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## GMAT

### Course: GMAT > Unit 1

Lesson 1: Problem solving- GMAT: Math 1
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# GMAT: Math 38

190-194, pg. 178. Created by Sal Khan.

## Want to join the conversation?

- I don's understand in question 192 who -1^2 = -1. In my opinion it should be 1. Can somebody please explain?(5 votes)
- I'm sure you have progressed through life just fine without having this question answered here, but someone else might wonder about it.

The question was to simplify the expression

(√2 + 1)(√2 - 1)(√3 + 1)(√3 - 1)

So you are asking about`-1²`

There is not any time during that part of the video where Sal squares -1

If he did, he would write it (-1)² and the result WOULD be positive 1. Meanwhile, the expression -1² always results in -1, due to PEMDAS. We do the square of the number that the exponent sits on, then we apply the negative multiplier in that expression.

If we read -1² the words are, "the negative of (1 squared)". It does not mean the square of -1, so if you ever want to convey "the square of -1" to someone, you should use the parentheses: (-1)²

Anyway, he was not squaring a -1 here at any point, he was using a super fast shortcut for FOILing a hairy set of expressions.

What he does do is use the pattern (a + b)(a - b) = (a² - b²) to quickly multiply the expressions.

Notice in a + b that both**a**and**b**are positive. Notice that in a - b, we are subtracting that positive**b**from a positive**a**. Finally notice that the result is b² subtracted from a²

So, the pattern is to take**a**, which is √2 and square that to get 2, THEN find the result of 1² to get 1, then subtract the results: 2 - 1 = 1

AND, for the second pair, again square**a**, which is √3 and square that to get 3, THEN find the result of 1² to get 1 again, and subtract that pair of results: 3 - 1 = 2

The last step was to multiply the 1 ∙ 2 for the grand finale of`2`

(1 vote)

- i found the first question difficult mind breaking it down(3 votes)
- I understand what Sal is saying in Q191 about maximising the radius (because that term is squared, but given the cylinder does't perfectly match the footprint of the box (i.e. there is waste), does that intuition necessarily follow? E.g. You could have 6 x 8 x 20 where r = 3 cylinder has greater volume than r = 4 one.(2 votes)
- The intuition that Sal refers to is the concept, "In order to maximize volume of a cylinder, do you want to have a lengthy height, or a hefty radius?"

The volume of the cylinder is pretty simple: the radius squared times the height (times π, of course). So the radius contributes more than the height, since the radius is squared. Yes, if you get a long enough cylinder, the height is going to make a big enough difference, but`if the numbers are fairly close,`

you can get a lot more volume by having a little more radius.

Let's look at the three possible side sizes for this 6 x 8 x 10 carton:

6 x 8 → the maximum possible diameter to fit within this side is 6, so the maximum radius is 3. The length of the cylinder is the remaining dimension of 10.

This volume will be 3²∙10π = 90 π

6 x 10 → the maximum possible diameter to fit within this side is**still**6, so the maximum radius is still 3. The length of the cylinder is the remaining dimension of 8.

This volume will be 3²∙8π = 72 π

8 x 10 → the maximum possible diameter to fit within this side is 8, so the maximum radius is 4. The length of the cylinder is the remaining dimension of 6.

This volume will be 4²∙6π = 96 π, the winner!

Since the test-makers are testing the concept that radius squared contributes more than a linear value, they probably will not have an example like yours. An interesting question is`how much bigger does the 3rd dimension have to be in order to cause the concept to break down?`

(1 vote)

- I had a question about Problem 191, I don't know if the person who wrote this problem has ever played with those stack-able shapes as a child, but you can't put something that is 8 inches, inside something else that is also 8 inches. Therefore the answer would have to be something that is less than a radius of 4. So the answer should be radius of 3 ( if there is not a choice like 3.9) right?(1 vote)

## Video transcript

We're on problem 190. If the operation star with a
circle around it is defined for all a and b by the equation,
a star with a circle around it b is equal to a
squared times b over 3, then what does 2 star with a circle
around it 3 star with the circle around it
minus 1 equal? So 3 star minus 1. That's equal to-- let's see,
I'll do this in magenta-- 3 star minus 1. That's equal to 3 squared times
minus 1 which is minus 3 squared, or minus 9. 3 squared, 9. Times minus 1. That's minus 9 divided by 3
which is equal to minus 3. So this reduces to 2 star minus
3 which is equal to 2 squared, which is 4,
times minus 3. All of that over 3. Minus 3 divided by 3. That's just minus 1. So you have minus 1 times
4 is equal to minus 4. And that's choice E. Problem 192. No, 191! Don't want to skip problems.
The inside dimensions of a rectangular wooden box
are 6 by 8 by 10. A cylindrical canister is to be
placed inside the box so it stands upright when the
closed box rests on one of its 6 bases. Of all such canisters that could
be used, what is the radius in inches of the one
that has maximum volume? OK, so they're going to put
one canister in this box. And how do you maximize
its volume? So let's think about
it this way. Let's think about what a volume
of a cylinder is. If this is my cylinder, its
volume is the area of the side times the height. If the area of the side is
equal to-- if this is the radius-- it would be pi r
squared times the height. Height times pi r squared. Now what's r going
to be equal to? r has to go all the
way around. So, r is going to be equal to
half-- well, it obviously equals half the diameter
of this circle. But if we look at the base
that's 6 and 8, if we use that as the base, and the 10 comes
out of the page, the radius can only be what? The radius can only be 3 because
the diameter of the circle can only be 6. We don't get to use the 8. So, this is the case if we
use the 6 and the 8 side. And so we'll kind of be-- I
guess the best way to think about it is we'll be wasting
this base over here. The other possibility, if we had
the 8 and the 10 side, if this was our base then we could
have a radius of 4. Let me think about it. Yeah, radius would be a 4. Because we would have
the height 8, we'd have a radius of 4. Now when we look at this
equation, if we had to maximize either the-- this is
going to be the volume of the cylinder-- if we had to maximize
either the height or the radius, because those are
the two things that we can deal with, which one, at least
to you, seems more important? Well, to me, the radius seems
more important because I'm squaring it. So, these, in my mind, are
the two contenders. The worst contender is where
we use the 6 and the 10 side as a base. Because then we use
a lot of space. But we can figure that
one out, too. Because then we waste a lot of
space and we have a height of 8 and the radius is still--. So, first of all, this one is
clearly better than this one. And why is that? They have the same area
of their base because the radius is 6. Sorry, the radius is 3,
the diameter is 6. But in here the height gets to
be 10, and here the height only gets to be 8. So this one is a non-option. So, really, we have to
compare these two. And I've drawn their bases. So, if you think about this one,
we have an area of the base of the radius is 4. So, 10 pi times the
height of six. That's the volume of
this cylinder. Here we have a radius--
we have a radius of 3. Half of the 6 diameter. So, the area is 9 pi. And we're going to multiply
that times the height, which is 10. So, here we have 90 pi. And here we have 6 times
16, which is 96 pi. So, this is the best
of our cases. And that was the correct
intuition. We're maximizing the radius. And when you maximize the
radius, we get a radius of 4. So what did they ask us
in the question again? What is the radius in inches
of the one that has the maximum volume? And so that radius is 4. Where one side is 8 and
one side is 10. And that's choice B. Next question. 192. Square root. OK, they have all sorts
of interesting things. They essentially want
us to simplify. The square root of 2 plus 1
times the square root of 2 minus 1 times the square root
of 3 plus 1 times the square root of 3 minus 1. So, this seems to be just an
application of-- I'm sure you've learned this in algebra--
that a plus b times a minus b is equal to a squared
minus b squared. You don't have to
memorize this. You could actually multiply it
all out and you'll get this. But it's a good thing
to know if you want to do things quickly. So, square root of 2 plus 1,
times square root of 2 minus one, is equal to square root of
2 squared, minus 1 squared. We're going to multiply that
times-- using the same principle-- square root of 3
squared minus one squared. Square root of 2 squared. That's 2 minus 1 times
3 minus 1. That's equal to 1 times 2. Which is equal to 2. And that is choice A. Problem 193. In a certain Calculus class,
the ratio of the number of math majors to the number of
students who are not math majors is 2:5. So, math to not math
is equal to 2:5. If 2 more math majors were to
enter the class the ratio would be 1:2. So if I have 2 more math majors,
then the ratio of them to the non-math majors
is equal to 1:2. How many students are
in the class? They want to know what m plus
nm, non-math majors, is. Let's see. Let's see if we can solve
for anything. We actually have two linear
equations and two unknowns. You might not see it yet,
but if we cross multiply we get 5m. I'll just call these
n, non-math majors. I don't want to keep
saying nm. 5m is equal to 2n. And here you get 2 times
m plus 2 is 2m plus 4 is equal to n. We could substitute that. If n is equal to that, then that
means that 5m is equal to 2 times n. n is equal
to 2m plus 4. So, you get 5m is equal
to 4m plus 8. Subtract 4m from both sides,
you get m is equal to 8. If m is equal to 8, we could
say 5m is equal to 2n. So, 5 times 8 is 40. And so n is equal to 20. Let's see, are they asking for
the-- if 2 more students were to enter the class? OK. So they want to know how many
are in the class right now. So they want to know m plus n. Well, that's equal to
8 plus 20 which is equal to 28 students. And that is choice D. Next question. I think I have time to
fit this one in. I'll continue in the
next if I don't. 194. What is the units digit of 13 to
the fourth times 17 squared times 29 to the third? So, here you just have to
realize that the units digit in any of these numbers, are
just the units digit of each of these numbers times
each other. And then you take the
units digit of that. So, we really just have to worry
about the units digits. The answer to this question
would be the exact same thing as if I said the units digit
of 3 to the fourth. We just have to worry about
the units digits. 3 to the fourth times 7 squared
times 9 to the third. It would be the exact
same thing. So, 3 to the fourth, what's
the units digit? 3 to the one is equal to 3. 3 squared is equal to 9. 3 to the third is 27, but
the units digit is 7. And then 3 to the fourth is 81
and the units digit is 1. And just so you can see this
principle that I just talked about-- and we could prove it if
I had more time-- is that I could just multiply 3 times 7
and the units digit is just 1. Or I could just multiply
3 times 9 and the units digit is 7. We could just ignore
everything else. So here the units digit is 1. 7 squared, what's
the units digit? Well, 7 squared, the units
digit is-- well, 7 squared is 49. The units digit is 9. And then 9 to the third,
what's the units digit? 9 to the 1 is 9. 9 squared. The units digit is 1. It's 81. And then 9 to the third, we
don't have to worry about anything else. When you multiply 81 times 9,
the units digit is going to be 9 times that 1 in 81. So, that equals 9. So, it's 1 times 9 times 9. And then 9 times 9 is 81. We want the units digit,
which is 1. So the answer is E. And I'll let you think about
that a little bit. Maybe you want to play around
and see why we only have to worry about the units digit
and why I was able to make this statement right here
to avoid doing a lot of harder math. Anyway, see you in
the next video.