Uploaded by khanacademy on 17.12.2008

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.