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Current time:0:00Total duration:10:46

We're on problem 129. If n is a positive integer, is
the value of b minus a at least-- OK, so let's see. So n is positive. n positive integer. And they're saying, is the value
of b minus a at least twice the value? So that's greater than or equal
to twice the value of 3 to the n minus 2 to the n. That's what we need to test. Statement 1 tells us, a is equal
to 2 to the n plus 1. And b-- this is still statement
1-- is equal to 3 to the n plus 1. So let's think about
this a little bit. So let's substitute this
into this equation. So you get 3 to the n plus 1
minus 2 to the n plus 1 is greater than or equal to-- and
let's multiply this out. 2 times 3 to the n minus
2 times 2 to the n. So what's 2 times 2 to the n? Well, that's just 2
to the n plus 1. 2 times 2 to the n is 2
to the n plus 1, and there's a minus sign. That's where I got it from. So interesting. Just like that, we have a minus
2 to the n plus 1 on both sides. So we can just add 2 to the n
plus 1 from both sides to cancel them out. So we get rid of that,
get rid of that. And then our statement boils
down to, is 3 to the n plus 1 greater than or equal to
2 times 3 to the n? We could rewrite 3 to the n plus
1 as 3 times 3 to the n. And we're testing whether
that's greater than 2 times 3 to the n. And they tell us that n
is a positive number. But that actually
doesn't matter. Because we have 3 to the
n on both sides. Even if you rose this to a
negative exponent, you would still have a positive number. So you could divide both sides
by 3 to the n and the statement boils down to 3 is
greater than 2, which is absolutely correct. So statement 1 is enough
information to say whether this statement is true or not. And it actually proves
that it's true. Let's look at statement
number two. Statement number two tells
us that n is equal to 3. So let's see. If we look at the original
statement, if n is equal to 3, then we have to say-- This is
what the statement reduces to. It reduces to, b minus a is
greater than or equal to 2 times 3 to the third power. So that's 27. Minus 2 to the third power. Well, 2 to the third is 8. So it reduces to, b minus a is
greater than or equal to-- 27 minus 8 is 19. Times 2 is 38. Well, I don't know. They don't tell us
anything else about b, or a, or anything. So this is a useless
statement. And actually, going through
this was a little bit of a waste of time. You could have just said, hey,
they told me what n is. I can come up with a number
here, but that still doesn't give me any information
over what b or a is. So statement 1, alone,
is sufficient to answer this question. And 2 is not. Let me go to sky blue. Problem 130. The inflation index for the
year 1989, relative to the year 1970, was 3.56. Indicating that, on average, for
each dollar spent in 1970 for goods, 3.56 had to
be spent for the same goods in 1989. Let me write that down. 1970, $1.00. In 1989, that same thing
would cost $3.59. Oh, $3.56. I'm getting my 6's
and 9's confused. $3.56 had to be spent. If the price of a model K mixer
increased precisely according to the inflation
index, what was the price of the mixer in 1970? So we want to know the
mixer K in 1970. So statement number one. The price of the model K
mixer was $102.4 more in 1989 than 1970. So let's call this, K. We're trying to figure
out K 70. That's what we're trying
to figure out. Mixer K in 1970. So they say the price of the
model K mixer was $102.40 more in 1989 than 1970. So this, essentially, tells us
that mixer K in 1989, minus mixer K in 1970, is
equal to $102.40. But we also have other
information. We know that mixer K, it went
exactly with inflation, the increase in inflation price. So mixer K in 1989, is equal to
3.56 times mixer K in 1970. Well, how do we know that? Because they told us that mixer
K, the price of it, increased directly with
this inflation. Pretty much anything that went
with inflation was worth 3.56 times more in 1989. So mixer K in '89 is
3.56 times more than it was in 1970. So now, we have this linear
equation, which was given in the problem statement. And we have this linear
equation. So we have two equations
and two unknowns. So we actually have enough
information just with statement 1 to solve
the problem. And I'm actually not even going
to go into it because it would be a waste of time. Hopefully, we're getting into
the rhythm of this. Two linear equations, two
unknowns, enough information. If they're quadratic or
something, then you'd have to dig a little bit deeper. Or if they're somehow parallel
equations, but I wouldn't worry about that too
much on the GMAT. Statement number two. The price of the model K mixer
was $142.40 in 1989. So they're telling us that
K in 1989 was 142.40 Well, this is definitely enough
information, because we already have this equation. The '89 price, which is 142.40,
is equal to 3.56 times the 1970 price. So you just divide this by 3.56
and you have your answer. So each statement,
independently, is enough to solve this problem. Problem 131. Is 5 to the k less
than a thousand? Statement number one tells us
that 5 to the k plus 1 is greater than 3,000. I already have a sense this
isn't going to help us, because it's essentially setting
a lower bound on 5. So let me just explain it. So 5 to the k plus 1. That's the same thing as
5 to the k times 5. So that's equal to 5
times 5 to the k. You could view this as
5 to the first power. Then, you would add the
exponents to get here. And that, they say, is
greater than 3,000. You divide both sides by 5 and
you get 5 to the k-- that's a k-- is greater than 600. That doesn't help us. 5 to the k could be 6,000. Well, 5 to the k could
be 700, in which case this would be true. 5 to the k could be 7,000. In which case, this
wouldn't be true. So statement 1, by itself,
isn't enough information. Statement number two tells us, 5
to the k minus 1 is equal to 5 to the k minus 500. This is interesting. Let's see what we can do here. Let's get all the
k's on one side. So this is the same thing as 5
to the k minus 1 minus 5 to the k is equal to minus 500. Well, this is the same
thing as 1/5, or 5 to the negative 1. Let me write that. 5 to the negative 1
times 5 to the k. Then, minus 5 to the k is
equal to minus 500. Let's factor 5 to the
k out of this. So you get 5 to the
k times-- 5 to the negative 1, that's 1/5. Minus 1-- we factored a 5
to the k out there-- is equal to minus 500. And let's see what we can do. 5 to the k. 1/5 minus 1. That's minus 4/5, right? Times minus 4/5 is equal
to minus 500. We can get rid of the minus
sign on both sides. Multiply both sides
by negative 1. And then, we can multiply
both sides by 5/4. So we get 5 to the k is equal
to 500 times what? 5/4. And that is what? That's 125. 2,500 divided by 4. 2,500 divided by 4. I want to say it's 475. Is that-- 5/4. No. So each fourth is 125 and
you have another. So it's equal to 625. Even if I just got my math
wrong, that doesn't matter. Because they just want to know
whether 5 to the k is less than a thousand. And it definitely is. Even if my math-- 500 times
5/4-- 1/4 of 500 is 125. 125 times 5. So we'll have one
more of those. 625. So statement 2, alone,
is sufficient to answer the question. Statement 1 was kind
of useless. I've run out of time. See you in the next video.