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Current time:0:00Total duration:11:05

We're on problem 80. If a, b, and c are integers,
is a minus b, plus c greater-- OK. So what they want to know is a
minus b, plus c greater than a plus b, minus c? Fair enough. And if you think about it,
there's an a on both sides of this equation. So what they're really wanting
to know is minus b plus c greater than b minus c? Or another way, they're just
trying to say-- I just took a from both sides of this
inequality, so it doesn't change anything. Really what they're asking, if
you just rearrange this, is c minus b greater than
b minus c? Which is another way of saying,
well, is one of them greater than the other one? Right? If c is greater than b,
this is going to be greater than this one. If b is greater than c,
this is going to be greater than this one. So let's look at our
statements now. Statement number one. b is negative. Well, that by itself doesn't
help us much because b could be negative, but c could
be even more negative. So c could still be smaller
than b, right? If b could be minus 5,
c could be minus 6. Or c could be plus 6. So it really tells us nothing
about c relative to b. Statement two says
c is positive. So this statement by itself,
once again, it tells us nothing about the value
of b relative to c. If c is positive, maybe
b is positive as well. Maybe b is more positive. Or maybe b is negative. Who knows? But this by itself
doesn't help us. But if we use both of these
together, if c is positive and b is negative, then we know, if
this is positive, this is negative, we know this is
going to be positive. And we know that this is
going to be negative. Essentially, we know that c is
larger than b, so this will be positive and this will
be negative. And we can answer
our question. If c is larger than
b, this is true. The left-hand side is greater
than the right-hand side, or this holds good. So we need both of the
statements together. Both statements together
are sufficient to answer this question. Switch colors. 81. If a certain animated cartoon
consists of a total of 17,280 frames on film, how many minutes
will it take to run the cartoon? So they want to know
how many minutes. So essentially, if we knew how
fast the frames have to go through-- if it's x frames per
second-- if we knew that number, we'd be able to figure
out how long the cartoon runs. So statement one is, the
cartoon runs without interruption at a rate of
24 frames per second. Well, that's exactly what
I just asked for. 24 frames per second. So using just that information,
how many seconds does it run? Well, it runs 17,280
frames divided by 24 frames per second. This will give you seconds, and
then if you divide this number by 60, you'll
get minutes. Divide seconds by 60,
you get minutes. So this is the answer. So one alone was enough to
answer the question. What does statement
two give us? I'll write it here. Statement two. It takes 6 times as long to run
the cartoon as it takes to rewind the film. And it takes a total of
14 minutes to do both. 6 times as long to run the
cartoon as it takes to rewind. Let's say 6 times to play. I want different letters. So to play the cartoon takes
6 times as long as to rewind the cartoon. And they say it takes a total
of 14 minutes to do both. So they say play plus rewind
is equal to 14. And what we want to know
is, how long does it take to play the movie? And we don't even have to use
that information alone, the 17,280, but just looking at
this, we have two linear equations and two unknowns, so
we can just solved for P. P is the playtime
of this movie. And if you wanted to figure it
out-- not that you would have to, because we would be
done at this point. We would say oh, one and two
are each independently sufficient to answer
this question. But just to show you that
we could figure it out. Say R is equal to 14 minus P. Then you substitute
that into that. You get P is equal
to 6 times R. So 6 times 14 is 60, plus 24. 84 minus 6P. You get 7P is equal to 84. P is equal to 12 minutes. And we'd be done. But actually, we didn't have
to do all of that. All of that is a
waste of time. I just want to prove it to you. 82. A box contains only red chips,
white chips, and blue chips. If a chip is randomly selected
from the box, what is the probability that the chip will
either be white or blue? So essentially, to answer this,
we just have to know what proportion of the chips in
the bag are white or blue. All right. So statement one says, the
probability that the chip will be blue is 1/5. So probability of blue
is equal to 1/5. That also tells us that 1/5 of
the chips in the bag are blue. But we don't know what
proportion are white, so we still can't answer that
top question. Statement two tells us, the
probably that the chip will be red is 1/3. The probability of red
is equal to 1/3. So, once again, just this
alone-- actually, just this alone is fine, right? I correct myself. The probability of getting
white or blue, that's probability of white-- we could
write, do set notation, white or blue-- that is
just equal to what? It's 1 minus the probability
that we get a red. Think about that. Either this is going to be
true, I'm going to pick a white or blue, or I'm
going to pick a red. Because there's only red, white,
and blue in the bag. So the probability of white or
blue is equal to 1 minus the probability of red, which is
equal to 1 minus 1/3, which is equal to 2/3. So statement two alone
is sufficient. And then if we have statement
number two, statement one alone is not sufficient. 83. See, this is tricky because, I
think, at least when I first read it, I was like, oh, I could
use this and this to figure out the probably
of getting a white. So I need both statements. But we don't even need to know
the probability of a white. We just need to know the
probably a white or a blue, which is just 1 minus the
probability of red. So statement two alone
gets us there. Let me do the next problem. 83. If the successive tick marks on
the number line are equally spaced, and if x and y are the
numbers designating the endpoints of the interval as
shown, what is the value of y? OK, so let me see if I can draw
what they have drawn. So they have a number line. And I probably have to draw the
exact same number of ticks as they did. So they have 0. And then let's see, there's 1
tick, 2 ticks, and then they have a third tick. And they say this is x. And then they have three more. 1, 2, 3. And then they have y. And then they have one more. And this keeps going
on and on. And they ask us, if the ticks
are equally spaced, and if x and y are the numbers designated
at the endpoints of intervals as shown, what
is the value of y? So they're equally spaced, but
they're not telling us that these are integers. We don't know that this is
1, this is 2, this is 3. We don't know that. But what we do know is y is
how much further than x? So let's see, x is
1, 2, 3 spaces. So we could say x is equal to 3
times, let's say, s, where s is the length of each interval,
or each tick. And y is 1, 2, 3, 4, 5, 6, 7. So y is equal to 7s. And so we could use this
to come up with a relation with x and y. This is before they've
done anything. This is just looking
at the diagram. So we could say that
s is equal to y/7. Or if we substitute this back
here, we could say that x is equal 3/7 y. And actually, you didn't have
to even do this s stuff. You could have just said, oh,
well, x is 3/7 as far as y is from 0. Because these are
equally spaced. So just the problem setup, just
looking at it, gives us that information. But we want to know what the
value of y is, and that by itself isn't enough. What does statement
one tell us? x is equal to 1/2. Well, that's enough
to figure out y. If x is equal to 1/2, you
just substitute there. 1/2 is equal to 3/7 y. Multiply both sides by 7/3. You get y is equal to what? 7/6. And you're done. You don't have to solve that. You just know that I have
this, I have that. I can solve it. So one alone is sufficient. Statement two tells us, y
minus x is equal to 2/3. So once again, we
have two linear equations with two unknowns. We can just substitute
and solve them. And just to prove it to you,
x is equal to 3/7 y. I can substitute that
right there. So I have y minus 3/7
y is equal to 2/3. So 7/7 minus 3/7. So that's just to go here. 7/7 y, so that's 4/7
y is equal to 2/3. y is equal to 2/3 times 7/4. Which is-- that's 1, this is 2--
which is 7/6 once again. See you in the next video. So once again, each of these are
independently sufficient to answer the question. See you soon.