GMAT: Math 51

We're on problem 237. A part-time employee whose hourly wage was increased by 25% decided to reduce the number of hours worked per week so that the employee's total weekly income would remain unchanged. By what percent should the number of hours worked be reduced? So before, he made his hourly wage times hours worked, right? This is what he would normally work in a week, right? So his wage per hour times hours per week. This is what he used to make. But now what he's done is since he's making 25% more of his wage, he's going to reduce his hours by some number, by some fraction. Let's call that x. So then the new wage times the new hours , and they're saying that he doesn't want to change the total amount that he makes in a week. So the new wage times the new hours will equal w times h. Let's divide both sides by w. This goes to 1, that goes to 1. Divide both sides by h. That goes to 1, that goes to 1. And you're left with 1 is equal to 1.25 times x. And I view that-- this is easier to think of as what? 5/4, right? 1.25-- right? 4/4 is 1, and then you add another 1/4. This is 1 and 1/4, so this is 5/4. So let's write that as 5/4x. Multiply both sides by 4/5, you get x is equal to 4/5, which is equal to 0.8. So another way to view it is he would reduce-- he's working 80% of the hours that he used to work, so he reduced by 20% the amount that he worked. And that will be choice B. Next problem, 238. OK, they've drawn us a figure, so let's draw the figure. Let me see if I could make it thicker. OK, so they drew a box in a box, and there's another box in that box, and it looks something like that. And then they shaded it in, so I'll shade it in. And they say that this side up here is 18, and they say that this side over here is 15. The shaded region in the figure above represents a rectangular frame with length 18 inches and width 15 inches. The frame encloses a rectangular picture that has the same area as this frame itself. So this area is equal to this area. Fascinating. If the length and width of the picture have the same ratio as the length and width of the frame, what is the length of the picture in inches? So they're saying that this is the length, so this is l. And this is the width. And they want us to solve for l. What is the length of the picture, right? They're calling-- I mean, I would call that width, but they originally say the frame with length 18 inches. So they want to know the length of the picture. OK, first of all, they tell us that the ratios are the same. So l:w of the picture is the same ratio as it is for the frame, is 18:15. You divide by 3. That's the same thing as 6:5, right? l:w is equal to 6:5. And the other thing they told us is that the area of the picture the same as the area of the frame. So the area the picture is l times w, and what's the area the frame? What's the area of this whole thing , which is 15 times 18, minus the area of the picture, right? Because the area of the frame is not this whole rectangle. It's just the part that doesn't have the picture in it, so minus l times w. So if we add lw to both sides, we get 2lw is equal to 15 times 18. I'm just going to leave that as 15 times 18. Let's see, we could already divide both sides by 2, so you get lw is equal to 15 times 9. And let's see, we have two equations with two unknowns. Let's see if we can solve. So we have l/w is equal to 6/5. We can cross-multiply, so we get 5l is equal to 6w. I ignored the stuff in the middle. And we want to solve for l, right? So let's substitute for w. So we get 5/6l is equal to w, and now let's substitute that into this equation. So we have l times w, which is 5/6l, is equal to 15 times 9. And let's see, let's multiply both sides by 6/5, and so you would get l squared is equal to 15 times 9 times 6 over 5. So let's see, we could do a little simplification. This becomes 3, this becomes 1, and then we have-- let's see, now we have to do some math. 3 times 9 is 27. 27 times 6 is what? 7 times 6 is 42. 2 times 6 is 12, plus 4 is 161. That's a strange number. So let's see if I can simplify this. Oh, I'm looking at choices, and they have all sorts of crazy square roots in it. So let's just say that l is equal to the square root of-- let's see, how can we rewrite this? We could rewrite the 6 as 3 times 2, right? So then we get 3 times 9 times 3, that's the same thing as 9 times 9. That's 81, right? 81 times 2. And we have a 1 in the denominator. So l is equal to 9, and I'll do one more step, because I think this confuses a lot of people. That's equal to the square root of 81 times the square root of 2, which is equal to 9 square roots of 2, which is choice A. Problem 239. Of the 200 students at College T majoring in one or more of the sciences-- I already feel a Venn diagram coming on. Of the 200 students at College T majoring in one or more of the sciences, 130 are majoring-- so let me draw the whole-- well, let me read the whole problem. 130 are majoring in chemistry and 150 are majoring in biology. If at least 30 of the students are not majoring in either chemistry or biology, then the number of students majoring in both chemistry and biology can be any number from-- OK, let me draw the Venn diagram. So that's chemistry, that's biology, and then there's a leftover, right? This is a whole pool. So they tell us there's 200 total, right? There's 200 total. And they say that 130 are majoring in chemistry, 150 in biology, and then at least 30 of the students are not majoring in either chemistry or biology, so we could say greater than or equal to 30. So this is driving a little spin here by making it an inequality. Then the number of students majoring in both chemistry and biology could be any number from-- OK, so let's say that this is 30, just to simplify our thinking. If this is 30, how many people are majoring in chemistry or biology? Well, then it's 170, right? So let's think of it this way. Let's think of this as the intercept-- this is both. Well, I used B for biology. So let's call this-- I don't know, let's call this-- I used C for combined. So let's call this I for the intersection. So if 30 people are majoring in-- well, actually, let's just call this x. So how many people are majoring in-- if I wanted to add these two numbers up, if I want to know the people who are in chemistry or biology, what would it be? It would be 130 plus 150, and remember, and we've done this a couple of times with these Venn diagram problems, if I counted these 130, and then I count these 150, I've counted this intersection twice. I counted it with the 130, and I counted with the 150. So I have to subtract out at least the intersection once. So I subtract it out once. So these are the number of people who are in chemistry or biology or both, right? So it's the intersection of chemistry or biology or both. And now we know that that has got to be equal to what? Well, 200 minus the people who are in nothing, right? So I'm going to call that x for now. 200 minus the people who are nothing. And let's see if we can solve for the intersection, which is what they're asking us in terms of x. So this becomes 280 minus I is equal to 200 minus x. And then they tell us to subtract 200 from both sides. You get 80 minus I is equal to minus x. Let's add I to both sides. So you get 80 is equal to I minus x, right? Or even better, let's add x to both sides. So you get I is equal to 80 plus x. And so, if x is greater than or equal to 30-- so we know that if x is 30, then we get what? We get 130 plus 80 is 110, right? So we know that I is going to be greater than or equal to 110. The intersection is at least 110. Now, what's an upper bound on it? How many people can be studying both? And if you look at the choices already, there's two choices that have 110 as the minimum for the number studying both. But what's the maximum that can be studying both? Well, it's 130, right? If all the chemistry people are also studying biology, then the intersection becomes 130, right? But you can't have 150 people studying both, because there's only 150 studying chemistry. So the intersection is greater than or equal to 110. And it's limited by the number of people studying chemistry, 130, which is choice D. And I'll see you in the next video.