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### Course: GMAT > Unit 1

Lesson 1: Problem solving- GMAT: Math 1
- GMAT: Math 2
- GMAT: Math 3
- GMAT: Math 4
- GMAT: Math 5
- GMAT: Math 6
- GMAT: Math 7
- GMAT: Math 8
- GMAT: Math 9
- GMAT: Math 10
- GMAT: Math 11
- GMAT: Math 12
- GMAT: Math 13
- GMAT: Math 14
- GMAT: Math 15
- GMAT: Math 16
- GMAT: Math 17
- GMAT: Math 18
- GMAT: Math 19
- GMAT: Math 20
- GMAT: Math 21
- GMAT: Math 22
- GMAT: Math 23
- GMAT: Math 24
- GMAT: Math 25
- GMAT: Math 26
- GMAT: Math 27
- GMAT: Math 28
- GMAT: Math 29
- GMAT: Math 30
- GMAT: Math 31
- GMAT: Math 32
- GMAT: Math 33
- GMAT: Math 34
- GMAT: Math 35
- GMAT: Math 36
- GMAT: Math 37
- GMAT: Math 38
- GMAT: Math 39
- GMAT: Math 40
- GMAT: Math 41
- GMAT: Math 42
- GMAT: Math 43
- GMAT: Math 44
- GMAT: Math 45
- GMAT: Math 46
- GMAT: Math 47
- GMAT: Math 48
- GMAT: Math 49
- GMAT: Math 50
- GMAT: Math 51
- GMAT: Math 52
- GMAT: Math 53
- GMAT: Math 54

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# GMAT: Math 20

106-109, pg. 166. Created by Sal Khan.

## Want to join the conversation?

- q.108;

i = true

ii = true

iii = true --> I agree with Sals logic, unless x = y, then we have a integer of 1 and iii = false.(4 votes)- @4:06, the question specifies that they have to be
**different**prime numbers, so you are safe with iii = true(1 vote)

- A gas station sells regular gas for $2.15 per gallon and premium gas for $2.65 a gallon. At the end of a business day 340 gallons of gas were sold, and receipts totaled $776. How many gallons of each type of gas were sold?(0 votes)
- 106. Was trying to reverse the equation to check my work but ran into a problem:

If there are 15 teachers at the end, there must be 375 students.

If the 'added' students and teacher are subtracted, there are 10 teacher and 325 students.

The ratio is 32.5/1, how did I mess this up?(0 votes)

## Video transcript

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.