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# GMAT: Data sufficiency 12

55-58, pg. 282. Created by Sal Khan.

## Want to join the conversation?

• At , why can't I used s+2s to denote single scoop and 2s to denote double scoop? Why is it s+2d? Please explain in detail. Thank you.
(1 vote) ## Video transcript

We're on problem 55. At a certain picnic each of the guests was served either a single scoop or a double scoop of ice cream. How many of the guests were served a double scoop of ice cream? So let's say that d is the number of guests that were served double scoop and that's what we're trying to figure out. So what are their statements? Statement number one says at the picnic 60% of the guests were served a double scoop of ice cream. So 60% percent of the total were served a double scoop. Let's see if we can write that algebraically. So 60% of the guests. So the total number of guests are the number of guests that were served double scoops plus the number of guests that were served single scoops. Then we have all of the guests, right? This is this is equal to the total number of guests. And then they tell us that 60% of them were served double scoop. So the number that were served double scoop is 60% of the total. So that statement number one is just an algebraic equation. And this alone though doesn't help me solve this equation. Maybe in the next statement they'll tell us how many total guests there are. What this is. And that'll help me but I don't know. Let's see, statement number two, they say a total of 120 scoops of ice cream were served to all the guess at the picnic. So once again, they're telling us that there are 120 scoops of ice cream, they're not telling us that there are 120 guests. So let's see, what are the total number of scoops. So there's the number of singles groups and each of those are-- the number of guests that got singles scoops are s and then the number of scoops is just 1 times that because they each got one scoop. And then you have the number of guests that got double scoops. But then the number of scoops they got is going to be 2 times that. So this is the total numbers scoops, right? The number of people that got single scoops times 1 plus the number that got double scoops times 2 and that is going to be equal to the total number of scoops. And this alone isn't going to help us solve how many double scoops were served, it depends on the singles scooped and all of that. But if you look at both of these statements we have two linear equations and two unknowns. So at this point you should immediately say, unless somehow these are the same equation or somehow these are two parallel lines, these should intersect and allow you to solve this equation. So immediately-- especially if you don't want to waste time-- you should say that both statements together are sufficient. And if you want me to prove this to you, let me see if I can solve this. So I'll do it in a different color. So let me do some substitution. So here we get s is equal to 120 minus 2d. And we can substitute this back here. So you get 0.6 times d plus s, which is 120 minus 2d, is equal to d. And so you get 0.6 times 120 minus d is equal to d. And what's 0.6 times 120? That's 72, right? 6 times 12 is 72. 72 minus 0.6d is equal to d. So you get 72 is equal to 1.6d and then 72 divided by 1.6 is equal to 45. So there, we solved the problem. d is equal to 45. 45 double scoops were served. And then we could substitute back into the equation to figure how many single scoops or how many total guests there were, but this is all that we had to figure out. And we were able to using both statements combined. To the next problem. 56. What is the value of xy? OK. Statement number one: they tell us that y is equal to x plus 1. And that's not going to tell me what xy is, I mean there's a bunch of x's and a bunch of different y's depending on what I choose. Two: they tell us y is equal to x squared plus 1. Once again, this by itself-- it completely depends on which xy's I pick, right? I can pick any combination and get different values. You could try that out yourself. But I have two equations and two unknowns. It's a little tricky because one of them's non-linear, but let's see if we can figure it out with substitution. If y equals this and y equals that, we could set these two equal to each other and see if we can solve for x. So we get x plus 1, that's the first one, should equal this one, because they're both equal to y. x squared plus 1. Let's see, we can subtract 1 from both sides and you get x is equal to x squared. And let's see, you could say-- well, you could divide-- if we assume that if x does not equal 0-- so there's two solutions here, right? x could be equal to 0. That's completely legitimate. They never told us that x isn't equal to 0. That would satisfy that equation. And then what other x would satisfy? x could be equal to 0 or x could be equal to 1. Either of those are completely legitimate answers here, right? Fair enough. If x is 0 this is true, 0 equals 0. If x is 1, 1 is equal to 1 squared. x can't be negative 1. Negative 1 is not equal to negative 1 squared. So these are the two solutions for x. So right now it seems like it's a little shady, but remember, we're not trying to solve for x or solve for y, we're trying to solve for xy. So maybe when we solve for y something, I don't know, interesting could happen. Let's solve for y. If you assume x is equal to 0 what's y? y is equal to 0 plus 1. y is equal to 1 in this case. That's one solution. And if you put x is equal to 0 here you also get y is equal 1. So this is one set of solutions. And in this reality xy is going to be equal to what? xy is going to be equal to 0, right? 0 times 1 is 0. Now if x is equal to 1, what is y? Well, if x is equal to 1, y is equal to 1 plus 1, y is equal to 2. Well, then xy is equal to 2. So using both of the statements, I still don't know whether xy is equal to 0 or 2. Someone would have to tell me that x is not equal to 0, x is equal to 0, and only then can I really solve this equation. So the answer to this, even though we have two equations and two unknowns, because one of them was a quadratic and had two solutions, we actually don't know what xy is. So the answer is that the statements together are not sufficient or E. That's interesting because they actually make you do a little bit more work than just eyeballing it and saying, oh, I have two equations to unknowns. You've got to be a little careful when the equations are non-linear. At least I think that's right. Alright, question 57. They want to know what 1/x plus 1/y-- some police sirens outside. I live in Palo Alto, there's not normally a high crime scene here but you never know. Anyway, so 1/x plus 1/y. So statement number one is x plus y is equal to 14. Actually let me try to simplify this a little bit. If I were to try to add these two, what could I do? Well, when you add two fractions you have to find a common denominator. An easy common denominator for x and y is just xy. So 1/x is what over xy? Well, it's y/xy plus-- right, this is the same thing as 1/x-- plus, and if I had xy here, what's 1/y? Well, that's x/xy. So that can be rewritten as y plus x over xy. And now when you rewrite this statement like this, then statement one looks a little interesting. They gave us the numerator at least. They haven't told us what xy is but they told us that y plus x, or x plus y is equal to 14. So statement one, it's kind of part of the puzzle but it's not completely helping us. Let's see what statement two tells us. I suspect, if this is solveable, they'll have to tell us what xy-- yep, they tell us what xy is. They tell us that xy is equal to 24. So then we're done, right? This simplifies to this and they're telling us what this is. x plus y is 14. So it equals 14/24-- not that we have to really solve for it-- but that is equal to 7/12. But as soon as we got-- you say, oh, that's that, that's that. You're done. You say, both statements together are sufficient. So C. 58. If d denotes a decimal is d greater than or equal to 0.5? They put a 0 here so you make sure you notice the decimal. d, greater than or equal to 0.5. Statement number one tells us that when d is rounded to the nearest 10th the result is 0.5. So let's see, it could be 0.52 and when you round it to the nearest 10th it becomes 0.5. But on the other hand, it could be 0.46 and when you're round it to the nearest 10th you also get 0.5. So since statement one says it could be either one of these, it really doesn't help us to know whether we're greater than or equal to 0.5. This one is greater; this one isn't. So statement one by itself doesn't help. Statement two: when d is rounded to the nearest integer the result is 1. Now this is interesting. If something is rounded to the nearest integer it has to be greater than 0.5, right? The smallest number that when you round it to the nearest integer equal to 1 is 0.5, right? So it has to be 0.5 or it has to be 0.51, because 0.499999, when you're round it to the nearest integer, is going to be equal 0. That's the only way it can get rounded to 1, if you're 0.5 or above. So actually statement two alone is all you need to know that d is greater than or equal to 0.5. So that is choice B. Statement two alone is sufficient. Alright, I've realized I've run out of time. See you in the next video.