Main content

### 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

© 2024 Khan AcademyTerms of usePrivacy PolicyCookie Notice

# GMAT: Math 36

180(simpler)-184, pgs. 176-177. Created by Sal Khan.

## Want to join the conversation?

- sorry, but am confused about q# 184... could anyone pls help me understand it? ty...(1 vote)
- Here is an easier way to solve this:

Let M = 10x + y (e.g. 15 = 1 x 10 + 5 x 1 right?)

So, N = 10y + x (e.g. 51 = 5 x 10 + 1 x 1)

So M + N = 10x + y + 10y + x = 11x + 11y = 11(x + y)

So, our answer MUST be a multiple of 11.

Now, the easiest way to check if a 3 digit number is a multiple of 11 is to check if the digit in the tens place (the middle one) is the sum of the one in the units place and the one in the hundreds place. e.g. given the three digit number ABC it will be a multiple of 11 if and only if B = A+C.

In option A this is not true. Whereas all other options are multiples of 11. So, our answer is A.(2 votes)

- i dont get what it is saying.(0 votes)
- well you have to multiply ok like this 10x4= its like this 4x1=4 then you add a 0 to the end of 4 so its like 10x4=40 so 10x22=220 do u get it now(1 vote)

## Video transcript

Before moving on, I want to
address problem 180 again. Because we got it right in the
last video, but then as soon as I finished the video, I
realized there was a much, much simpler way
of doing that. They said in a nationwide
poll, n people were interviewed. 1/4 of them answered yes
to question one. And of those, 1/3 answered
yes to question two. So then you have another 1/3. So what fraction of the entire
population said yes to both? Well, it would be 1/4
of-- or 1/3 of 1/4 which is equal to 1/12. 1/12 of the population
said yes to both. Or you could say 1/12 of
n said yes to both. And they ask us which of the
following expressions represents the number of people
interviewed who did not answer yes to both? So it's everybody else. These are the people who
said yes to both. So essentially you just
subtract that from 1. So 1 minus 1/12 is equal to
12/12 minus 1/12, which is equal to 11/12 of the
population did not say yes to both. And so 11/12 and then the
population is n, so it's 11n over 12. And that's how we
got choice E. Problem 181. The ratio of two quantities
is 3:4. So, let's say one is x, one
is y, and it equals 3:4. If each of the quantities is
increased by 5, what is the ratio of these new quantities? x plus 5 to y plus 5. Well, it really depends
on what multiple these are of 3 and 4. Well, I already have a sense
that it's choice E. It cannot be determined from
the information given. And let's prove it. I'm going to show you two
different x's and y's, and when you add 5 to both you're
going to get a completely different answer. So you cannot determine it
from the information. So they snuck in a data
sufficiency question into the problem solving. So, x could be 3. If x is equal to 3 and y is
equal to 4, that definitely satisfies this condition. Or we could have x is equal
to 6 and y is equal to 8. Right? That also satisfies
the 3 to 4 ratio. What happens when we add
5 to both of these? x plus 5 would be equal to 8. And y plus 5 would
be equal to 9. So, the new ratio becomes 8:9. What about this case? You would have x plus
5 is equal to 6 plus 5, which is 11. And then y plus 5 is equal
to 8 plus 5, which is 13. And 8/9 is a very different
fraction than 11/13. It's not like you can reduce
one into the other. So I can find an x and a y to
satisfy this, but when I add 5 to the top and bottom I get
two different answers. So you cannot determine this
with the information given. So the choice is E. Sneaky data sufficiency
problem. 182. If the average arithmetic mean
of x and y is 60-- so essentially you're saying x plus
y over 2 is equal to 60. The average of them is 60. And the average of
y and z is 80. So, y plus z over 2
is equal to 80. What is the value
of z minus x? So let's see if we
can do that. So let's rewrite this. Let's solve for x
in terms of y. I'll do it in a different
color. We get x plus y is
equal to 120. Just multiplied both
sides by 2. You get x is equal
to 120 minus y. Now let's solve for z
in terms of y here. Multiply both sides by 2, you
get y plus z is equal to 160. z is equal to 160 minus y. So what's z minus x? So z is 160 minus y, minus
x, 120 minus y. So that is equal to 160 minus
y, minus 120 plus y. And it's good that the y's
are canceling out. Whenever they give you something
like this, it's usually a safe guess that if
you just go forth with the algebra that nice things
will happen like this. So these y's cancel out and
we're left with 160 minus 120 which is 40. And that is choice B. Problem 183. If 1/2 of the air in a tank is
removed with each stroke of a vacuum pump-- so we have
1/2 tank per stroke. I'm guessing this is going
to be a rate of change. What fraction of the original
amount of air has been removed after four strokes? Oh no. This is interesting. OK. So, after one stroke, what
fraction of the original amount of air has
been removed? Let's just figure out how
much has been left. We could do both actually. Left and then removed. So, after one stroke-- 1/2 of
the air in the tank is removed with each stroke. So after one stroke, you
have 1/2 left and you have 1/2 removed. But, after two strokes, it takes
out half of the air. Right. So it takes out 1/2
of this half. So if you take out 1/2 of this
half you have 1/4 left over. And you've also removed
another 1/4. Right? Because these add up to
what was left before. After three strokes, you
take out 1/2 of this. So you have 1/8 left over. And you took out another 1/8. And then after four strokes,
you have 1/16 of the air left over. Actually I didn't even have
to do this column. After four strokes you have
1/16 of the original air. But they want to know what
fraction of the original amount of air has
been removed. So this is what you have left. So what's been removed
is 1 minus this. So, 1 minus 1/16. Well, that's 16 over
16 minus 1 over 16. Which is equal to 15 over 16. Choice A. I didn't even have to worry
about that column. Problem 184. If the 2 digit integers M and
N are positive and have the same digits, but in reverse
order, which of the following cannot be the sum of M and N? All right, same digits
in reverse order. So M could be AB. And N could be BA. So which cannot be the sum? I'm just experimenting. Let's see you if you
add AB and BA. So, if B plus A-- I guess
there's two assumptions. If B plus A is less 10,
let's do this one. B plus A is less than 10. Then you would have-- this
digit would be B plus A. And then this digit over here
would also be B plus A. So, definitely choices
D and E work. So we're left with choices
A, B, and C. Because none of these fit this
paradigm where we have the same digit twice. So, all of those assume that B
plus A is greater than 10. So, if B plus A is greater than
10, then what happens? A plus B. B plus A. So let's think about
it a little bit. If B plus A, you'll have the
ones digit from B plus A. And then you'll have
a 1 up here. You add 1 plus A plus B. So, you'd have B
plus A plus 1. And so this will be a
two-digit number. And if you think about it-- if
we do the three digits of the number, this is going
to be the ones digit from B plus A right? This is going to be the ones
digit from B plus A plus 1. So this number is going to be
one more than this number. That's an interesting problem. This one has to be one
more than that one. And so both B and C satisfy
that, right. 165. 6 is one more than 5. And it should work out. We should be able to figure
out a B and an A that satisfy that. And then C also satisfies that. 121. 2 is one more than 1. I want to make sure you
understand that because it's actually pretty interesting. Deductive reasoning,
A is the answer. But if you think about it, we're
assuming B plus A is greater than 0. So, here, you'd have the ones
digit of B plus A and you'd carry a 1. Now when you add A plus B and
that 1, the ones digit is going to be 1 greater. This is going to be the ones
digit of B plus A plus 1. And that's why I'm saying that
the tens digit has to be 1 greater than the ones. That was an interesting
problem. Anyway, see you in
the next video.