Tip:
Highlight text to annotate it
X
We're on problem 13.
A company sells boxes of balloons in which the balloons
are red, green, or blue.
Luann purchased-- that doesn't sound like a
name related to me.
Luann purchased a box of balloons in which 1/3
of them were red.
If there were 1/2 as many green balloons in the box as
red ones-- so green is equal to 1/2 red.
Fair enough.
Green is equal to 1/2 red.
There's 1/2 as many green balloons in the box as red
ones, and 18 balloons were blue, how many balloons were
in the box?
OK, so this is fascinating.
So 1/3 are red.
And then the green is 1/2 as many as the red, right?
So if 1/3 of them are red, how may are going to be green?
Well, 1/2 of the 1/3, right?
So we know that 1/6 are going to be green.
How did I get 1/6?
Let's say there's 10 red, right?
And 10 is 1/3, then there's going to be-- let's say
there's 30 balloons.
1/3 is red, so then it's 10.
So green's going to be 1/2 of that, so it'd be 5/30 or 1/6.
So whatever the number is, 1/3 are red, 1/2 of
that, or 1/6, are green.
And there are 18 blue.
So what can we do now?
Well, let's figure out what fraction
would have to be blue.
We know that 1/3 are red 1/6 are green.
What's left over?
So what's 1 minus 1/3 minus 1/6?
This will tell us how many blue, what fraction of the
balloons have to be blue.
So that equals-- let's make 6 the common denominator, right?
1 is equal to 6/6 minus 1/3.
That's equal to 2/6 minus 1/6.
So that's 6 minus 2 minus 1.
So that's 3/6 are blue, right?
And I just subtracted the fraction that are red and I
subtracted the fraction that are green from the whole.
And so I get 3/6, or 1/2, are blue.
So 1/2 of the balloons in the container, or in
the box, are blue.
And they want to know how many balloons are in the box?
Well, there are 18 blue, and that's 1/2 of all of the
balloons, right?
So 18 is equal to 1/2 of all of the balloons.
Multiply both sides by 2, you get 36 is equal to x.
And you know that.
18 is 1/2 of something, then the total number
of balloons is 36.
That's our answer.
Next problem, 14.
The three distinct points P, Q, and R lie on line L, OK?
The four distinct points S, T, U, V lie on a different line
that is parallel to L.
What is the total number of different lines that can be
drawn so that each line contains exactly two of the
seven points.
OK, I see what they're saying.
So let's draw the first line, P, Q, and R.
So let me just draw the two lines first. So the first line
is line P, Q, and R.
And then the second line is S, T, U, and V.
So then we have S, T, U, and V.
And this line is parallel to this line, right?
They're parallel, so they're never going to
intersect each other.
OK, what is the total number of different lines that can be
drawn so that each line intersects exactly two of the
seven points?
So you can't even intersect three of this.
Well, you can, only if you go through that line.
So what is the total number of different lines that can be
drawn so that each line contains exactly two of the
seven points?
Right.
So that's interesting.
So you can't count these lines, right?
Because these lines have three.
This line has three of the seven points, and this line
has four of the seven points.
So those can't be it, because it says exactly two.
You can't have even three of the points.
Sp what are they?
Well, P can go to four points, right?
I mean, I could just count them out.
P can go to four points, Q can be connected with four points,
and R can be connected with four points, so
it should be 12.
And if you don't know what I'm saying, let me
just draw it out.
P could be this line, one, two.
Sorry, that one wasn't drawn well.
One, two, three, four.
One, two, three, four, and then one, two, three.
Well, that's kind of a nice-looking shape there.
So 4 plus 4 plus 4 is 12, so there's 12 possible lines that
intersect exactly two of these seven points.
Next problem, problem 15.
I'm still using the line tool.
OK, problem 15.
If 2 to the x plus 2 to the x plus 2 to the x plus-- how
many of these are there?
There's four of them-- plus 2 to the x is equal to 2 the
seventh, what is the value of x?
So how many of these are there?
There's one, two, three, four, which I had to figure out
while I was drawing it.
So there's four.
We're essentially adding to the x four times, so this is
the same thing as saying 4 times 2 to the x, right?
Because we have 2 to the x four times:
one, two, three, four.
So this is the same thing as this: 4 times 2 to the x.
And that equals 2 to the seventh.
What's 4 written as base 2?
That's the same thing as 2 squared, right?
Whenever you see these problems and you have two
different bases, try to see if you can convert them all to
the same base.
So that's 2 squared plus-- 2 squared times 2 to the x is
equal to 2 to the seventh.
2 squared times 2 to the x, that's the same thing as 2 to
the x plus 2.
That equals 2 to the seventh.
So x plus 2 must equal 7.
x plus 2 is equal to 7.
x is equal to 5.
And we are done.
Next problem, problem 16.
Each of five people had a blank card on which they wrote
a positive integer.
If the average of these integers is 15, what is the
greatest possible integers that can be
on one of the cards?
This is fascinating.
So essentially, they're saying you have five positive
integers and their average is 15.
What is the greatest possible integer that could be among
these cards?
So think of it this way: the sum of the five integers is
going to be what?
So let's say it's x1 plus x2 plus x3
plus x4 plus x5, right?
All of them over 5 is equal to 15, right?
That's what they told us.
The average of the t numbers is 15, so the sum x1 plus x2
plus x3 plus x4 plus x5 is equal to what?
5 times 15, that's 75.
5 times 10 is 50 plus-- OK, that's 75.
So the sum of the integers are going to be 75.
And so we want to know what the largest one of these, the
greatest possible integer here.
So let's just say that this is what we're
trying to figure out.
Let's try to maximize this number here, x5.
If we want this number to be as large as possible, these
numbers have to be as small as possible, right?
And you could subtract these numbers from the other side.
You could say x5 is equal to 75 minus x1 minus x2 minus x3
minus x4, right?
And we're going to try to maximize this number.
And then we're going to say that's going to be the
largest.
So if this is the largest, we want to subtract as small a
number as possible here, here, here.
And what are the constraints?
They have to be positive integers.
So each of these numbers have to be greater than zero and
have to be integers.
So we want them to be as small as possible, so
let's make them 1.
So let's say it's 75 minus 1 minus 1 minus 1 minus 1.
That's 75 minus 4, which equals 71.
So that's the greatest possible value
of one of the integers.
And I have a minute left and do problems. No, I'd better do
it in another video.
So I'll see you in the next video.