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## Rozwiązywanie problemów — film z polskimi napisami

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# Matematyka GMAT: 34

## Transkrypcja filmu video

We're on problem 173. The probability is 1/2 that a
certain coin will turn up heads on any given toss. Fair enough. If the coin is to be tossed
three times, what is the probability that on at least
one of the tosses the coin will turn up tails? So probability, at
least one tails. That might seem really
hard for you. How do I know which tails
it is and all of that? But let's think about the
opposite circumstance. This is equivalent to saying
what is the probability that I don't get all heads? These are equivalent things. At least one tails means that
I don't get all heads. What is the probability that
I don't get all heads? Well, that's 1 minus
the probability that I get all heads. Figuring out the probability
that you get all heads isn't that hard. Because you have 1/2 probability
of heads on the first time, times 1/2
probability because you have to get a heads on the second
time, times 1/2 probability because you have to get a
heads on the third time. That's equal to 1/8. So 1 minus 1/8 is equal
to, 8/8 minus 1/8, which is equal to 7/8. So that is choice D. A lot of these probability
questions, you have to get at least 1 of something. If you think of it that
way, it's very hard. You'll have to start thinking of
combinatorics and binomial coefficients and all of that. But if you just think of the
opposite problem, what would have to be true? That I don't get all heads. What's the probability that
I do get all heads? And then that's a much easier
thing to solve for. You just take that from 1, and
you get the probability that it doesn't happen. Next question, 174. Of the final grades received by
the students in a certain math course, 1/5 are A's, 1/4
are B's, 1/2 are C's, and the remaining 10 grades are D's. What's the number of the
students in the course? So if we know what fraction 10
is, then we know the number of students in the course. So, essentially, what do these
fractions add up to? So let's find a common
denominator. Looks like a common denominator
between 4, 5, and 2 is 20. So, 1/5 get A's. So that's four 20's plus 1/4. So, that's five 20's
plus one 1/2. That's ten 20's. So the A's, B's, and C's are
equal to-- what is this? 4 plus 5 is 9 plus 10 is equal
to 19/20 of the class. They're saying that the
remaining 10 grades are D's. So there are no F's
in this class. So the remaining grades, that's
1/20 of the class. So that's equal to 1/20. Right? Because if the A's, B's and
C's are 19/20, and all the other grades are D's, then all
the other grades are going to be 1/20 of the class because
that's what's left over. So if I say 1/20 of the class is
equal to 10, multiply both sides by 20. You get the class is equal
to 200 students. That's choice D. Next question, 175. As x increases from 165 to 166,
which of the following must increase? Interesting. So 1, 2x minus 5. Well sure, as I go from 2 times
165 to 2 times 166, this is obviously going to increase,
and this is going to stay the same. So this is going to increase. You can substitute the numbers
in there if you don't believe me, but I think that should
make some intuition. Statement 2, 1 minus 1 over x. So, 1/165 is greater
than 1/166. This is a larger fraction
than that is. 1 over 1 million is much
smaller than 1/2. So 1/65 is greater than 1/66. So you might say, as I go from
165 to 166, this term right here gets smaller. Right? That term does get smaller. But we're subtracting
that term. So if we're subtracting a term
that's getting smaller, the whole value is actually
going to get larger. We're subtracting a smaller
number as we go to 166. 1 minus 1/166 is a larger number
than 1 minus 1/165. So this, also, gets larger. Statement 3, 1 over
x squared minus x. So the question is, what happens
to x squared minus x as we go from 165 to 166? If you think about it, we are
well beyond any type of x squared, any kind of
vertex in this. If you know a little bit of
calculus, you could figure out the vertex really fast. So, this
is going to be some type of a parabola. This x squared minus
x is going to look something like this. Right? Its vertex is actually-- 2x
minus 1 is equal to 0-- its vertex is at x is
equal to 1/2. It's a very small number. So we're in the point
of this curve that's increasing upwards. Right? So when we go from 165 to 166,
this right here is definitely increasing. This denominator is definitely
going to increase from 165 to 166. You might say, we're
going to be subtracting a larger number. But that's more than offset by
squaring the larger number. So this denominator is
going to increase. But if the denominator is
increasing, then the whole fraction, which is the inverse
of the denominator, that's going to decrease. So this one is not increasing. So, 1 and 2 only. So that's C. Next question, 176. I'll do it in a different
color. A rectangular box is 10 inches
wide, 10 inches long, and 5 inches high. So, it's 10 inches, 10 inches,
and then 5 high. Fair enough. That's the box. What is the greatest possible
distance in inches between any two points on the box? So, this is a bit
of an old trick. So, the longest distance in
this box is kind of a three-dimensional diagonal. So, a normal diagonal goes
along one surface. But if you go from this point
back here-- and I'll do that in a different color-- to this
point over here, that's the longest dimension in the box. If you know how to figure out
distance in three dimensions, then that's all you really
need to know how to do. It's going to be the square
root of 10 squared plus 10 squared plus 5 squared. So that's the fast
way to do it. I'll show you the logic
of it in a second. So, 10 squared plus 10 squared
plus 5 squared. If we had a fourth dimension,
which is hard to visualize, we would at that distance squared
and take the square root of everything. So this is equal to the
square root of what? 100? 225? 225, which is equal to 15. That's choice A. Let me give you the logic. So if I'm taking the distance
from this point to this point, how do we figure that out? Well let's see, the easiest way
to think about it is can we figure out this distance
right here? So this is going along
the top of the box. Let me do that in the magenta. Well sure, if we look at the box
straight down, what I just drew-- if this is 10 and
10-- what I just drew goes like that. Right? It's going along the
top of the box. So, this is going to be the
square root of 100 squared, the square root of 10
squared plus the square root of 10 squared. So, this just the Pythagorean
theorem. This distance right here is
the square root of 200. I just used the Pythagorean
theorem. I just said this is 10, this is
10, so this has to be the square root of 200. Now we have to do a little
bit of visualization. To figure out that green line
that's the longest diagonal in this box, think about
it this way. We still have a right triangle
although maybe it's a little harder to visualize. Well, maybe it isn't. So, the right triangle looks
like this, where the side is the square root of 200,
that's this up here. This side right here is 5. They tell us that. So the side right here is 5. Then the diagonal that
we're trying to solve for is in green. That's the longest diagonal. Remember this line is the same
thing, if you view from the top, as this line. It's just that you can't
see what's below it. So here we just use the
Pythagorean theorem. So, we could say c squared--
let's call this c-- is equal to 25 plus this squared
plus 200. Then you get to the same answer
that we just got to. So c is equal to the
square root of 225, which is equal to 15. What problem was I on? That was choice A if I haven't
already said it. Anyway. See you in the next video.