-
We're on problem 42.
-
And they've drawn this little,
looks like a pie graph.
-
And they say, in the circular
region with center o shown
-
above, the two unshaded sections
constitute 3/7 and
-
1/3 of the area of the
circular region.
-
Fair enough.
-
The shaded section constitutes
what fractional part of the
-
area of the circular region?
-
So the shaded section is just
the whole area minus these two
-
fractions, right?
-
So if you said what fraction of
the whole area is the whole
-
area, you would say that's 1.
-
And you would subtract out these
two areas to get the
-
shaded area.
-
So 1 minus 3/7 minus 1/3 is
equal to the fraction of the
-
totally area that this
shaded area is.
-
And let's just add or subtract
those fractions.
-
The least common
multiple is 21.
-
1 is the same thing as 21/21.
-
Minus 3/7.
-
Let's see, 7 goes
into 21 3 times.
-
So 3 times 3 is 9.
-
So this is the same thing
as minus 9 over 21.
-
And minus 1/3 is the same thing
as minus 7 over 21.
-
So this is equal to 21
minus 16 over 21.
-
And that's 5/21, which
is choice D.
-
Next question.
-
0.3.
-
[SNEEZES]
-
Excuse me.
-
My apologies.
-
0.3 to the 5th over
0.3 to the 3rd.
-
Well, anything to the 5th
divided by anything to the
-
3rd, you could essentially say
divide the top and the bottom
-
by 0.3 squared.
-
Well, actually, you could divide
the top and the bottom
-
by 0.3 cubed.
-
You could say this is the same
thing as 0.3 to the 5th times
-
0.3 to the minus third.
-
That's just another
way of doing this.
-
And so if you're dividing these
two numbers, you would
-
subtract the exponents.
-
But now we're multiplying.
-
We're adding the exponents.
-
But either way, it becomes
0.3 squared.
-
And that is equal to 3
times 3, which is 9.
-
And you're going to have two
numbers behind the decimal
-
points, right?
-
0.3 times 0.3.
-
Two numbers behind the
decimal points.
-
1, 2.
-
So two numbers behind
the decimal point.
-
So 0.09.
-
Or another way of saying it
is 30% of 0.3 is 0.09.
-
And that is choice C.
-
44.
-
In a horticultural experiment--
this is sounding
-
interesting already-- 200 seeds
were planted in plot 1.
-
So plot 1 got 200 seeds.
-
And 300 were planted
in plot 2.
-
So plot 2 got 300 seeds.
-
If 57% of the seeds in plot 1
germinated, and 42% of the
-
seeds in plot 2 germinated, what
percentage of the total
-
number of planted seeds
germinated?
-
So the total number of planted,
what percent?
-
So how many total seeds
germinated is going to be 200
-
times 0.57.
-
That's how many in plot
1 germinated.
-
Plus 300 times 42%, or 0.42.
-
That's how many in plot
2 germinated.
-
All of that divided by 500.
-
Right?
-
And how do I know 500?
-
Because there were a
total of 500 seeds.
-
So just to simplify the math,
we could just divide
-
everything by 100, right
from the get go.
-
So if you divide the bottom by
100 and the top by 100, you
-
have to do both terms by 100.
-
So you get 2 times 0.57 plus
3 times 0.42 divided by 5.
-
2 times 0.57.
-
That is what?
-
Let's see, this is
1.14 plus 1.26.
-
Is that right?
-
3 times 4 is 12.
-
3 times 2 is 6.
-
Right?
-
1.26.
-
All of that over 5.
-
This becomes what?
-
This is equal to 2.4
divided by 5.
-
And so 5 goes into 2.4.
-
Let's see, goes into
4, 4 times 5 is 20.
-
48.
-
So 0.48.
-
So the answer is 48%
or 100 times 0.48.
-
And that's choice C.
-
Question 45.
-
Let's switch to a more
interesting color.
-
3 and 8 are the lengths of two
sides of a triangular region.
-
Which of the following can be
the length of the third side?
-
OK.
-
So let's think about
it a little bit.
-
3 and 8 are the lengths of two
sides of a triangular region.
-
Let me write their
choices down.
-
Choice one is 5.
-
I can easily imagine
a triangle that has
-
sides 3, 8, and 5.
-
That seems completely
reasonable.
-
I'm just experimenting.
-
I don't know where
this is going.
-
8.
-
Well, sure, that's just
an isosceles triangle.
-
You can easily have a triangle
that has 8, 8, and 3.
-
Choice three.
-
11.
-
Now, this is interesting.
-
Let me ask you a question.
-
Can I have a triangle that
looks like this?
-
11 and then 3 and then 8.
-
Is this possible?
-
Well, no, because
3 plus 8 is 11.
-
So the only way you're going to
get 11 is if you push this
-
side all the way flat.
-
That's the only way you're going
to get the length of
-
this third side to be 11.
-
In fact, 11 is the upper
bound on what this
-
third side could be.
-
Because imagine this.
-
Imagine if I made the triangle
really flat, I made this angle
-
right here really wide, as
close to 180 as I can.
-
And I make it really flat.
-
Right?
-
If this length plus this length,
or this length plus
-
this length is equal to 11,
this length is going to be
-
shorter than it.
-
This length right here has to
be shorter than this length
-
plus this length, right?
-
Because it's kind of a
straight-line distance between
-
this point and this point.
-
So 11 is the upper
bound, right?
-
The only way to get 11 is if
you completely flatten out
-
this triangle, at which point
that's not a triangle anymore.
-
It'll be a line.
-
So it can't be choice three.
-
So the only possibility.
-
They say, which of the following
can be the length of
-
the third side?
-
So it's only choices
one and two.
-
And that is choice C.
-
Next problem.
-
46.
-
How many integers n are there
such that 1 is less than 5n
-
plus 5, which is less than 25?
-
OK.
-
So they say how many integers n
are there so that 5n plus 5.
-
So they didn't say positive
integers, right?
-
So that's an interesting
thing to keep in mind.
-
So let's just try to simplify
this a little bit.
-
Let's subtract 5 from all
sides of this double
-
inequality.
-
So if you subtract 5 from
everything you get minus 4 is
-
less than 5n, which
is less than 20.
-
Right?
-
So another way you could say
it, is let's just divide
-
everything by 5.
-
So because 5 is positive you
don't have to change the
-
inequalities.
-
So you get minus 4/5 is less
than n, which is less than 20
-
divided by 5, is 4.
-
So now the question becomes
a lot simpler.
-
How many integers n are
there such that this?
-
How many integers are there
between minus 4/5 and 4?
-
And it's not equal
to any of those.
-
So 0 is an integer.
-
1, 2, and 3.
-
So there are 4 integers.
-
So that is B.
-
OK, next problem.
-
47.
-
A car dealer sold x used cars
and y new cars during May.
-
So number used is equal to x.
-
Number new is equal to y.
-
During May.
-
If the number of used cars sold
was 10 greater than the
-
number new cars, which of the
following expresses this
-
relationship?
-
So the number of used cars, x,
was 10 greater than the number
-
of new cars.
-
So it's 10 greater than y, so
it equals y plus 10, right?
-
This says that the number of
used cars is 10 more than the
-
number of new cars.
-
So we just have to look for
that. x is equal to y plus 10.
-
That's choice D.
-
I think we have time
for one more.
-
48.
-
If a 10% deposit that has been
paid toward the purchase of a
-
certain product is $110,
how much more
-
remains on the product?
-
So essentially they're saying,
110 is 10% of what number?
-
That's the first thing
you have to say.
-
So 110 is equal to 0.1
times what number?
-
So that's the price.
-
Let's call that the original
price of the product.
-
So the original price of the
product is going to be what?
-
It's going to be 110 divided
by 0.1, which is
-
just this times 10.
-
Which is 1,100, right?
-
Just add a 0.
-
So that's original purchase
price of the product.
-
The deposit is $110 and
they want to know
-
what do you have left.
-
So you're going to put
$110 deposit on it.
-
This was the original
purchase price.
-
You put $110 deposit.
-
Let's see, 1,100 minus 100 would
be 1,000, but then we
-
have another 10.
-
So it would be 990.
-
So that is choice B.
-
And you could do it
the other way.
-
You could just do a little
bit of borrowing.
-
Anyway, you get the idea.
-
And you would get choice
B, which is 990.
-
And I'm out of time.
-
See you in the next video.