استراتژی QC - عملیات تطبیق
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QC Strategies - Matching Operations
Matching operations. So to begin I’ll say there’s only so much about QC’s we can discuss right now because I realize you’re at the beginning of the math module, and there’s quite a bit of math that you haven’t reviewed yet. So after the math content modules, we’ll have a much deeper module called advanced QC strategies, in which we will discuss how the different content areas show up on the quantitative comparison questions.
It doesn’t really make sense right now, for example, to talk about quantitative comparisons in geometry, because we haven’t reviewed geometry yet. So all of those videos are at the end of the math content modules. So right now, I just wanna give you one strategy that will be very helpful, because in your practice you’re gonna start to encounter quantitative comparison questions, and I want you to at least have some strategies for attacking them.
So I’m gonna describe a quick trick that one can use in a wide variety of quantitative comparison questions. For now, I’ll call this trick matching operations. In any QC, we are trying to determine the relationship between two quantities. We could say that, between the two quantities, there’s some unknown relationship.
It might be always greater than, always less than, always equal, or it might be some combination or indeterminate. It’s important to recognize there are mathematical steps we might take to simplify the problem that do not change this underlying relationship at all. We could add the same number to both quantities. We could add a positive or negative number to both quantities.
We could subtract the same number from both quantities, or we could multiply or divide both quantities by the same positive number. Very important, can’t multiply or divide by a negative, but we can multiply or divide by any positive number. Why we can do these steps and not others, for example, divide by a negative, we will discuss in the advanced videos, after we’ve discussed all the math content.
Right now, simply practice this as a strategy, because when you are used to thinking this way, this strategy can enormously simplify many QC problems. Here’s a practice problem. Pause the video, and then we’ll talk about this. So first of all, I realize that if you haven’t reviewed fractions yet, it might be a little bit challenging to think about all this.
And so if you’re rusty with fractions, don’t worry, we have a whole module with fractions coming up soon. I’m just gonna talk about some of the very basics here. Now you may notice, for example, that 35 divided by 8, that’s 4 and some kind of remainder. And 13 divided by 3, that’s also 4, and some sort of remainder.
In other words, 4 equals 32 divided by 8, or 12 divided by 3. So, both of those are reasonably close to what we have here. So we could subtract 4 from both of these and then in each case we’re gonna find the appropriate common denominator. So with the 8s we’re gonna turn the 4 into 32 over 8 and with the 3rds we’re gonna turn it into 12 over 3, and then we have a common denominator, so we can subtract in the numerators, and we get three-eights and one-third.
Well, we have a few different ways that we can approach this. One way is to find a common denominator of the two fractions now, so multiply the left one by 3 over 3, multiply the right one by 8 over 8. And we get nine-twenty-fourths compared to eight-twenty-fourths. And so we see quantity is bigger. Incidentally, when we got the three-eighths compared to one-third we also just could have cross-multiplied.
And then we’d get nine on the left, eight on the right. And again we’d see quantity A is bigger. So it turns out, none of those steps change the relationship at all. And so quantity A has to be bigger, even in the original. Here’s another practice problem. Pause the video and then we’ll talk about this.
This can be a tricky problem for a variety of reasons. This involves a little bit of algebra, and again if you’re rusty with algebra don’t worry the algebra here is very simple. We have a couple of modules coming up with a lot of algebra in it for your review. So, what we’re gonna do here, some very basic algebraic simplification. First of all, we’re just gonna subtract 3 from both sides.
So we get x plus 7 and 3x then we’re gonna subtract x from both sides. Then we’re gonna divide both sides by 2, and now we can see that this comparison between quantity A and quantity B. This is identical to comparing x to 3.5. So this is interesting. If x is bigger than 3, how does it compare to 3.5?
Well, if you’re stuck thinking about integers, well, then if you think greater than 3, well, that would be 4, 5, 6, 7. All those are bigger than 3.5. But part of this is one of QC’s favorite tricks, don’t get stuck thinking about integers because there are all kinds of fractions and decimals. For example, 3.1 is bigger than 3, but it’s less than 3.5.
And so we can pick numbers that are all greater than 3, but some of them could be greater than 3.5, some of them could be less than 3.5. So we have no way to determine and so the answer is D. Here’s another practice problem, pause the video and then we’ll talk about this. So machine A can produce 11 nails in 14 seconds. And so, each single nail is produced in fourteen-elevenths of a second.
And similarly, B each nail is produced in nine-sevenths of a second. So these are the two fractions we’re comparing, not very nice fractions. Well, notice that I could multiply both sides by 7. Now I could multiply both sides by 11. And essentially what I’ve done here is I’ve cross multiplied. Well now, 7 times 14 that’s 98, 11 times 9 that’s 99.
99 is greater than 98. And so the time it takes machine B to produce one nail is marginally larger than the time that it takes A to produce a nail and so, the answer is B. In summary, one strategy we can use to simplify many quantitative comparison questions is matching operations. We can add the same number to both quantities, subtract the same number from both quantities, or multiply or divide both quantities by the same positive number.
These operation preserve the underlying relationship between the quantities, so we can come to a simplified answer and that relationship will be the same as the relationship in the original problem.
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