07 Challenging Word Problem

Ali and John had 550 sticks of satay altogether. Ali had 50 more sticks than John. When John sold 1/3 as many sticks of statay as Ali, he was left with twice as many sticks as Ali. How many sticks did John sell?

The challenge is to “simplify” or “reduce” the model into some common parts that we can find the value of the common part. There might be different parts in the model and the trick here is to express the different parts into 1 common part that we can find. In algebraic terms, we are trying to express the multiple variables into 1 variable so that we can solve the equation. In this example, I am going to draw the model systematically. For the better students, they can skip a few steps if they can “visualize” the initial models in their hard.

Ali and John had 550 sticks of satay altogether. Ali had 50 more sticks than John. 

Most students should be able to draw this model without much prompting. Possible questions to ask : Who had more? How do I represent 550 in the model?

When John sold 1/3 as many sticks of statay as Ali, he was left with twice as many sticks as Ali.

There are two parts to this sentence. I shall tackle the 2nd part first as the 2nd part seems more easier (After all, “twice as many” does not seem as difficult as “1/3”).  Always tell the pupils to draw models that they find it more easier first. Possible questions to ask : Who had more now? How are the two “after bars” related? 

 When John sold 1/3 as many sticks of statay as Ali, he was left with twice as many sticks as Ali.

Now, for the first part. This is more complicated than the 2nd part as it involves “1/3” and “sold”. Possible questions to ask : What does the word “sold” mean? How can I show the “sold” in the after model? How can I show “1/3 sold”?

The model drawn below has captured all the information in the question. Get the students to see that it is impossible to solve for any of the parts. There is a need to “reduce” the model to some common part.  Possible questions to ask : Are all the parts in the model the same? Can I find any of the part? If not,  what must I do?

The trick is to relate the cyan parts and the blue parts together. Possible questions to ask : Is there anyway that the cyan parts and blue parts are related ? 
Some students might see the relationship. We can use MaPS to rearrange the parts so that the relationship is clearer.

Get the students to articulate the relationship

2 cyan parts = 1 blue part  + 50

Using this relationship, we can attempt to “simplify” the model. Emphasize the aim is now to convert the model into common part (i.e. I choose blue part). Get the students to articulate why they choose that as “common part”. It is okay if students choose cyan part to be the “common part”.

This is how the model looks like now:

We are still not too happy as the cyan parts stick out like a sore thumb :-) Possible questions to ask: How can I “get rid” of the cyan parts?  How can I get “2 cyan parts” as I know the relationship? Is there another way to draw the model? 

Now, get the students to redraw the model horizontally.

Now, the model looks much more “doable”. Possible questions to ask: Does the model looks more “friendly” now? Can you express the 2 cyan parts into blue parts?”

Finally, we get to the “solvable” model.

The battle is 75% won. The students just need to solve for the blue part. And make sure the final answer is what the question is asking.

 

 

Leave a Reply

Your email address will not be published.