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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 30

121-124, pg. 288. Created by Sal Khan.

## Want to join the conversation?

- Question 123: Since Sal points out on statement number two, that knowing the coordinates of point Q we can tell tha OQ is 13, he's assuming that O has cordinates 0.0. Which leads to my question:

Can't statement number 1 be sufficient? Knowing the coordinates of point O and P I can calculate the lenght of OP. Using this info and the hight I can then calculate the area of the left portion of the triangle PNO (let's call PN the hight of the triangle). The area is 24.

Now, since we know OP<PQ, we know that the right half (PNQ) will have an area bigger than triangle PNO, whose area is 24. Which means that the two areas combined should be bigger than 48, which is exactly what the question is asking. Hope I was not too confusing!(4 votes)- The area of the left part of the triangle must be smaller than the right part. This is because QP>OP. If we call the area on the left area L, and the right part area R, then L<R, and L=24, then area L+R>48. So statement 1 is sufficient.(2 votes)

- 122. If W=2 then V again has to be zero. This is the only way to make VW=V^2. ( Vx2=V^2 ) ( Ox2=O ) There is no other option than V to be zero. So both 1 and 2 are correct. right?(1 vote)
- Question 123: can you assume point O has the coordinance 0,0?(1 vote)

## Video transcript

We're on 121. It says, if x/2 is equal to
3/y, is x less than y? So this is essentially saying
that x times y is equal to 6. I just cross multiplied. Or you could do that
step by step. Multiply both sides by 2. Then, multiply both
sides by y. And you get xy is equal to 6. And they want to know whether
x is less than y. Statement 1 says, y is greater
than or equal to 3. If y is greater than or equal
to 3, what does that mean? So let's think about this. x is equal to 6 divided by y. If y is greater than or equal
to 3-- so if it's 3, that's the largest that x
can be if y is 3. If y is any larger number, then
x will only be smaller. So x will be less than
or equal to 6/3. Or x will be less than
or equal to 2. And just, I want you to
understand why I said that. Think about this. It says, y is greater
than or equal to 3. If y is 3, x is going to be 2. If y is 4, x is going
to be 1 and 1/2. If y is 6, x is going to be 1. So the highest value for x
happens when y is equal to 3. And then, as y gets larger and
larger, x just gets smaller. So y being greater than or equal
to 3 says that x is less than or equal to 2. So statement 1, alone, is
sufficient to say that x is less than y. If x is less than 2 and y is
greater than 3, then x is definitely less than y. Statement 2 tells us, y is
less than or equal to 4. So this is interesting. It's really just the inverse of
this statement right here. Now, you're saying, instead of y
being greater than a certain number, y is less than
a certain number. So if you think about it
this way-- so let's go back to this equation. If y is 4, x is going to be--
let's do a little table. y, x. If y is equal to 4,
x is equal to 1.5. If y is equal to 3,
x is equal to 2. If y is equal to 2, then
x is equal to 3. So actually, this constraint,
I can pick a bunch of different y's that meet
this constraint. But it still doesn't tell me
about anything whether x is greater than or less than y. Here, x is less than y. And here, x is greater than y. So this statement is useless. So statement 1, alone,
is sufficient to answer this question. Statement 2 doesn't
help us much. 122. If v and w are different
integers, does v equal 0? Different integers. Remember, it says, different. They say that vw is equal
to v squared. Now, let's think about this. This doesn't tell me for sure
that v is equal to 0. And, in fact, as long as-- well,
let's think about this. The only way that this could be
true is if v is equal to 0, that's one case. Now, if v does not equal to 0--
So if v does not equal to 0, then this case
implies what? That soon. w equals v. Think about it. If v doesn't equal 0, we can
divide both sides by v. Because we know we're
not dividing by 0. You divide both sides by
v, you get, on the left-hand side, w. Because this v cancels out. And then v squared divided
by v, you get v. And so you are left with-- Well, I want to be careful
here, though. Because w would have
to equal v. And they already told
us that w and v are two different numbers. So we know that this situation--
if w equals v, then v does not equal 0. But they tell us in the problem
statement that w and v are two different integers. They tell us that w
does not equal v. So we know that the only way
that this could be satisfied, if w does not equal v, is
if v is equal to 0. So statement 1, alone,
is sufficient to answer this question. Statement 2. w is equal to 2. Well, that doesn't do
anything for me. I mean w could be 2
and v could be 0. Or v could be 3. All they told us was that v and
w are different integers and that they don't
equal each other. So this doesn't help me at all
in determining whether v is equal to 0. So statement 1, alone,
is sufficient. I just want to make sure I got
that first part right. I mean the only way that this is
true, this is true if v is equal to 0. Or it's true if w
is equal to v. Either of those cases
are true. That's the only way it works. And we know w does
not equal v. So v has to be equal to 0. What are we on? 123. What is the value of 36,500
times 1.05 to the nth power? All right, statement number 1
tells us, n squared minus 5n plus 6 is equal to 0. So we could solve for n. There will be two solutions. So this is n minus 2
times n minus 3. I just factored it. I said, what two numbers, when
I multiply them, equal positive 6? So I said, minus 2, minus 3. And when you add them
become minus 5. That's why I picked the minuses,
as opposed to plus 2 and plus 3 or some other
combination. Equals 0. So statement 1 tells us that
n is equal to 2 or n is equal to 3. So statement 1, by itself,
isn't enough. Because I don't know whether
I should put a 2 here or a 3 here. So I can't determine this
value just yet. Statement 2 tells us, n minus
2 does not equal 0. This statement, by itself,
is fairly useless. Because it just tells me that
n does not equal 2. So n could be any
other number. So I don't know what
number to put here. But used in conjunction, these
two statements-- statement 1 says that it is either 2 or 3. Statement 2 says, well,
it's not 2. So if you use both of these
together, we know that n is equal to 3. And then we can, obviously,
solve this problem. Taking 1.05 to the third power
and multiplying by 36,500, which I won't do because it
would be a waste of time. 124. In the rectangular
coordinate system above-- let me draw that. They have the y-axis. They have the x-axis. Then, they draw on a triangle. I'll do that in a
different color. That and like that. And the letters are
O, P, you and Q. In the rectangular coordinate
system above, if OP is less than PQ-- so OP is
less than PQ. Let me write that down. OP is less than PQ. Is the area of OPQ
greater than 48? So we want to know, is
area greater than 48? OP is less than PQ. So the way I drew this
actually isn't right. It's probably a little
bit shifted to the left a little bit. But fair enough. We can visualize. We can imagine that. So let's see. Statement number 1. The coordinates of point
P are 6 comma 8. So this is 6 comma 8. Let's see. That tells us the height. When we figure out the area of
a triangle, we need to know the base times the height. And then multiply
that times 1/2. The height of this triangle is
the y-coordinate, which is 8. So the altitude of this
triangle is 8. So we have the height, but we
still don't know the distance from O to Q. So statement 1, by itself,
isn't much information. So that tells that the
point was 6, 8. That P was 6, 8. Statement 2 says, the
coordinates of point Q are 13 comma 0. Well, this is helpful. 13 comma 0. So this distance right
here is 13. So by itself, that's useless. Because if I didn't have the
height, I still wouldn't be able to figure out the area
of this triangle. But now, if I use both
statements, I know the base times the height. So the area is going to
be the base times the height times 1/2. So 8 times 1/2 is 4. Times 13 is 52. And so 52 is clearly
greater than 48. Let me make sure. Because I didn't use
this information. I didn't use that information
at all. Let's see. Let me think about it. Actually, no. I think I'm wrong. I think statement 1, alone,
might be enough. Let's assume that we don't
have-- these are tricky. Let's assume that we don't
have statement 2 there. Let me see if I can figure out
whether the area is greater than 48 without it. Let me see. Yeah, I think I'm wrong. I caught myself. OK, so if this is 6 comma 8,
the area of this triangle right here, well, this
distance right here is going to be 6. This distance right here is 6. 6 comma 8. Right. OK, so the area of this
triangle is 6 times 8. 48 times 1/2. So the area of this
triangle is 24. But they told us that
OP is less than PQ. That this distance is less
than this distance. Maybe I should draw it
like that, so you can kind of see it. And the only way that this
distance is less than this distance is if this distance
is less than this distance. This is a tricky problem. This has to be less than this. Q has to be further away. Because that's the only way
you're going to get this altitude to be longer
than this altitude. Because you're assuming that
Q is along the axes. So that's the only thing that
you can shift back and forth. So if this distance is larger
than this distance, then the area of this triangle is going
to be at least 24. So the area is going to
be greater than 48. So actually, all you needed
was statement 1 to solve this problem. That's all you needed. Statement 1 gave you
enough information. Statement 2 does not, by
itself, give you enough information because you don't
know how high P is. It could be really low down
here, in which case the area is going to be less than 48. It could be really high, in
which case the area would be greater than 48. But anyway, see, I almost
got that one wrong. But I'll leave you there. I'll see you in the
next video. Statement 1, alone, was
sufficient for this one. See