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GMAT: Math 7

36-41, pgs. 156-157. Created by Sal Khan.

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Video transcript

We're on problem 36. They've drawn a little graph here. So let's draw it up for ourselves. So that's the y-axis. That is the x-axis. And then they have this point. Let's see, minus 1. They say, 1, 2, 3. Let's see. 1, 2. And then they have this line segment PQ that goes from 0, minus 1. Let me draw that in a different color. It goes from 0, minus 1 to 3 comma 2. So it goes from there, so 3, so roughly there. I think that's about as good as I can do. And the way they drew it, it intersects 1 here. And so this would be 3. This is the point 3 comma 2. And they say in the figure above, the point on segment PQ, so they say this is point P, this is point Q. The point on PQ that is twice as far from P as from Q. So the point is on this segment, so this point is someplace around here. We could find points that are away from the segment that are twice as far from P as from Q, but they say it's on this segment. So let's think about it a little bit. If we just look at how do we get from P to Q, we go 3 to the right, right? x increases by 3 and y increases by 3, right? So if we wanted to go 2/3 along the way, what if we just increased x by 2 to get to x is equal to 2 and increased y by 2, right? So the x's and y's only increase by 2/3 of the way to Q. So then that would be the point. x is 2 and then y is 1. That would be the point 2 comma 1, which is B. Next question. 37. If a positive integer n is divisible by both 5 and 7, than n must be divisible by which of the following? Let's see, they give us a couple of choices: I, II and III. So they say it has to be divisible by 12. No, that's not the case. 35 is divisible by 5 and 7. It's not divisible by 12. 35. Well, sure. If you're divisible by 5 and 7, that means it's equal to some constant times 5 times 7, right? And this could be a bunch of numbers multiplied together. But that means it's equal to 35 times some other integer, and that integer could have a bunch of other factors. So, right. It has to be divisible by 35. And statement III is 70. Well, no, 35 is divisible by 5 and 7, but it's not divisible by 70. So statement III is not right. So it's II only. So that's C. Question 38. If 4 is one solution to the equation x squared plus 3x plus k is equal to 10, so 4 is one solution. So that means if I were to factor this, it'll be x minus 4 times x plus or minus something else, because we know that x minus 4 equals 0, which tells us that x is equal to 4 is one of the solutions. So what are they going to ask us? They're going to say, where k is a constant, what is the other solution? So let's think about it a little bit. Well, actually, let me rewrite it a little different. I didn't realize that they said this is equal to 10. So let's subtract 10 from both sides. We get x squared plus 3x-- you essentially want to find the roots of this quadratic-- plus k minus 10 is equal to 0. And you want to factor it, right? You could factor any quadratic. It's going to be x plus a times x plus b. And they're telling us that 4 is one of the solutions, right? 4 is one of the solutions, so minus 4 could be a. So this would be x minus 4 times some other number, x plus b, is equal to 0. And I said positive 4 in the solution. Why did I put a minus 4 here? Because think about it. If we factor this quadratic and we get this, that's tells us either x minus 4 is 0 or x plus b is 0. If x minus 4 is 0, then the solution is x is equal to 4, so that's why I wrote x minus 4 there. But anyway, let's figure out what b could be. So if we multiply this out to try to pattern match it with this, we get x times x is x squared minus 4x plus b times x, and then you have minus 4 times b, so minus 4b. So then this is equal to x squared, and then we could write it plus b minus 4 times x. I just factored an x out of these two terms, right? And so you get b minus 4 minus 4b is equal to 0. So now you just pattern match it. You say b minus 4-- x squared equals x squared. b minus 4x has to be equal to 3x, right? We're just pattern matching. That's the only first-degree x term there. So b minus 4 has to be equal-- x has to be equal to 3x. We don't even care about k. So b minus 4 has to be equal to 3, or b is equal to 7. Oh, right, right. So they want to know what the other solution is. They don't want to know what b is. And I was looking and was like, oh, wow, there isn't a 7 there. There's a negative 7. So that tells us that the factorization of this problem-- I would have gotten that wrong if they put a 7 there, because I wasn't careful-- is x minus 4 times x plus 7 is equal to 0, right? So we already said we could say x minus 4 is equal to 0. And that's where we get the 4 solution. Or we could say x plus 7 is equal to 0. Subtract 7 from both sides. The other solution is x is equal to minus 7. That was a careless mistake I almost made, but I got saved by the choices. So that's choice A. Next problem. 39. If x is equal to negative 3, what is the value of minus 3x squared? So I don't know, maybe they're trying to test your ability to do order of operations or something. So this is equal to minus 3 times minus three squared, right? x squared. So this is equal to minus 3 times positive 9. Minus 3 squared is positive 9, which equals minus 27, which is choice A. Next question. 40. 29 squared plus 29, over 29 is equal to? Well, let's just divide the numerator and the denominator by 29. What's 29 squared? This is the same thing as 29 squared over 29 plus-- I'm just going to separate these two terms out-- plus 29/29. 29 squared divided by 29. x squared divided by x is x. 29 squared divided by 29 is 29. 29 divided by 29 is 1, so it equals 30. So that is choice E. Problem 41. If x is equal to 1 minus 3t and y is equal to 2t minus 1. So I'm going to write it like this. minus 1 plus 2t. I just want to put the t's under each other, right? 2t minus 1. For what value of t does x equal y? OK, so let's just set x equal y and solve for t. So we could just say if x is equal to y, we get 1 minus 3t has to equal this, which is 2t minus 1. Let's add 3t to both sides. So you get 1-- 3t goes away from there-- equals 5t, minus 1. Add 1 to both sides. You get 2 is equal to 5t. Divide both sides by 5. You get t is equal to 5/2, which is choice A. And I'm almost out of time. The next one has a diagram, which always takes me some time to draw, so I'll see you in the next video.