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## Rozwiązywanie problemów — film z polskimi napisami

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# Matematyka GMAT: 6

## Transkrypcja filmu video

We're on problem 30. Let's see, they've drawn
a circle there. If o is the center of the circle
above, what fraction of the circle is shaded? Let me draw that. They draw a circle. And then they have two lines
that go through the center, essentially. And one line looks like that. Another line looks something
like that. So they tell us that this
is the center, o. They tell us this
is 150 degrees. So the question is, what
fraction of the circular region is shaded? And the shaded region is this. This is the shaded region. So we just have to figure out
what percentage of the total degrees of the circle
are shaded. So let's think about
it this way. If this is 150 degrees, what
is this angle right here? Well, this angle and
150 degrees are supplementary, right? So this angle plus this angle
is going to be equal to 180. So 30 plus 150 is
equal to 180. And then you could either say,
oh, these are opposite angles, so they're equal. Or you can make the same
argument here, that 150 plus this angle has to be equal to
180, so this angle is also equal to 30. So what is the total angle
measure of the shaded region? You have 30 degrees here. And you have 30 degrees here. So a total of 60 degrees. And the whole circle is
how many degrees? If you run it all the way around
the circle, how many degrees is that? Well, that's 360 degrees. So 60/360. That's 1/6 of the circle
is shaded. And that is choice C. Question 31. This was question 30. Question 31. If Juan takes 11 seconds to run
y yards-- so y yards for every 11 seconds. Well, another way you could
think about it is, y/11 yards per second. Fair enough. How many seconds will
it take him to run x yards at the same rate? So he is going at y
yards per second. So how long will it take
him to go x yards? Well, if I said that I'm going
at 5 miles per hour, how long will it take me to
go 10 miles? Let's say I go 5
miles per hour. Or let's say 5 meters
per hour. How long will take me
to go 10 meters? You would say 10 divided
by 5 is equal to 2. 2 hours. So likewise, we have
his speed. That's his speed we're talking
about, how fast he runs. His speed is y/11 yards
per second. And we want to know how
long will it take him to run x yards. So we could just take x divided
by the speed, divided by y/11, and we'll get the
answer in seconds. And you could work out the
units, but x divided by y/11, that's the same thing as x
times 11/y, equals 11x/y. Another way to view it is this
is equal to the rate. And we know, from probably the
first day in physics, that distance is equal to
rate times time. This might be an easier way
for you to think about it. However it best forms
connections in your brain. Distance is equal to
rate times time. They tell us that the distance
is x, and that the rate is y/11 yards per second. So y/11 times t is equal to x. So if you want to find t, you
do x divided by y/11. Anyway, and you get 11x/y,
which is choice A. Question 32. John has 10 pairs of
matched socks. If he loses 7 individual socks,
what is the greatest number of pairs of matched
socks he can have left? So he loses 7. He loses 7 socks. So think about it. If you wanted to optimize the
number of matched pairs, what's the worst
case scenario? The worst case scenario is if
every one of these socks came out of a different pair. Because then he'd be left with 7
socks that had their brother socks, or whatever you want
to call it, their twin socks are gone. So that's the worst case. But we want to think of what
the best case-- what's the best case if you want to
preserve the number of pairs you have left? If you want to preserve the
number of pairs you have left, the 7 should have as many
pairs as possible in it. So if 7 could be made of 3
pairs and 1 lone socks. This is kind of the best case
scenario if he wants to have the greatest number of
matched socks left. So 7 could be 3 pairs and 1 lone
socks, in which case he'd be left with what? So these 3 pairs
would be gone. Remember, he has 10 pairs. Let's keep in mind, those
aren't 10 socks. He has 10 pairs. Let's say that he just
loses 3 pairs. Then he'll just have 7
pairs left, right? But then if he loses 1 more
sock, then he's only going to have 6 pairs left. He's going to have 6
pairs and one sock. So he's going to have 6 matched
pairs left, and that's choice B. Question 33. The worst case, if the question
was the other way-- maybe I'm just overly
fascinated with this question-- but if the question
was the other way, what's the worst case if-- what's the
minimum number of pairs that he has left? You could say, well, each of
these socks exactly destroy 1 pair, so losing 7 socks
destroys 7 pairs. So at the worst case, if each of
these were a different sock from a different pair, he
would have 3 pairs left. But anyway, that's not
what they ask, so I shouldn't focus on it. 23. What is the lowest positive
integer that is divisible by each of the numbers 1 through
7, inclusive. 1 through 7, including
1 and 7. So let's think of
a number that's divisible by all of these. So it has to be divisible
by 7, right? 7's a prime number, so it has
to be divisible by 7. It has to be divisible by 6. And 7 and 6 share no common
factor, so the number is going to be at least 7 times 6. It has to be divisible by 5,
because 5 is a prime number, and because that doesn't
share any factor with any of these numbers. It has to be divisible
by 4, right? Think about it. 6 is 2 times 3. So to make it divisible by 4,
you only have to multiply it by another 2. So it has 2 times 2 in its
prime factorization. Maybe that's how I should
be writing it. It has to be divisible by 7. It has to be divisible by 6,
which is the same way as saying it has to be divisible
by 2 times 3. It has to be divisible by 5. It has to be divisible by 4. But we already have one
of 4's factors here? It's 2, so we just have to
multiply it by another 2. So this is to say it's
divisible by 4. It has to be divisible by 3. Well, it's already
divisible by 3. It has to be divisible by 2. It's already divisible by
2, actually, twice. It has to be divisible by 1, and
that's already the case. So this is the number. And notice, I was very
careful not to have extra prime factors. If I just did 7 times 6, times
5, times 4, times 3, times 2, times 1, I would have gotten a
number that is divisible by all of them, but it wouldn't
have been the smallest one. So anyway, what is
this number? Let's do the small numbers
first. Well, I'll do the big number first. 7 times 5 is 35. 2 times 3 is 6, times 2 is 12. So 35 times 12. 35 times 2 is 70-- 0,
1 times 35-- is 350. And then I have 0. 7 plus 5 is 12. 420. That is choice A. And you can feel good about
it, because it's also the lowest number on their list. Next question. 34. What percentage of 30 is 12? So 12 is equal to some
decimal times 30. And we could convert
that decimal, eventually, to a percent. We just multiply it by 100. So x is equal to 30-- sorry,
is equal to 12/30, which is the same thing as 4/10,
which equals 0.4. And if you wanted to write a
decimal as a percent, you just multiply it by 100. That is equal to 40%. And that is choice D. And these can be confusing, so
it's very important to write it that way. What percentage of 30 is 12? So 12 is what percentage
of 30? It's going to be a
smaller number. You can kind of think, 12
is what fraction of 30? Or you could say 12/30 equals
what as a percentage? It equals x percent. You could think of
it that way, too. It's all the same thing. The important thing is
to be able to parse the language correctly. Anyway. 35. If 1.5 over 0.2 plus x is equal
to 5, then x is equal to-- let's just solve it. Multiply both sides of this
equation by 0.2 plus x. On the left-hand side,
it cancels out. We just have 1.5. On the right-hand side,
what's 5 times 0.2? It's 1, right? 0.2 is the same thing
as 20% or 1/5. So it's 5 times 1/5. That's 1-- you can think of it
a bunch of ways-- plus 5x. Subtract 1 from both sides. You have 0.5-- 1.5 minus 1
is 0.5-- is equal to 5x. So x is equal to 0.5
divided by 5. Well, that's just
equal to 0.1. The answer is B. See you in the next video.