Picking number strategies. So in VIC problem, variables in the answer choices, we can always use an algebraic approach. But sometimes that's time consuming or it can be hard to see how to begin the algebra. So sometimes, it's better to pick numbers and I'll say right away that there's an art to picking number well.

That's what we're going to talk about here. I'll distinguish two stages. First of all, what I'll call the low-hanging fruit stage and what I mean by that is eliminating answer choices that are very easy to eliminate. This will seldom eliminate all four answer choices, but it often will eliminate some and it just makes your job much easier.

So it's a very quick way to simplify the problem. For example, if one of the variables is zero, it may be very easy to compute the answer of the question in the prompt and it may be easy to test the value of zero quickly. So zero is an excellent next on value to use in the low-hanging fruit stage. This strategy is most efficient if there's only one variable in the answer choice.

But if there's a small number of variables or if there's one key variable that we can set equal to 0, then it's helpful also. If there's only one variable on the problem, that variable's a percent of something, then it may be very easy to figure out the prompt answer when this variable is either 0 or 100. If there's more than one variable, typically it would not be as easy to apply this strategy, unless it's a rare case in which it might be possible when one variable is zero, the other has to be zero or something like that.

We've already looked at this problem. But here, we're gonna talk about it. We're picking numbers and see if you can just use the low-hanging fruit approach to simplify things a bit here. Pause the video and then we'll talk about this. Okay, we'll suppose that A=0.

So in other words, the first leg was zero miles. So really, there was no first leg. And so it would take no time at all and that would mean the second leg was the whole trip. Well, if the second leg was the whole trip, the average speed for the second leg would be the average speed for the whole trip, V.

So if A=0, the whole expression should simplify to V. So what we will do here is we will plug-in A=0 and then we see A and D still equal V. B, C and E don't. And so right away just with one simple plug-in, we've simplified. Well, now we could plug-in other numbers and we only have two answer choices to test, because we've limited three answers.

Also if you're running out of time and needed to guess, your odds of guessing well or guessing the correct answer would be very high. Because you're only guessing now between two answer choices rather than five. Now, of course, there's something a little unrealistic about that. In a real world scenario, of course, we would not talk about a two-leg trip if the first leg were zero miles.

That would be a little duplicitous. We'd be trying to cheat somebody by talking about it that way, but don't worry about that. The low-handing fruit plug-in stage is often about pushing things to an unrealistic extreme to see the mathematical consequences. Now, of course, it's exceedingly rare that you can 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. It's purely about simplifying the problem. Also, question writers have this trick in mind. Sometimes, designs multiple answer choices that cannot be eliminated using this trick.

Sometimes, not even a single answer choice can be eliminated by using a low-hanging fruit approach. So the more you practice, the more you'll develop a sense when you can simplify and when you can't. All right, supposed were done with step one. We picked zero.

We've eliminated a few answer choices. We've eliminated whatever low-hanging fruit that we could. And now, we actually need to plug-in answers in an attempt to determine which one is correct. Here is some guidelines for picking number. First of all, if you're not in the low-hanging fruit stage, avoid one or zero.

Because one or zero, first of all, one times anything is just that thing. And so you don't know, suppose the correct factor were g squared. Well, one answer choice had a g, one had a g squared, one had a g cubed. All of them would equal the same thing if you plugged in g equals 1. And so one is a very bad choice. And for similar reasons, zero is a very bad choice.

Pick numbers different for different variables and different from those given in the problem. And so that way, they will be much easier to see if you have a factor of three, a factor of seven that you plug-in, then it's easy to see. Well, I wound up with this in the numerator and this in the denominator. It will be easy to see which numbers led to which consequences.

Don't pick numbers that are multiples of each number, picking different prime numbers is a good idea. Keep the answers small and pick, so that the answer to the prompt question is very easy to calculate. The ideal would to be to pick one set of numbers and have the answer choices equal five different values when we plug-in.

So that it would be very easy to say, okay, this is the value we want and the other four are wrong. Here's a practice problem. Pause the video, think about this. And even if you can do it with algebra, put the algebra on hold right now and see if you can do this with picking numbers.

Okay, well, let's talk about this. We have the vendor sells just two products. One for $6, one for $21. Q is a percent of the product sold. How many items are sold and T is a percent of revenue. And so we want to express Q in terms of T.

First, we don't know how to approach the problem algebraically. Now clearly, when Q=0, T=0 and when Q = 100, T=100, those would be good low hanging fruit guesses. But unfortunately, all five choices are designed to be consistent with those choices. So here the test writer, all ready anticipated.

People might try to use the low-hanging fruit approach. And so they created answer choices where nothing could be eliminated. All right, let's think about what it would mean for T to equal 60, okay. So in other words, 50% of the revenue is coming from each. 50% on revenue is coming from product A and 50% other revenue is coming from product B.

All right, and now we have to think about this a bit. Product B is $21, product A is $6. And so we'd need a certain number of product Bs and a certain number of As, and we'd need those to equal the same amount. In other words, we would need a common multiple of 21 and 6. And so finding the least common multiple.

If this is a skill that is foreign to you, if this is something that you have not studied, let me suggest that you should go back to integer properties at some point and watch the videos there when we talk in-depth about the common multiple. I'll just say here that 21, of course is 2 times 7 and 6 is 2 times 3. Both have a three in them.

And so we just multiply 2 times 3 times 7, then that will help B something divisible by 21, divisible by 6. And in fact, that is 42. And so 42 is the common multiple. In fact, it's the least common multiple. So let's just say that we sell two of product B.

So in other words, it would be $42 of product B and we'll sell seven of product A and that will be $42 of product A. So now we're getting $42 from product B, $42 from product A, so that does mean that T equals 50. 50% of the revenue is coming from product B. Now what would Q be?

Q is a percent of the item sold. We sell two of product B, seven of product A And so how many of B are sold Well, they're nine items sold all together, 2 plus 7. So two-ninths are product B. And so Q would equal two-ninths times 100 and we can change that into that decimal, but it's actually easier to think about it in this form here, two-ninths rimes 100.

And so now what we're gonna do is plug-in T equals 50 and see if we get two-ninths times 100. So this one, I'm gonna factor out. I'm gonna write it this way. And clearly, we're getting really, really big numbers here. This is not gonna be a number smaller than 100.

This is gonna be something gigantic. 5 times 15 is 75, 5 times 50 is 750. So clearly, we have a number in the hundreds times 50. This is a gigantic number, way over 100. This can't be a percent, so we can eliminate answer choice A. Answer choice B, plug-in T equals 50.

In the numerator, I get 750. In the denominator, I get 615-300 or 750 over 315. Well, that's some number between two and three. And so it can't be correct. So eliminate answer choice B.

Answer choice C, plug-in, multiply out 1,050 over 321. So that's somewhere between three and four, so that can't be the right answer. So we eliminated A, B and C with this answer choice of T equals 50. D, plug this in and we get 200 in the numerator. We get 90 in the denominator and that simplifies to two-ninths times 100.

So that's exactly what we want. Okay, that's promising. That is one solution that works. For E, multiply, we use doubling and halving in the numerator. So now, we get three-elevenths times 100.

Three-elevenths times 100 is not the same as two-ninths times 100. It's close, but its not the same. So this is not correct. We eliminate E and the only possible answer is D. So this was a case where picking one good choice of the number eliminated four answer choices.

And so we were done with just one choice. We were able to narrow things down to a single correct answer choice with only one value of the variable. So that we were lucky in some ways that only one picked number eliminated all the answer choices. Another point to be aware, don't pick numerical choices that are too obvious.

This is especially true in percent problems in which say, a price increases or decreases in some given percent. If the price is the variable 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=100. And of course, the test-maker anticipating this will design all kind of trap answers with that choice in mind.

If you want to excel on the test with extraordinary results, rule number one is not always to do the predictable things that everyone else does. So if you're just as predictable as everyone else, it's gonna be easy for the test-maker to trap you the way the test-maker traps everyone else. Here with percents, if you have to pick a number for some quantity of which you will take a percent instead of 100, pick 200 or 300 or 400 or 1,000.

Some number for which it is exceptionally easy to calculate percent, but not the completely obvious choice that everyone else will make. And so just a shade different from obvious is exactly what you want. Here's a practice problem. Pause the video and then we'll talk about this. Okay, let's talk about this problem.

It actually wouldn't be too hard to use algebra in this problem, but let's just put the algebra on a hold. After the video, you could go back and try and solve the problem with algebra. Here though, let's practice picking numbers. And of course, the most obvious choice for that dress price is gonna be 100. So I'm not gonna use that.

I'm gonna pick dress price equals 200, which means the shoes are 40. So let's think about this. The dress increases by 20%, by 40%. So 200 plus 40% would be 80, that would be 280. The shoes increase by 50%. So 40+20=60.

They combine to a price of $340. Now, Roberta gets a 30% discount on that price. Well, 10% would be 34. 30% would be 3 times 34. Well, 3 times 30 is 90. 3 times 4 is 12.

90 plus 12 is 102. And so what does Roberta pay? She pays 340-102, which is 238. So that's the price she pays. And so the choices of D=200 and H=40 should produce the answer to 38. So we go to the answer choices.

Answer choice A is 240. Answer choice B is 239. Answer choice C is 280. So those are not right. All right, what about 5.95 times H? Well, 5.95(40), how does this work?

Well, 5.95, I'm gonna write that as 6-0.05. Well, 6 times 40 is 240. So 40 times 0.05, that would be the same as 4 rimes 0.5, move the decimal of 1 to the left, the decimal of 1 to the right. So it's 0.5 times 4, which is 2. And so it's basically 240-2, which is 238.

So this actually works, choice D actually works. Now we have to check choice E. Have we eliminated both answers or do both D and E work? Let's see. For E, we get 1.21(200). And of course, that's bigger than 1.2(200) and 1.2(200) is 240.

And so this is gonna be something that equals something bigger than 240, so it's not gonna equal 238. So this is incorrect and the only correct answer here is answer choice D. Again, we were lucky to eliminate four answers with one choice of numbers. The more conventional and predictable the numbers you pick, the more likely it is that the test-writer has designed trap answers to match the choices.

It can work very much to your advantage to make a choice slightly different from what you think most people would choose, because the trap is gonna be designed for what most people choose. And so if you're only slightly different from that, you're gonna avoid the trap. Finally, what's a good standard by which to decide whether to use an algebraic solution and when to pick numbers?

Well, here really, the best thing to do is practice, practice, practice. If you're the type of person who is good with algebra, it's natural for you to do algebra, then it was definitely the choice. If you see how to do the algebra, then that is a wonderful way to do the problem. If algebra is not really a strong suit for you and picking numbers just makes more sense, then by all means try picking numbers.

And in fact, really, the best thing to do is when you get a practice problem, do it one way when you are in practice. But before you look at the answer, try and do it the other way. And get a sense of which one was harder for me, which one took more time. The more you can develop a sense of yourself, the more comfortable you'll be making this choice when you're in a test situation.

In summary, picking numbers is an important strategy in problems that have variables in the answer choices. Picking very easy choices to eliminate low-hanging fruit can be a helpful way to begin. So that's just a way very quickly to knock off a few answer choices to simplify the problem.

Pick numbers that are prime, different from one another, easy to work with, and not entirely predictable. In general, an algebraic approach is a good default, but get to know your own style and what works best for you.

Read full transcriptThat's what we're going to talk about here. I'll distinguish two stages. First of all, what I'll call the low-hanging fruit stage and what I mean by that is eliminating answer choices that are very easy to eliminate. This will seldom eliminate all four answer choices, but it often will eliminate some and it just makes your job much easier.

So it's a very quick way to simplify the problem. For example, if one of the variables is zero, it may be very easy to compute the answer of the question in the prompt and it may be easy to test the value of zero quickly. So zero is an excellent next on value to use in the low-hanging fruit stage. This strategy is most efficient if there's only one variable in the answer choice.

But if there's a small number of variables or if there's one key variable that we can set equal to 0, then it's helpful also. If there's only one variable on the problem, that variable's a percent of something, then it may be very easy to figure out the prompt answer when this variable is either 0 or 100. If there's more than one variable, typically it would not be as easy to apply this strategy, unless it's a rare case in which it might be possible when one variable is zero, the other has to be zero or something like that.

We've already looked at this problem. But here, we're gonna talk about it. We're picking numbers and see if you can just use the low-hanging fruit approach to simplify things a bit here. Pause the video and then we'll talk about this. Okay, we'll suppose that A=0.

So in other words, the first leg was zero miles. So really, there was no first leg. And so it would take no time at all and that would mean the second leg was the whole trip. Well, if the second leg was the whole trip, the average speed for the second leg would be the average speed for the whole trip, V.

So if A=0, the whole expression should simplify to V. So what we will do here is we will plug-in A=0 and then we see A and D still equal V. B, C and E don't. And so right away just with one simple plug-in, we've simplified. Well, now we could plug-in other numbers and we only have two answer choices to test, because we've limited three answers.

Also if you're running out of time and needed to guess, your odds of guessing well or guessing the correct answer would be very high. Because you're only guessing now between two answer choices rather than five. Now, of course, there's something a little unrealistic about that. In a real world scenario, of course, we would not talk about a two-leg trip if the first leg were zero miles.

That would be a little duplicitous. We'd be trying to cheat somebody by talking about it that way, but don't worry about that. The low-handing fruit plug-in stage is often about pushing things to an unrealistic extreme to see the mathematical consequences. Now, of course, it's exceedingly rare that you can 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. It's purely about simplifying the problem. Also, question writers have this trick in mind. Sometimes, designs multiple answer choices that cannot be eliminated using this trick.

Sometimes, not even a single answer choice can be eliminated by using a low-hanging fruit approach. So the more you practice, the more you'll develop a sense when you can simplify and when you can't. All right, supposed were done with step one. We picked zero.

We've eliminated a few answer choices. We've eliminated whatever low-hanging fruit that we could. And now, we actually need to plug-in answers in an attempt to determine which one is correct. Here is some guidelines for picking number. First of all, if you're not in the low-hanging fruit stage, avoid one or zero.

Because one or zero, first of all, one times anything is just that thing. And so you don't know, suppose the correct factor were g squared. Well, one answer choice had a g, one had a g squared, one had a g cubed. All of them would equal the same thing if you plugged in g equals 1. And so one is a very bad choice. And for similar reasons, zero is a very bad choice.

Pick numbers different for different variables and different from those given in the problem. And so that way, they will be much easier to see if you have a factor of three, a factor of seven that you plug-in, then it's easy to see. Well, I wound up with this in the numerator and this in the denominator. It will be easy to see which numbers led to which consequences.

Don't pick numbers that are multiples of each number, picking different prime numbers is a good idea. Keep the answers small and pick, so that the answer to the prompt question is very easy to calculate. The ideal would to be to pick one set of numbers and have the answer choices equal five different values when we plug-in.

So that it would be very easy to say, okay, this is the value we want and the other four are wrong. Here's a practice problem. Pause the video, think about this. And even if you can do it with algebra, put the algebra on hold right now and see if you can do this with picking numbers.

Okay, well, let's talk about this. We have the vendor sells just two products. One for $6, one for $21. Q is a percent of the product sold. How many items are sold and T is a percent of revenue. And so we want to express Q in terms of T.

First, we don't know how to approach the problem algebraically. Now clearly, when Q=0, T=0 and when Q = 100, T=100, those would be good low hanging fruit guesses. But unfortunately, all five choices are designed to be consistent with those choices. So here the test writer, all ready anticipated.

People might try to use the low-hanging fruit approach. And so they created answer choices where nothing could be eliminated. All right, let's think about what it would mean for T to equal 60, okay. So in other words, 50% of the revenue is coming from each. 50% on revenue is coming from product A and 50% other revenue is coming from product B.

All right, and now we have to think about this a bit. Product B is $21, product A is $6. And so we'd need a certain number of product Bs and a certain number of As, and we'd need those to equal the same amount. In other words, we would need a common multiple of 21 and 6. And so finding the least common multiple.

If this is a skill that is foreign to you, if this is something that you have not studied, let me suggest that you should go back to integer properties at some point and watch the videos there when we talk in-depth about the common multiple. I'll just say here that 21, of course is 2 times 7 and 6 is 2 times 3. Both have a three in them.

And so we just multiply 2 times 3 times 7, then that will help B something divisible by 21, divisible by 6. And in fact, that is 42. And so 42 is the common multiple. In fact, it's the least common multiple. So let's just say that we sell two of product B.

So in other words, it would be $42 of product B and we'll sell seven of product A and that will be $42 of product A. So now we're getting $42 from product B, $42 from product A, so that does mean that T equals 50. 50% of the revenue is coming from product B. Now what would Q be?

Q is a percent of the item sold. We sell two of product B, seven of product A And so how many of B are sold Well, they're nine items sold all together, 2 plus 7. So two-ninths are product B. And so Q would equal two-ninths times 100 and we can change that into that decimal, but it's actually easier to think about it in this form here, two-ninths rimes 100.

And so now what we're gonna do is plug-in T equals 50 and see if we get two-ninths times 100. So this one, I'm gonna factor out. I'm gonna write it this way. And clearly, we're getting really, really big numbers here. This is not gonna be a number smaller than 100.

This is gonna be something gigantic. 5 times 15 is 75, 5 times 50 is 750. So clearly, we have a number in the hundreds times 50. This is a gigantic number, way over 100. This can't be a percent, so we can eliminate answer choice A. Answer choice B, plug-in T equals 50.

In the numerator, I get 750. In the denominator, I get 615-300 or 750 over 315. Well, that's some number between two and three. And so it can't be correct. So eliminate answer choice B.

Answer choice C, plug-in, multiply out 1,050 over 321. So that's somewhere between three and four, so that can't be the right answer. So we eliminated A, B and C with this answer choice of T equals 50. D, plug this in and we get 200 in the numerator. We get 90 in the denominator and that simplifies to two-ninths times 100.

So that's exactly what we want. Okay, that's promising. That is one solution that works. For E, multiply, we use doubling and halving in the numerator. So now, we get three-elevenths times 100.

Three-elevenths times 100 is not the same as two-ninths times 100. It's close, but its not the same. So this is not correct. We eliminate E and the only possible answer is D. So this was a case where picking one good choice of the number eliminated four answer choices.

And so we were done with just one choice. We were able to narrow things down to a single correct answer choice with only one value of the variable. So that we were lucky in some ways that only one picked number eliminated all the answer choices. Another point to be aware, don't pick numerical choices that are too obvious.

This is especially true in percent problems in which say, a price increases or decreases in some given percent. If the price is the variable 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=100. And of course, the test-maker anticipating this will design all kind of trap answers with that choice in mind.

If you want to excel on the test with extraordinary results, rule number one is not always to do the predictable things that everyone else does. So if you're just as predictable as everyone else, it's gonna be easy for the test-maker to trap you the way the test-maker traps everyone else. Here with percents, if you have to pick a number for some quantity of which you will take a percent instead of 100, pick 200 or 300 or 400 or 1,000.

Some number for which it is exceptionally easy to calculate percent, but not the completely obvious choice that everyone else will make. And so just a shade different from obvious is exactly what you want. Here's a practice problem. Pause the video and then we'll talk about this. Okay, let's talk about this problem.

It actually wouldn't be too hard to use algebra in this problem, but let's just put the algebra on a hold. After the video, you could go back and try and solve the problem with algebra. Here though, let's practice picking numbers. And of course, the most obvious choice for that dress price is gonna be 100. So I'm not gonna use that.

I'm gonna pick dress price equals 200, which means the shoes are 40. So let's think about this. The dress increases by 20%, by 40%. So 200 plus 40% would be 80, that would be 280. The shoes increase by 50%. So 40+20=60.

They combine to a price of $340. Now, Roberta gets a 30% discount on that price. Well, 10% would be 34. 30% would be 3 times 34. Well, 3 times 30 is 90. 3 times 4 is 12.

90 plus 12 is 102. And so what does Roberta pay? She pays 340-102, which is 238. So that's the price she pays. And so the choices of D=200 and H=40 should produce the answer to 38. So we go to the answer choices.

Answer choice A is 240. Answer choice B is 239. Answer choice C is 280. So those are not right. All right, what about 5.95 times H? Well, 5.95(40), how does this work?

Well, 5.95, I'm gonna write that as 6-0.05. Well, 6 times 40 is 240. So 40 times 0.05, that would be the same as 4 rimes 0.5, move the decimal of 1 to the left, the decimal of 1 to the right. So it's 0.5 times 4, which is 2. And so it's basically 240-2, which is 238.

So this actually works, choice D actually works. Now we have to check choice E. Have we eliminated both answers or do both D and E work? Let's see. For E, we get 1.21(200). And of course, that's bigger than 1.2(200) and 1.2(200) is 240.

And so this is gonna be something that equals something bigger than 240, so it's not gonna equal 238. So this is incorrect and the only correct answer here is answer choice D. Again, we were lucky to eliminate four answers with one choice of numbers. The more conventional and predictable the numbers you pick, the more likely it is that the test-writer has designed trap answers to match the choices.

It can work very much to your advantage to make a choice slightly different from what you think most people would choose, because the trap is gonna be designed for what most people choose. And so if you're only slightly different from that, you're gonna avoid the trap. Finally, what's a good standard by which to decide whether to use an algebraic solution and when to pick numbers?

Well, here really, the best thing to do is practice, practice, practice. If you're the type of person who is good with algebra, it's natural for you to do algebra, then it was definitely the choice. If you see how to do the algebra, then that is a wonderful way to do the problem. If algebra is not really a strong suit for you and picking numbers just makes more sense, then by all means try picking numbers.

And in fact, really, the best thing to do is when you get a practice problem, do it one way when you are in practice. But before you look at the answer, try and do it the other way. And get a sense of which one was harder for me, which one took more time. The more you can develop a sense of yourself, the more comfortable you'll be making this choice when you're in a test situation.

In summary, picking numbers is an important strategy in problems that have variables in the answer choices. Picking very easy choices to eliminate low-hanging fruit can be a helpful way to begin. So that's just a way very quickly to knock off a few answer choices to simplify the problem.

Pick numbers that are prime, different from one another, easy to work with, and not entirely predictable. In general, an algebraic approach is a good default, but get to know your own style and what works best for you.