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Current time:0:00Total duration:11:03

We're on problem 68. And they want to know, what
is the average arithmetic mean of j and k? So essentially, if we knew what
they were, you could take j plus k, and divide by 2, and
you'd know their average. What is statement one? Statement one tells us the
average of j plus 2, and k plus 4, is 11. So this is interesting. So means that j plus 2, plus
k plus 4, over 2-- I'm just averaging the two numbers--
they're telling us that that is equal to 11. And remember, the question
they're asking is the average of j and k. So if we could just figure out
what j plus k, over 2 is equal to, we're done. Or, if we just knew what j
plus k is, we were done. Maybe we could figure
that out from this. Let's see if we can simplify. This simplifies to j plus
k, plus 6, over 2, is equal to 11. This simplifies to j plus k,
over 2, plus 3, right? So plus 6 over 2. I'm just taking the
6 over 2 out. So plus 6 over 2,
is equal to 11. So then we get j plus k,
over 2 is equal to 8. And we're done. The average of j and k is 8. We're done. Statement one alone
is sufficient. Let's see what statement
number two does for us. Statement number two says the
average of jk and 14 is 10. I suspect we're going
to be able to do the exact same thing. So the average of j plus k,
plus 14-- so now we're averaging three numbers--
is equal to 10. And here, we should be
able to figure out what j plus k is again. We don't have a convenient 2 in
the denominator anymore, so let's just solve for j plus k. So then you get j plus k, plus
14-- multiply both sides by 3-- is equal to 30. And then you get j plus
k is equal to what? Subtract 14 from both sides,
is equal to 16. And then if you wanted to figure
out the average of the two, you just divide
both sides by 2. And you get j plus k over
2 is equal to 8. So each statement alone
was sufficient to solve this problem. Next problem. I feel a sneeze coming on. This is not happening. All right. Where was I? Problem 69. Let me move the scroll
bar up all the way. Paula and Sandy were among those
people who sold raffle tickets to raise money
for club x. If Paula and Sandy sold a total
of 100 of the tickets, how many of the tickets
did Paula sell? OK, so essentially they're
telling us Paula plus Sandy sold 100 tickets. And they want us to figure out
how many did Paula sell? Where P is the number Paula
sold, and S is the number Sandy sold. Problem number one. Sandy sold 2/3 as many of the
raffle tickets as Paula did. So Sandy is equal to
2/3 times Paula. Well, this alone
is sufficient. We have one equation of two
unknowns, and now we have another equation of the
same two unknowns. These are both linear
equations. So we have two equations
of two unknowns. We can easily now solve
for S and P. And maybe I'll do it, but you
should just already recognize that this is sufficient
if you're on the GMAT. And we'll solve it, just
to do it after this. Sandy sold 8% of all
the raffle tickets sold for club x. All right. Now, this is a little
different. This is Sandy is equal to 8%
of not all the tickets that Sandy and Paula sold,
she sold 8% of the total that club x sold. So this is a different number,
because there could have been-- I might have been in club
x selling raffle tickets. So we don't know how
many I sold. So we don't know what this
total number is. So there's not much we can do
with it, because there's no way for us to figure out
this total number. So this is not that useful. So statement one alone
is sufficient. And just to prove the point to
you-- we're trying to figure out what Paula sold. So let's just substitute
this back in. So you have P plus--
for S I'll write 2/3P-- is equal to 100. And so this is 5/3P
is equal to 100. Just added 1 plus 2/3,
or 3/3 plus 2/3. And then you have P is equal
to 3/5 times 100, which is equal to 60. So that's how many she sold. So we would definitely be able
figure it out, but this would've been a waste of
time if you were taking the GMAT for real. You should have just recognized
two linear equations and two unknowns. I'm done. Next problem, 70. Is ax equal to 3 minus bx? Who knows? [? Problem ?] statement number one. Let me scroll down
a little bit. Statement number one tells
us, x times a, plus b is equal to 3. Well, that's essentially the
same thing as this top equation, right? Let me show you. x times a, that's ax, plus
bx is equal to 3. And then subtract bx
from both sides. You get ax is equal to 3 minus
bx, which is exactly what we were trying to prove. So if this is true, then this
is definitely true. Statement one alone
is sufficient. Statement two tells us, a equals
b, equals 1.5, and x is equal to 1. Well, let's see if
this is true. So a is-- you have
1.5 times 1. So you have 1.5 is equal to
3, minus 1.5 times 1. So 3 minus 1.5. Well, this is true. 1.5 is equal to 3 minus
1.5, is 1.5. So two alone is also
sufficient. So the answer is D. Each statement alone
is sufficient to solve this problem. 71. Let me switch colors. A number of people each wrote
down one of the first 30 positive integers. Were any of the integers written
down by more than one of the people? All right. That's interesting. So are there any repeats? A number of people. So clearly if we had more than
30 people and they all had to pick one of the first 30
integers, then you're going to have repeats. But let's see what
they tell us. The number of people who wrote
down an integer was greater than 40. Well, there you go. So 40 people each have to pick
a number between-- one of the first 30 positive
integers, right? So they only have a pool
of 30 to pick from. So if 40 people have to pick
numbers from a pool of 30 numbers, there's definitely
going to be repeats, right? So even if the first 30 people
all picked different numbers, which we cannot guarantee by
any means, the next 10 are going to have to repeat
with somebody. Because all of the first 30
would've already been picked. And you could very easily
have even more repeats. So statement number one
is very sufficient. Statement number two. The number of people who
wrote down an integer was less than 70. Well, that's useless. I mean, by this statement, maybe
only one person wrote down an integer, right? And if only one person wrote
down an integer, we definitely don't have any repeats because
there's no one to repeat with. So this is a useless
statement. So statement one alone is
sufficient and statement two is useless. 73. I think there should be a third
option, where you can rate the degree of uselessness
of a statement. Some of them borderline on
almost being useful, some of them are ridiculous,
like this one. OK, 73. 72. In the figure above, is
CD greater than BC? So they drew us a number
line, or whatever you want to call it. Some type of a line. Let me do another color. This end we have A. Then we have B. Then we have C. And then we have D. In the figure above, is
CD-- so this is CD-- greater than BC? So they want to know is this
greater than this? So is CD greater than BC? All right. So number one, they tell us
that AD is equal to 20. So this whole thing
is equal to 20. So that doesn't help me much. Statement number one tells
us AD is equal to 20. It just tells me the
whole length. Maybe I can get some
more information. Statement number two. AB is equal to CD. So this is interesting. They're telling us that
this is equal to this. But this still doesn't
help us, right? I can imagine a situation where
both of these are-- let's say that both
of these are 5. That could be 5, that
could be 5. And then this would
be 10, right? 5 plus 10, plus 5. In which case, CD would not
be greater than BC, right? So this would be a situation
that meets all of the conditions where CD would
not be greater than BC. But then I can construct one
where it is greater than BC. I can make CD equal to--
what if that was 8? And then this one also
has to be 8. And then we would only
have 4 left. Where CD is greater than BC. So in this case, both statements
together are still not sufficient. So the answer is E. And I've run out of time. I'll see in the next video.