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Friday, July 21 2017 @ 09:34 PM CDT
 Forum Index >  Test Paper Related >  Primary 6 Matters
 P6-Speed
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By: fanny6573 (offline)  Wednesday, July 11 2012 @ 09:33 PM CDT (Read 1945 times)  
fanny6573

Anyone please help to solve the question.

Andy, Benjamin and Charles travel from Town X to Town Y.
They leave Town X at the same time.
Andy drives at 56km/h , Benjamin walks at 8km/h and Charles walks at 7km/h.
Andy gave Benjamin a lift at the start of journey to travel a part of the journey and
then comes back to pick up Charles.
They arrived at Town Y at the same time. If Benjamin walks 2.8km, how far does Charles walks?

Thank you.

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By: MathIzzzFun (offline)  Wednesday, July 11 2012 @ 09:46 PM CDT  
MathIzzzFun

Quote by: fanny6573

Anyone please help to solve the question.

Andy, Benjamin and Charles travel from Town X to Town Y.
They leave Town X at the same time.
Andy drives at 56km/h , Benjamin walks at 8km/h and Charles walks at 7km/h.
Andy gave Benjamin a lift at the start of journey to travel a part of the journey and
then comes back to pick up Charles.
They arrived at Town Y at the same time. If Benjamin walks 2.8km, how far does Charles walks?

Thank you.



http://tinyurl.com/mathizzzfun-Speed-01

cheers.

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By: fanny6573 (offline)  Thursday, July 12 2012 @ 11:07 PM CDT  
fanny6573

Thank you for the answer but my daughter still cannot understand, anyone can explain further about the working step of this question?

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By: echeewh (offline)  Saturday, July 14 2012 @ 10:11 PM CDT  
echeewh

Hey Fanny,

Trust that you understand the Distance / Time model given.

The method used by MathizzzFun is based on Speed , Distance proportion / ratio technique.

This Speed , Distance proportion / ratio technique is applicable given that Time is constant.
In this case, Time taken by A (from P -> Q -> Y) = Time taken by C ( from P -> Y)

S(A) : S(B )
56 : 8
D(A) : D(B )
56 : 8

8u --> 2.8 km
56u --> (2.8 / 8) x 56 = 19.6 km

***********************
<Alternative method>
For this, usual DST formula can also be used as follows:

Time taken by B (P -> Y) = 2.8 ÷ 8 = 0.35 hr
So Dist travelled by A (P -> Q -> Y) = 56 x 0.35 = 19.6 km

***********************

Dist PQ = (19.6 - 2.8) / 2 = 8.4 km

Time taken by A (from X -> P -> Q) = Time taken by C ( from X -> Q)
Using the same technique above, we have ...

S(A) : S(C)
56 : 7
D(A) : D(C)
56 : 7

So dist travelled by A is 8x that of C ( --> 8 XQ) and this is the dist travelled by A from X -> P -> Q , which is , XQ + QP + PQ ( from model, dist travelled by A to go to P with B, and returned back to fetch C ).

Hence, 8 XQ = XQ + 2 PQ
7 XQ = 2 PQ = 2 x 8.4 = 16.8
XQ (dist travelled by C) = 16.8 ÷ 7 = 2.4 km

======================================================

Trust the above helps to clarify further. Btw I used same method but simplifying the ratios further; thus working with smaller numbers Smile

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

Cheers,
Edward

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By: fanny6573 (offline)  Tuesday, July 17 2012 @ 10:02 PM CDT  
fanny6573

Thank you Edward.

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