Ho Soo Thong

The problem solving approach to the following problem illustrates a bar modelling approach to mathematical deductions.

Buying Apples  and Oranges

Four pupils bought a total of 7 apples and oranges each.
Kim paid $ 6.60 for 5 apples and 2 oranges. Ann paid $ 6.0 for 2 apples and 5 oranges.
(a) Lee bought 4 apples and 3 oranges, how much did Lee paid for?
(b) Mei bought 7 oranges, how much did Mei paid for each orange ?

Solution

Comparing the bar models for the situations of Kim and Ann, Kim have 3 more apples than Ann and Kim paid $0.603 more than Ann.
We deduce that an apple costs $ 0.20 ( = $0.60÷3 ) more than an orange.

(a) Comparing the situations of Ann and Lee, Lee has 2 apples more than Lee and we deduce that Lee paid 2× $ 0.20 = $ 0.40 more than Ann.

(b) Finally, Mei bought two apple less than Ann and we deduce that she paid 2 × $ 0.20 = $ 0.04 less than Ann.

Mei paid $ 5.60 ÷7 = $0.80 for each orange

 

苹果与橙子分配问题 -杆模型方法 

接下来,我们將注重于如何应用分配法 (Distributive Law)来处理同等分配情况 (Distributive Property Situations),并进行数学推理来解答问题。 

苹果与橙分配问题

王原花了6.60元买了5个苹果和2个橙子。陳安6元买了2个苹果和5个橙子。

(a)李进买了3个苹果和4个橙子,李进付了多少钱?

(b)杨兰花了多少钱买了7个橙子?

题解

第一个步骤 :

我们注意到数学情景:苹果和桔子的总数是7。

我们构建条形模型,展示他们的各自情景。

第一个步骤 

首先, 比较 王原与陳安的 情境, 推断 ( =>) 出一个结果: 一个苹果比一个橘子貴0.20元。

接下来,以同样推断 , 得到李进和杨兰的花費,如杆模型所示。

备注: 

分配律 也是解答鸡兔同笼问题和盈亏问题的基本数学。