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GMAT: Data sufficiency 25

107-109, pg. 287. Created by Sal Khan.

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  • blobby green style avatar for user sujatha.kaipa
    I am truly benefiting with the videos for GMAT Problem Solving and Data Sufficiency. Could you please let me know if you are planning on uploading GMAT videos for verbal (Reading Comprehension, Critical Reasoning and Sentence correction)? Thank you for your time.
    (20 votes)
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  • leaf blue style avatar for user samroodbar
    About Q.107; based on your argument that (any # / 4 ) results into a finite decimal, can we (mostly me) deduce that (any # / any even #) results into a finite decimal? is the opposite true? (any # / any odd # ) = repeating decimal?
    It seems to be straight forward but I just wanted to make sure. Thanks
    (1 vote)
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    • hopper cool style avatar for user Dhaval Furia
      To determine whether r/s is terminating (where r and s are integers), it is enough to check the prime factorization of s.
      If the prime factorization of s contains only a bunch of 2s and/or 5s, then r/s is terminating.
      If the prime factorization of s contains any number besides 2 and 5, then r/s is non-terminating.
      Hope this helps.
      (2 votes)

Video transcript

We're on problem 107. Any decimal that has only a finite number of non-zero digits is a terminating decimal. For example, 24-- in the example they give us-- 24, 0.82, and 5.096. So they're saying if you just have a finite number of these non-zero things, at some point you have to end, and just have zeroes just left over. So that's what they mean by a terminating decimal. If r and s are positive integers, and the ratio r divided by s is expressed as a decimal, is r divided by s a terminating decimal? So essentially it wouldn't work if r and n-- that this is 1/3. That's not going to work. Because you'll have 0.3 forever. So they're saying, does it at some point end? This is interesting. So statement 1 tells us that 90 is less than r which is less than 100. That statement by itself doesn't help me. Now let's say that r was-- let me think of a good r. Let's say r was 93. If r is equal to 93 and then s is equal to-- well, let's just say that s, for the sake of argument, is 3 times r. Actually, I don't have to pick an r. Whatever r is, I can just pick an s to be 3 times r. And I'm not going to get a terminating decimal. On the other hand, if I pick s to be 2 times r, then I'm going to have a terminating decimal. So statement 1 by itself isn't going to help me. I hope you understand that. I can pick any s to make it terminating or not at this point. No matter whether r is constrained between 90 and 100. Statement 2 tells us that s is equal to 4b. Now, this is interesting to me. So where did this b come from all of a sudden? Any decimal that has only a finite number of non-zero digits is a terminating decimal. If r and s are integers and the ratio s is equal to 4b. Where did this come from? Let me look it up. What is this? I think this is a typo. Because where did b come from all of a sudden? Oh, yeah, when I look at the solution they say s is equal to 4. So that's definitely a typo. 4b. s is equal to 4. OK that makes a lot more sense. All right, so if s is equal to 4. Pretty much any number, when you divide any number by 4, you're going to have a terminating decimal. It's either going to be divisible by 4 or it's going to be-- the worst case, if you have 1 divided by 4, that equals 0.25. 5 So no matter what, when you divide something by 4, you're always going to have a terminating decimal. It's never going to repeat forever. I mean you can try it out with every-- no matter what number you divide, if you view it as a fraction, or kind of your fourth grade remainder problems, the remainder is either going to be 1, 2 or 3. If the remainder is 1, the decimal is going to be 0.25. If the remainder is 2, the decimal is going to be 0.5. And if the remainder is 3, the decimal is going to be 0.75. In any case, as long as s equals 4, you have a terminating decimal. So statement 2 alone is sufficient and statement 1 really doesn't tell us much. And I'm annoyed that they had that b typo there. Next problem. 108. OK, they've drawn a triangle right there. Let me see if I can draw it properly. There you go. And let me label it. This is B. That's A. That's C. This is y degrees. And this is D and this is x degrees. In the figure above, what is the value of x plus y? So x degrees plus y degrees. x plus y is equal to what? So they don't tell us anything else. They don't tell us if this is an equilateral triangle. They don't tell us that these angles are equal. They don't tell us it's symmetrical. They don't tell us anything about this triangle so far. So statement number 1 tells us that x is equal to 70. Well, just by x equalling 70, that still doesn't help me. I still don't know what y can be. I mean, you could draw this interior triangle in a bunch of different ways and you could see that the angle could change depending on how much you lift up D or compress D. This y angle can change irrespective of what x is. So x is equal to 70 does not in any way give me any information on what y is. Statement 2. Triangle ABC and triangle ADC are both isosceles. So isosceles actually doesn't mean necessarily that those two sides are the same. I mean, that kind of tends to be someone's immediate assumption, when we say isosceles. Oh maybe that side is equal to that side. But no, it just means that two of the sides are the same. So, likewise on this triangle. Although you can, at least, here-- can I make an argument? No, you can't make an argument which two sides are going to be the same. So, even if you assumed that both of those sides are the same, you're still not going to-- because you don't know. My gut instinct would be to say, oh, that side's the same as that side. But no, I can't make that assumption. Because it could be that this side is equal to that side. That would still make it isosceles. And this side is kind of viewed as like the base side. And it's a very big difference. Because if these are the two equal sides, and if this is x, this is going to be x. But then if I take the other scenario, where if these are two sides and if this is x, then both of these are really be 180 minus x divided by 2, right? Because this would be-- The bottom line is, this still doesn't get me anywhere. Even if I know that this is 70 degrees. So I'm going to go with, E, not enough information to solve this problem. 109. Are positive integers p and q both greater than n? They told us positive integers. OK. They say p minus q is greater than n. Well I'll give you a case right now where that doesn't satisfy this. What if p is equal to 10, q is equal to 1, and n is equal to 2. Then 10 minus 1 is 9 is definitely greater than 2. But look, q is not greater than 2. 1 is not greater than 2. So q wouldn't be greater than n. Or I could say, it could be, 110 and 108. Would equal-- and when you subtracted you'd get 2. And in this case q is bigger than 2. So this doesn't give me enough information. I can think of two combinations of p's q's and n's. One that says that q is not greater than n and one that says that q is greater than n. So one, by itself, isn't enough. Statement 2. q is greater than p. That still doesn't help me. Both of these statement actually cannot be true simultaneously. Because if these are both positive integers, which they told us-- well, greater than n. Both greater than n. They didn't tell us necessarily that n is positive. So n doesn't have to be positive. My logic for statement 1 still holds. But let's see, if q is greater than p, this by itself doesn't help me much. Because I can think of a similar combination to this. I can say q is equal to 10, p is equal to 9, and I could make n is equal to 2. Or I could say n is equal to 15. This in no way constrains what n is. So it, by itself, doesn't solve it. But what if we were to use both constraints. This is the interesting case. What if we were to say that p minus q is greater than n. And q is greater than p. So, we know that both p and q are positive. They told us that in the problem statement. If p and q are both positive, and q is greater than p, this is going to end up becoming a negative number. These are both positive. And this is the larger of the two positive numbers based on the second constraint. So this is going to be a negative number. And so if a negative number is greater than n, then we can be sure, we can be positive, that both p and q are greater than n. Because they told us that p and q are positive. So actually both of these statements imply that n is negative. Because we have a negative number here being greater than n. And if n is negative, then both p and q are definitely greater than n. So both statements combined are sufficient to answer this question. And the answer is actually true. I'll see you in the next video.