We're on problem 106.
The present ratio of students to teachers at a certain
school is 30:1.
So the ratio of students to teachers is equal to 30:1.
If the student enrollment were to increase by 50 students,
and the number of teachers were increased by 5, the ratio
of students to teachers would then be 25:1.
So if we increase the students by 50 students, and increase
the teachers by 5, the new ratio is going to be 25:1.
What is the present number of teachers?
Let's just simplify these a little bit.
We could cross-multiply.
We could just write s over T.
30 over 1 is equal to 30, or s is equal to 30T.
And let's substitute that into this equation, because we're
going to have to solve for the present number of teachers,
which is just T.
So substitute this here, so we get 30T plus 50, over T plus
5, is equal to-- well, 25 over 1, that's just 25.
And then let's multiply both sides of this equation by T
plus 5, and you get 30T plus 50 is equal to 25T, plus--
what's 25 times 5?-- 125.
Let's subtract 25T from both sides.
You get 5T plus 50 is equal to 125.
Subtract 50 from both sides, you get 5T is equal to 75.
And then T is equal to 15.
15 times 5 is 50, plus 25, right, 75.
T is 15.
So the answer is E.
Next question.
I don't want to run out of space, so I'll just do it
right here.
Problem 107.
What is the smallest integer, n, for which 25 to the n is
greater than 5 to the 12th?
Whenever you see some type of an equation or inequality
where the variables are under the exponent, and you're
comparing numbers of different bases, your first thing that
should register in your brain is how can I get these to be
in the same base?
And it should be pretty obvious here, because 25 is
the same thing as 5 squared, so let's rewrite it that way.
This statement is equivalent to saying 5 squared to the n
is greater than 5 to the 12th.
And 5 squared to the n, that's the same thing as 5 to the 2n,
is greater than 5 to the 12th.
And this right here is going to be true when 2n is greater
than 12, when the exponent up here is greater than the
exponent up here.
That's when it's going to be true.
You divide both sides by 2, and you get is greater than 6.
So what is the smallest integer, n, for which 25 to
the n is greater than 5 to the 12th?
So n has to be greater than 6, so we can't pick choice A,
which is 6, because if we pick 6, then we would have
5 to the 12th here.
And then it would be an inequality, but this is
greater than.
So the smallest integer, n, now has to be greater than 6,
so n has to be 7.
So that's choice B.
And don't get tricked there.
When you get 6 as your answer, you might want to pick A, but
n has to be greater than 6.
So it's essentially the smallest integer
greater than 6, is 7.
Problem 108.
If x and y are different prime numbers, each greater than 2,
which of the following must be true?
Each greater than 2, they're both prime numbers.
So statement number 1: x plus y cannot equal 91.
Let me think about that.
If x are y are different prime numbers, each greater than 2,
which of the following must be true?
x plus y cannot equal 91.
Let me think about that a little bit.
That means that I can't have two prime numbers that, when I
add them together, equal 91.
Well, let's think about it.
They're greater than 2, and then they're prime.
So what do we know about all prime numbers greater than 2?
Well, we know that they're both odd.
All prime numbers are odd.
2 is the only even prime number.
So both of these have to be odd.
So if I add an odd plus an odd, what do I get?
One way to think about it, this might be second nature to
you at this point, but an odd number can be written in the
form, 2 times some integer plus 1.
It's some even integer, some even number plus 1.
So I could write the first odd number like that.
Maybe I'll say this is x.
I'm just saying that x is odd.
I'm not saying that x is prime, but if you're larger
than 2 and you're prime, you're going to be odd.
Otherwise, you're divisible by 2.
And let's say y is equal to 2 times some other
integer, plus 1.
So if you add two odd numbers together, you get 2 times k,
plus m, plus 2, which is an even number.
When you add two odd numbers, you get an even number.
So if you add two prime numbers greater than 2, you're
essentially adding two odd numbers greater than 2, so
they have to be equal to an even number.
So they can't be equal to 91.
So statement 1 must be true.
Statement 2, let's think about it.
Statement 2 tells us, x minus y is even.
So once again, x and y are definitely
going to be odd numbers.
When you add them, you get an even number.
What happens when you subtract them?
If I were to subtract y from x, what would I get here?
If I were to subtract it, I would get 2 times k, minus m,
and then the 1 minus 1.
But this k minus m is going to still be an integer.
These are two integers, so this is still
going to be an integer.
And they're different prime numbers, so k and m are going
to be different, so this isn't going to be equal 0.
So the difference has got to be an even integer.
It might be a negative even integer, because we don't know
that x is necessarily larger than y, but we know that it
has to be an even integer.
So statement 2 is also correct.
Now statement 3.
x divided by y is not an integer.
Well, we know that x is prime, which means that it only has
two factors, 1 and itself.
And we know that x isn't 2.
We know that y is another prime number.
Now, if this statement right here-- if x divided by y-- if
that were an integer, then that means that y divides
evenly into x.
Which would mean that y is a factor of x.
Which would mean that x has a factor other than 1 and
itself, which would mean that x is not prime.
But they tell us it x is a prime number, and y isn't just
the number 1.
They're both prime numbers and they're each greater than 2,
so y is definitely not 1.
So in order for this to be an integer, x could not be prime.
Since x is prime, and y is some number greater than 2, we
know that this is not an integer.
So all three of these statements have to be true.
And that's statement E, answer E.
Problem 109.
Let me do it in a different color.
All of the following have the same value except-- OK, so
we're just going to try to do the math real fast here.
1 plus 2, plus 3, plus 4, plus 5, all of that over 3.
1 plus 2 is 3, 3 plus 3 is 6, 6 plus 4 is 10, 10 plus 5 is
15, divided by 3, this is equal to 5.
B, 1/3, times 1, plus 1, plus 1, plus 1, plus 1.
That's 1/3 times-- this is 5, so this is equal to 5/3.
So after we see the next answer we should know where
we're going.
So the next answer, we have 1/3 plus itself five times.
1/3 plus 1/3, plus 1/3, plus 1/3, plus 1/3.
Well, that's five 1/3s, so that's also equal to 5/3.
So we should immediately be able to recognize that this is
different than the other two.
And assuming there's only one answer-- they say all the
following have the same value except-- I think we've done
enough math, because we know that B and C are
the the same value.
And A is already different, so A is probably
going to be our choice.
And if we kept going, D and E are going to
have the same value.
And just eyeballing it, it does look
like that's the case.
It definitely looks like that's the case, because if
you look at the other two problems, you still get 5/3.
And E looks confusing, but the fact that A is already
different than B and C tells you that A is the answer.
And I'll see you in the next video.