## Challenging Problems

##### Articles for Learning

The following short articles show solutions to some challenging problems for learning effective Bar Model Method with Basic Algebra.

**Bar Model Method**

This article aims to illustrate how to solve problems involving mixtures of two or more different compositions following steps:

1. use the ratio approach as a heuristic system in the Bar Model Method; and

2. use the *Unitary Method* for simple *Algebra*.

The first step uses bar models with identical bars to depict the ratio situations for a clearer overview of the problem. This allows for the effective use of the Unitary Method for simple algebra employment in the second step.

For comparison, we present the traditional approach vis-à-vis all solutions.

**A Mixture of Liquid Solutions**

A chemist has a solution with an acidic concentration of 20% and another with an acidic concentration of 25%. What amount of each solution should be used in order to make 300 ml of a 22% acidic solution?

*Solution*

As 20% = 1/5 and 25% = 1/4 , the bar model solution for this problem is as follows:

Note : The solution shows the heuristic approach in the bar model method and the need to understand the problem before solving it.

*Alternatively***,** a traditional approach involves solve the simultaneous equations

*x* + *y* = 300

20% *x* + 25%*y* = `22%×300

Next, we will solve a problem which involves a mixture of cupcakes.

**A Simple Cupcake Mixture Problem**

Mrs. Lim baked some cupcakes to sell at a fun fair. 3/5 of them were chocolate cupcakes and the rest were blueberry cupcakes. Mrs. Lim had sold 100 blueberry cupcakes and 5/6 of the chocolate cupcakes. 1/6 of the cupcakes remained unsold.

How many unsold cupcakes?

*Solution*

Since 1/6 of the chocolate cupcakes were unsold and 1/6 of the total cupcakes remained unsold, we deduce that 1/6 of the blueberry cupcakes remained unsold. We begin our arithmetic operations with the blueberry cupcakes situation.

Note : The solution shows the heuristic approach in the bar model method and the need to understand the problem before solving it.

*Alternatively*, we consider the chocolate cupcakes and begin with a common unit U and proceed with a simple algebraic equation.

We can use the *traditional algebraic approach* in which we let the total number of cupcakes be *x* and equate the sum of the unsold cupcalkes of chocolate and blueberry to 1/6 of* x*.

Finally, we solve a challenging problem.

**A Complex Cupcake Mixture Problem**

Mrs. Lim baked some cupcakes to sell at a fun fair. 3/5 of them were chocolate cupcakes and the rest were blueberry cupcakes. Mrs. Lim had sold 4/5 of the chocolate cupcakes and 106 blueberry cupcakes. 1/6 of the cupcakes remained unsold, How many unsold cupcakes were there?

*Solution*

First, we apply the *Greatest Common Unit Procedure* to the number of chocolate cupcakes.

Again we can use the traditional algebraic approach in which we let the total number of cupcakes be x and equate the sum of the unsold cupcalkes of chocolate and blueberry to 1/6 of x.

*Remarks*

We have illustrated the simple algebraic approache to employing the Bar Model Method.

• The bar model approach involves variables of small values and simple algebraic manipulation. Find the greatest common unit and perform simple algebraic manipulation (Unitary Method).

• In the traditional approach, we derive complex fractional equations, which involves complex algebraic manipulations of fractions.

Intriguing Speed Problems

The bar modelling view of motion in kinematics helps to realise the vast potential and capacity of Mathematics.

‘The bicycles and the Fly’ is a well-known speed problem imagined by Martin Gardner and this kinematic problem gives rise to a learning attitude of intrigue towards the basic concepts of time and distance in kinematics.

Similar examples and adapted problems of this nature can be seen in our earlier works (see [1], [2], [3])).

Here, we focus on a challenging kinematic problem at the Primary Olympiad level.

Problems such as the below example will be henceforth known as ‘Hitch-hikers’ Problems’.

A motorist wants to ferry some hitch-hikers travelling from one place to another.

He only has one back seat to carry one passenger at a time.

How can he plan the journey so that all the hitch-hikers will arrive at their destination at the same time?

Let us begin with a simpler two-hikers problem.

** A Hitch-Hikers’s Problem**

Two hitch-hikers, A and B, walk at different speeds. A walks at a speed of 5 km/h, while B walks at a speed of 6 km/h. A passing motorist, M, travels at 30 km/h on a motorcycle, with space for only one passenger. A and B then engage Motorist M to ferry them for a distance of 11.4 km, from Town X to Town Y.

The motorist decides to pick A up at X for part of the journey and returns to pick B up for the remaining journey, so that both of them can reach Y at the same time. What will be the total travelling time?

*Solution*

First, we construct a time model to depict and relate the travelling times of M, A and B as shown below.

Referring to the overall *Time-Model* , we apply the relation

*Distance = Speed × Time*

to construct a corresponding *Distance Model* and obtain the algebraic relations as shown.

Using 2*a* = 3*m*, 7*m* = 5*b* for the distance travelled,

30*a* + 5*m *+ 5*b* = 11.4

45*m* + 5*m *+ 7*m* = 11.4

*m* = 0.2

Hence, the total travelling time is

* a + m + b* = 0.3 + 0.2 + 0.28 = 0.78 hours = 46.8 minutes = 46 minutes 48 seconds.

*A Traditional Approach*

First, we construct a *Distance model* to depict the situations and compute the travelling times of A B and M as shown below.

Comparing the travelling times of A, B and M, we have

5*x* = 4*y* = 2×(11.4 −*x* − *y*)

Solving the simultaneous equations, we have

*x* = 2.4 and *y* = 3

A Summary of the problem-solving strategy for Speed Problems

This approach begins with a time model representing the scenarios, followed by the application of the distance-speed-time relation to construct a distance model, depicting and visualising all complex kinematic situations taking place.

The solution is derived by applying the Euclidean Algorithm for a greatest common unit and performing direct calculations.

A similar approach can be applied for more complex problems at the Primary Olympiad Level.

*References*

[1] Ho Soo Thong, Ho Shuyuan, *Bar Model for Speed Problems – A Ratio Approach*, barmodelhost.com

[2] Ho Soo Thong, Ho Shuyuan, *Speed Problems – A Ratio Approach*, Mathematical Medley, Vol 37, No. 1-2,

Dec 2011.

[3] Ho Soo Thong, Ho Shuyuan and Leong Yu Kiang Problem *Solving methods for Primary Olympiad Mathematics*, AceMath, 2012.

**A Chicken-Rabbit Approach**

The following classic word problem – involving chickens and rabbits in a cage – was posed in the ancient Chinese book Sunzi Suanjing :

*There are chickens and rabbits in a cage.*

*Look at the top of the cage – there are 35 heads.*

*Look at the bottom of the cage – there are 94 legs.*

*How many chickens and how many rabbits are there in the cage?*

A problem-solving strategy is to create a hypothetical or supposed situation, and compare it with the actual situation using mathematical deduction from the *Distributive Law*.

*Actual situation* : There are 35 heads and 94 legs.

*Supposed situation* : There are 35 chickens and 70 ( = 2×35 ) legs.

We construct a comparison bar model for the two situations:

Therefore we have 12 rabbits and 23 chickens.

The above problem solving approach is known as the Chicken-Rabbit Approach.

Next, we proceed to apply the *Chicken-Rabbit Approach* to some challenging problems.** **

**A Coins Distribution Problem**

There is a total of 33 coins of 10¢, 20¢ and 50¢, and they have a total value of $12.10.

The number of 20¢ coins is 3 more than that of 10¢ coins.

How many 50¢ coins are there?

*Solution*

By adding 3 more 10¢ coins, we create an adjusted situation there are equal number of 10¢ and 20¢ coins.

Originally,, there are 8 20¢ coins, 5 20¢ coins and 20 50¢ coins.

Notes:

1. Our solution involves the simple algebraic manipulation of one unknown.

2. In this solution, we break down a situation into simpler situations for easy computations.

3. For practice, pupils may reduce the number of 20¢ coins by 3 and work with n 10¢ coins.

In a *Traditional Approach***,** we assume that there are *x* 10¢ coins, *y* 20¢ coins and z 50¢ coins and then proceed to obtain and solve the simultaneous equations:

*x* + *y* + *z* = 33

10*x* + 20*y* + 50*z* = 1210

*x* + 3 = *y*

Finally, we solve a complex problem with heuristic bar modelling approach.

**A Weight Distribution Problem**

A total of 31 teachers and pupils plan for a two-nights excursion.

Teachers prepare 94.5 kg of food and 57 kg of camping equipment.

Each teacher will carry 3.5 kg of food. Among the pupils, each boy will carry 3 kg of food together with 2.5 kg of equipment and each girl will carry 3 kg of food together with 1.5 kg of equipment.

How many boys and how many girls are there in the excursion?

*Solution*

We solve the problem with a pictorial view in three steps.

1. Construct a comparison bar model for the “Food” situation and the “Equipment” situation.

2. Add a supposed Food situation and apply the Chicken-Rabbit Approach for the number of teachers and pupils

3. Add a supposed equipment situation and find the number of boys and the number of girls.

Note:

This heuristic problem-solving solution involves simple algebra for two *Chicken-Rabbit Situations*.

A *traditional approach* involves relating three unknowns: *t* (number of teachers), *b* (number of boys) and *g* (number of girls), to solve the simultaneous equations:

*t* + *b* + *g* = 31

3.5*t* + 3*b* + 3*g* = 84.5

2.5*b* + 1.5*g* = 57

*Remark*

Similar approach can be applied for problems involving three or more Chicken-Rabbit situations.