Main content

## 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

© 2023 Khan AcademyTerms of usePrivacy PolicyCookie Notice

# GMAT: Math 40

198-200, pg. 179. Created by Sal Khan.

## Want to join the conversation?

- 1+1+1+1+1+1+1+1+1+1+1+1×0+1=?(2 votes)
- It's 12 — maybe it'd help to rewrite it with parenthesis :

1+1+1+1+1+1+1+1+1+1+1+(1*0)+1 =

1+1+1+1+1+1+1+1+1+1+1+0+1 = 12(2 votes)

- How did you get x=12, at .49(2 votes)
- Set up a common denominator for the fractions 1/2, 1/3, and 1/4 on the left-hand side before adding them, then cross multiply to solve for x. Hope this answers your question!(3 votes)

- if 84/of a book consists of 420 pages. find total number of pages in the book?(2 votes)

## Video transcript

We're on problem 198. If 1/2 plus 1/3, plus 1/4 is
equal to 13/x, which of the following must be an integer? Fascinating. So let's see what
they're saying. I don't know. I'm tempted to just
work through this. So if I get a common denominator
here, the common denominator is going to be 12. 1/2 is 6/12 plus 4/12,
plus 3/12. 6 plus 4, plus 3, is 13. So, that equals 13/12. So that tell us that
x is equal to 13. I'm sorry. That tells us that
x is equal to 12. It's the denominator. OK, so they give us a
couple of choices. Choice one, x/8. Is that an integer? No, 12 divided by 8
is not an integer. It's 1 and 1/2. Choice two, x divided by 12. Is that an integer? Yeah, sure. I mean 12 divided by
12 is equal to 1. So, that works. Choice three, x divided by 24. No, x is 12. So, x divided by 24 is 1/2. So, two only is the answer,
and that is choice B. Problem 199. And they have drawn a graph,
which I will draw. So that's a y-axis. And there's an x-axis. And that's y. That is x. See, in the rectangular
coordinate system above, the line, y equals x is the
perpendicular bisector of segment AB, not shown. OK, so let me draw this
y equals x graph. Start there. Nope. I thought I was using the
line tool, but I'm not. Let me use the line tool. That's the y equals x graph. y is equal to x. And they draw point A. The place they drew it, it looks
like it's roughly at the point 2, 3. I don't know if that's going
to become relevant. 2, 1, 2, 3. So A looks like it's
at the point 2,3. I don't know if we
can assume that. The line, y equals x, is a
perpendicular bisector of AB, and they don't show us B. And the x-axis is the
perpendicular bisector of BC. All right. Not shown. If the coordinates of point A
are 2 comma 3-- OK, we were right; A is 2 comma
3-- what are the coordinates of point C? OK, perpendicular bisector means
there's some line that's exactly perpendicular to
this line, where it splits it in half. So if you think about it,
point A is at 2, 3. So where would B have to be in
order for the line between A and B to be perpendicular to
this line, and to be split directly in half? Well, it would be at the point--
I mean, you could just eyeball it-- it'd be at
the point 3 comma 2. If we're at the point 3 comma 2,
there's a lot of, I guess, intuitive reasons why you could
think that this would be the perpendicular bisector. We know y equals x is a
perpendicular bisector. The first is, you can see that
the slope of this line-- you could calculate it if you want--
the slope of this line is negative 1. And I don't know if you remember
in calculus class, but if you have the slope of
one line-- so the slope of this line is 1-- the slope of
a perpendicular line will be the negative inverse
of this slope. That's something that,
hopefully, you learned in calculus-- oh, in algebra
a long time ago. So, the negative inverse
of 1 is negative 1. So this line has a slope
of negative 1. And you could make a bunch
of arguments for why it's equidistant. Both of these points are on
the same line, and they're equidistant from the
point 2 comma 2. So that's why, you know, that
it's the bisector, that this distance is equal to
this distance. And, frankly, on the GMAT,
you're doing it fast. You don't have to do a
rigorous proof. You can eyeball it. You could say, oh, you know,
those are going to be equal. I'm just going to take the
mirror image and go on the other side. So this is point B. And they tell us that the x-axis
is a perpendicular bisector of segment BC. OK, so x is going to be perpendicular to this new line. So this new line has to be
parallel to the y-axis. So this new line is going
to go like that. And whatever the distance is
between B and the x-axis has to be the distance between
x-axis and C. So B is two above the x-axis, so
C is going to be two below the x-axis. So that's where C is. And x, at that point,
is equal to 3. And so this y would be
equal to negative 2. So what are the coordinates
of point C? 3, negative 2. And that is choice D. Interesting problem. Problem 200. A store currently charges
the same price for each towel that it sells. If the current price of each
towel were to increase by $1, 10 fewer of the towels could
be bought for $120, excluding sales tax. What is the current price
of each towel? So let's say the current
price of each towel. So if we take 120 and we divide
it by the current price of each towel, we could buy,
I don't know, n towels. Now, they're saying that if we
had $120, and if we were to increase the price by $1-- so
now the new price is p plus 1; now I can't buy n towels
anymore-- they're saying I can only buy 10 fewer towels. So I can only buy n
minus 10 towels. Now we have two equations
and two unknowns. Well, they're not linear
equations, but I think we should be able to solve them
one way or the other. So here we get p times
n is equal to 120. And here, let's see, we can
multiply-- we have 120 is equal to p plus 1,
times n minus 10. So that's equal to p times n,
minus 10p, plus n, minus 10. And so both of these are equal
to 120, so we could set them equal to each other. Both of these are
equal to 120. So we could use both
of these equations. Say pn is equal to pn minus
10p, plus n, minus 10. So let's see if we can make
any headway here. I could subtract pn
from both sides. So I get 0 is equal to minus
10p, plus n, minus 10. And let's see, I could add 10p
to both sides, and I get 10p is equal to n minus 10. Well, what's n? Actually, I shouldn't have
substituted like this. I'm kind of doing it
a little messily. But we can do another
substitution. n is equal to 120/p, so
that should help us. The real way I should have done
it is-- well, actually, no, this is I think the
easiest way to do it. 10p. n is 120/p. A lot of times you just have to
go forth with the algebra and see what happens. Minus 10. And let's multiply both sides
of the equation by p. So you get 10 is equal-- sorry,
10p squared is equal to 120 minus 10p. Let's divide both sides by 10
just to get rid of them. So you get p squared is
equal to 12, minus p. Let's take these on to
the left-hand side. So you get p squared plus p,
minus 12, is equal to 0. We have a quadratic. We can just factor it. That is p plus 4, times p
minus 3 is equal to 0. What two numbers, when you
add them, equal plus 1? The coefficient there. When you multiply them they
equal negative 12. That's just practice you got
in algebra doing that. So that's p plus 4,
times p, minus 3. So p could be minus 4, or
p could be equal to 3. And p can't be might minus 4
unless he was giving away the towels, he was paying people
to take the towels, which isn't the case. We're assuming that
price is positive. So the price had to be $3,
which, luckily enough, after doing all that algebra,
is one of the choices. Choice C. Next question. 201. I think I have time
to do this one. 201. If the sum of n consecutive
integers is 0, --this is my ad hoc notation-- which of the
following must be true? n consecutive integers is 0. n must be an even number. So let's think a lot about,
like, our even and odd type of talk. So we know that an even
plus an even is even. And we know that an odd plus
an odd is also even. And you could prove
it to yourself. You could try a bunch
of numbers. Or if you just write an odd as
2k plus 1, and you add it to another 2m plus 1, 2k-- that
becomes 2 times k, plus m, plus 2. So, that's an even number. It's divisible by 2. So an plus an odd is even. And so the only way you're going
to get an odd number is if you have an even
plus an odd. Now, the only way you're going
to have an even plus an odd-- so we're summing n consecutive
numbers. So, I don't know, let's call
it number 1, plus number 2, plus number 3. Let's think about it a couple
of different ways. If this number is odd, then this
number is obviously even, and then this number is odd. Now, if we're taking the sum
of an even number, if n is even, we're going to get the
same number of evens and odds. So n could be 2. If n could be 2, we're going to
have one even and one odd no matter what, right? This could be odd. This could be even. Or this could be even and
this could be odd. And if n is 2, which is an
even number, we can very easily-- you know an odd plus
an even is an odd number. So, they say if the sum of n
consecutive integers is 0-- oh, you know what? I completely missed
that point. Let me continue this
in the next video.