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## GMAT

### Course: GMAT > Unit 1

Lesson 1: Problem solving- GMAT: Math 1
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- GMAT: Math 49
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# GMAT: Math 27

138-142, pg. 171. Created by Sal Khan.

## Want to join the conversation?

- Around the2:20mark, Hal explains how in problem 138 you can divide the top and bottom to get 1/1 + 3/5, but I thought you couldn't do that when dealing with fractions in which either the numerator or denominator has more than one number in which you add or subtract. In this case, the denominator has 1 + 3/5 so why was he able to change the Wd(Bd) in the numerator and denominator?(2 votes)
- Are you confused about the fraction in the denominator ? First of all think of the Denominator as a separate item for a moment. Think of it as an entity in its own right, and forget the Numerator. I explain this to my son as " Work on the Top and Botton however you want. Just make sure you can regain the original fraction from any answer you end up with". So as long as you preserve the laws of arithmetic in the Top and Bottom and you can get back the original numbers in the Top and Bottom, your arithmetic is correct. Think of fractions as Ratios, everthing is relative in fractions. The guy in the video is simply reducing a fractional expression to its simplest form(2 votes)

- 142. I get 22.2% 3600(1/3) = 1200, 1200(1/3) = 400 1200-400 = 800, 800/3600 = .222

22.2% Anyone with me?(2 votes)- If you watch the video again, Sal explains you need to subtract 400 from the total workers as well. So it's 800/3200 which equals 1/4.(2 votes)

## Video transcript

We're on problem 138. At a loading dock, each worker
on the night crew loaded 3/4 as many boxes as each worker
on the day crew. So that means that the boxes per
night crew worker is equal to 3/4 the number of boxes
per day crew worker. If the night crew has 4/5 as
many workers as the day crew-- the number of night crew workers
is going to be 4/5 times the number of
day crew workers. What fraction of all the boxes
loaded by the two crews did the day crew load? So, how many boxes will
the day crew load? The day crew is going to load
the number of day crew workers times the boxes per
day crew worker. And they want to know what
fraction of all of the boxes loaded is this? So that'll be the numerator,
and the denominator will be all the boxes loaded. So all the boxes loaded are
going to be the boxes loaded by the day crew. So that's the number of workers
times the boxes per worker, plus the number
of boxes loaded by the night crew. That's going to be the number
of night crew workers times the number of boxes per
night crew worker. And to simplify this, if we
could just substitute these variables with things that look
a lot more like that. And lo and behold, we have easy
substitutions to tell us what b sub n is, the boxes
per night crew worker. And they tell us what the number
of night crew workers are in terms of the
day crew workers. So let's just substitute. So this is equal to day crew
workers times boxes per day crew worker, divided by day crew
workers times boxes per day crew workers, plus,
what is the number of night crew workers? It's w sub n, but that's the
same thing as 4/5 w sub d. And then what are the
number of boxes per night crew worker? That's 3/4 times the boxes
per day crew worker. The 4's cancel out. Actually, we could
simplify this. We can divide the top and the
bottom by this w sub d times b sub d, and you get 1 over 1
plus-- all of these cancel and you're just left with 3/5. That's equal to 1 over 8/5,
which is equal to 5/8. And that is choice E. Next problem, 139. A restaurant meal costs $35.50
and there is no tax. If the tip was more than 10% but
less than 15% of the cost of the meal, then the total
amount must have been-- So what's the range for the tip? 10% of $35.50. That's just going to be $3.55. And 15%-- there's a couple ways
you can think about it. You could say it's half
more than this. What's half of this? Half of this is $1.77 1/2. I think that's right. soon. Really, you can't
get a half-cent. Is that right? 2 times that, you get 2. And 2 times 70 is $1.40. So this would be 5%. I just took half of this That's
how people do it. Half of 10%. If where you live, the tax
is 10%, but I don't want to get into that. So if you add this to this, you
get 4, 5, 12, 13, and 5. So the tip is going to be
between $3.55 and $5.33, I guess we could say. And so the total bill, that's
what they want to know, we just add both of these. So at minimum, you're going to
pay what's this plus $35? So $35.50 and $3.55. 5, 10, 9. $39.05. And at the high end,
you have $35.50. And then you have $5.33. And that is equal to
3, 8, 0, 1, 4. So your total bill is going
to be between $39 and $41. And that is exactly choice B. They just round up. They say $41 is here
and $39 is here. If it's between these 2 numbers,
it's definitely between $39 and $41. And that's choice B. Problem 140. In a weightlifting competition,
the total weight of Joe's two lifts was 750. So l1, his first lift plus the
second lift was 750 pounds. If twice the weight of the
first lift was 300-- If 2 times the first lift was equal
to 300 more than the weight of his second lift, what was
the weight in pounds of his first lift? So we just want to
solve for l1. So this top equation we can
rewrite as l2 is equal to 750 minus l1. Then we can substitute
that here. So then you get 2 times l1 is
equal to 300 plus l2, which is this, plus 750 minus l1. Add l1 to both sides. You get 3 l1 is equal
to 300 plus 750. That's 1,050. And 3 goes into 1,050-- 3 times
3 is 9, 15, 3 goes into 15 5 times, so it's 350. So his first lift
was 350 pounds. And that's choice D. Problem 141. A club collected exactly
$599 from its members. If each member contributed
at least $12, what is the greatest number of members
the club could have? So the more that each member
contributed, the smaller you the number of people
in the club. If each member contributed at
least $12-- so what is the greatest number of members
the club could have? So if they all contributed
exactly $12-- 12 goes into 600 50 times. So if this number were $600,
you could say if $600 were collected in increments of at
least $12, then you could have at most 50 people. But we don't have $600. $600 would be 50 times 12. We have $599. $599 can be done by at
most not 50 people, but at most 49 people. Let's think about it. Let's make sure that
that's right. So my answer is 49,
which is C. And let's think about how much
money would be left over, so 49 times 12. This is just a reality check. 49 times 12. 2 times 9 is 18. 2 times 4 is 8 plus 1 is 9. 0, 49. 8, 18, 588. So if you have 49 people, and
they all contribute $12 exactly, you'll raise $588. And then you could have had a
couple of other people, you could have had 11 other
people who instead of $12, they paid $13. And that's how you get
to $599, because they say at least $12. You can't have 50 people,
because 50 people, if they contribute at least $12,
you would have to raise at least $600. So it can't be 50, so
the answer is 49. And you might say why did you
go immediately to $600? And a lot of times on the GMAT,
where they have these numbers that are 12 and 599--
those are strange numbers. You say 599 is awfully
close to 600. Let's see if I can use that
information to deduce something interesting
about this problem. Anyway, next problem, 142. Of the 3,600 employees of
Company x, 1/3 are clerical. So clerical is equal
to 1/3 times 3,600, which is equal to 1,200. If the clerical staff were to
be reduced by 1/3, what percentage of the total number
of remaining employees would then be clerical? We're going to reduce the
clerical staff by 1/3, so how many people are we going
to get rid of? So 1/3 of 1,200 is what? It's 400. So the clerical staff is now
going to be 1,200 minus 1/3, minus 400, which is
equal to 800. So what proportion is this of
the remaining employees? So clerical is 800, and
how many remaining employees are there? Are there 3,600? No, we've gotten rid of 400
employees who were clerical. So now we have 3,200
employees. 3,600 minus 400, so that
is equal to 1/4. And that's A, 25%. Problem 143. Actually, I'm out of time. I'll continue this in
the next video. See