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 Forum Index >  Test Paper Related >  Primary 6 Matters
 p6 msths - speed
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By: fanny6573 (offline)  Sunday, August 10 2014 @ 05:28 AM CDT (Read 1375 times)  
fanny6573

Hi please help on the question below:
Jean started his journey from town x to y at a constant speed. Sometime later, Michelle started her journey from town x to town y at the same speed as jean.
At 11 a.m, Jean travelled 4 times the distance. Michelle had travelled.
At 11.42 a.m, Jean finished the journey while Michelle travelled half of the distance from town x to town y.
At what time did Jean start his journey?

Thank you.

Regards
Fanny

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By: echeewh (offline)  Sunday, August 10 2014 @ 10:37 PM CDT  
echeewh

Hello <fanny6573>

Following is my worked solution :

*** - for alignment purpose


********************** <11am> * <11.42am>
J |-----|-----|-----|-----|-------------|
M |-----|-------------|
******* <11am> * <11.42am>

By 11am,
J completed 4u distance ; M completed 1u of same distance.

By 11.42am,
J completed the entire distance; M completed half the distance.

From above models and given that both J, M speeds are same and constant,

Time taken by M to complete half distance
= 1u (distance) + 42 mins

Referring to J's model,
Time taken by J to complete full distance
= 4u (distance) + 42 mins

Since both speeds are same and constant,
J would also take [ 1u (distance) + 42 mins ] to complete half distance.

This infers the following :
J would need [ 1u (distance) + 42 mins ] to complete the 3u (distance)

i.e. J would take 42 mins to cover 2u (distance)

Hence, 1u (distance) --> 42 ÷ 2 = 21 mins
4u (distance) --> 4 × 21 = 84 mins = 1 hr 24 mins

Time (when J started journey):
1 hr 24 mins before 11am

= 11 00 - 1 24
= 9 36

J started journey at 9.36am

==========

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: fanny6573 (offline)  Monday, August 11 2014 @ 12:31 AM CDT  
fanny6573

Thank you Edward.

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