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GMAT: Math 28

143-147, pgs. 171-172. Created by Sal Khan.

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Video transcript

Problem 143. 3.003 divided by 2.002 is equal to-- well, this already looks suspicious. It looks like 3/2, but just to confirm that, this is the same thing as 3 times 1.001. And this is the same thing as 2 times 1.001. So you cancel those out, and you get 3/2, which is the same thing is 1.5. And that's choice E. Problem 144. If 4 minus x over 2 plus x is equal to x, what is the value of x squared plus 3x minus 4? Let's see if we could simplify this little bit. Let's multiply both sides of this equation times 2 plus x, so you get 4 minus x is equal to 2 plus x times x is 2x plus x squared. Let's subtract x from both-- no, add x to both sides, so you get 4 is equal to 3x plus x squared. Subtract 4 from both sides, and you get 0 is equal to minus 4 plus 3x plus x squared, which is just rearranging that, which was the same thing as x squared plus 3x minus 4, so that is equal to 0, which is C. Next problem, 145. They've drawn us a trapezoid. Let me see if I can draw the trapezoid as well. So they have the base, one side, and then they have another side that looks something like that, and they connect the two lines like that. And then they labeled this side as 2 feet, this side as 5 feet, and they say this is a, and this is b. All right. The trapezoid shown in the figure above represents a cross-section of the rudder of a ship. If the distance from a to b is 13 feet, what is the area of the cross-section of the rudder in feet? So let's think about this a little bit. Let's see if we can figure out this area. So this area right here is easy to figure out. If this is 13, this is 5, what is this side going to be equal to? We can use the Pythagorean theorem. If we call this x, we could say x squared plus this side squared, plus 25, is equal to 13 squared. 13 squared is what? 169? I want to say that, but let me check. I haven't memorized my 13 times tables. 3 times 13 is 39. 130. Yeah, 169. Subtract 25 from both sides, you get x squared is equal to 144. That's tells us we're on the right track. We got a perfect square. So x is equal to 12. So it's very easy to figure out the area of this part of the trapezoid, right? It's just the base times the height times 1/2. So it'd be 12 times 5, which is 60, times 1/2, so this would be 30, this area here. But this is a little bit trickier. Well, let's think about it this way. Let's extend this trapezoid into a rectangle. So if you did that, then you can view the area of the rudder, so the original problem that I have in magenta there as the area of this entire rectangle minus this green area right here. So let's do that. First of all, what's the area of this green area? So this side is what? If this is 2, this side is 3. And we already figured out this width is 12. So the green area is 3 times 12 times 1/2, that's the area of a triangle. So 3 times 12 is 36 times 1/2 is 18. Do it in a different color. So the area of this green area is 18. So the area of our rudder, this whole magenta area that I'm shading n, is equal to the area of the rectangle minus 18. So the area of the rectangle is 12 times 5, which is 60. And then let's subtract out the green area minus 18 is equal to 60 minus 10, that's 42. And that's choice C. Problem 146. If 0 is less than or equal to x, which is less than or equal to 4, and y is less than 12, which of the following cannot be the value of xy? Let me going to write down all the choices. A, minus 2. and y is less than 12, I don't see why not. x could be 2 and y could be minus 1. y could be arbitrarily small. So that's not true, it's not A. B, 0. Well, x and y actually can both be zero, so if you multiple them, no reason why you can't get zero. C, 6. Well, if x is 2, and y is 3, I can easily get 6. And they don't even tell us that x and y have to be different. They don't even tell us they have to be integers. 24, Well, the largest number that x can be equal to is 4, and x can't be a negative, so if x is 4, times y is 6, you get 24, so I'm already guessing the answer is E without having looked at it. E, 48. Because the largest value of x is 4, and so you'd have to multiply 4 times 12 to get to 48, but you can't get there. So the answer is E. Problem 147. Turn the page. I'll do it in blue since it seems to be dealing with water. 147: In the figure above-- let me draw the figure above. So I have a V-axis, then I have this other axis, and then they've drawn what looks like water, at least before reading the problem. I've drawn that. They say this is V. They say that this distance is 5 feet. They say that this distance is 10 feet, and they label us a couple of points. It's R and that's S. In the figure above, V represents an observation point at one end of a pool. From V, an object that is actually located on the bottom of the pool at point R, so an object that is actually here, appears to be at point S. The light goes from here, but then refracts, and goes like that, so if you just follow the path, it would go straight to point S. If VR is equal to 10 feet-- what? Oh, VR is equal to 10 feet, so let me draw VR. VR is equal to 10 feet. What is the distance RS, in feet, between the actual position and the perceived position of the object? So they want to know this distance right here. Well let's think about it a little bit. If we can figure out this distance, then we subtract that distance from 10, we'll have this distance. So what's this? Let's call this x. This is a right triangle, so you have x squared plus 5 squared, plus 25, is equal to this hypotenuse squared, is equal to 100, which is the Pythagorean theorem. x squared is equal to 75, so then x is going to be equal to-- let's see, square root of 75. That's someplace between 8 and 9, right? 8 is 64, 9 is 81, so x is greater than 8 and less than 9. Oh, actually, they write it in terms of square root of-- let me simplify this a little bit more. So I can write that as-- the square root of 75, I can write as 3 times 25. Square root of 25 is 5, so that becomes 5 square roots of 3. So this distance right here is 5 square roots of 3. So if I want to figure out this distance, I just subtract that from 10. So 10 minus 5 square roots of 3, and that is choice A. See you in the next video.