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## GMAT

### Course: GMAT > Unit 1

Lesson 1: Problem solving- GMAT: Math 1
- GMAT: Math 2
- GMAT: Math 3
- GMAT: Math 4
- GMAT: Math 5
- GMAT: Math 6
- GMAT: Math 7
- GMAT: Math 8
- GMAT: Math 9
- GMAT: Math 10
- GMAT: Math 11
- GMAT: Math 12
- GMAT: Math 13
- GMAT: Math 14
- GMAT: Math 15
- GMAT: Math 16
- GMAT: Math 17
- GMAT: Math 18
- GMAT: Math 19
- GMAT: Math 20
- GMAT: Math 21
- GMAT: Math 22
- GMAT: Math 23
- GMAT: Math 24
- GMAT: Math 25
- GMAT: Math 26
- GMAT: Math 27
- GMAT: Math 28
- GMAT: Math 29
- GMAT: Math 30
- GMAT: Math 31
- GMAT: Math 32
- GMAT: Math 33
- GMAT: Math 34
- GMAT: Math 35
- GMAT: Math 36
- GMAT: Math 37
- GMAT: Math 38
- GMAT: Math 39
- GMAT: Math 40
- GMAT: Math 41
- GMAT: Math 42
- GMAT: Math 43
- GMAT: Math 44
- GMAT: Math 45
- GMAT: Math 46
- GMAT: Math 47
- GMAT: Math 48
- GMAT: Math 49
- GMAT: Math 50
- GMAT: Math 51
- GMAT: Math 52
- GMAT: Math 53
- GMAT: Math 54

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# GMAT: Math 47

221-225, pgs. 182-183. Created by Sal Khan.

## Video transcript

We're on problem 221. And let's see, they have--
is all of that in the denominator? Well, I guess it is. Let's see, they wrote 1 over--
and I'll write it the way they did it. 1 over-- and they have this--
you know, I think it's got to be 1 plus. I think it's 1 plus. The way they wrote
it is weird. 1 plus 1 over 1 plus 1/3. This looks a lot like another
problem that we did. But anyway, I'll do this one. So let's see. If we do 1 plus 1/3, that's 3/3,
plus 1/3, so that's 4/3. So then that turns into 4/3,
1, 1 plus, 1 over that, and then you have 1 plus that. And so then this equals
1 plus 1 over 1 plus-- what's 1 over 4/3? Well, that's the same thing as--
1 divided by 4/3 is the the same thing as 1 times 3/4,
and that's equal to 3/4. And what's 1 plus 3/4? Well, that's 4/4 plus
3/4, right? 4/4 plus-- so that's 7/4. So then that becomes 1 plus 1
over 7/4, which is equal to 1 plus 4/7, and that equals 7/7
plus 4/7 is equal to 11/7. And that's choice D. The way they wrote
it was weird. They wrote it like this,
1 plus down here. But I assume that this is
what they wanted, what I originally had done. Problem 222. In the rectangular coordinate
system above-- well, let me draw it. That's the y-axis, that's
the x-axis. y, x. And they have a point out here
called point p, and they say that's at 4. x equals
4, y equals 0. In the rectangular coordinate
system above, if point r, not shown, lies on the positive
y-axis, so it lies someplace over here, and the area of
triangle orp , so this is point o, and r is going to be
some place on this, is 12, what is the coordinate
of the point r? So the coordinate of the point
r, let's just call it-- well, the x-value is going
to be 0, right? And this y-value, I don't
know, let's call it x. No, let's call it y. That's a better variable
name for the y-value. So they essentially want to
know what the area of this triangle right here is. Or actually, they want to know
what y is, given that the area of that triangle, they
tell us, is 12. The area of this
triangle is 12. So what's the area
of this triangle? It's the base times the
height times 1/2. The base is 4, right? This is x is equal to 4. So it's 4 times the height. Well, the height is going
to be y, right? Because the coordinate
of r is 0 comma y. So 4 times y times 1/2-- if you
don't do the 1/2, you're getting the area of the whole
rectangle-- is equal to the area of the triangle
is equal to 12. So you get 2y is equal to 12. y is equal to 6. So the coordinate of
r is 0 comma 6. And that is choice-- oh, they
just ask what is the y-coordinate of point r. So that's 6, and that's
choice B. Problem 223. Car A is 20 miles
behind car B. So let me see if I
can draw that. So this is A, this is B-- and
it's 20 miles behind it; I don't know if that's relevant--
which is traveling in the same direction along
the same route as car A. Car A is traveling at a constant
speed of 58 miles per hour, so car A is at 58 miles
per hour, and car B is traveling at a constant speed
of 50 miles per hour. Fair enough. How many hours will it take
for car A to overtake and drive 8 miles ahead of car B? So they don't want just
car A to catch up. They want car A to catch up and
go ahead 8 miles, right? And drive 8 miles ahead. So another way to think about
it is how long will it take for car A to go 28 miles
further than car B? And the easiest way to think
about it is just what their differential is, right? Car A is going 8 miles an
hour faster than car B. So the relative velocity
of A relative to B is 8 miles an hour. And you could say that, well,
I don't want to use fancy notation, but car A, if you're
sitting in car B and you looked at car A, you'd say,
oh, it's going in my same direction, but I'm stationary
and it's going at 8 miles per hour. So the question essentially says
if I'm going at 8 miles per hour, how long does it take
to go 28 miles, right? Because I have to catch up to
20 miles with B, and then I have to go 8 more miles. You could just pretend like you
are in car B and you think that you are stationary. You ignore the fact that the
world is speeding behind you. And you just see car A. You say, oh, car A must be
moving at 8 miles an hour in the direction in which
my car is pointed. So, anyway, if you have to go 28
miles, and you're going at 8 miles per hour, distance
divided by rate is equal to time. That's equal to 3 and 1/2 hours,
which is choice E. Next question, 224. For the past n days, the average
daily production at a company was 50 units, so for n
days, 50 unit per day average, if today's production of 90
units, so we have 90 units, and that raises the average to
55 units per day, what was the value of n? OK, so the old average would
have been n times 50-- and this is kind of obvious
-- divided by n is equal to 50, right? Now, if I want to average in
this 55, what happens is I get to n times 50, right? That's the sum of the n days
before, the sum of the total production for the
n days before. And what I want to do is I want
to have the production for today, and that is 90, plus
90, and now I'm going to divide by n plus
1 days, right? And now the average is 55. So we just have to
solve this for n. So let's see what we can do. I'll do it in a different
color. We get 50n plus 90 is equal to--
multiply both sides by n plus 1-- 55n plus 55. So let's subtract 50n
from both sides. You get 90 is equal
to 5n plus 55. Subtract 55 from both sides, 55
to 35 is equal to 5n, n is equal to 7. And that's choice E. Problem 225. OK, They write x plus 1 over
x minus 1 squared. And they say if x does not equal
0 and x does not equal 1-- and that's good, because
if x equaled 1, we'd be dividing by 0 here-- and if x is
replaced by 1/x everywhere in the expression above-- so
x is replaced with 1/x everywhere in the expression
above-- then the resulting expression is equivalent
to-- I don't know, let's just do it. So if we get 1/x-- we're just
replacing x with 1/x-- plus 1 over 1/x minus 1 quantity
squared, and that equals -- let's see. This is the same thing as
1/x plus x/x, right? 1 is the same thing as x/x, And
this 1 is the same thing as x/x, right? That's an x/x. So then this equals 1 plus
x/x over 1 minus x/x, everything squared. And this inside of the brackets,
that's the same thing as 1 plus x/x-- instead
of being divided by this, we could multiply by the
reciprocal-- times x over 1 minus x. I just reciprocated this
and multiplied. And this is all still
in the squared . Now these x's cancel out, and lo
and behold, we get 1 plus x over 1 minus x squared, which
is the same thing as x plus 1-- well, let's think
about it. x plus 1 over 1 minus x squared. And see, that's not one
of the choices, right? But let's think about it. 1 minus x, that's the
same thing as-- just let me rewrite it. This is the same thing as x plus
1 divided by-- now this 1 minus x is the same
thing as x minus 1 times negative 1, right? And we're going to square
all of this. Well, this over here, this is
the same thing as x plus 1 over x minus 1 squared
times 1 over negative 1 squared, right? If you take the square of a
product, you could-- this is the same thing as that times
1 over negative 1. So that turns into this, and
this, obviously if I'm squaring another number,
this just becomes 1. So I'm left with x plus 1
over x minus 1 squared. And that's exactly what we
started off with, and that's choice A. And that's the trick, to realize
that this is just a negative of minus 1. And you can take a negative 1
out, but we're going to square it anyway, so it's going to
turn it back positive. Next question. In choice E, actually, in that
one, they tempt you by taking the negative outside, but you
can't take the negative outside the square. You're squaring it, so the
negative will disappear. Anyway, problem 200. Oh, I'm all out of time. See you in the next video.