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 P6 Challenging Maths-Pls help!
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By: geniuskids (offline)  Tuesday, February 05 2013 @ 07:07 AM CST (Read 1485 times)  
geniuskids

Hi there, please help solve this question:-

Adam and Rahmad shared some beads. If Adam gave 1/3 of his share to Rahmad, Rahmad would have 70 more than Adam. If Adam gave 1/5 of his share to Rahmad, Rahmad would have 10 more than Adam. How many beads does Adam have at first?

Urgent!
Thanks

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By: echeewh (offline)  Tuesday, February 05 2013 @ 06:57 PM CST  
echeewh

Hey there,

Following pls find my worked solution:

<After>
A |---------|
R |---------|<--70-->|

<process1>
Since A had (2/3) of his share left, split this portion of A into 2 equal parts (2p). Working backwards, A would have 1p added to his 2p, and R had 1p taken from his, leaving (1p + 70) as shown below <before> model.

<before>
A |----|----|----|
R |----|<--70-->|

<process2>
A gave (1/5) of his share. Using common multiple method, Split his 3p into 15 equal units (u), i.e. 1p = 3u. R would also have his 1p split into 5u.

A = 15u; R = 5u + 70;

Since A gave (1/5) of this (or 3u) to R, A would be left with 12u. And R would have 3u more to his share as shown below.

<After>
A: 12u
R: 8u + 70

Given that R had 10 more than A in the end, we have ...

R - A = 10
8u + 70 - 12u = 10
12u - 8u --> 70 - 10
4u --> 60
1u --> 15

A (at first) = 15u --> 15 × 15 = 225

==========

Do let me know again if this is different from your Answerkey or if theres further clarification.

Cheers,
Edward

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