
By: ycthun (offline) Saturday, July 28 2012 @ 10:12 PM CDT (Read 1318 times)



ycthun 
Hi.
I need help with this question. Can anyone help. Thanks
At 0900hr, John started cycling from point A to point B. At the same time, Mary started cycling from Point B to point A using the same route. Cycling at a speed of 4km/hr faster than Mary, John met up with Mary 300m from the midpoint.
a. At what time did the two meet?
b. if John took another 3 mins to reach point B, at what time will Mary reach point A?
Ted

Newbie
Registered: 06/23/09 Posts: 8





By: echeewh (offline) Sunday, July 29 2012 @ 10:00 PM CDT



echeewh 
Hey Ted,
Following pls find my worked solution :
P  Midpoint
X  where J,M would meet
A **************** P ****** X ******* B

> *************** <300m> ***** <
J ************************* > ***** M
S(J)=S(M) + 4 ********* <
************************ (J1,M1)
Notes:
If J were to cycle at the same speed as M, both would be 300 m away from the midpoint P, i.e. distance apart/separating each other = 600 m.
The extra speed of 4 km/hr by J would take him to meet M at X, which is 300 m from P. (refer distancetime model (J1,M1))
(a)
Time taken to meet at X = (Distance apart) ÷ (Speed Difference)
= 0.6 ÷ 4 = 0.15 hr = 9 mins
As both J,M started at 0900, both met at 0909
(b)
Given that J took another 3 mins to reach B,
Dist (XB ) covered by J = [(S(M) + 4) x (3/60)]
This is the same distance travelled by M in 9 mins, which is ...
Dist (BX) coverd by M = S(M) x (9/60)
So we have ...
[(S(M) + 4) x (3/60)] = S(M) x (9/60)
3 S(M) + 12 = 9 S(M)
9 S(M)  3 S(M) = 12
6 S(M) = 12
S(M) = 2 km/h
S(J) = S(M) + 4 = 2 + 4 = 6 km/h
Dist AB = 6 x (12/60) = 1.2 km
Time taken by M to reach A = 1.2 ÷ 2 = 0.6 hr = 36 mins
Time at which M reached A = 0936
===========================
Trust this helps.
Do let me know again if these do not agree with your Answerkey or if there's any clarification.
Cheers,
Edward

Active Member
Registered: 04/21/11 Posts: 625





By: ycthun (offline) Sunday, July 29 2012 @ 10:13 PM CDT



ycthun 
Thank you very much for the prompt and succint answer.
Ted

Newbie
Registered: 06/23/09 Posts: 8



