A Direct Bar Model Approach to Concentration Problems

“The esscence of mathematics lies in its freedom”

- Georg Cantor

A Direct Bar Model Approach to Concentration Problems

Ho Soo Thong

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Copyright © June 2015 AceMath, Singapore. 

Abstract
This article illustrates the ratio approach in the Bar Model Method with pictorial view of complex situations in two challenging problems at Primary Olympiad Level.

Bar Modelling of Concentration Scenarios of Alcohol Solutions

A homogeneous alcohol solution is a mixture that alcohol and water are uniformly distributed throughout the mixture.
Figure 1 shows a homogeneous 25% = ( 1/4) alcohol solution, a mixture of alcohol and water in the ratio 1 : 3 depicted in a bar model.

1

Figure 2 gives more examples.

2

Figure 3 shows the splitting of a 100 cc of 25% ( = 1/4) alcohol solution.

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Figure 3 illustrates the result that splitting a homogeneous alcohol solution leads to solutions with the same concentration,

 

Two Challenging Problems

In the next example, we will see how Steve, a laboratory technician, manage to solve problems using the above result.

Example 1

Steve was to mix 300 ml of 25% alcohol solution with a bottle of unknown alcohol solution to obtain a 18% alcohol solution.
Steve made a mistake by pouring 300 ml of 15% alcohol solution with the unkown solution. But he added a 200 ml of alcohol solution to the mixed solution to obtain the desired 18% alcohol solution.
What is the concentration of the added 200 ml of alcohol solution?

Solution

In Figure 4 and Figure 5, we construct bar models for the two scenarios A and B respectively:
A : 300 ml of 25% alcohol solution was mixed with the unknown solution for a desired 18% alcohol solution.

4re

B : 300 ml of 15% alcohol was mixed with the unknown solution and then another 200 cc of alcohol water was added to obtain a 18% alcohol solution.

5

Then Steve performed a splitting process:
He poured out 200 ml of 18% alcohol solution in B and the remaining solution is the same as the scenario A as shown in Figure 6.

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Equating the volumes of alcohol in the scenarios A and B* , we have

45 + Y − 36 = 75
Y = 66

Therefore, the concentration of the added 200 ml alcohol solution is 33% ( = 66/200 ).

 

Example 2

Bottle A contains 1 litre of pure alcohol and Bottle B contains 1.5 litres of water. Steve mixes the two solutions for a 62.5% alcohol solution in A and a 25% alcohol solution in B.
How many litres of the solutions will be there in A and B?.

Solution

Figure 7 shows two mixing processes:

A to B : For a 25% ( = 1/4 ) alcohol solution, the alcohol-water ratio is 1 : 3 and so we mix 0.5 litres of alcohol from bottle A with 1.5 litres of water in the bottle B as depicted by the 2nd bar model.

B to A : Steve split the 25% alcohol solution in bottle B. He mixed 4U of the 25% alcohol solution in bottle B with the pure alcohol in bottle A so that the bottle A contains a 62.5% alcohol solution.

Noting that 62.5% = 5/8, the alcohol-water ratio in bottle A is 5 : 3.

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Therefore,the bottle A contains 1 litre of 62.5% alcohol solution and the bottle B contains 1.5 litres of 25% alcohol solution.

Notes :
An alternative way is to obtain a 62.5% alcohol solution in bottle A first and then 25% alcohol solution in bottle B.
Pupils may try this approach as an exercise.

An Algebraic Approach.

In Figure 8, bar models help obtaining the simultaneous equation with two unknowns:

5U + V = 1.0
3U + 4V = 1.5.

8

 

[1] Ho Soo Thong and Ho Shuyuan, Bar Model Method for PSLE and Beyond, AceMath, 2011
[2] Ho Soo Thong, Advanced Bar Model Method for Counting Method, Asia Pacific Mathematics Newsletter,
[3] Ho Soo Thong & Ho Shuyuan, Bar Model Method for Speed Problem, barmodelhost.com, 2014
[4] 严军 冯惠愚 小学数学奥赛优化解题题典 吉林教育出版社 2007
[5] 朱华伟 多功能题典小学数学 竞赛 华东师范大学出版社 2007