Albert Einstein once said
55 minutes thinking about the problem and 5 minutes thinking about solutions.”
Recently, George Polya proposed a problem solving strategy in his book “How to Solve It “ with four steps:
- 1Understand the Problem
- 2Devise a Plan
- 3Carry Out the Plan
- 4Look Back
We will follow a similar idea for a bar modelling problem solving strategy, Read More, with the following steps.
- 1Understanding the problem (mathematical situations)
- 2Looking for an appropriate mathematical model for the related mathematical situations
- 3Constructing a bar model with the related mathematical situations for a pictorial view.
- 4With the pictorial view, perform calculation, algebraic manipulations and deductions for solutions.
- 5Practise with variants of the problems
The following two examples illustrate the basic steps for solving two recent PSLE questions.
Bar Model Approach to a 2019 PSLE Question (15/01A/19)
The effective Ratio Approach is illustrated
Bar Model Approach to a 2019 PSLE Problem (14/02/19)
It is a Distributive Property – Percentage Problem which involves two mathematical situations:
1. Kelvin spent $61.20 for an unknown number of egg tarts.
2. With 15% discount ,Julie also spent $61.20 and she got 6 more egg tarts than Kelvin.
A comparison bar model is needed to depict the two distributive situations.
Construct a bar model for the two situations
On the bar model, we perform simple calculations for
1. the percentage situation and compute the price of each egg tart with discount
2. the number of egg tarts that Julie bought
3. the price of each egg tart without discount as shown.
The Problems & Solutions includes the following Examples and an Exercise for practice.
Four pupils were to buy a total of 9 apples and oranges each.
Kim paid $ 6.40 for 5 apples and 4 oranges. Ann paid $ 5.80 for 2 apples and 7 oranges.
(a) Lee bought 4 apples and 5 oranges, how much did Lee pay for?
(b) Mei paid $ 7.00, how many apples did Mei pay for?
The bar model depicts the difference that Kim have 3 more apples and Ann has 3 more oranges. Then we deduce that an apple costs $ 0.20 ( = $ 0.60÷3) more than an orange as shown.
(a) Lee had one apple less than Kim we deduce that Lee paid $ 0.2 less than Kim. Lee paid $ 6.40 – $ 0.20 = $ 6.20.
(b) Finally, Mei paid $ 0.60 more than Kim and so she had 3 more apples than Kim. Mei bought 8 apples and an orange.
The above approach can be applied to solve a recent PSLE question (17/02/15 ).
There were a total of 85 pupils sharing some apples.
A package of three apples for four girlsA package of two apples for three boys
The girls and the boys shared equal number of apples.
How many apples were there?
First, we constructs bar models to depict the sharing situations.
Next, we construct bar model to find that 12 ( = 2×6) is the least possible number of apples.
Now, we construct a corresponding bar model for corresponding number of pupils as a group of 17 (= 2×4 + 3×3) as shown.
There are 85÷17 = 5 groups of pupils and so there are 5×12 = 60 apples.
Note : Following the approach, you can easily solve the PSLE question (16/02/2015).
The heuristic approach to the problem can be applied to more challenging problem. A sample of challenging problem is given below.
There were a total of 92 pupils sharing some apples.
A package of three apples for four girls A package of two apples for three boys
The ratio of the number of apples shared by the boys to the number of apples shared by the girls is 3 : 5.
How many apples were there?
We begin with the bar modelling of the sharing situations.
There are 92 ÷ 23 = 4 groups of pupils and so there are 4×16 = 64 apples.