If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains ***.kastatic.org** and ***.kasandbox.org** are unblocked.

Main content

Current time:0:00Total duration:9:54

All right. I re-read the question,
and now I realize the error in my logic. I misread the question. I had misread it and I said,
oh, there are 60 siblings. No, there are 60
sibling pairs. So that means in each class
there-- so 60 siblings pairs, that means there are
120 siblings. But that tells us that in each
class there aren't 30, there are 60 members of a sibling
group, right? So let's go back through
our logic again. Let's say I pick someone from
the junior class first. What is the probability that
I even pick someone from a sibling pair? That's going to be 60/1,000,
which is the same thing as 6/100, which is the same
thing as 3/50, right? Fair enough. So there's a 3 in 50 chance that
I pick someone who's a member of a sibling pair. Now, in order for me to have a
complete sibling pair when I go to the senior class and pick
someone, I have to pick this junior sibling, and that
junior only has 1 sibling in the senior class. So there's only a 1
in 800 chance that I pick their sibling. It doesn't matter that-- the
other 59 siblings in that are useless, because they're
not this guy's sibling. So I have a 1 in 800 chance
of doing that. You multiply them,
you get 3/40,000. And that's choice A. If you don't see your choice,
make sure to read the question again. Almost out of space so let
me clear all of this. Let me clear all of that. And we're on problem 218. Which of the following cannot
be the median of the three positive integers x, y, and z? So the median, the
middle number. They say A is x. Well, no, that's not true. What if this was the order?
x y, z, then x would be the median. So it's not A. B, z . No reason why z can't
be the median. The order can be y, z, x. It's not B. C, x plus z. Well, let's see, could x
plus z be the median? x plus z, could that
be the median? Let me think about it. If we have x, z, and y, I mean,
by definition one of these numbers has to
be the median. We have three numbers,
one of them has to be the middle number. Under what circumstances--
because x plus-- I mean, if x plus z is the median, that
implies that neither x nor z is the median, right? Do they tell us that
those are-- right, they're positive integers. None of these are zero. So in order for x plus z to be
the median, that tells us that neither x nor z is the
median, right? And if neither x nor z
is the median, then y has to be the median. And then that tells us that x
plus z has to be equal to y. But if y is the median, the
order is this: x, y, z. And there's no way that I can
add x and z and get a smaller number than z, because
all of these numbers are positive, right? Yeah, they said three
positive integers. So I go with choice C. That cannot happen. The median cannot be x plus z. Problem 219. What is the 25th digit to the
right of the decimal point in the decimal form of 6/11? And whenever they ask you for
something ridiculous, there must be some type of pattern we
can exploit to figure out what the 25th digit is. Let's just start dividing. 11 goes into 6, so we're
going to have to have some decimals there. 11 goes into 60 five times. 5 times 11 is 55,
and remainder 5. And you're going to
50, you get a 4. 4 times-- I already see
the pattern-- 44. Oh no, never mind. Wait, let me make sure
that's right. 55, you get 50 remainder, 11
goes into 50 four times. You get 40. So then you get 60. 11 goes into 60 five times,
and you repeat so on and so forth. I just want to make sure. Let's see, the first digit
to the right is 25. The second digit to
the right is 4. The third digit to
the right is 5. And so, you know, we
could keep going. The fourth digit to
the right is 4. So every odd digit to the
right of this is what? It's 5. So they want to know the 25th
digit to the right, and that's an odd number, 25. So it has to be 5 as well. So that's choice C. Next question. Oh, I thought I had
to erase that. OK, question 220. John and Mary were each paid x
dollars in advance to do a certain job together. Fair enough. John worked on the job for 10
hours, so John worked 10 hours, and Mary worked 2
hours less than John. So Mary worked 8 hours, right? John worked on the job for 10
hours and Mary worked for 2 hours less than John, so
she worked 8 hours. If Mary gave y dollars of her
payment so that they would have received the same hourly
wage, what was the dollar amount in terms of y that John
was paid in advance? OK, John and Mary were each paid
x dollars in advance to do a certain job together. OK, so how much money did they--
so John got x dollars and Mary got x dollars. They both got x dollars,
right? So if you think about
it, let's ignore who did what, right? We had how many total
hours of labor? We had 18 hours of labor,
and they were paid how many dollars? They were paid 2x dollars
for 18 hours of labor. So the average wage should be--
you could kind of do it the fair-- you know, per hourly
wage should be-- let's see, we could divide by 9, you
get a 1 and you get a 9. You get x/9 dollars per hour. That's the fair way. So let me re-read
the question. If Mary gave John y dollars of
her payment so that they would have had the same hourly wage,
what was the dollar amount in terms of y that John was
paid in advance? Actually, let's think
of it this way. So Mary pays John-- let
me ignore this. I just thought of a better
way of doing it. So they were each paid
x hours in advance to do a certain job. John worked on the job for 10
hours and Mary worked 2 hours less than John. If Mary paid John y dollars of
her payment so that they would have received the same hourly
wage-- so after Mary pays John y, what-- so that means John got
a total of y dollars and Mary has paid y dollars. So now John has x plus y, and
Mary has x minus y dollars. And so, if Mary paid of her
payment so that they would receive the same hourly wage,
what was the dollar amount in terms of y that John was
paid in advance? So let me think about
this a little bit. So they got the same-- they
received the same hourly wage. So Mary should have 8/10 the
money of John, right? So x minus y should be equal
to 0.8 times x plus y. Am I thinking about that right
or is there an easier way to think about it? Well, let's think about
it this way. John worked 2 hours more
than Mary, right? So Mary, essentially, if she
pays him for 1 hour-- no, yeah, this is the right thing.
x minus y is equal to 0.8 times x plus y, but I'm
going to still end up with an x here. So you get x minus y is equal
to 0.8x plus 0.8y, right? And all the logic here I just
said is-- well, Mary's total money, if we're going to be
fair, should be 0.8 of John's total money, because she worked
0.8 as much hours. And so, let's see. Oh, actually, they want to know
what John was paid in advance, so that's x. So actually, OK, we're fine. So they just want to know
what x is in terms of y. So we're cool. So let's subtract 0.8x
from both sides. So you get 0.2x minus
y is equal to 0.8y. And then add y to both sides. You get 0.2x is equal to 1.8y. And then multiply both sides by
5, you get x is equal to-- what is that? 5 times 18 is 50 plus 40. Add a decimal point is
90 is equal to 9y. And that is choice E I got a little confused with
the wording of it when they say what was the dollar amount
in terms of y that John was paid in advance? They should've said
what was x? And that would've been
a little bit easier. Anyway, I'm all out of time. Sorry for bumbling through
that problem. See you in the next video.