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Course: GMAT > Unit 1
Lesson 1: Problem solving- GMAT: Math 1
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GMAT: Math 35
177-180, pg. 176. Created by Sal Khan.
Want to join the conversation?
- at 10.00, it might be easier to think of solving it by using the complement, i.e. 1-(1/4)*(1/3) = 11/12(3 votes)
- For Q180 I presume you can simply work out 1 - P(YY) as Sal had done in the previous video (i.e. 1 - 1/4*1/3)?(2 votes)
- how do you do $14.99 20% out of the $14.99 plus a 50 cent tax?(1 vote)
Video transcript
My wife is looking at me kind
of strange right now. So forgive me if I have a
problem doing some of these problems. But anyway,
problem 177. She said she wanted to observe
me making videos. Let's see how this goes. Do you want to say hi? No, no, OK. No, I'm not going
to record it. A company accountant estimates
that airfares next year for business trips of 1,000 miles or
less will increase by 20%. So less than 1,000 miles. Airfares next year going
are going to go up 20%. And for all other business
trips, it will increase by 10%. So essentially greater than or
equal to 1,000 miles, airfare is going to increase by 10%. This year total airfares for
business trips of 1,000 miles or less were $9,900. So that's the total for
1,000 miles or less. And airfares for all
other business trips were 13,000 miles. According to the accountant's
estimate, if the same number of business trips will be made
next year as this year, how much will be spent for
airfares next year? I have a problem with what they
say, because they say the same number of business trips. But we really, I'm thinking,
have to assume that not only are the same number of business
trips made, but the allocation between less than
1,000 and greater than 1,000 is also going to have
to be the same. But anyway, let's just do what
I think they're assuming. So essentially you just have
to take the $9,900 and increase it by 20%. 9,900 times 1.2. And then add that to the 13,000
increased by 10%. So 1.1 times 13,000. That's a 13, not a B. So let's see. Let's just write 9,900
times 1.2. 2 times 0 is 0. 2 times 0 is 0. 2 times 9 is 18. 2 times 9, 18, plus 1 is 19. Add a 0 and then
you have 9,900. 0, 0, 9, 9-- scroll down--
0, 0, 8, 9 plus 9 is 18, 2 plus 9 is 11. And we have one number
behind the decimal. So that's 11,880. And then 1.1 times 13,000. So it's 13,000 times 1.1. 1 times 13,000 is 13,000, 0,
1 times 13,000 is 13,000. 0, 0, 0, 3, 4, 1. And we have one number
behind the decimal. So it's 14,300. And if we were to add
these two together. Let's see, if I add 11,880
to this, I get 0, 8. 3 plus 8 is 11. So, 26,180. And that's choice B. Next problem. Problem 178. If x star y is equal to xy minus
2 times x plus y, for all integers x and y,
then what does 2 star minus 3 equal? So we essentially just have
to do pattern matching. Everywhere we see an x here,
we substitute with 2. And everywhere we see a y, we
substitute with minus 3. And just to be clear, this isn't
an equal, that's a minus right there. So it's 2 times minus 3, minus
2 times 2, plus minus 3. 2 times minus 3 is minus 6. Minus 2 times 2 plus
minus 3 is minus 1. That's the same thing
as 2 minus 3. And let's see, you have a minus
and a minus, so we can make those both plus. So you have minus 6 plus 2. When you multiply it by 1,
that's the same thing. So minus 6 plus 2 is
equal to minus 4. And that is choice C. Problem 179. I'll switch colors to
ease the monotony. 179, the table above shows the
number of students in the 3 clubs at [? McCullough ?] School. So let me write the table. So they have the club. And they say the number
of students. And they have chess, drama,
and math clubs. The chess club has 40 students,
drama has 30, and math has 25. The table above shows the number
of students in 3 clubs at [? McCullough ?] School. Although no student is in all
3 clubs, 10 students are in both chess and drama. I feel a Venn diagram
coming on. 10 students are in both
chess and drama. 5 students are in both
chess and math. And 6 students are in
both drama and math. How many different students
are there in the 3 clubs? Yeah, I think we need to
do a Venn diagram. So let's draw the chess students
first in yellow. Actually I'm not going to do
circles, because I think I have to draw it a little
bit different. Because there's no students
in all 3 clubs. So let's say that this
is the chess club. There's 40 people in that. And let's do drama next. There are 30 students
in the drama club. And then how many people
are in both? Well they told us that
there are 10 people in chess and drama. So this is right here is 10. And then let's draw math. So math will intersect
with both of these. It looks like that. Math has a total
of 25 students. And their overlap with
math and drama is 6. And the overlap with math
and chest is 5. So the main question we have to
ask is, if we were to add up all of these-- if we were to
add up the 40 plus the 30 plus the 25-- how many
times are we counting each of these sections? So let's just try that, because
we just want to know the total number of students
in 3 clubs. So let's take the 40. We're already counting this 10
once, right, because they're in that 40. So let's add the 30. So 40 plus 30. If we add the 40 plus the 30,
we're double counting this 10 that's in both clubs. So if we were to subtract one
of them, this number right here is the number of students
in the chess and drama clubs, right? So that is-- what is that-- that
is 40 plus 30 minus 20. That's 60 students. So 60 students in
chess and drama. And now if we add 25 students--
so we're adding the math club to that-- who are
we double counting? Now we're double counting these
6 that we've already counted right? Because we added them in the
30 from the drama club. And we're also double counting
these 5, which we added when we first took into account the
40 from the chess club, right? So since we've already counted
those, we need to subtract those from the 25. So minus 5 and minus 6. This problem is all about
not double counting. So let's see, 60 plus
25 is 85, minus 5 is 80, minus 6 is 74. So 74 total students. So that's choice C. Problem 180. In a nationwide poll, N people
were interviewed. So question number one. That's in the question,
question number 1. So it said for question number
1, 1/4 said yes. And of those, 1/3 answered
yes to question 2. Which of the following
expressions represents the number of people interviewed
who did not answer yes to both questions? OK, so first of all, you have
the people who answered no to question 1. So that was 3/4 of the
people, right? No on question 1. And who cares what they answered
on question 2, right? Because they definitely said
no to one of the questions. Because they want to
say who did not answer yes to both questions? So these guys fall into
that category. They said no to one
of the questions. And then you want to say, what
percentage of the total people said yes on the first question
and then no on the second? That combination would be if 1/3
of the people who said yes on the first question, said
yes on the second, then 2/3 said no. And what percentage, or what
fraction is this of the entire population? Well it's 2/3 of 1/4. So that is 2/3 times 1/4
is equal to, 1, 2. That's equal to 1/6. 1/6 of the entire population
said yes on the first and no on the second. So if we were to add these two
up, these are people who said no in either or both questions,
you get 1/6 plus 3/4 is equal to-- let's give a
common denominator of 12-- so that's 2/6 plus 3/4,
that is 9/12. So that is 11/12. Oh, there were N people
interviewed. So 11/12 of the people did not
answer yes to both questions. And so if they want to know
the number of people, you would say 11/12 times the total
population, which is N. So 11N over 12. And that's choice E. And my wife is still
staring at me. See you in the next video.