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## GMAT

### Course: GMAT > Unit 1

Lesson 1: Problem solving- GMAT: Math 1
- GMAT: Math 2
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- GMAT: Math 27
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- GMAT: Math 44
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- 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 51

237-239, pg. 185. Created by Sal Khan.

## Want to join the conversation?

- At 6.46 Sal receive answer 9*sqrt(2) but actually if we rewrite 5/6 L = W as 6/5 L = W which is actually the same because we do not know which is bigger L or W.

Then if we substitute in L * W = 9 * 15 we will receive

L * 6/5 L = 9 * 15

L^2 = ( 9 * 15 * 5 ) / 6

L^2 = ( 3 * 15 * 5 ) / 2

L^2 = ( 15^2 ) / 2

L = 15 / sqrt (2)

W = 6/5 L = 6/5 * 15/sqrt(2) = (6*15)/5 * sqrt(2) = 18/ sqrt(2)

So, we have two different answers which are fit the problem.(1 vote) - Sal, correction 27 X 6 = 162 and not 161 (5:57)(1 vote)

## Video transcript

We're on problem 237. A part-time employee whose
hourly wage was increased by 25% decided to reduce the number
of hours worked per week so that the employee's
total weekly income would remain unchanged. By what percent should
the number of hours worked be reduced? So before, he made his
hourly wage times hours worked, right? This is what he would normally
work in a week, right? So his wage per hour times
hours per week. This is what he used to make. But now what he's done is since
he's making 25% more of his wage, he's going to reduce
his hours by some number, by some fraction. Let's call that x. So then the new wage times the
new hours , and they're saying that he doesn't want to change
the total amount that he makes in a week. So the new wage times the new
hours will equal w times h. Let's divide both sides by w. This goes to 1, that
goes to 1. Divide both sides by h. That goes to 1, that
goes to 1. And you're left with 1 is
equal to 1.25 times x. And I view that-- this is easier
to think of as what? 5/4, right? 1.25-- right? 4/4 is 1, and then you
add another 1/4. This is 1 and 1/4,
so this is 5/4. So let's write that as 5/4x. Multiply both sides by 4/5,
you get x is equal to 4/5, which is equal to 0.8. So another way to view it is he
would reduce-- he's working 80% of the hours that he used to
work, so he reduced by 20% the amount that he worked. And that will be choice B. Next problem, 238. OK, they've drawn us a figure,
so let's draw the figure. Let me see if I could
make it thicker. OK, so they drew a box in a box,
and there's another box in that box, and it looks
something like that. And then they shaded it in,
so I'll shade it in. And they say that this side up
here is 18, and they say that this side over here is 15. The shaded region in the figure
above represents a rectangular frame with length 18
inches and width 15 inches. The frame encloses a rectangular
picture that has the same area as this
frame itself. So this area is equal
to this area. Fascinating. If the length and width of the
picture have the same ratio as the length and width of the
frame, what is the length of the picture in inches? So they're saying that this is
the length, so this is l. And this is the width. And they want us
to solve for l. What is the length of
the picture, right? They're calling-- I mean, I
would call that width, but they originally say the frame
with length 18 inches. So they want to know the
length of the picture. OK, first of all, they tell us
that the ratios are the same. So l:w of the picture is the
same ratio as it is for the frame, is 18:15. You divide by 3. That's the same thing as 6:5,
right? l:w is equal to 6:5. And the other thing they told
us is that the area of the picture the same as the
area of the frame. So the area the picture is
l times w, and what's the area the frame? What's the area of this whole
thing , which is 15 times 18, minus the area of the
picture, right? Because the area of the frame
is not this whole rectangle. It's just the part that doesn't
have the picture in it, so minus l times w. So if we add lw to both
sides, we get 2lw is equal to 15 times 18. I'm just going to leave
that as 15 times 18. Let's see, we could already
divide both sides by 2, so you get lw is equal to 15 times 9. And let's see, we have two
equations with two unknowns. Let's see if we can solve. So we have l/w is
equal to 6/5. We can cross-multiply, so we
get 5l is equal to 6w. I ignored the stuff
in the middle. And we want to solve
for l, right? So let's substitute for w. So we get 5/6l is equal to w,
and now let's substitute that into this equation. So we have l times w, which is
5/6l, is equal to 15 times 9. And let's see, let's multiply
both sides by 6/5, and so you would get l squared is equal to
15 times 9 times 6 over 5. So let's see, we could do
a little simplification. This becomes 3, this becomes
1, and then we have-- let's see, now we have to
do some math. 3 times 9 is 27. 27 times 6 is what? 7 times 6 is 42. 2 times 6 is 12,
plus 4 is 161. That's a strange number. So let's see if I can
simplify this. Oh, I'm looking at choices,
and they have all sorts of crazy square roots in it. So let's just say that l is
equal to the square root of-- let's see, how can
we rewrite this? We could rewrite the 6
as 3 times 2, right? So then we get 3 times 9 times
3, that's the same thing as 9 times 9. That's 81, right? 81 times 2. And we have a 1 in
the denominator. So l is equal to 9, and I'll
do one more step, because I think this confuses
a lot of people. That's equal to the square root
of 81 times the square root of 2, which is equal
to 9 square roots of 2, which is choice A. Problem 239. Of the 200 students at College
T majoring in one or more of the sciences-- I already feel
a Venn diagram coming on. Of the 200 students at College
T majoring in one or more of the sciences, 130 are majoring--
so let me draw the whole-- well, let me read
the whole problem. 130 are majoring in
chemistry and 150 are majoring in biology. If at least 30 of the students
are not majoring in either chemistry or biology, then the
number of students majoring in both chemistry and biology can
be any number from-- OK, let me draw the Venn diagram. So that's chemistry, that's
biology, and then there's a leftover, right? This is a whole pool. So they tell us there's
200 total, right? There's 200 total. And they say that 130 are
majoring in chemistry, 150 in biology, and then at least 30
of the students are not majoring in either chemistry
or biology, so we could say greater than or equal to 30. So this is driving a little
spin here by making it an inequality. Then the number of students
majoring in both chemistry and biology could be any number
from-- OK, so let's say that this is 30, just to simplify
our thinking. If this is 30, how many
people are majoring in chemistry or biology? Well, then it's 170, right? So let's think of it this way. Let's think of this as the
intercept-- this is both. Well, I used B for biology. So let's call this-- I don't
know, let's call this-- I used C for combined. So let's call this I for
the intersection. So if 30 people are majoring
in-- well, actually, let's just call this x. So how many people are majoring
in-- if I wanted to add these two numbers up, if I
want to know the people who are in chemistry or biology,
what would it be? It would be 130 plus 150, and
remember, and we've done this a couple of times with these
Venn diagram problems, if I counted these 130, and then I
count these 150, I've counted this intersection twice. I counted it with the 130, and
I counted with the 150. So I have to subtract out at
least the intersection once. So I subtract it out once. So these are the number of
people who are in chemistry or biology or both, right? So it's the intersection of
chemistry or biology or both. And now we know that that has
got to be equal to what? Well, 200 minus the people who
are in nothing, right? So I'm going to call
that x for now. 200 minus the people
who are nothing. And let's see if we can solve
for the intersection, which is what they're asking
us in terms of x. So this becomes 280 minus I
is equal to 200 minus x. And then they tell us to
subtract 200 from both sides. You get 80 minus I is
equal to minus x. Let's add I to both sides. So you get 80 is equal
to I minus x, right? Or even better, let's
add x to both sides. So you get I is equal
to 80 plus x. And so, if x is greater than
or equal to 30-- so we know that if x is 30, then
we get what? We get 130 plus 80
is 110, right? So we know that I is
going to be greater than or equal to 110. The intersection is
at least 110. Now, what's an upper
bound on it? How many people can
be studying both? And if you look at the choices
already, there's two choices that have 110 as the minimum for
the number studying both. But what's the maximum that
can be studying both? Well, it's 130, right? If all the chemistry people
are also studying biology, then the intersection
becomes 130, right? But you can't have 150 people
studying both, because there's only 150 studying chemistry. So the intersection is greater
than or equal to 110. And it's limited by the number
of people studying chemistry, 130, which is choice D. And I'll see you in
the next video.