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## GMAT

### Course: GMAT > Unit 1

Lesson 1: Problem solving- GMAT: Math 1
- GMAT: Math 2
- GMAT: Math 3
- GMAT: Math 4
- GMAT: Math 5
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- GMAT: Math 7
- GMAT: Math 8
- GMAT: Math 9
- GMAT: Math 10
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- GMAT: Math 12
- GMAT: Math 13
- GMAT: Math 14
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- GMAT: Math 17
- GMAT: Math 18
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- GMAT: Math 27
- GMAT: Math 28
- GMAT: Math 29
- GMAT: Math 30
- GMAT: Math 31
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- GMAT: Math 39
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- GMAT: Math 41
- GMAT: Math 42
- GMAT: Math 43
- GMAT: Math 44
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- 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

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# GMAT: Math 28

143-147, pgs. 171-172. Created by Sal Khan.

## Want to join the conversation?

- @4:51problem 145: I learned in grade school the equation for a trapezoid as (1/2)*(h1+h2)*b

Intuitively, you are taking the average height of the trapezoid and then multiplying that value times the base.(5 votes) - How many different ways 36 cents change can be obtained by using quarter, dime, nickle and pennies.

Please give a technique/formula. It takes a long time to compute the problem in a regular way.(2 votes) - Problem 145: instead of calculating the area of the square you can also take the top triangle and apply the same logic used for the bottom triangle namely base =2 height = 12 thus 2*12*1/2=12 and then 30+12=42(1 vote)
- Hey guys you want a sprite cranberry(1 vote)
- You also do trapezoid area = 1/2 (a+b)h(1 vote)
- At1:00, Number 144, Why doesn't answer choice D work. I solved this problem by just plugging in the answer choices to see which one fulfills both equations. To my appearance, if you plug in 1 to both equations it works...(1 vote)

## Video transcript

Problem 143. 3.003 divided by 2.002 is equal
to-- well, this already looks suspicious. It looks like 3/2, but just to
confirm that, this is the same thing as 3 times 1.001. And this is the same thing
as 2 times 1.001. So you cancel those out, and you
get 3/2, which is the same thing is 1.5. And that's choice E. Problem 144. If 4 minus x over 2 plus x is
equal to x, what is the value of x squared plus 3x minus 4? Let's see if we could simplify
this little bit. Let's multiply both sides of
this equation times 2 plus x, so you get 4 minus x is equal
to 2 plus x times x is 2x plus x squared. Let's subtract x from both--
no, add x to both sides, so you get 4 is equal to
3x plus x squared. Subtract 4 from both sides,
and you get 0 is equal to minus 4 plus 3x plus x squared,
which is just rearranging that, which was the
same thing as x squared plus 3x minus 4, so that is
equal to 0, which is C. Next problem, 145. They've drawn us a trapezoid. Let me see if I can draw
the trapezoid as well. So they have the base, one
side, and then they have another side that looks
something like that, and they connect the two lines
like that. And then they labeled this side
as 2 feet, this side as 5 feet, and they say this
is a, and this is b. All right. The trapezoid shown in the
figure above represents a cross-section of the
rudder of a ship. If the distance from a to b is
13 feet, what is the area of the cross-section of
the rudder in feet? So let's think about
this a little bit. Let's see if we can figure
out this area. So this area right here
is easy to figure out. If this is 13, this is
5, what is this side going to be equal to? We can use the Pythagorean
theorem. If we call this x, we could say
x squared plus this side squared, plus 25, is equal
to 13 squared. 13 squared is what? 169? I want to say that,
but let me check. I haven't memorized my
13 times tables. 3 times 13 is 39. 130. Yeah, 169. Subtract 25 from both
sides, you get x squared is equal to 144. That's tells us we're
on the right track. We got a perfect square. So x is equal to 12. So it's very easy to figure out
the area of this part of the trapezoid, right? It's just the base times
the height times 1/2. So it'd be 12 times 5, which
is 60, times 1/2, so this would be 30, this area here. But this is a little
bit trickier. Well, let's think about
it this way. Let's extend this trapezoid
into a rectangle. So if you did that, then you
can view the area of the rudder, so the original problem
that I have in magenta there as the area of this entire
rectangle minus this green area right here. So let's do that. First of all, what's the area
of this green area? So this side is what? If this is 2, this side is 3. And we already figured
out this width is 12. So the green area is 3 times 12
times 1/2, that's the area of a triangle. So 3 times 12 is 36
times 1/2 is 18. Do it in a different color. So the area of this
green area is 18. So the area of our rudder, this
whole magenta area that I'm shading n, is equal
to the area of the rectangle minus 18. So the area of the rectangle
is 12 times 5, which is 60. And then let's subtract out
the green area minus 18 is equal to 60 minus
10, that's 42. And that's choice C. Problem 146. If 0 is less than or equal to x,
which is less than or equal to 4, and y is less than 12,
which of the following cannot be the value of xy? Let me going to write down
all the choices. A, minus 2. and y is less than 12, I don't
see why not. x could be 2 and y could be minus 1. y could be arbitrarily small. So that's not true,
it's not A. B, 0. Well, x and y actually can
both be zero, so if you multiple them, no reason
why you can't get zero. C, 6. Well, if x is 2, and y is
3, I can easily get 6. And they don't even
tell us that x and y have to be different. They don't even tell us they
have to be integers. 24, Well, the largest number
that x can be equal to is 4, and x can't be a negative, so
if x is 4, times y is 6, you get 24, so I'm already guessing
the answer is E without having looked at it. E, 48. Because the largest value of x
is 4, and so you'd have to multiply 4 times 12 to get to
48, but you can't get there. So the answer is E. Problem 147. Turn the page. I'll do it in blue since
it seems to be dealing with water. 147: In the figure above-- let
me draw the figure above. So I have a V-axis, then I have
this other axis, and then they've drawn what looks like
water, at least before reading the problem. I've drawn that. They say this is V. They say that this distance
is 5 feet. They say that this distance is
10 feet, and they label us a couple of points. It's R and that's S. In the figure above, V
represents an observation point at one end of a pool. From V, an object that is
actually located on the bottom of the pool at point R, so an
object that is actually here, appears to be at point S. The light goes from here, but
then refracts, and goes like that, so if you just follow the
path, it would go straight to point S. If VR is equal to
10 feet-- what? Oh, VR is equal to 10 feet,
so let me draw VR. VR is equal to 10 feet. What is the distance RS, in
feet, between the actual position and the perceived
position of the object? So they want to know this
distance right here. Well let's think about
it a little bit. If we can figure out this
distance, then we subtract that distance from 10, we'll
have this distance. So what's this? Let's call this x. This is a right triangle, so
you have x squared plus 5 squared, plus 25, is equal to
this hypotenuse squared, is equal to 100, which is the
Pythagorean theorem. x squared is equal to 75, so
then x is going to be equal to-- let's see, square
root of 75. That's someplace between
8 and 9, right? 8 is 64, 9 is 81, so
x is greater than 8 and less than 9. Oh, actually, they write it in
terms of square root of-- let me simplify this a
little bit more. So I can write that as-- the
square root of 75, I can write as 3 times 25. Square root of 25 is 5,
so that becomes 5 square roots of 3. So this distance right here
is 5 square roots of 3. So if I want to figure out this
distance, I just subtract that from 10. So 10 minus 5 square roots of
3, and that is choice A. See you in the next video.