H    [B][B][B][G][G]   <—–(3/4) x (3/5) = 9/20  => 9u blue stickers given away, total parts = 20u

       <-12u–><-8u->

S     [       B         ][       B         ][       G        ]

(a)  From H, stickers left = 20u-9u =11u.

11u = 165 => u = 15

Blue stickers that H has given away = 9u = 9 x 15 = 135

(b)  S’s stickers      Blue               Green

         Before               2s                      s

       From H           +135 

         After                  7p                    3p

s = 3p

2(3p) + 135 = 7p

p = 135

Hence Sherman has 10p = 1350  stickers in the end.        

My goodness.  I tried looking at this problem and went googly-eyed.

Wonder where the question was from.  I find it difficult to believe it is a PSLE question.

Hi Chief

It’s from Anglo-Chinese school prelims 2021.

Chief,

While I agree it looks tough but actually this kind of question is good in helping kids to learn how to break complex problems down into smaller parts and tackle them step by step.  It’s part of building the mind to solve problem systematically.   Having said this, I also recognise that sometimes it is hard to do within exam conditions, within a time limit.   So for exams, it is really about practice and that’s what exam is good for….it makes us do enough that we can remember some important stuff for life.