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## GMAT

### Course: GMAT > Unit 1

Lesson 2: Data sufficiency- GMAT: Data sufficiency 1
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- GMAT: Data sufficiency 21 (correction)
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# GMAT: Data sufficiency 25

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

## Want to join the conversation?

- Sal,

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) - 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)- 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.