GMAT: Math 23

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.