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Word Problem Strategies. So, so far throughout the word problem module, we have talked about the very straightforward algebraic solution to problems, and always those work that's always one way to solve the problem, but it's not the only way,. And in this set of videos at the end we're gonna be talking about some alternative strategies that you can use.

The first strategy we're gonna talk about is backsolving. If all five answer choices are numerical, we get five numbers, then we can use the strategy known as backsolving. Backsolving means that we start by assuming one of the numerical answers is the answer to the question, we pick one and we say, okay, let pretend that this is the answer.

From this we work backwards and see if it actually makes sense. If the answer doesn't work, we often can tell that we needed a larger or smaller number for the end result and this allows us to eliminate more than one answer at a time. A relatively simple way to approach this is always begin with answer choice C. So this is the most straightforward backsolving strategy.

If you start with C, that will be correct 20% of the time and if it's not correct, you will be able to eliminate either the two larger answers or the two smaller answers. So, just in one trial, even if you don't get the right answer, you can usually eliminate three answers at once, and then all you have to do is test one of the remaining two.

There's an alternate strategy, so this is a little more sophisticated but you can certainly use this if you like, alternate strategy would be to pick B first, there's a 20% chance that it's the official answer and another 20% chance that is too big which would make A the official answer,. That gives you a 40% chance of being right on the first guess, and of course, if B doesn't work then you'd use D next, and then that would narrow everything down to a single answer.

So that's a slightly more sophisticated approach but statistically it gets you to the right answer on average a little bit faster. So certainly, it's something to consider. Okay, here's a practice question, pause the video and then we'll talk about this. Okay, this is a relatively easy practice question. In a certain state, schools pay 2% tax on food and 8% tax on stationery.

The school placed a combined order of $500 on food and stationery, and paid $19 on tax on the order. How much of that money was spent on food? So we could solve this, we could set up algebra and solve this. Instead of doing that, we're gonna explore a backsolving approach. So we're just gonna pick C, we're just gonna say, pretend 300 is the answer that means they spent 300 on food so they must've spent 200 on stationery.

There's a 2% tax on food, so 2% of 300 gives us $6. There's an 8% tax on stationery so that gives us $16, the total tax there is $22 and that's too much. They paid $19 in tax, so 22 is too much and so what this means is, to pay less in taxes the school must have spent more on food and less on stationery, cuz food is taxed at a much lower rate.

So if they spent more on food they would be paying less in taxes. So right away, we can eliminate A, B and C, so, this is great. Even if we were running out of time, we could guess from the remaining two we would have a very high likelihood of, of guessing by chance the right answer because three answers have already been eliminated. But that was pretty quick, so we probably would have time still to do another round of backsolving.

So another round we could pick either D or E. I'm just gonna pick E because it's divisible by 100, it's just slightly more convenient so that's the only reason I'm gonna pick it. Let's say E is the answer. Well that means that food costs $400 that stationery costs $100. The food tax 2% of 400 is $8, 8% of 100 of course is $8 and that would give us a total of $16 in tax which is too little.

So A, B, and C are too big, E is too little as far as the amount of tax, is that leaves the only right answer, the only possible answer would be D. So after two choices we are done, we know that D has to be the answer. Here's another practice question, pause the video and then we'll talk about this. Okay, this is a mixture problem.

If you haven't seen the video on, on mixture questions it may help to go back and watch that video before talking about this particular problem because this particular problem depends on knowledge of that video. A chemical supply company has 60 liters of 40% of some chemical, that chemical happens to be nitric acid. You don't need to worry about that.

How many liters of pure undiluted acid must the chemists add so that the resultant solution is a 50% solution? So, we're gonna add solute to increase the concentration and we want to increase it to 50%. So C is a nice round number so we'll just pick C as our answer. Before we even pick it, notice that the amount of solute in the beginning solution 40% of 60 is 24 liters.

There are 24 liters of the acid in the initial solution. So, now, let's pick C. We add 20 more liters of acid, and so that means that we have a total concentration of 44 liters that is the amount of acid in the solution and of course, the total solution the volume of the solution is gonna be 80 liters and so we wanna know, do we have the right concentration?

Well, we don't have to do the division. We can see that 44 is more than half of 80. So, we don't even note, need to know the exact number. All we need to know is the concentration would be over 50%, so, this is too high. We've added too much acid. So right away.

We can eliminate C, D and E, all of those add to much acid. So now we have to pick either A or B, it doesn't matter. I'll just pick A. Suppose we add 12 liters of acid. Well, now the total concentrate is 24 plus 12, 36 liters of acid. So that's how many liters of acid total would be in the solution and at that point, the volume, the total volume of the solution will be 72, and notice that 36 is exactly half of 72, so this would be a 50% solution, this answer actually works.

So we pick the right answer here, the answer is A. In summary, when all five answer choices are numbers, one alternative strategy is to solve by backsolving. So the very straightforward approach is start with answer C. Try this as the answer to the prompt and see if it works in the scenario. If C doesn't work, the information about too big or too small will allow us to eliminate other answers, and remember, an alternate strategy would be to pick, for example, B and then you could, you might be able to determine that either B or A is the answer, depending on whether it was just right, or too big, or too small and if it didn't work.

Then you could pick D, and that second choice would allow you to narrow everything down.

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