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GMAT: Data sufficiency 27

113-115, pg. 287. Created by Sal Khan.

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  • orange juice squid orange style avatar for user Chris Kelley
    one question 114, I believe it should be c. If you have a set of 8 values and each of those values is between 1 and 25 inclusive, the probability of x being less than or equal to 20 for any of the items in the set of 8 is 20/25 or 4/5. It doesn't matter which member of the set is chosen, they all have the same probability 20/25. so it should be c. idk
    (1 vote)
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    • mr pants teal style avatar for user Ben
      The key here is that the set T that we're drawing from is a specific set of numbers.

      From statements 1 and 2 we know that there are 8 numbers between 1 and 25, but we're not drawing 8 numbers from the range (1,25). We're drawing 8 numbers from the set T.

      T could be {1,2,3,4,5,6,7,8} (in which case P(x<=20) = 1), or T could be {3,5,7,11,21,22,23,24} (in which case P(x<=20) = 0.5).
      (1 vote)
  • blobby green style avatar for user Brian Whitehead
    The data sufficiency problems seem very simple when Sal works through them. How equivalent are these to what I'm going to see on the GMAT in July this year (2020)?

    I like Sal's approach to thinking through them, I will definitely use this method during the test. I worry about time constraints though.

    I am wondering, is there an instructor willing to help me with GMAT studying?
    (1 vote)
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Video transcript

All right we're on problem 113 on page 287. So they've drawn this thing. Let me see if I could draw it. So we have a right triangle in the center. So that's the side. That's the other side, and then the base. Right triangle. And then they drew some squares on every side of that triangle. Let me see if I can do that. That's one square. Well, I don't know how I'm going to draw the other square because I don't know how to draw an angled one. I'll just draw the lines. That's the other square. And the last one. I'll just draw in some lines something like this. It comes up like that, like that and like that. So that's what they have. And they call this A, this B, C, D, and they tell us that this is a right angle. They say, in the figure above, if the area of triangular region D is 4, what is the length of a side of square region A? So all of these are-- well, they don't tell us that this and this are a square region just yet, but they tell us this is a square region. So the length of this side-- if we know the length of this side and that side, then we can figure out the length of the hypotenuse of this right triangle, so that's all we need. And let's see, let's call this length right here little b. And let's call this length little c. So we already know that, if we know the area of this triangle is 4, we know that little b times little c, so height times base times 1/2, is equal to 4. But that also tells us that b be times little c is equal to 8. I don't know that does for us. Let's read the statements. Statement number 1. The area of square region B is 9. So B is equal to 9. Well, that's useful. If the area of this region, and it's a square, is 9, what's each of the sides going to be? The sides are going to be the square root of that. Little b is going to be equal to 3. If little b is equal to 3, then the area of this is 3 times 3. We can do that because it's a square, which would be 9. So if little b is 3, then what is C? Well, then we get that C is equal to 8/3 using this information. And we got that, remember, just based on the fact that the area of this triangle right here is 4. So little b times little c times 1/2 would equal to 4. So, if we know what little b and we know what little c is, then we can solve for-- let's call that little a. And little a is what the whole point was about. What is the length of a side of the square region A? So we could just solve for b squared plus c squared is equal to a squared, right? This squared soon. plus this squared is equal to the hypotenuse squared. And we just solve for a and we'd be done. So statement one alone is sufficient. And this is really messy, especially taking an 8/3 squared and all that, so I'm not going to do it. And it would be a waste of time on the GMAT anyway. Let's look at statement number two. The area of square region C is 64/9, so C is equal to 64 over 9. So this is kind of very similar to the information they gave us in statement one. If the area of this is 64/9 and this is a square region, then the length of each of its sides, or in this case, little c, the length of little c is just going to be the square root of that. So little c is going to be 8/3, which is exactly what we figured out here. And then we could use this information that we got from the problem statement that b times c is going to be equal to 8, and we got that from the fact that the area of this triangle is 4. So we use that and we can figure out that b times 8/3 is equal to 8. And then you get b is equal to 3, and then you apply this again to solve for a. So each of the statements one and two are each individually sufficient to answer this question. Next problem. 114. If x is to be selected at random from set T, what is the probability that 1/4x minus 5 is less than or equal to 0? Let's simplify this real fast. That means that what's the probably that 1/4x is less than 5? And that means what's the probability? Let's multiply both sides by 4. I just added 5 to both sides here. And let's multiply both sides by 4. And we don't have to change the inequality since 4 is a positive number. So it's the same thing as what's the probability that x is less than 20? And I like this a lot more because it's just simple and it lets my brain grasp the problem better. Now, what are the statements they give us? Statement one: T is a set of 8 integers. T has 8 integers. That by itself doesn't help me because I don't know what those eight integers look like. I mean, maybe all eight could be larger than 20, in which case, it would be impossible to find one that's less than or equal to 20, or maybe all 8 are less than 20, in which case, you would have 100% chance. So statement one by itself is fairly useless. Statement two. T is contained in the set of integers from 1 to 25. So that also doesn't help me. So T within 1 to 25. So they're essentially telling us that every number in T is between this range. But if we just take statement two alone, they're not telling us, first of all, how many numbers are in T. Statement one tells us that, but statement two doesn't tell us. And then even if we knew how many numbers were in T, we don't know where they're distributed. Maybe all of the numbers in T are less than 20, in which case, there is 100% chance that you would select one that was less than 20. But maybe 4 of them are greater than 20. you Maybe you have 21, 22, 23, 24. And four of them are less than 20, or maybe something in between. So given that, you still don't know where T is distributed relative to 20. So I say that we still don't have enough information, even with both of these statements, to solve this problem. Next question. 115. It takes the most confidence to choose that E, to say that one cannot solve this problem. 115. I'm always second-guessing myself. I'm like, well, let me make sure I haven't missed some piece of information. No, but I'm pretty confident here. All right. 115. If Sarah's age is exactly twice Bill's age, what is Sarah's age? They want to know Sarah's age. S for Sarah's age. Let me scroll down a little bit. Problem statement number one says four years ago, Sarah's age was exactly three times Bill's age. So Sarah's age four years ago, so Sarah minus 4. S is Sarah's age today. So Sarah minus 4 was exactly three times Bill's age, so was three times Bill's age then, so B minus 4, right? Sarah's age four years ago was three times Bill's age four years ago. Fair enough. So just with this and this, can I solve for S? Well, sure. I have two linear equations and two unknowns, so I can solve for either variable. And to prove it to you, I'll do it, although this, once again, is a waste of time from the point of view of the GMAT. S is equal to 3B minus 8. And then B-- from there we get B is equal to S/2, so S is equal to 3, right? I just divide 2 into both sides. So 3S/2 minus 8. What can I do here? So I have 1 and 1/2 S. I could subtract it from both sides. I'll have minus-- I'm doing it fast just to-- minus 1/2S is equal to minus 8. And then multiply both sides by minus 2, and you get Sarah is 16 years old. And you're done. But you didn't have to do any of this. That was all a waste of time. You just had to see that you had two linear equations and two unknowns, which is enough to solve for either of the variables. Statement two tells us, eight years from now, Sarah's age will be exactly 1.5 times Bill's age. So, same thing. Sarah's age eight years from now is going to be equal to 1 and 1/2 times Bill's age eight years from now. Same thing. We have two linear equations, one that they gave in the problem statement and one they gave in statement two, and two unknowns. No reason why you can't solve for both variables. And you might want to do it for yourself just as an exercise. But you'd do it the same way you did it right here. I always find substitution to be the easiest, especially when one of the equations is really easy to manipulate. So each statement individually is sufficient to solve this problem. Next problem. 116. Oh, I just realized I'm out of time. I'll do 116 in the next video. See