Ho Soo Thong   Ho Shuyuan

This write-up illustrates the bar modelling approach to a variant of a recently highlighted 2017 PSLE question. The problem solving strategy involves the use of Euclidean Division Algorithm to depict the two “remainder” situations.

A Ribbon Problem

Jane, a sales supervisor, has to prepare 150 ribbons for gift wrapping. Each ribbon is of length 160 cm. The ribbons are cut out from tapes of length 20 m each.
How many tapes are needed to prepare all 150 ribbons?

Solution
For the number of ribbon that can be made from each tape of length 20 m = 2000 cm,we construct a bar model with Euclidean Division Algorithm.

12 ribbons can be made per tape and the remaining 80 cm of tape has to be discarded.

For 150 ribbons, we again apply the Euclidean Division Algorithm and the bar model shows that one more tape is needed for the remaining 6 ribbons.

Therefore, Jane needs 12 + 1 = 13 tapes.

Remarks
An earlier PSLE question (15/02/2014 ) also involves a non-discarded “leftover“ situation. This was solved with bar model and Euclidean Division Algorithm. This also highlight the use for a pictorial view of the problem to give students a clearer understanding of the situation described in the problem.

 

 

红丝带问题 – 杆模型方法

这篇文章展示了如何使用欧几里德分割算法来描述两个“剩余”情况。

红丝带问题

销售主管,小呢,必须准备185条红丝带用于礼品包装。每条红丝带长度为160厘米,红丝带是从每卷长度20米的红丝带上剪下。

小呢需要多少盒 丝带?

题解 (分二个阶段)

第一阶段 : 每卷丝带能剪成几条丝带.

我们先用一个杆模型展示如何应用欧几里德分割算法把20 m = 2000 cm 长的丝带裁剪成12条160cm 的红条丝带 ,并丢弃剩余80厘米的丝带。

第二阶段 : 需要多少卷丝带才能剪成185条丝带.

我们再次应用欧几里德分割算法,杆模型显示剩余的5条色带需要再多一卷丝带。

因此,小呢需要15 + 1 = 16卷丝带。

备注: 

这問题涉及二个不同结果的“剩余”情境。于杆模型让学生更清楚地了解问题中描述的情境.

早期的PSLE问题(15/02/2014)也常 涉及不同的“剩余”情境。

練習題 : PSLE(15/02/2014)