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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 49
- GMAT: Math 50
- GMAT: Math 51
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- GMAT: Math 54

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# GMAT: Math 50

232-236, pgs. 184-185. Created by Sal Khan.

## Want to join the conversation?

- The explaination for 234 is still a little strange.

If every red bead is worth 7points/bead then how can it be that only 2 beads were removed? Since all 4 types of beads need to add up to 147000 points. It is true that 7 can be factored out twice, which gives us the 2, but that doesn't mean only 2 beads were removed.

If the question is how many beads were removed, shouldn't it be a lot more than just 2 beads?(2 votes)- 9 Beads in total but the question only asked for how many red beads, ie how many 7's.(2 votes)

## Video transcript

We're on problem--
where are we? Problem 232. Only a few left. If 1/x minus 1 over x plus 1 is
equal to 1 over x plus 4, then x could be which
of the following? And you might have a temptation
to just invert both sides, but you can't
do that just yet. Because if you took the inverse
of this, it's not like you just get x minus x plus 1. You actually have to subtract,
or you could say add these two fractions together. So let's do that. So if we simplify this left-hand
side, a common denominator would be
x times x plus 1. And then a numerator would be,
let's see-- 1/x is the same thing as x plus 1 over this
thing, because that cancels out and you get 1/x, minus-- 1
over x plus 1, that's the same thing as x/x times x plus 1. The x's would cancel out,
you get 1 over x plus 1. So this left-hand side, the x's
cancel out and you get 1 over-- well, we could
just make that x squared plus x, right? So now this simplifies to-- the
left-hand side is 1 over x squared plus x, and the
right-hand side is 1 over x plus 4. And now we can invert both
sides or just set the denominators as equal
to each other. But let's just invert both. It's the same thing, actually. So if you take the inverse of
both sides and you get-- and I'll just switch colors just
randomly-- x squared plus x is equal to x plus 4. Let's subtract x from
both sides. You get x squared
is equal to 4. So x is equal to plus
or minus 2. And they say then x could be--
well, choice C is minus 2, so that's one of the values
that x could be. Next question, 233. All right, they have 1/2 to the
minus 3 power times 1/4 to the minus 2 power times
1/16 to the minus 1. The first thing I do when I see
negative exponents, just so they don't confuse
me, is that I just invert all of the numbers. So this is the same thing. This is the exact same thing as
equal to-- this is the same thing as 2 to the third
power, right? When you take something to a
negative exponent, that's the same thing as the inverse of
that thing to the positive exponent, so this is 2
to the third power. This is times 4 squared
times 16. And so 2 to the third is 8
times 16 times 16, right? That's 8 times 256. And then-- oh, wait. I'm going down the wrong path,
because they have it in completely different format. So actually, maybe I should look
at the choices before I continue these problems.
So let me rewrite it. So they all have it as -- you
know, 1/2 to some power, 1/8 to some power, so let
me rewrite this. I went down the wrong road. So let's write all of these
as powers of 1/2. So 1/2 to the minus 1/3, fine. That's still 1/2 to
the minus 1/3. 1/4, what is that? 1/4 is the same thing
as 1/2 squared. We're raising that
to the minus 2. And then 1/16, that's the same
thing as 1/2 to the third power is 1/8, so it's
to the fourth power. We're taking that to
the negative 1. And then this is-- we'll just
use what we know about exponent rules, and we've proven
this in other Khan Academy videos. 1/2 squared to the minus 2. You can just multiply these
exponents, so that's equal to 1/2 to the minus 4. And then 1/2 to the 4 to the
minus 1, you multiply them and that's equal to minus 4. And now we have the same base,
and we're multiplying them, so we can just add the exponents. This is equal to 1/2 to the
minus 3 plus minus 4, minus 7, minus 4, is equal to minus 11. So it's 1/2 to the negative 11th
power, which is choice B. That was a good lesson there. Look at your choices before you
go down some path because I was actually going to evaluate
this thing, which would have taken a little
bit of extra time. Problem 234. In a certain game, a large
container is filled with red, yellow, green-- so let's see,
there's red, yellow, green, and blue beads, worth
respectively-- so respectively means the same order; we said
the colors are in the same order that they're worth -- so
worth respectively 7, 5, 3, and 2 points each. OK, a number of beads are then
removed from the container. If the product of the point of
values of the removed beads is 147,000, how many red
beads were removed? So this is interesting. So there's something that
hopefully will jump out at you about these numbers. They're all prime
numbers, right? So if you look at any number,
you can tell how many of a certain prime number there is in
that number by just doing a prime factorization. So let's do that for 147,000. And you'd be surprised
how quickly a number can get smaller. So let's see. How many times-- let's
see, does 3 go 147? I think we've done
this before. Sure, 3 times -- let me
just do it up here. 3 goes into 147 4,
12, 27, 49 times. So you get a 3 and you
get a 49, right? 49,000, I should say. You could do a 7 and a 7,000. And then you can get another
7, and you can get a 1,000. And then you could do
2 a bunch of times. So you get a 2, you get a --
well, the main thing, just so we don't waste time is they want
to know-- what color do they want to know? They want to know how many red
beads were removed, right? So red corresponds to 7. So all the beads, you
multiply them together and you get 147,000. So we've essentially factored
out all the 7's that we can fit into 147,000. 7 does not go into
1,000 at all. 7 is not a factor. So we've already solved
our problem. There are exactly two 7 prime
factors in 147,000. You could keep going
with this. You know, you get a 2. 2 times 500. You get another 2. 2 times 250. There's a bunch of
2's in here. And then you'd probably get-- I
don't know what else you're going to get. You'd probably get a couple
of 5's in here, and then you'd be done. But these are all the 7's. There are two 7's in 147,000. Two 7 prime factors, so there
would be 2 red beads. And that is choice D. Problem 235. Switch colors. 235. If 2 over 1 plus 2 over
y is equal to 1, then what does y equal? Let's take the inverse
of both. Let's multiply both sides of the
equation times 1 plus 2/y. And so we get 2 is equal
to 1 plus 2/y. And let's multiply both sides
of the equation by y. So you get 2y. And actually, you could probably
already eyeball this. Actually there's
an easier way. Let's subtract 1 from
both sides. You get 1 is equal to 2/y. Multiply both sides by y and
you get y is equal to 2. And that is choice D. And you could've actually
eyeballed it, because, say, 1 plus 1 is equal to 2,
so this has to be 1. So how am I going to
make that a 1? That equals 2, but sometimes
it's easier just to mechanically go through the
algebra, because sometimes, you can't eyeball it,
and it's good to just keep moving forward. Problem 236. If a, b, and c are consecutive
positive integers, and a is less than b, which is less than
c, which of the following must be true? Consecutive positive, so, you
know, we could say that b would be equal to a plus 1, and
c would be equal to a plus 2 or b plus 1. And, of course, a
is equal to a. I don't know if that helps,
I just felt like writing that down. So in statement 1, they say
c minus a is equal to 2. Sure. They're consecutive positive
integers, so c is going to be 2 more than a. So this is right. Statement 2, they tell
us abc is even. Well, instead of going into some
algebraic thing where I represent even and odd integers
and all of that, let's just try it with two
cases, one where a is odd and one where a is even. So let's say a is even. So the numbers would be 2--
sorry, would be 2 times 3 times 4, right? So then we would get 8. Well, you know what? This has to be even because one
of these numbers are going to be even, right? And so if you're divisible, and
by definition, if one of these numbers are even,
it's divisible by 2. So the whole thing, when you
multiply it together, it's got to be divisible by 2, right? Either a and c are even, in
which case this whole thing has to be even, or b is even, in
which case this whole thing is even, right? If you multiply a string of
numbers and at least one of them is even, then that whole
number is going to be even, because that number that's even
is a factor of that whole string of numbers. But anyway, you could try it
with 2, 3, 4, or you could try it with 3, 4, 5. Either way, you're going
to get an even number. So this is also right. Statement 3. a plus b plus c over
3 is an integer. Well, let's think about it. I don't know. Let's say if we were to
take a plus b plus c. I don't know, even if we were to
take-- let's see, 3 plus 4 plus-- no, that works. 2 plus 3-- no, that works. Let's write it as-- so this
is a plus a plus 1, right? B is a plus 1, plus a plus
2 over 3, right? And so this is equal to 3a plus
3 over 3. b is just a plus 1, c is a plus 2. You add them all together. Sure enough, you know, you could
divide the top and the bottom by 3 and you
get a plus 1. So yes, statement 3
is also correct. This is always going to evaluate
to a plus 1, which is an integer. So all three are true,
so that is choice E. And I'm all out of time. See you in the next video.