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## Problem solving

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# GMAT: Math 45

## Video transcript

We're on problem 214. Of the 50 researchers in a
work group, 40% will be assigned to team A. So let's call that team A. Of the 50 researchers working,
40% percent will be assigned to team A. So 40% of 50, that's equal
to 20 go to team A. And the remaining 60%
go to team B. So B gets the remaining 60%,
which is 30 people, right? 60% percent of 50 is 30. However, 70% of the researchers
prefer team A, and 30% prefer team B. So 70%, which is what,
35, prefer A, and 30% prefer team B. So the remainder, which is 15,
right, or 30% of 50, prefer B. What is the lowest possible
number of researchers who will not be assigned to the
team they prefer? So there's a lot of
pseudo-double negatives here. What is the lowest possible
number of researchers who will not be assigned to the team
that they prefer? So this is -- I have to read
that a couple of times. What is the lowest number of
researchers who will not be assigned to the team
they prefer? So essentially they're saying
this is the case where most people get the team they
prefer, right? Where as many people as possible
to get the team they prefer, and then we want the
lowest number of people who didn't get the team
they prefer. So what's the situation? Well, the best situation is --
let's see, 35 prefer A, so let's say that 20
of them got A. And then 15 fall into
this upset category. And likewise, out of these 15
that prefer B, well, I don't know, let's say that
all of them got B. All of these got B, but then we
have to put another 15 in B, and that's these 15 people. So the best case scenario in
which all of these people get B, and 20 of these people get
A, but then you've got to stick 15 people of these in B,
so those 15 people are going to be upset. And so that is the lowest
possible number of researchers who will not be assigned to
the team they prefer, so that's these 15. And that's choice A. Problem 215. If m is the average of the first
ten positive multiples of 5, m is equal to average of,
I don't know, let's call it multiples of 5, and M is the
median of the first ten -- and this is the first ten --
and M-- big M-- is equal to the median of the same thing,
what is the value of big M minus small m? So let's just write them out. I think that's the simplest
way to do it. The first ten multiples of 5. 5, 10 -- the first ten positive
multiples of 5. Right, so we won't count 0. 5, 10, 15, 20, 25, 30, 35 --
actually, let me write it another way, I know this
is a little confusing. So this is 5, 10, 15, 20, 25,
30, 35, 40, 45, and 50. And I just thought of writing
it that way for a reason. So, first of all, they want a
little m, which is the mean. Little m is equal to the
average of this. Let me ask you a question. If you average-- so I paired
these numbers from the lowest to the highest, second lowest to
second highest. If you take the sum of this one and
this one, you get 55. This one and this
one, you get 55. This one and this
one, you get 55. So I think you get the point. The sum of any two
of them are 55. So if you take the average of
any of these, the average of 5 and 50 is 27.5. And that's going to actually be
the average for the whole set, because you can
pair them up. The average of 25
and 30, 27.5. The average of 20
and 35 is 27.5. And if you don't believe me, I
mean, there's a bunch of ways you can think about this. You could say, oh, this is
the same thing as 55. If you wanted to figure out the
sum of this, you can say if I sum these, it would
be 55 times 5. So you'd have-- the sum of all
of them would be 55 times 5 divided by the total of all of
them, which is 10, which is 55 times 1/2, which is 27.5. Anyway, there's a bunch of ways
to think about it, but I just wanted to show you a quick
way of figuring out the average of the first
ten multiples of 5. And then they want to know
what is the median of the first ten positive
multiples of 5? So the middle numbers--
so there's two middle numbers, these two. So when you have two middle
numbers, you take the average of them to get the median. Well, guess what? The average of those
two is 27.5. So big M is equal to 27.5. So big m minus small m is
0, and that is choice B. Next question, 216. If m is greater than 0, and x is
m% of y-- so x is equal to m/100 times y, right? If m is 6%-- and this'll
turns into 0.06, right? x is m% of y, then in terms of
m, y is what percent of x? OK, so let's multiply both
sides of this by 100/m. So you get 100/m x, and 100/m
times m/100 is just 1 is equal to y, right? Let me switch the order to so it
makes a little more-- y is equal to 100/m times x. So if I wanted to write this
as a percentage, this right here is going to be some
type of decimal number. And so if I have a decimal and
I want to multiply, and if I want to express it as a
percentage, I mean, if this right here is 1, if I wanted
express it as a percentage, it would be 100%. If this right here ends up being
0.2, I would multiply it by 100 and get 20%. So to express this as
a percentage, I'd multiply it by 100. So 100/m times 100 is
equal to 10,000/m. And that is choice E. 10,000 over m%, which
is the same thing as 100/m as a decimal. Next question, 217. I'll use green. A certain junior class has 1,000
students, so the junior class has 1,000 students,
and a certain senior class has 800 students. So some people dropped out. Maybe they didn't. Maybe it started bigger
or smaller. Among these students there
60 sibling pairs. 60 siblings-- I already sense
a Venn diagram coming-- each consisting of 1 junior
and 1 senior. If one student is to be selected
at random from each class, what is the probability
that the two students selected will be a sibling pair? If one student is to be selected
at random from each class, what is the probability
that the two students will be a sibling pair? OK, so how many sibling
pair members are there in each class? Well, there have to be
30 in each class. So there's 30 sibling pair
members in that class and there's 30 sibling pair
members in that class. OK, so if one student is to be
selected at random from each class, what is the probability
that two students will be a sibling pair? OK, so let's say we pick a
student from the junior class first, right? So first of all, what's the
chances that they're even part of a sibling pair? Well, that's going to be--
there's 30 of them that are part of a pair, and there's
1,000 in the class, so the probability of that happening is
30/1,000, which is the same thing as 3/100, right? That's the probability that
we got a part of a pair. Now we're going to go and pick
someone from the senior class. If we want to find this guy's
sibling, we can't just find another sibling pair, we've
got to find this guy's actual sibling. So actually, only one person in
the entire senior class is going to be-- whoever this
person we pick from the junior class, if we did happen to pick
one, the only way that we got an actual sibling pair is if
we pick that one person in the senior class who is
this person's sibling. So there's actually only
a 1 in 800 chance that we picked them. So if we multiply that out,
that is equal to 3 over-- let's see, what is it? 80,000? 3/80,000, which is not
one of the choices. So what did I do wrong? Here. So I see 3/40,000, and for some
reason, I think maybe I'm not counting two scenarios. There's probably two scenarios
in which is this can occur, right? I could have picked-- if I go to
the junior class first and there's a 30 in 1,000 chance
that I pick the correct sibling, so that's 3 in 100, and
then the odds that I take his or her sibling from the
senior class is 1 in 800. But maybe then-- let me see if
I'm missing some logic here. Maybe I could have picked--
well, instead of me stumbling because I'm already over time,
let me continue this in the next video.