# A Direct Bar Model Approach to Concentration Problems

##### Ho Soo Thong

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.

Figure 2 gives more examples.

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

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

*respectively:*

**B****: 300 ml of 25% alcohol solution was mixed with the unknown solution for a desired 18% alcohol solution.**

*A***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.

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.

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.

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.

[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