Working Backwards: Heuristic for Problem Solving

Working Backwards is a non-routine heuristic that all pupils learn in primary schools. Many pupils learn this heuristic as early as when they were in primary two. You can easily find this heuristic being included in one of the topics in assessment books.

Though common, this heuristic is in fact one of the toughest in primary Mathematics syllabus. Problem sums that can be solved using Working Backwards are usually wordy with a sequence of events taking place which make them complex. You need endurance and clarity of mind to follow through a series of events that are unfolding in sequence, not to mention in a backward manner.

Without further ado, let’s delve into one typical upper primary problem sum to find out how we can tackle this category of problem sums using the heuristic of Working Backwards.

Example

Xiaoming and Ali were playing a card game using 96 pokemon cards. In the first game, Ali lost $\frac{1}{5}$ of his cards to Xiaoming. In the second game, Xiaoming lost $\frac{1}{3}$ of his cards to Ali. After the second game, both boys had the same number of pokemon cards. How many pokemon cards did Xiaoming have at first?

1. Study and Understand the Problem

   In this problem sum, there are only two variables – Xiaoming and Ali.

   Cards are transferred between Xiaoming and Ali in a series of games, but the total number of cards between them is still the same – internal transfer.

2. Think of a Plan

   There are contextual clues of a typical “Working Backwards” problem sum:

   • Final information is given on how a situation ends and you need to find the answer in the beginning.
   • There is a focus on sequence of events.

3. Act on the Plan

   Reverse the solution steps by working backwards in a systematic manner using a table. A table will help to organise your working in a more orderly manner and track the steps in sequence.

   End

   Final number of cards each person had = 96 ÷ 2 = 48

   	XM	A
   End	96 ÷ 2 = 48	96 ÷ 2 = 48

   Second game

   After Xiaoming lost $\frac{1}{3}$ of his cards to Ali, he was left with 2 units as 1 unit was won by Ali (Refer to the numerator and denominator). Thus, 2 units = 48

   	XM	A
   Second game	2 units = 48 1 unit = 24 3 units = 72 XM had 72 cards before second game	96 – 72 = 24 Ali had 24 before second game

   First game

   After Ali lost $\frac{1}{5}$ of his cards to Xiaoming, he was left with 4 units as 1 unit was won by Xiaoming (Refer to the numerator and denominator). Thus, 4 units = 24

   	XM	A
   First game	96 – 30 = 66 XM had 66 cards at first	4 units = 24 1 unit = 6 5 units = 30 Ali had 30 cards at first

4. Reflect on my Answer

   Work forward with the answer you have.

   At first

   XM → 66
   A → 30

   First game

   A → of 30 = 6
   30 – 6 = 24
   XM → 66 + 6 = 72

   Second game

   XM → of 72 = 24
   72 - 24 = 48 (correct!)
   A → 24 + 24 = 48 (correct!)

More Examples of Problems Sums Involving Working Backwards

Try to pick out the contextual clues that tell you that Working Backwards can be used.

P3 Math question

Alan bought some fish. One day, 6 of his fish died. After that, he bought the same number of fish as those which were still alive. He gave away all his fish equally among 8 friends and each friend had 4 fish. How many fish did Alan have at first?

P6 Math question

A MRT train left Bugis station with some passengers. At Lavender station, no passengers alighted and the number of passengers who boarded the train was $\frac{1}{4}$ of the original number of passengers in the train. At Kallang station, $\frac{2}{5}$ of the passengers alighted and 51 passengers boarded the train. At Aljunied station, $\frac{2}{3}$ of the passengers alighted and 24 passengers boarded the train. At Paya Lebar station, all 122 passengers alighted from the train. How many passengers were there when the train left Bugis Station?

Conclusion

Whenever possible, use a table or draw boxes to help you solve Working Backwards questions in an orderly and systematic manner.

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