Read : Parents up in arms again over PSLE Mathematics paper.
My two cents – Firstly, if all the smart students can solve all the problems then who will be the smarter and who will be the smartest ? Secondly, the question mentioned in the article is ‘nothing new’ as similar ones can be found in exam papers from ‘top ten schools’. (The most recent – ACS Prelim 2009 Q16)
In my classes, we call this type of questions, ‘Before – After’. Typically, an initial ratio, percentage, fractions is given (Before). Then something happen – sweets eaten, marbles lost, wateva! Finally, the end ratio, percentage, fractions is given. (After).
For my students, I teach them both methods but I highly recommend the ‘X-method’ (Algebra) for solving ‘Before-After’ questions. Modals can be time consuming!
There are two ways to solve the problem.
1) Modal Method (Step-by-Step Solution)
Step 1 : Jim bought some chocolates and gave half of it to Ken. Ken bought some sweets and gave half of it to Jim.
Step 2 : Jim ate 12 sweets and Ken ate 18 chocolates.
Step 3 : The ratio of Jim’s sweets to chocolates became 1:7 and the ratio of Ken’s sweets to chocolates became 1:4
Step 4 : Focus on the two BLUE parts. Both the BLUE parts (Jim’s Sweets & Ken’s Sweets) are the same because from the question, ‘Ken bought some sweets and gave half of it to Jim’. This means Ken and Jim have equal share of the sweets.
The modal can be redrawn to look like this:
Step 5 : Focus on the GREEN parts. Both GREEN parts (Jim’s chocolates and Ken’s chocolates) are the same because from the question, ‘Jim bought some chocolates and gave half of it to Ken’. This means Ken and Jim have equal share of the chocolates.
7u = (12 + 1u) + (12 + 1u) + (12 + 1u) + (12 + 1u) + 18
7u = 48 + 4u + 18
7u – 4u = 48 + 18
3u =66
1u = 22
Ken bought : (12 + 1u) + (12 + 1u) = (12 + 22) X 2 = 68
Ans : 68 sweets.
2) Algebra
Step 1 : Jim bought some chocolates and gave half of it to Ken. Ken bought some sweets and gave half of it to Jim.
This means Jim and Ken both has 1 unit of chocolate and 1 unit of sweets.
Jim Ken
1s : 1c 1s : 1c
Step 2 : Jim ate 12 sweets and Ken ate 18 chocolates
Jim Ken
1s – 12 : 1c 1s : 1c – 18
Step 3 : The ratio of Jim’s sweets to chocolates became 1:7 and the ratio of Ken’s sweets to chocolates became 1:4
Jim Ken
1s – 12 : 1c 1s : 1c – 18
X X (This is what I coined the ‘X’-method, a common lingo in my class)
1 : 7 1 : 4
Equation 1 : 7s – 84 = 1c
Equation 2 : 4s = 1c – 18 or 4s + 18 = 1c
Solving the two equations simultaneously
7s – 84 = 4s + 18
7s – 4s = 18 + 84
3s = 102
s = 34
2s = 34 x 2 = 68
Ans : 68 sweets


