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GMAT: Data sufficiency 15

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We're on problem 68. And they want to know, what is the average arithmetic mean of j and k? So essentially, if we knew what they were, you could take j plus k, and divide by 2, and you'd know their average. What is statement one? Statement one tells us the average of j plus 2, and k plus 4, is 11. So this is interesting. So means that j plus 2, plus k plus 4, over 2-- I'm just averaging the two numbers-- they're telling us that that is equal to 11. And remember, the question they're asking is the average of j and k. So if we could just figure out what j plus k, over 2 is equal to, we're done. Or, if we just knew what j plus k is, we were done. Maybe we could figure that out from this. Let's see if we can simplify. This simplifies to j plus k, plus 6, over 2, is equal to 11. This simplifies to j plus k, over 2, plus 3, right? So plus 6 over 2. I'm just taking the 6 over 2 out. So plus 6 over 2, is equal to 11. So then we get j plus k, over 2 is equal to 8. And we're done. The average of j and k is 8. We're done. Statement one alone is sufficient. Let's see what statement number two does for us. Statement number two says the average of jk and 14 is 10. I suspect we're going to be able to do the exact same thing. So the average of j plus k, plus 14-- so now we're averaging three numbers-- is equal to 10. And here, we should be able to figure out what j plus k is again. We don't have a convenient 2 in the denominator anymore, so let's just solve for j plus k. So then you get j plus k, plus 14-- multiply both sides by 3-- is equal to 30. And then you get j plus k is equal to what? Subtract 14 from both sides, is equal to 16. And then if you wanted to figure out the average of the two, you just divide both sides by 2. And you get j plus k over 2 is equal to 8. So each statement alone was sufficient to solve this problem. Next problem. I feel a sneeze coming on. This is not happening. All right. Where was I? Problem 69. Let me move the scroll bar up all the way. Paula and Sandy were among those people who sold raffle tickets to raise money for club x. If Paula and Sandy sold a total of 100 of the tickets, how many of the tickets did Paula sell? OK, so essentially they're telling us Paula plus Sandy sold 100 tickets. And they want us to figure out how many did Paula sell? Where P is the number Paula sold, and S is the number Sandy sold. Problem number one. Sandy sold 2/3 as many of the raffle tickets as Paula did. So Sandy is equal to 2/3 times Paula. Well, this alone is sufficient. We have one equation of two unknowns, and now we have another equation of the same two unknowns. These are both linear equations. So we have two equations of two unknowns. We can easily now solve for S and P. And maybe I'll do it, but you should just already recognize that this is sufficient if you're on the GMAT. And we'll solve it, just to do it after this. Sandy sold 8% of all the raffle tickets sold for club x. All right. Now, this is a little different. This is Sandy is equal to 8% of not all the tickets that Sandy and Paula sold, she sold 8% of the total that club x sold. So this is a different number, because there could have been-- I might have been in club x selling raffle tickets. So we don't know how many I sold. So we don't know what this total number is. So there's not much we can do with it, because there's no way for us to figure out this total number. So this is not that useful. So statement one alone is sufficient. And just to prove the point to you-- we're trying to figure out what Paula sold. So let's just substitute this back in. So you have P plus-- for S I'll write 2/3P-- is equal to 100. And so this is 5/3P is equal to 100. Just added 1 plus 2/3, or 3/3 plus 2/3. And then you have P is equal to 3/5 times 100, which is equal to 60. So that's how many she sold. So we would definitely be able figure it out, but this would've been a waste of time if you were taking the GMAT for real. You should have just recognized two linear equations and two unknowns. I'm done. Next problem, 70. Is ax equal to 3 minus bx? Who knows? [? Problem ?] statement number one. Let me scroll down a little bit. Statement number one tells us, x times a, plus b is equal to 3. Well, that's essentially the same thing as this top equation, right? Let me show you. x times a, that's ax, plus bx is equal to 3. And then subtract bx from both sides. You get ax is equal to 3 minus bx, which is exactly what we were trying to prove. So if this is true, then this is definitely true. Statement one alone is sufficient. Statement two tells us, a equals b, equals 1.5, and x is equal to 1. Well, let's see if this is true. So a is-- you have 1.5 times 1. So you have 1.5 is equal to 3, minus 1.5 times 1. So 3 minus 1.5. Well, this is true. 1.5 is equal to 3 minus 1.5, is 1.5. So two alone is also sufficient. So the answer is D. Each statement alone is sufficient to solve this problem. 71. Let me switch colors. A number of people each wrote down one of the first 30 positive integers. Were any of the integers written down by more than one of the people? All right. That's interesting. So are there any repeats? A number of people. So clearly if we had more than 30 people and they all had to pick one of the first 30 integers, then you're going to have repeats. But let's see what they tell us. The number of people who wrote down an integer was greater than 40. Well, there you go. So 40 people each have to pick a number between-- one of the first 30 positive integers, right? So they only have a pool of 30 to pick from. So if 40 people have to pick numbers from a pool of 30 numbers, there's definitely going to be repeats, right? So even if the first 30 people all picked different numbers, which we cannot guarantee by any means, the next 10 are going to have to repeat with somebody. Because all of the first 30 would've already been picked. And you could very easily have even more repeats. So statement number one is very sufficient. Statement number two. The number of people who wrote down an integer was less than 70. Well, that's useless. I mean, by this statement, maybe only one person wrote down an integer, right? And if only one person wrote down an integer, we definitely don't have any repeats because there's no one to repeat with. So this is a useless statement. So statement one alone is sufficient and statement two is useless. 73. I think there should be a third option, where you can rate the degree of uselessness of a statement. Some of them borderline on almost being useful, some of them are ridiculous, like this one. OK, 73. 72. In the figure above, is CD greater than BC? So they drew us a number line, or whatever you want to call it. Some type of a line. Let me do another color. This end we have A. Then we have B. Then we have C. And then we have D. In the figure above, is CD-- so this is CD-- greater than BC? So they want to know is this greater than this? So is CD greater than BC? All right. So number one, they tell us that AD is equal to 20. So this whole thing is equal to 20. So that doesn't help me much. Statement number one tells us AD is equal to 20. It just tells me the whole length. Maybe I can get some more information. Statement number two. AB is equal to CD. So this is interesting. They're telling us that this is equal to this. But this still doesn't help us, right? I can imagine a situation where both of these are-- let's say that both of these are 5. That could be 5, that could be 5. And then this would be 10, right? 5 plus 10, plus 5. In which case, CD would not be greater than BC, right? So this would be a situation that meets all of the conditions where CD would not be greater than BC. But then I can construct one where it is greater than BC. I can make CD equal to-- what if that was 8? And then this one also has to be 8. And then we would only have 4 left. Where CD is greater than BC. So in this case, both statements together are still not sufficient. So the answer is E. And I've run out of time. I'll see in the next video.