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Wednesday, November 22 2017 @ 01:20 AM CST
 Forum Index >  Test Paper Related >  Primary 6 Matters
 p6 2012 nanyang prelim
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By: tchtby (offline)  Friday, August 09 2013 @ 05:46 AM CDT (Read 1097 times)  
tchtby

pls help me solve these 2 qn

thks

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By: Lisa Wong (offline)  Saturday, August 10 2013 @ 12:59 AM CDT  
Lisa Wong

Hope I'm of some help. Pls see attachment for solution. I'm also a P6 anxious mum.Smile

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By: echeewh (offline)  Saturday, August 10 2013 @ 02:43 PM CDT  
echeewh

Hey <tchtby>

Following is my worked solution for Q10:

For T, J to pass / meet each other (from the opp direction), the duration / time taken has to be the same.

Applying the <Common Time> method, where the Distance ratio is the same as the Speed ratio, we have ...

sT : sJ
160 : 200
4 : 5

dT : dJ
4 : 5

This distance of 9u (dT + dJ) is based on their initial starting points on the track (i.e. T and J were 200 m apart).

1u distance = 200/9 m

Based on the given circular track of 400m, T and J would meet for a 2nd time after running / completing a combined / joint distance (dT + dJ) of 600m.

dT = 4u × 3 = 12u
12u distance = 12 × (200/9) = 266 2/3 m

=========

Trust this helps.

Do let me know again if there's further clarification.

Cheers,
Edward

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