
By: Rohana (offline) Tuesday, April 23 2013 @ 07:36 AM CDT (Read 1085 times)



Rohana 
Hi, can anyone help me with this question.
1. Two sisters, Melissa and Nicole, started briskwalking at 6.45a.m frothier house to a nearby park.
Melissa walked at an average speed of 90m/min while Nicole's average speed was 60m/min.
After arriving at the park Melissa rested for 10min and then walked back home. If Melissa met Nicole
At 7.40a.m on her return journey, what was the distance between their house and the park?
Thankyou in advance!

Junior
Registered: 01/24/12 Posts: 18





By: echeewh (offline) Tuesday, April 23 2013 @ 01:44 PM CDT



echeewh 
hi Rohana,
Following pls find my worked solution:
(Revised: 24 Apr, 8:30 am)
***  for alignment purpose
H ******************************* P

> ****************************** >
M ****************************** M1(rested 10mins)
S(M)=90
> ****************** >
N ****************** N1
S(N)=60 ********* <
t1=0645 ********** M2
****************** t2=0740
t(NN1) = 0740  0645 = 55mins = t(MM1M2) {inclusive of M's rest time of 10mins}
d(NN1) = 60 × 55 = 3300m
d(MM1M2) = 90 × 55 = 4950m { since Average Speed = (Total Distance) / (Total Time) }
d(M1M2) = [4950  3300] ÷ 2 = 825m {since N1M1 = M1M2}
Hence, d(MM1) [house to park] = 3300 + 825 = 4125m = 4.125km
===========
Trust this helps.
Do let me know again if this is different from your Answerkey or if there's further clarification.
Cheers,
Edward

Active Member
Registered: 04/21/11 Posts: 623





By: agnesphuah (offline) Wednesday, April 24 2013 @ 08:15 PM CDT



agnesphuah 
Hi Rohanna/Edward
Edwards approach is correct except Edward forgotten to factor in the 10mins rested.
as such the working should be edited as such.
t(NN1) = 0740  0645 = 55mins
d(NN1) = 60 × 55 = 3300m
d(MM1M2) = 90 × 45 = 4050m { 55mins  10mins rest time }
d(M1M2) = [4050  3300] ÷ 2 = 375m {since N1M1 = M1M2}
Hence, d(MM1) [house to park] = 3300 + 375 = 3675m = 3.675km

Newbie
Registered: 08/30/11 Posts: 1





By: echeewh (offline) Thursday, April 25 2013 @ 05:48 PM CDT



echeewh 
Hello Agnes (and all),
I understood the question clearly and was aware of M's 10 mins rest time. But the speed used in this calculation is given as the Average Speed.
Using your solution and putting it back into the question, we shall calculate the average speed for M.
Total Distance = d(MM1) + d(M1M2)
= 3675 + 375
= 4050 m
Total Time = 55 min { from 0645  0740 }
Average Speed = 4050 ÷ 55 ~ 74 m/min ( rounded to whole number )
This Average Speed , however, does not agree with that given in question of 90 m/min.
As for my solution,
Total Distance = 4125 + 825 = 4950 m
Average Speed = 4950 ÷ 55 = 90 m/min
==========
Trust the above clarifies.
Cheers,
Edward

Active Member
Registered: 04/21/11 Posts: 623



