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### Course: GMAT > Unit 1

Lesson 2: Data sufficiency- GMAT: Data sufficiency 1
- GMAT: Data sufficiency 2
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- GMAT: Data sufficiency 6
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- GMAT: Data sufficiency 21 (correction)
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# GMAT: Data sufficiency 14

63-68, pgs. 282-283. Created by Sal Khan.

## Want to join the conversation?

- Problem #67: Statement no. 1: AB+CD+AC+BD=6

AB+BD=6-CD-AC => B(A+D)=6-C(D+A) => B(A+D)=6-C(A+D) => B=6-C =>B+C=6

someone please correct me if I am wrong, but isn't statement no.1 alone sufficient or did I solve the equation wrong?(4 votes)- You forgot to divide 6 on the right hand side by A+D.(4 votes)

- how to download your lessons(3 votes)
- To download a video, go to Info (underneath the video) > Download this lesson.(2 votes)

- At time5:40, question 65 part 2. Sal identifies X as odd. Why did Sal decide odd, why not even? Can you designate X as even? If not, why not? if you can, the statement would be insufficient...Help.(3 votes)
- Question 65 states that M/2 is not an even integer. This means that M/2 could be an odd integer, could be a fraction, could be an irrational number, in fact it could be anything EXCEPT for an even integer. Therefore the only thing that M can't be is an even integer, because any even integer divided by two will be an even integer. Therefore by deduction M must be an odd integer.

The ONLY way for M/2 to be an even integer is if M is an even integer. Therefore because M/2 is not an even integer M has to be non-even which means it has to be odd.

Anyone agree with this logic??(2 votes) - in #66 sal says that triangle ABC could be derived from #1 the problem is that without #2we cannot know if ABD and CBD are symmetrical?(1 vote)
- Isn't 0 an even integer o-3 =-3.(0 votes)
- Question 64, if you solve for x and y, you can actually get that x is 6 and y is 4.(1 vote)

## Video transcript

We're on problem 63. Carlotta can drive from her home
to her office by one of two possible routes. If she must also return by one
of these routes, what is the distance of the shorter route? So essentially she has a long
route to go to her office, that's her home, that's
her office. And then she has a short route,
with fewer curves, that's my guess. And she can go and come
back by one of two-- Statement number one: when she
drives from her home to her office by the shorter route
and returns by the longer route, she drives a total
of 42 kilometers. So when she goes this, way
shorter route this way and then longer route back,
it's 42 kilometers. So shorter plus longer
is equal to 42. And that by itself doesn't let
me know what the distance of the shorter route is. Statement number two: when she
drives both ways from her home to her office and back by the
longer route, she drives a total of 46 kilometers. So if she goes by the longer
route and then by the longer route again, so 2 times a longer
route is equal to 46 kilometers. Well, this by itself helps me
figure out what the longer route is, but they want to
know the distance of the shorter route. So this, I could just eyeball
this, the longer route is 23 kilometers. But if you want to know the
distance of the shorter route you would need this information
as well. The shorter route plus
23 is equal to 42. So the shorter route
is 19 kilometers. So you need both of
the statements. Both statements are sufficient,
but one by themselves are not enough
to figure out what the shorter route is. If they ask for the longer
route, just statement two could have been good enough. Next question. 64. Is x greater than y? My computer's updating
something. It's a little bit-- Is
x greater than y? Statement number one tells us
that x is equal to y plus 2. Well, let me think about that. No matter what y is, x is
going to be 2 more. Even if y is a negative million,
x is still going to be a bigger number. It's going to be 2 more than
negative a million. If it's y is negative 10,
this will be negative 8. No matter what x will
be greater than y. So statement one
is sufficient. Statement number two: x/2
is equal to y minus 1. OK. This doesn't look
as clear there. So x is equal to 2y minus 2. So we just have to come up with
a condition, one y where y is less than x and another y
where y is greater than x and then we would have
proven our point. So let's think about it. When y is 0-- this is using
statement number two-- when y is 0, what's x? x is minus 2. So that's a statement where
y is greater than x. And then when y is equal
to 10, what's x? x is 18. So this is a statement where
y is less than x. So statement number two actually
tells us nothing about whether x is greater than
y or y is greater than x. So statement number
two is useless. Statement number one, alone,
is sufficient. Next problem. 65. If m is an integer is m odd? Statement number one: m divided
by 2 not-- they write it real big and bold--
not even integer. Well, that doesn't
tell me much. I mean that doesn't tell
me that this is odd. In fact, it's not an even
integer but if this was an odd integer then that means that m
is an even integer, right? Anything that's divisible by 2
and results in an integer-- right, that's essentially saying
that it's divisible by 2-- by definition is
going to be even. So for example, what
if m divided by 2 is equal to 3, right? 3 is not an even integer but
still m would equal 6. And then m would be even. And I mean they're not
telling us much here. m divided by 2 could be equal
to 2.5 in which case m would be equal to 5 so that would
be an odd integer. So this by itself doesn't give
me much information. Statement number two: m minus
3 is an even integer. Well, this is useful, right? If you subtract an odd number
from a number and you get an even number, this number
is going to be even. And let's think about that. If I have an even number--
let me just write a bunch of numbers. Let me write x, y, and z. And let's say they're
consecutive, right? Actually let me write
it even better. Let's say that x is even--
let's say x is odd. Then x plus 1 is going
to be what? It's going to be even, right? The next number will be even. Then x plus 2 is going
to be odd. And then x plus 3 is
going to be even. And then if you go the other
way-- actually I should have gone to the way first-- x
minus 1 would be even. x minus 2 would be odd. And x minus 3 would
be even, right? If you have an odd number and
you subtract out an odd number and you get an even number
then this number is going to be odd. So this statement two by itself
is actually sufficient. And play around with some
numbers if that intuition doesn't make sense. Statement two, by itself, is
sufficient to solve this problem and you don't need
statement one at all. Next problem. 66. what is the area of the
triangular region ABC above? Well, they've drawn us a little
triangle so I will draw us a triangle. Let's see, we have our
base and then, it looks pretty symmetric. I'll try to draw it the
way they drew it. One end, that's the other end,
and then this is the altitude. Good enough. And then they label
it A, B, C, D. This is a right angle. This is x degrees. Alright, what is the area of
the triangular region? So area of a triangle is base
times height times 1/2. So if we can do that then
we're all done. One: the product of
BD and AC is 20. So BD times AC-- oh no, sorry,
BD times AC, oh that's it. So BD is the height. So area, let me just write it,
area is equal to 1/2 times this altitude, BD, times the
base, times AC, right? Statement number one, they tell
us what BD times AC is. They tell us that BD times
AC is equal to 20. So that immediately follows, if
this thing is equal to 20 then the area is equal to 10. And you didn't even have
to figure that out. So statement number one, by
itself, is sufficient. Statement number two: x is
equal to 45 degrees. This one's a little bit
more interesting. x is equal to 45 degrees. Well, I mean we could do a
little angle game, we could say, OK, this is a 90 degree
angle and we could say if this is 90, this is 45, and this is
going to be 45-- actually that's about where
we could stop. And this tells us no information
about what any of the sides are so you could
imagine if this is 45 degrees, but this could be a million mile
high triangle or it could be a nanometer high triangle. We don't know. It could be of any size. You could scale it up or down. This just tells us the degree
and the angle information. So two, by itself, is useless. So for this one, statement one,
alone, is sufficient to answer this problem, A. Next problem. 67. What is the value of b plus c? Alright, number one: they tell
us that ab plus cd plus ac plus bd is equal to 6. And this by itself doesn't help
us much because there's all these a's and d's here and
they could be anything. So we can't solve
for b plus c. But I'm already suspecting that
we could factor out the a's and d's somewhere. So we can factor out the b's
plus c's, then we might be able to figure out the numbers
if we're given more information about a and d. Statement two, and they give
us more information about a and d. a plus d is equal to 4. So let me see if I can simplify
this top one. Let me see if I can factor out
a bunch of b plus c's. So if we think about
it, let me think. So if you take this term and
that term and you factor out the a you get a-- let me do it
in a different color-- so this term and that term is equal to a
times what? b plus c, right? You can multiply that out
and you get ab plus ac. And then plus, and let's
do the same thing with this term, d. That's d times c plus b. But we can switch the order,
b plus c-- and if you don't believe me that this equals
this, multiply it out. All I did is I grouped the ab
and the ac terms and then factored out the a and then I
grouped the cd and the bd terms and factored out the d. And then now we can factor
out a b plus c. We can do the reverse
distributive property again. So now this equals a plus
d times b plus c, right? And if you don't believe me,
just multiply b plus c times a and then b plus c times d
and you get that, right? That's just the distributive
property. But now we're ready to solve
because they told us that this thing's equal to 6 and they
tell us that this thing is equal to 4. So b plus c is going to be equal
to 6/4 or 1 and 1/2. And we're done. You need both statements
together. Both statements, together,
are sufficient to solve this problem.