# GMAT: MathÂ 23

## Video transcript

We're on problem 120. David has d books, which is 3
times as many as Jeff, and 1/2 as many as Paula. So d is equal to 1/2 the number
of books Paula has. How many books do the
three of them have together in terms of d? So they want to know how
much d plus j plus p is, in terms of d. So what's j in terms of d? j is equal to-- divide both
sides by 3-- is equal to d/3. And what's p in terms of d? If we multiply both sides by 2,
you get p is equal to 2d. So this turns into d plus
j, which is d/3, plus p, which is 2d. Let's find the common
denominator. Well, 3. 1d is the same thing
as 3d over 3. d over 3, well, that's still
just going to be d/3. 2d, that's the same thing
is 6d over 3. That's equal to 3, 4, plus 6 is
10d over 3, or 10 thirds d. And that's choice C. 10 thirds d. Problem 121. There are 8 teams in a certain
league, and each team plays each of the other teams
exactly once. If each game is played by 2
teams-- it's a traditional game-- what is the total
number of games played? Let's think about
it a little bit. They each play each other
exactly once. So this turns into that grid
problem that we did last time. If I have 8 teams, 1,
2, 3, 4, 5, 6, 7, 8. And I'm going to write
the same thing. 1, 2, 3, 4, 5, 6, 7, 8. They can't play each other. 1 is never going to play 1. 2 is never going to play 2. 3 is never going to play 3. 4 is never going to play
4, et cetera, all the way down the diagonal. You can't play yourself. And 1 is going to play
2 exactly 1 game. That's 1 game. And we're not going to write
2 played 1, because 1 already played 2. They only play each
other once. So you don't want to count
this one down here. You don't want to
count this one. This is when 1 plays 3. This is when 2 plays 3. So, essentially, you want to
figure out all of these spots that are above the diagonal. Let's think about how many
spots there are. If we have an 8 by 8 square,
there are 8 times 8 squares on this grid, if I actually
drew them as squares. I did that a couple
of problems ago. If I actually drew it like that,
then you might see the grid a little bit better. But I think you get the idea. You have 8 by 8 spots on this
grid, because it's 8 spots high, 8 spots wide. So you have 64 spots. Now, you're not going to have
one of the games along the diagonal, because that would
be a team playing itself. So how many spots are
along the diagonal? There's 8. There's 1 against 1,
2 and 2, 3 and 3. So that's each team
playing itself. So there's 8 spots along
the diagonal. So 64 minus 8 is 56. And so 56 would give us the
number of squares that aren't in the diagonal. But if we counted all of them,
we would be double counting. We'd be counting 1 playing 2
up here, and 2 playing 1. So we only want to count half
of the remaining squares. So if you divide this
by 2, you get 28. So there'd be a total
of 28 games. And that's choice C. Question 122. An operation theta is defined
by the equation a theta b is equal to a minus b over a plus b
for all numbers a and b such that a does not equal minus b. That ensures that the
denominator does not equal 0. So they're defining
the domain. If a equaled minus b, then it
would be undefined, because you'd have something
divided by 0. If a does not equal negative c,
and a theta c is equal to 0, then c is equal to-- So
they're telling us right now that a does not equal minus c. So let's just write it down. a
theta c is equal to a minus c over a plus c. And then, that we want
to set equal to 0. Because they're telling us
a theta c is equal to 0. They're telling us that a
does not equal minus c. So that's essentially saying
if a was minus c, then the denominator would be 0,
and we would have something really strange. So they are telling us the
denominator does not equal 0. So if the denominator does not
equal 0, and this fraction equals 0, then the numerator
of this fraction has to be equal to 0. Or another way you could say it,
a minus c over a plus c is equal to 0. We know a plus c is
not equal to 0. So let's multiply both sides
of the equation times it. And you get a minus c is equal
to 0 times a plus c. That's just 0. In other words, if a fraction
is equal 0, the denominator is not 0. That would make it undefined. Then the numerator has
to be equal to 0. So then, just solving, you
get a is equal to c. Add c to both sides. So they want to know
what c is equal to. So c is equal to a. And that is choice E. Next question. Let me switch colors. 123. The price of a lunch for 15
people was $207, including a 15% gratuity for service. So that was plus 15%. What was the average
price per person, excluding the gratuity? So essentially, we want to
figure out the price of 15 before gratuity times-- so
this is after gratuity. So I'll call this, price
before gratuity. I don't want to get too
complicated with my notation-- Price before gratuity times
1.15, because we're going to increase by 15, is
equal to $207. So the price before gratuity of
the 15 people is equal to 207 divided by 1.15. And we want to figure out the
average price per person. So we want to figure out
this divided by 15. This looks, at its heart,
kind mathy. But let's just chug through it
and see where we can go. So this is the same thing
is 207/1.15 times 1/15. And that's equal to 207 divided
by 15 times 1.15, the denominator. Let's figure out the
denominator first. Well, maybe the numerator--
207 divided by 1.15. 1.15 goes into 207. Add some decimals. Push this two to the right,
two to the right. The decimal goes there. 115 goes into 207 one time. 1 times 115 is 115. Then, it's 85 plus 7. 85 plus 7 is 92. 920. And then then this should
go roughly 8 times. 8 times 115. 8 times 5 is 40. 8 times 1 is 8 plus 4 is 12. Carry the 1. 8 times 1 is 8, plus 1 is 9. It actually equals
exactly 920. Great. So before gratuity,
it was $180. Before gratuity, our
price was $180. And now, we have to divide
that by 15 to get the average price. So 15 goes into 180
how many times? 1 times 15, 30. 15 goes into 30 two times. So the average price before
gratuity was exactly $12, B. And I'm out of time. See you in the next video.