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VICs - Picking Numbers

In a problem that has variables in the answer choices, we always can use an algebraic approach, but sometimes that’s time consuming, or difficult to begin. An alternative strategy is to pick numbers for the answer choices. It’s very important to keep in mind that there is an art to picking numbers well. Picking numbers does not mean just plugging in the first numbers that come into your head.

We really have to be very strategic to do picking numbers well, and that’s what we’re going to talk about in this video. The first stage of picking numbers is to go after any, what I will call low hanging fruit. That is very easy to get to kind of stuff. This means eliminating answers that are very easy to eliminate.

This will seldom eliminate all four in, incorrect answer choices, but it will often eliminate some. For example, one of the variables is 0. It may be very easy to compute the answer of the question in the prompt, and it may be very easy to test values of zero quickly. So, this whole thing about testing low hanging fruit, this is something that you really should be able to do within say fifteen or twenty seconds.

This should not be long at all. This should be something very quick. And, this is most efficient if there is only one variable in the answer choice. If there is only one variable in the answer choice and that variable is a percent of something, then it may be very easy to figure out the prompt answer when the variable is either 0 or 100.

If there is more than one variable, typically it would not be as easy to apply this strategy unless it’s a rare scenario in which it might be possible for two variables to be zero at once, or a scenario in which is setting one variable to zero enormously simplifies the whole problem. So, for example, this question is one we actually solved algebraically in the previous video.

And so, if you’re not familiar with it, pause the video, and reread it here. All right, so we’re going to use a very simple trick for picking numbers here, to eliminate some low hanging fruit. If A equals 0, the distance of the first leg equals 0. Then the first leg took no time at all, and the second leg was the whole trip. Therefore, the average speed for the second leg, which is what the question is asking, that would just be the average speed for the whole trip, because the whole trip would be the second leg.

So, if A equals 0, the prompt answer would be V. If we replace A with 0, the correct answer will equal V. Well, here are the answer choices. I’m going to replace A with 0, and then I’m just going to see what happens. Well, two of the answer choices equal V. Two of them do not.

The two that do not are wrong, so immediately just by plugging this in, and here I was kind of slowing it down to show you all the details. But, if you can go through this very fast in your head, this is an enormously beneficial way just to knock off some answers right away. So,that it, it simplifies your job. So, that is the goal of the low hanging fruit stage.

You should not spend a long time on the low hanging fruit stage. In other words, the idea is that very quickly you’re gonna be able to plug in something very simple and eliminate a couple answers. And here, we’ve eliminated two of the answer choices. Of course, there’s something a little unrealistic about that.

In a real world scenario, we wouldn’t talk about a quote two leg trip if the first leg were actually zero. Don’t worry about that. The low-hanging fruit plug-in stage is often about pushing things to an unrealistic extremes to see the mathematical consequences. So, that’s what we’re doing in the low-hanging fruit stage of plugging in numbers.

Now, of course, it’s exceedingly rare that you could eliminate down to a single answer choice with this trick. This is about knocking out a few choices right away, so that on subsequent plug-ins, you have fewer choices to check. Also, question writers have this trick in mind, and sometimes design multiple answer choices that cannot be eliminated using this trick.

Sometimes, not a single answer choice can be eliminated! Depending on how well the question is written. Okay, that should just be a very short part of plugging in the low hanging fruit stage. Suppose we have eliminated whatever low-hanging fruit that we could, now we need to plug in answers in an attempt to determine the answer to the question.

Here’s some guidelines for picking numbers. Avoid 1 or 0. So, 0 or 1, those can be very useful in the low hanging fruit stage but once you’re done with that, when you actually want to determine a unique answer, avoid 1 or 0. Pick different numbers for different variables, and numbers different from those given in the problem.

Don’t pick numbers that are multiples of each other. Picking different prime numbers is a very good strategy. And keep the numbers small, and pick so that the answer to the prompt question is very easy to calculate. It might be useful to have one of the values equaling 10, or something like that, so that it’s easy to multiply.

The ideal would to be pick, would be to pick one set of numbers and have the five answer choices equal five different values when we plug in. See, that’s the ideal, because then you can distinguish among the answer choices if, if they equal five different things. Only one of them is going to equal the thing that it should, and that would have to be the answer.

So, that would make the, the picking numbers a very efficient strategy if you’re able to accomplish that. So, here’s a practice problem. Pause the video. You don’t have to solve this, just think about this problem, then we’ll talk about it.

Suppose we don’t know how to approach the problem algebraically. Clearly, when Q equals 0, then T equals 0, when Q equals 100, then T equals 100. So, those would be good low hanging fruit choices. But unfortunately, all five answer choices are designed to be consistent with those answer choices.

So, plugging in zero or a hundred is not going to help us. All five answer choices work for those choices. Let’s think about what it would mean for T to equal 50. In other words, 50% of your revenue is gonna come from each product. This would meant that 50% would come from Product B, which is $21, and 50% would come from Product A, which is $6.

All right, so let’s think about this. If the revenues are the same, it means that we have some multiple of $21 equals some multiple of $6. And, so here, we need to find a least common multiple. This is a skill we talked about in the integer properties videos. If this is something unfamiliar to you, you might want to go back and watch the integer property videos, as it turns out the least common multiple of 21 and 6 is 42.

So, if we sold 2 of product B, that would bring in $42. And if we sold 7 of product A, that would bring in $42, the same. And, so we’d have equal revenue from both the products. Well, 2 plus 7, 9, that means we’re selling 9 total products. And therefore, 2/9 of the units sold are product B. So, Q would equal 2/9 written as a percent.

And, that happens to be about 22%. It actually is more useful just to leave it in the form of the fraction, 2/9 times 100. So that’s our target answer. That’s the answer we want. So now, we are gonna go through the answer choices.

Plug in T equals 50, and what we want is 2/9 times 100. So, I plug it into answer choice A. First, I am gonna factor this out for easier plug in. I don’t actually have to multiply this out. What I see here, is that we’re gonna get something that is 1500 minus 700. That’s gonna be something in the hundreds, times 150.

This is gonna be way over 100. This is not gonna be a percent at all. So, this is wrong. Answer choice B, plug in 50. Now, I don’t have to do the exact division, I get down to something like 750 over 315.

Well, 315 times 2 is smaller than 750, 315 times 3 is bigger than 750. So this is between two and three so it can’t be right, because we want something that would be more around 20. So, something between two and three is not correct. So, this is wrong. Answer choice C, again, I’m gonna plug in 50, gonna multiply a bit, do the subtraction, well, 321, let’s see 321 times three is something in the 900’s.

That’s smaller than 1050. 321 times 4, that would be something in the 1200s, that would be bigger. So, it means that this fraction is something between 3 and 4. And again, if it’s between 3 and 4, it’s not gonna be close to, the, the range around 22, which is what we’re looking for. So, this can’t be right.

So, we’ve eliminated this. Answer choice D, plug in fifty. I can cancel these factors of ten. And, I get 2/9 times 100. Okay, so this is actually what we want. So, this is a possible solution.

We don’t know that this is right, but at least this is one answer that works. So, this is promising. Now, we have to check E. Here, we’ll plug in 50, we’ll multiply, we’ll cancel the factors of 10, we get 3/11ths times 100. Well, 3/11ths is not equal to 2/9ths, as it turns out it’s a little bit bigger, but it’s a different value.

What we want is 2/9ths times 100. 3/11 times 100 is just a different value. So, it’s close, but it doesn’t work. We eliminate that. And, the only possible answer is D. In this problem, we were lucky in that only one picked number eliminated all the incorrect answer choices.

Another point of which to be aware, don’t pick new, numerical choices that are too obvious. What do we mean by this? This is especially true in percent problems in which say, a percent, a price increases or decreases by some given percent. If the price variable is P, what value do you think every single test-taker on the planet is going to pick for P?

Of course, the universal choice will be P equals 100. The price is $100. Everyone’s gonna pick that. And, of course, the test makers, knowing that everyone’s gonna pick that, will design trap answers with precisely that choice in mind. So, if you pick 100, you will not be able to eliminate all the answer choices because several answer choices will work for that value, guaranteed.

If you want to excel on the test with extraordinary results, rule number one is not to do, not to always do the predictable things that everyone else does. If you have to pick a number for some quantity in which you will take the percent, pick 200, or 300, or 400, or 1000, some number for which it is exceptionally easy to calculate percents.

But, not the complete obvious choice that everyone else will make. Here’s a problem. Pause the video, work on this, and then we’ll talk about this. Okay, before January, the price of a dress was D and the price of matching shoes was H.

In January, the price of the dress increased by 40%, the price of the shoes increased by 50%, and didn’t change again in the following months. In March, Roberta bought both items with a 30% discount. If D = 5H, so D is 5 times H, which of the following represents the amount that Roberta paid? So, first of all, what choice do you think everyone is going to make for this problem?

Everyone is going to pick dress equals 100, and then the shoes are 20. That’s going to be the really obvious choice. Several of the answer choices are designed to work for that particular choice. So, we’re not gonna make that choice. So, in fact, it wouldn’t actually be hard even to do this with algebra, but we are gonna pick numbers here, and I’m gonna pick 200 rather than 100.

I’m gonna pick 200, which is a less obvious choice. So, the dress increases by 40%. So, it goes from 200, so 20% of 200 would, would be, 10% would be 20, so 40% is 80, so it goes up to 280. The shoes increase by 50%, so they go from another 20 up to 60. That means the total ensemble goes up to $340.

And, then Roberta buys it at a 30% discount. Well 10% of 340 is 34. Multiply that by 3, 3 times 34 is 102. That’s her discount. So 340 minus 102 is 238. She pays $238.00.

So, the answer choices, we’re going to plug in D equals 200, and H equal 40, and that should produce the answer 238. So, when I plug this into A, I get 240, that doesn’t work. When I plug it into B, I get 239, that doesn’t work. That’s close but it doesn’t work. When I plug it in to C, I get 200 plus 80, 280, that doesn’t work.

When I plug in to D, okay, here we have to multiply out a little bit. Let’s see, 595 times 40, I’m gonna write this as 6 minus 0.05 times 40. Well, 6 times 40, that’s easy, that’s 240. And, 0.05 times 40, that we can multiply out, that’s gonna be 2. And, so this is 240 minus 2 is 238. So, that actually works, that actually produces the answer we want.

And, then when we do 200 times 240, we’ll notice that this is gonna be bigger than 1.2 times 200, and 1.2 times 200 is 240 so whatever this is gonna be larger than 240. So, this is not gonna work and with this one choice, this non-obvious choice, we’ve been able to eliminate four answers and we’ve got the answer. Again, we were lucky to eliminate four answers with one choice of numbers.

The more conventionable, conventional and predictable the numbers you pick are, the more likely it is that the test writer has designed trap answers to match such choices, and therefore, you will not be effective at eliminating answers if you make very conventional choices. It can work very much to your advantage to make a choice slightly different from what you think most people would choose.

Finally, what’s a good standard by which to decide when to use an algebraic solution and when to pick numbers? As a general rule, if you see how to do the algebra, do the algebra. Tend to use that as your default approach because the algebra is always efficient. Get to know yourself. And, this is really important.

If you solve a problem one way during a practice section, take the time to solve it the other way after the session, and take note. Which is easy for you? Which is faster for you? Which is, which do you prefer? What qualities of a problem make one method, or the other method preferable to you?

You see, you really have to learn your, about yourself. For any given problem, it’s not that the problem is better doing it one way or the other way, it’s about you yourself with your particular comfort level. It will be easier for you to do it one way or another way, that’s the thing you need to discern by practicing and experimenting over time. In summary, picking numbers is an important strategy when we have variables in the answer choices.

Picking numbers, picking very easy choices to eliminate low-hanging fruit can be a helpful way to begin. Remember, you don’t want to spend a lot of time on that. That should be a very quick process if you can eliminate a couple things quickly. Pick numbers that are prime, different from one another, easy to work with, and not entirely predictable.

And, in general, an algebraic approach is a good default. But, get to know your own style, and what works best for you.

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