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GMAT: Data sufficiency 18

80-83, pg. 284. Created by Sal Khan.

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Video transcript

We're on problem 80. If a, b, and c are integers, is a minus b, plus c greater-- OK. So what they want to know is a minus b, plus c greater than a plus b, minus c? Fair enough. And if you think about it, there's an a on both sides of this equation. So what they're really wanting to know is minus b plus c greater than b minus c? Or another way, they're just trying to say-- I just took a from both sides of this inequality, so it doesn't change anything. Really what they're asking, if you just rearrange this, is c minus b greater than b minus c? Which is another way of saying, well, is one of them greater than the other one? Right? If c is greater than b, this is going to be greater than this one. If b is greater than c, this is going to be greater than this one. So let's look at our statements now. Statement number one. b is negative. Well, that by itself doesn't help us much because b could be negative, but c could be even more negative. So c could still be smaller than b, right? If b could be minus 5, c could be minus 6. Or c could be plus 6. So it really tells us nothing about c relative to b. Statement two says c is positive. So this statement by itself, once again, it tells us nothing about the value of b relative to c. If c is positive, maybe b is positive as well. Maybe b is more positive. Or maybe b is negative. Who knows? But this by itself doesn't help us. But if we use both of these together, if c is positive and b is negative, then we know, if this is positive, this is negative, we know this is going to be positive. And we know that this is going to be negative. Essentially, we know that c is larger than b, so this will be positive and this will be negative. And we can answer our question. If c is larger than b, this is true. The left-hand side is greater than the right-hand side, or this holds good. So we need both of the statements together. Both statements together are sufficient to answer this question. Switch colors. 81. If a certain animated cartoon consists of a total of 17,280 frames on film, how many minutes will it take to run the cartoon? So they want to know how many minutes. So essentially, if we knew how fast the frames have to go through-- if it's x frames per second-- if we knew that number, we'd be able to figure out how long the cartoon runs. So statement one is, the cartoon runs without interruption at a rate of 24 frames per second. Well, that's exactly what I just asked for. 24 frames per second. So using just that information, how many seconds does it run? Well, it runs 17,280 frames divided by 24 frames per second. This will give you seconds, and then if you divide this number by 60, you'll get minutes. Divide seconds by 60, you get minutes. So this is the answer. So one alone was enough to answer the question. What does statement two give us? I'll write it here. Statement two. It takes 6 times as long to run the cartoon as it takes to rewind the film. And it takes a total of 14 minutes to do both. 6 times as long to run the cartoon as it takes to rewind. Let's say 6 times to play. I want different letters. So to play the cartoon takes 6 times as long as to rewind the cartoon. And they say it takes a total of 14 minutes to do both. So they say play plus rewind is equal to 14. And what we want to know is, how long does it take to play the movie? And we don't even have to use that information alone, the 17,280, but just looking at this, we have two linear equations and two unknowns, so we can just solved for P. P is the playtime of this movie. And if you wanted to figure it out-- not that you would have to, because we would be done at this point. We would say oh, one and two are each independently sufficient to answer this question. But just to show you that we could figure it out. Say R is equal to 14 minus P. Then you substitute that into that. You get P is equal to 6 times R. So 6 times 14 is 60, plus 24. 84 minus 6P. You get 7P is equal to 84. P is equal to 12 minutes. And we'd be done. But actually, we didn't have to do all of that. All of that is a waste of time. I just want to prove it to you. 82. A box contains only red chips, white chips, and blue chips. If a chip is randomly selected from the box, what is the probability that the chip will either be white or blue? So essentially, to answer this, we just have to know what proportion of the chips in the bag are white or blue. All right. So statement one says, the probability that the chip will be blue is 1/5. So probability of blue is equal to 1/5. That also tells us that 1/5 of the chips in the bag are blue. But we don't know what proportion are white, so we still can't answer that top question. Statement two tells us, the probably that the chip will be red is 1/3. The probability of red is equal to 1/3. So, once again, just this alone-- actually, just this alone is fine, right? I correct myself. The probability of getting white or blue, that's probability of white-- we could write, do set notation, white or blue-- that is just equal to what? It's 1 minus the probability that we get a red. Think about that. Either this is going to be true, I'm going to pick a white or blue, or I'm going to pick a red. Because there's only red, white, and blue in the bag. So the probability of white or blue is equal to 1 minus the probability of red, which is equal to 1 minus 1/3, which is equal to 2/3. So statement two alone is sufficient. And then if we have statement number two, statement one alone is not sufficient. 83. See, this is tricky because, I think, at least when I first read it, I was like, oh, I could use this and this to figure out the probably of getting a white. So I need both statements. But we don't even need to know the probability of a white. We just need to know the probably a white or a blue, which is just 1 minus the probability of red. So statement two alone gets us there. Let me do the next problem. 83. If the successive tick marks on the number line are equally spaced, and if x and y are the numbers designating the endpoints of the interval as shown, what is the value of y? OK, so let me see if I can draw what they have drawn. So they have a number line. And I probably have to draw the exact same number of ticks as they did. So they have 0. And then let's see, there's 1 tick, 2 ticks, and then they have a third tick. And they say this is x. And then they have three more. 1, 2, 3. And then they have y. And then they have one more. And this keeps going on and on. And they ask us, if the ticks are equally spaced, and if x and y are the numbers designated at the endpoints of intervals as shown, what is the value of y? So they're equally spaced, but they're not telling us that these are integers. We don't know that this is 1, this is 2, this is 3. We don't know that. But what we do know is y is how much further than x? So let's see, x is 1, 2, 3 spaces. So we could say x is equal to 3 times, let's say, s, where s is the length of each interval, or each tick. And y is 1, 2, 3, 4, 5, 6, 7. So y is equal to 7s. And so we could use this to come up with a relation with x and y. This is before they've done anything. This is just looking at the diagram. So we could say that s is equal to y/7. Or if we substitute this back here, we could say that x is equal 3/7 y. And actually, you didn't have to even do this s stuff. You could have just said, oh, well, x is 3/7 as far as y is from 0. Because these are equally spaced. So just the problem setup, just looking at it, gives us that information. But we want to know what the value of y is, and that by itself isn't enough. What does statement one tell us? x is equal to 1/2. Well, that's enough to figure out y. If x is equal to 1/2, you just substitute there. 1/2 is equal to 3/7 y. Multiply both sides by 7/3. You get y is equal to what? 7/6. And you're done. You don't have to solve that. You just know that I have this, I have that. I can solve it. So one alone is sufficient. Statement two tells us, y minus x is equal to 2/3. So once again, we have two linear equations with two unknowns. We can just substitute and solve them. And just to prove it to you, x is equal to 3/7 y. I can substitute that right there. So I have y minus 3/7 y is equal to 2/3. So 7/7 minus 3/7. So that's just to go here. 7/7 y, so that's 4/7 y is equal to 2/3. y is equal to 2/3 times 7/4. Which is-- that's 1, this is 2-- which is 7/6 once again. See you in the next video. So once again, each of these are independently sufficient to answer the question. See you soon.