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 P6 revision test on Speed
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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 brisk-walking 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!

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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

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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

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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

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