Welcome to Test-paper.info
Tuesday, January 16 2018 @ 03:08 PM CST
 Select a Forum » General Chat » Primary 1 Matters » Primary 2 Matters » Primary 3 Matters » Primary 4 Matters » Primary 5 Matters » Primary 6 Matters » Question, Feedback and Comments » Education Classified
 Forum Index >  Test Paper Related >  Primary 6 Matters p6 2012 nanyang prelim
 | Printable Version
 By: tchtby (offline)  Friday, August 09 2013 @ 05:46 AM CDT (Read 1134 times)
tchtby

pls help me solve these 2 qn

thks

Newbie

Registered: 10/15/12
Posts: 13

 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.

Newbie

Registered: 05/09/10
Posts: 1

 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

Active Member

Registered: 04/21/11
Posts: 627

 All times are CST. The time is now 03:08 pm.
 Normal Topic Locked Topic Sticky Topic
 New Post Sticky Topic w/ New Post Locked Topic w/ New Post
 View Anonymous Posts Able to Post HTML Allowed Censored Content