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Tuesday, September 26 2017 @ 04:51 PM CDT
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 Branching Method 2
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By: shirley wee (offline)  Monday, January 07 2013 @ 08:29 AM CST (Read 765 times)  
shirley wee

Hi everyone! I am stuck with this question. Can you help me? Please show your workings clearly.

1. 27 Primary 5 pupils and 38 Primary 6 pupils took part in a marathon. The number of Primary 6 girls was 80% of the number of Primary 5 girls, and there were 12 more Primary 6 boys than Primary 5 boys.
a. How many Primary 5 boys took part in the marathon?
b. How many more Primary 6 boys than Primary 6 girls took part in th marathon?

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By: echeewh (offline)  Monday, January 07 2013 @ 11:51 PM CST  
echeewh

Hi Shirley

I dont quite understand why the name "Branching method" is used for the title, where the method is not suitable for this kind of question.

It will also be much appreciated if you can provide the Answerkey , wherever possible. Thank you.


Following pls find my worked solution:

*** - for alignment purpose


80% = (80 / 100) = (4 / 5)

P6G : P5G
4 : 5 --> 4p of P6G and 5p of P5G

***** <-1u->
P6B |------|<12>| --> 1u + 12
P5B |------| ***** --> 1u

Given there were 27 P5 and 38 P6 pupils, applying <parts / units method> (or Simultaneous method), we have ...

P6G + P6B = 38
4p + 1u + 12 --> 38
4p + 1u --> 38 - 12 = 26 -- (1)

P5G + P5B = 27
5p + 1u --> 27 -- (2)

(2)-(1):
5p - 4p --> 27 - 26
1p --> 1

(a)
From (2)
P5B = 1u = 27 - 5p --> 27 - 5 = 22

(b)
P6B = 1u + 12 --> 22 + 12 = 34
P6G = 4p --> 4

Hence, P6B - P6G = 34 - 4 = 30

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

Trust the above 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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