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All right we're on problem
113 on page 287. So they've drawn this thing. Let me see if I could draw it. So we have a right triangle
in the center. So that's the side. That's the other side,
and then the base. Right triangle. And then they drew some
squares on every side of that triangle. Let me see if I can do that. That's one square. Well, I don't know how I'm going
to draw the other square because I don't know how
to draw an angled one. I'll just draw the lines. That's the other square. And the last one. I'll just draw in some lines
something like this. It comes up like that, like
that and like that. So that's what they have. And they call this A, this B,
C, D, and they tell us that this is a right angle. They say, in the figure above,
if the area of triangular region D is 4, what is the
length of a side of square region A? So all of these are-- well, they
don't tell us that this and this are a square region
just yet, but they tell us this is a square region. So the length of this side-- if
we know the length of this side and that side, then we can
figure out the length of the hypotenuse of this
right triangle, so that's all we need. And let's see, let's call this
length right here little b. And let's call this
length little c. So we already know that, if we
know the area of this triangle is 4, we know that little b
times little c, so height times base times 1/2,
is equal to 4. But that also tells us that
b be times little c is equal to 8. I don't know that does for us. Let's read the statements. Statement number 1. The area of square
region B is 9. So B is equal to 9. Well, that's useful. If the area of this region,
and it's a square, is 9, what's each of the sides
going to be? The sides are going to be
the square root of that. Little b is going to
be equal to 3. If little b is equal to
3, then the area of this is 3 times 3. We can do that because it's a
square, which would be 9. So if little b is 3,
then what is C? Well, then we get that C is
equal to 8/3 using this information. And we got that, remember, just
based on the fact that the area of this triangle
right here is 4. So little b times little c times
1/2 would equal to 4. So, if we know what little b and
we know what little c is, then we can solve for-- let's
call that little a. And little a is what the
whole point was about. What is the length of a side
of the square region A? So we could just solve for b
squared plus c squared is equal to a squared, right? This squared soon. plus this
squared is equal to the hypotenuse squared. And we just solve for
a and we'd be done. So statement one alone
is sufficient. And this is really messy,
especially taking an 8/3 squared and all that, so
I'm not going to do it. And it would be a waste of
time on the GMAT anyway. Let's look at statement
number two. The area of square region
C is 64/9, so C is equal to 64 over 9. So this is kind of very similar
to the information they gave us in statement one. If the area of this is 64/9 and
this is a square region, then the length of each of its
sides, or in this case, little c, the length of little c is
just going to be the square root of that. So little c is going to be 8/3,
which is exactly what we figured out here. And then we could use this
information that we got from the problem statement that b
times c is going to be equal to 8, and we got that from the
fact that the area of this triangle is 4. So we use that and we can figure
out that b times 8/3 is equal to 8. And then you get b is equal to
3, and then you apply this again to solve for a. So each of the statements one
and two are each individually sufficient to answer
this question. Next problem. 114. If x is to be selected at random
from set T, what is the probability that 1/4x minus 5
is less than or equal to 0? Let's simplify this real fast.
That means that what's the probably that 1/4x
is less than 5? And that means what's
the probability? Let's multiply both
sides by 4. I just added 5 to
both sides here. And let's multiply
both sides by 4. And we don't have to change the
inequality since 4 is a positive number. So it's the same thing as what's
the probability that x is less than 20? And I like this a lot more
because it's just simple and it lets my brain grasp
the problem better. Now, what are the statements
they give us? Statement one: T is a
set of 8 integers. T has 8 integers. That by itself doesn't help me
because I don't know what those eight integers
look like. I mean, maybe all eight could
be larger than 20, in which case, it would be impossible to
find one that's less than or equal to 20, or maybe all 8
are less than 20, in which case, you would have
100% chance. So statement one by itself
is fairly useless. Statement two. T is contained in the set of
integers from 1 to 25. So that also doesn't help me. So T within 1 to 25. So they're essentially telling
us that every number in T is between this range. But if we just take statement
two alone, they're not telling us, first of all, how many
numbers are in T. Statement one tells us that,
but statement two doesn't tell us. And then even if we knew how
many numbers were in T, we don't know where they're
distributed. Maybe all of the numbers in T
are less than 20, in which case, there is 100% chance that
you would select one that was less than 20. But maybe 4 of them are
greater than 20. you Maybe you have
21, 22, 23, 24. And four of them are less
than 20, or maybe something in between. So given that, you still don't
know where T is distributed relative to 20. So I say that we still don't
have enough information, even with both of these statements,
to solve this problem. Next question. 115. It takes the most confidence to
choose that E, to say that one cannot solve this problem. 115. I'm always second-guessing
myself. I'm like, well, let me make
sure I haven't missed some piece of information. No, but I'm pretty
confident here. All right. 115. If Sarah's age is exactly twice
Bill's age, what is Sarah's age? They want to know Sarah's age. S for Sarah's age. Let me scroll down
a little bit. Problem statement number one
says four years ago, Sarah's age was exactly three
times Bill's age. So Sarah's age four years
ago, so Sarah minus 4. S is Sarah's age today. So Sarah minus 4 was exactly
three times Bill's age, so was three times Bill's age then,
so B minus 4, right? Sarah's age four years ago
was three times Bill's age four years ago. Fair enough. So just with this and this,
can I solve for S? Well, sure. I have two linear equations
and two unknowns, so I can solve for either variable. And to prove it to you, I'll
do it, although this, once again, is a waste of time from
the point of view of the GMAT. S is equal to 3B minus 8. And then B-- from there we get
B is equal to S/2, so S is equal to 3, right? I just divide 2 into
both sides. So 3S/2 minus 8. What can I do here? So I have 1 and 1/2 S. I could subtract it
from both sides. I'll have minus-- I'm doing it
fast just to-- minus 1/2S is equal to minus 8. And then multiply both sides by
minus 2, and you get Sarah is 16 years old. And you're done. But you didn't have
to do any of this. That was all a waste of time. You just had to see that you had
two linear equations and two unknowns, which is
enough to solve for either of the variables. Statement two tells us, eight
years from now, Sarah's age will be exactly 1.5
times Bill's age. So, same thing. Sarah's age eight years from now
is going to be equal to 1 and 1/2 times Bill's age
eight years from now. Same thing. We have two linear equations,
one that they gave in the problem statement and one
they gave in statement two, and two unknowns. No reason why you can't solve
for both variables. And you might want to do it for yourself just as an exercise. But you'd do it the same way
you did it right here. I always find substitution to
be the easiest, especially when one of the equations is
really easy to manipulate. So each statement individually
is sufficient to solve this problem. Next problem. 116. Oh, I just realized
I'm out of time. I'll do 116 in the next video. See