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 By: eqkt (offline)  Saturday, January 12 2013 @ 09:11 AM CST (Read 1068 times)
eqkt

Hi! Can anyone help me to solve this question? Thank you!

(No bead was to be divided in any way)

Junior

Registered: 03/18/10
Posts: 18

 By: echeewh (offline)  Saturday, January 12 2013 @ 08:09 PM CST
echeewh

Hi there

This is a re-post of my solution here last year dated 31/5/12 (refer title of post - Nanyang Primary School CA1 Q11 & 12 ).

It will be much appreciated if you are doing any of the past years' questions , you may want to review the questions posted here earlier as your question could have already been answered . Thank you.

Following pls find my worked solution:

*** - for alignment purpose

Apply <Branching method>

*******|--> 3/4 + 3/4 b (S)
Total -|******************|--> 3/4 + 3/4 b (Y)
*******|--> 1/4 - 3/4 b - |
**************************|--> 1/4 - 3/4 b = 1 b (A)

In other words,

From Total, S = 3/4 + 3/4 b
Remainder (R) = 1/4 - 3/4 b
From R, Y = 3/4 + 3/4 b
Remainder (R1) = 1/4 - 3/4 b

Given that A received the last bead in the bag, then ...
R1 = 1/4 - 3/4 b = 1 b

So, 1/4 = 1 3/4 b

Working Backwards, we have ...
R = 4/4 = (7/4) x 4 = 7 b

So we have ...
R = 1/4 - 3/4 b = 7 b
1/4 = 7 3/4 b

S = 3/4 + 3/4 b
= 3 x (31/4 b) + 3/4 b = 93/4 b + 3/4 b
= 96/4 b = 24 b

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

Trust this helps.

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

Cheers,
Edward

Active Member

Registered: 04/21/11
Posts: 627

 By: eqkt (offline)  Saturday, January 12 2013 @ 11:06 PM CST
eqkt

Junior

Registered: 03/18/10
Posts: 18

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