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## GMAT

### Course: GMAT > Unit 1

Lesson 2: Data sufficiency- GMAT: Data sufficiency 1
- GMAT: Data sufficiency 2
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- GMAT: Data sufficiency 21 (correction)
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- GMAT: Data sufficiency 41

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# GMAT: Data sufficiency 41

154-155, pg. 290. Created by Sal Khan.

## Want to join the conversation?

- Keep up the good work Mr Khan!(4 votes)
- Sal, I wonder what I am doing wrong, but I don't follow you on problem 154/2.

If I add 1 to both sides as you did, to get X^2<1, then X < +/- 1, then - 1 > x < 1 not -1 < x < 1.

When I try different approach to cross check, I break down X^2 -1 < 0 into (x+1)(x-1) < 0, which again shows that X< -1 or x<1 . in either case I show that X is less then -1. Where am I going wrong?(1 vote)

## Video transcript

We're on problem 154. They ask us, is x negative? So the question is
x less than 0. Statement number 1 tells us--
let's see, they say x to the third times 1 minus x squared
is less than 0. Now let's simplify it. Let's see, x to the third
minus x to the fifth is less than 0. We could add x to the fifth to
both sides. x to the third is less than x to the fifth. Let's divide both sides
by x squared. That won't change the inequality
since x squared is definitely going
to be positive. Any integer squared, even if
it's negative, is a positive. So you divide both sides by x
squared, you get x is less than x to the third. Now let's see, does this
answer that question? Let's see what x's would
satisfy this condition. So we can already say that x
cannot be equal to 0, right? Anything where this
would have been an inequality doesn't work. So x can't be equal 0, it can't
be able to 1, because 1 is equal 1 to the third,
and it can't be equal to negative 1. So we know that immediately. We know that for regular
positive numbers, for numbers greater than 1, taking it to
the third power is always going to be greater than
the number itself. If you took 2 to the third
power, that's greater than 2. 3 to the third power
is greater than 3. 1.5 to the third power
is greater than 1.5. So we know that this statement
right here implies that x is greater than 1. Now let's see, is there
a situation where x can be negative? Well, let's see. First off, x is between
0 and 1. 1/2 is not less than 1/8. So it doesn't work
between 0 and 1. But what if x were between
negative 1 and 0? Let's try an example. Negative 1/2 is less
than negative 1/8. It's more negative. Negative 1/2 is less than
negative 1/8, right? You could try that with
any fraction. Because when you take it to the
third power, it becomes a smaller fraction,
but it becomes a smaller negative fraction. So x will be less than it,
because it's more negative. So this is true. If this is true, then x is
either greater than 1, or it's between negative 1 and 0. Let's see, does it work if x
is less than negative 1. Negative 2, is negative 2
less than negative 8? Nope. Right. So these are the only ranges
where it works. This statement implies this. But this still isn't enough
information to tell us whether x is less than 0. x could be greater than 1, which
is definitely greater than 0, or x could
be less than 0. So we don't know yet just
from statement 1. Statement 2 tells us x squared
minus 1 is less than 0. So add 1 to both sides, that
tells us that x squared is less than 1. So when could something squared
be less than 1? Well, it's going to
have to be between negative 1 and 1, right? It's essentially going to have
be a positive or a negative fraction less than 1. That's the only way that when
you square it, you get a number less than 1. If it was greater than 1, you're
definitely going to get a number greater than 1 and
it's going to be positive either way. If it's 1, it's going to be 1,
and then 0 is in our range. So statement 2 tells us this,
but it includes both positive and negative numbers. So it doesn't answer our
question whether x is less than 0. But if we use both statements
combined, what is the intersection? What is the overlap
of this and this? Well, if x has to be between
negative 1 and 1 and it has to be one of these, the only
one that it overlaps with is this one. You cannot have x being greater
than 1 and x being in this range. You can have x being in this
range and x being in this range, since this is
a subset of this. So this is actually the
more restrictive. And so since we go to that, that
actually tells us that x has to be negative. So both statements combined are
sufficient to answer this question, but individually
they're not. The last question, 155. Marsha's bucket can hold
a maximum of how many liters of water? So we want to know the capacity
of her bucket. Statement number 1 says, the
bucket currently contains 9 liters of water. That's useless. We don't know how-- that
could be the bucket, 9 liters could be that. Or maybe 9 liters is
the full capacity. That tells us nothing about
what the capacity of the bucket is. If I told you I had a bite of
the sandwich, that tells you nothing as far as how large
of a sandwich I can eat. So statement 1 doesn't
seem that useful. Statement 2, if 3 liters of
water are added to the bucket when it is 1/2 full of water. So 1/2 times the capacity,
right? When it's 1/2 full of water,
you add 3 liters. So if 3 liters of water are
added to the bucket when it is 1/2 full of water, the amount
of water in the bucket will increase by 1/3. To increase something by 1/3,
that is like multiplying it by 1 1/3 That's 1 1/3 times
the capacity. That's what they're saying. The amount of water in the
bucket will increase by 1. Sorry. You're going to have 1 1/3 times
what you started off with, which was half
your capacity. That make sense? We say we're starting with 1/2
of our capacity and we're adding 3 liters. They're saying that that will
increase the amount we have by 1 1/3. Let me write that. Let's say we start with x, and
we're adding 3 liters. They're saying that will equal
1 1/3 or 4/3 times x, right? This shows that x is
increasing by 1/3. You could view this as
1 plus 1/3 times x. This is an increase of 1/3. They tell us that our starting
point is 1/2 of capacity. So that's where I get the 1/2
c-- that's our starting point-- plus 3 is equal to 4/3
times the starting point, or where we're adding water
to, 1/2 times c. We could easily solve
for c here. So you get a 2, this
cancels out. Then you get 1/2 c plus
3 is equal to 2/3 c. Let's see, we could subtract 1/2
c from both sides, and you get 3 is equal to
2/3 minus 1/2 c. Then you can just do that
fraction, and you get c is equal to 3 over-- well,
what is this? 1/2-- so if you do it over a 6,
4 minus 3, so this is 1/6. So this turns into 1/6. So c is equal to 18. Wait a minute. You didn't have to
do all that. You should have just realized
that as soon as you can write this algebraic equation down,
it's a linear equation with one unknown, and that one
unknown is what we're trying to solve for, you have
enough information to solve the problem. So statement 2, alone,
is sufficient to solve this problem. We're all done with
data sufficiency. See you all in the
next section.