vicky said:
4 years ago
How 3 came? Please explain.
(7)
Prerna23 said:
5 years ago
Time reached -- time taken -- Speed
2 p.m. -- t -- 10 km/hr
12 noon -- t-2 -- 15 km/hr
1 p.m. -- t-1 -- ?.
Distance is constant.
Writing in terms of time and speed, t x 10 = (t-2) x 15 which gives t = 6.
Distance is 60 km, so speed required to reach in t-1= 5 hrs i.e. at 1 p.m. is 60/5 = 12 km/hr.
2 p.m. -- t -- 10 km/hr
12 noon -- t-2 -- 15 km/hr
1 p.m. -- t-1 -- ?.
Distance is constant.
Writing in terms of time and speed, t x 10 = (t-2) x 15 which gives t = 6.
Distance is 60 km, so speed required to reach in t-1= 5 hrs i.e. at 1 p.m. is 60/5 = 12 km/hr.
(7)
Anushree Kulkarni said:
3 years ago
@Farzana.
Best and neat explanation, Thank you.
Best and neat explanation, Thank you.
(7)
Aasa sri said:
10 months 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.
(6)
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