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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
- GMAT: Math 6
- GMAT: Math 7
- GMAT: Math 8
- GMAT: Math 9
- GMAT: Math 10
- GMAT: Math 11
- GMAT: Math 12
- GMAT: Math 13
- GMAT: Math 14
- GMAT: Math 15
- GMAT: Math 16
- GMAT: Math 17
- GMAT: Math 18
- GMAT: Math 19
- GMAT: Math 20
- GMAT: Math 21
- GMAT: Math 22
- GMAT: Math 23
- GMAT: Math 24
- GMAT: Math 25
- GMAT: Math 26
- GMAT: Math 27
- GMAT: Math 28
- GMAT: Math 29
- GMAT: Math 30
- GMAT: Math 31
- GMAT: Math 32
- GMAT: Math 33
- GMAT: Math 34
- GMAT: Math 35
- GMAT: Math 36
- GMAT: Math 37
- GMAT: Math 38
- GMAT: Math 39
- GMAT: Math 40
- GMAT: Math 41
- GMAT: Math 42
- GMAT: Math 43
- GMAT: Math 44
- GMAT: Math 45
- 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 52

240-244, pgs. 185-186. Created by Sal Khan.

## Want to join the conversation?

- What are the steps in solving math problems like this 3-(-52)(3 votes)
- When you have a minus sign next to a negative number, you make it a plus sign: 3+52.

3+52 is the same as 52+3, which is 3 numbers after 52: 52,53,54,-55-.**55**is our answer.(4 votes)

- Mixture X+Y contains some bluegrass in it, which originates from X, therefore the % of mixture X in X+Y is the %ryegrass from X + the% bluegrass from X / everything.

Why are we ignoring the fact X+Y contains some bluegrass from X. We are in fact answering percent of the ryegrass in X /ryegrass in mixture.

I have no issues with algebra, it´s the wording which I feel is incorrect(2 votes)- Try to forget what is ryegrass, bluegrass and fescue. Those specifications are just trying to confuse you. What's more helpful is if you try to reduce the number of variables.

Here's an alternative way to think about it.

40% of X is rye (R) and 60% of X is other (O).

25% of Y is rye (R) and 75% of Y is other (O).

When you mix X and Y, you get 30% R + 70% O.

Multiply all percents by 100 to get numbers that are more manageable. This info helps you set up the following system of equations:

40X + 25Y = 30

60X + 75Y = 70

Now solve for X by eliminating Y. To do so, find the Greatest Common Factor (GCF) for 25 and 75. In this case, GCF = 75, so all you need to do is multiply the first equation by -3 so you can cancel out the Y.

-120X - 75Y = -90

60X + 75Y = 70

Y cancels out and you are left with:

-60X = -20

X = 1/3 = 33 1/3%(4 votes)

## Video transcript

We're on problem 240. Seed mixture x is 40% rye
grass and 60% bluegrass. Let's see, x is 40% rye grass
and 60% bluegrass. Fair enough. Seed mixture y is 25% rye
grass-- so it's 0.25 rye grass-- and 75% fescue. So I should look up
what that is. 75% fescue. If a mixture of x and y contain
30% rye grass, what percent of the weight
of the mixture is x? OK. Well, let's say that A is equal
to the percent of weight of mixture that is x, what
they're asking for. So how much rye grass-- what
percentage of that mixture will be rye grass that
comes from x? Well, we have A%, or you know,
we could call that-- fraction A of the mixture is from x. And then x is 40% rye grass. So if we multiply the amount
that we're getting from x times 40%, times 0.4, this'll
tell us the amount of rye grass from x. So r from x, or the proportion
of rye grass from x. And now, how much rye grass are
we going to get from y? So first of all, what's
our proportion of y? If this mixture is A of
x, there's going to be 1 minus A of y. If this is 1/4, this is going
to be 3/4, right? Makes sense. So this is how much y we have,
and y is 25% rye grass. So our rye grass from y will
be the amount of y we have times how much y is rye grass. So this is rye grass from y. And they tell us that this
is equal to a mixture. The mixture should have
30% rye grass. So let's just solve for A. We get 0.4 times A plus-- let's
see, multiply-- 0.25 minus 0.25A-- just distributed
the 0.25-- is equal to 0.30. 0.4A minus 0.25A-- 40 minus
25-- that's 0.15A. And then subtract 0.25 from both
sides, is equal to 0.5. So A is equal to 0.5 divided by
0.15, is the same thing as 5 over 15, which is equal to
1/3, which is equal to 33 and 1/3%, Which is choice B. Next problem. Problem 241. If the integer n has exactly
three positive divisors, including 1 and n-- so it's
divisors are, I don't know, 1, some number, and n-- how many
positive divisors does n squared have? So one thing, I don't know if
you've ever thought about this, but how can any
number have an odd number of divisors? Right? Because normally, if you take
a number, let's take the number 6-- and you shouldn't
do this on the GMAT, it's a waste of time. But its divisors are 1 times
6, or 2 times 3. So every divisor tends to have
a corresponding divisor. So how can you have an odd
number of divisors? Well, only if one of the
divisor's corresponding divisors is itself. If it's 1 times n, and x times
x, which implies that if you have an odd number of divisors,
you're dealing with a perfect square. Anyway, I just wanted
to point that out. But that's actually not that
relevant to that problem. It's just something, you know,
it is useful sometimes. But anyway. They just want to know how many
divisors n squared has. Let's just find a number
that has this property. Well, we know any perfect
square-- well, a small perfect square will have
this property. So let's see. Number 4, it has its property. It's factors are 1, 2, and 4. So how many factors does
4 squared have? Well, 4 squared is 16. Its factors are 1 and 16,
2 and 8, and 4 and 4. So it has five factors,
and that is choice B. And you could do it
mathematically. You could say all the
combinations of when I multiply all the factors of
this times itself, and you could say, oh, they have
five combinations. But in this case, it's just way
easier to think of, OK, let me think of a small perfect
square that's going to have three factors. A perfect square just tells you
it has an odd number of factors, but a small one
will have three. And let me just square it and
just count its factors. Fastest way to do that
problem, I think. Unless there is-- I
don't know, maybe there is a faster one. 242. If n is a positive integer, then
n times n plus 1, times n plus 2 is-- OK, they have
a bunch of things. What happens when n-- let's see,
even when n is even, the even only one is odd,
odd [? 1n ?] is odd. OK, so they want us to
see what happens when n is even or odd. So let's see what happens
when n is even. Let's say that n-- if n is even,
that means it's equal to 2 times some integer k. We don't know, some positive
integer k. So let's substitute that in. So then this breaks down
to 2k times 2k plus 1, times 2k plus 2. Interesting. And let's see, it looks like
this is saying that e should be-- right. Because let's see, we could
factor 2 out of this one. So this becomes 2k times
2k plus 1, times 2, times k plus 1. Right? I just factored the
2 out of there. And let's multiply these two. So you get 4k times 2k plus
1, times k plus 1. So this one is divisible by 4. And I just glanced at statement
e, they talk about divisibility by 4, so
I think we're done. This simplifies to this if we
can assume that n is even. So if n is even, then we are
definitely divisible by 4. 4 can definitely go into
the number evenly. So statement E is right. Divisible by 4 when n is even. Statement E. Next problem. We're on the last page, 243. A straight pipe 1 yard in
length was marked off in fourths and also in thirds. Maybe I'll give some
dimensionality to our straight pipe. So let's say the pipe looks
something like that. It's a straight pipe, and
it was marked off in fourths and thirds. So if I mark it off in fourths,
let's see, that's the middle, and that's a
fourth, and that's another fourth, roughly. And if I mark it off
in thirds, it looks something like this. Right? All right. If the pipe was then cut into
separate pieces at each of these markings, which of the
following gives all the different lengths of the pieces
in fraction of a yard? Fascinating. OK, so we're going to cut it,
actually, at here, and here, and here, and here,
and at here. And actually, if we go halfway,
then we can just use symmetry to say, OK, this piece
is going to be the same as this piece, and this piece
is going to be the same as this piece, and that piece
is going to be the same as that piece. So we really just have to
figure out up until the halfway point. So let's think about it. So this whole thing is 1 yard,
so this distance right here, that distance right
here is 1/4. That distance is 1/4. We know that this whole
distance is 1/3. This distance is 1/4, so this
distance right here is going to be 1/3, because that's
1/3, minus 1/4. Common denominator over 12. 1/3 is 4/12, minus 3/12
is equal to 1/12. So this distance is
going to be 1/12. And then what's going
to be this length? Let me get a different color. I want to make sure that
you see that 1/12. Well, this distance is 1/2, and
this distance is 1/3, so it's going to be 1/2 minus 1/3
is equal to common denominator of 6, is equal to 3/6 minus
2/6, is equal to 1/6. So this piece is 1/4. This piece is 1/12. This piece is 1/6. And then you could just say, oh,
well, it's going to happen the same thing on
the other side. This is 1/6, 1/12, and
then this is 1/4. And so if I say-- let's see,
1/6, 1/12, and 1/4. Right, that is choice D. 1/6, 1/12, and 1/4. Choice D. Those are the different
piece sizes we have. Problem 245. If 0.0015 times 10 to the m,
over 0.03 times 10 to the k, is equal to 5 times 10 to the
seventh, then what does m minus k equal to? And that's important to
read, I was about to solve-- m minus k. So they want to know
what m minus k. So if we immediately divide
these, we should be able to come up with something
interesting, right? Because this is the same thing
as 0.0015/0.03 times 10 to the m divided by 10 to the k. Well, that's just times 10 to
the m minus k, is equal to 5 times 10 to the seventh. And I don't know, let's see if
we can do anything interesting to this decimal number. Let's see, if we were to--
let's just divide it. So if we just say 0.03, instead
of multiplying by 10 and doing-- going into 0.0015. Let's add some 0's. OK. If we take this decimal two to
the right, we're going to take this decimal two to the right. So then we get 3 goes into 1
zero times, 3 goes into 15 five times. So we get 0.05. So this whole thing simplifies
to-- and I'll write it in magenta-- so that simplifies
to 0.05 times 10 to the m minus k, is equal to 5 times
10 to the seventh. So let's divide both
sides by this 0.05. What is 0.05 goes into 5? Let's add some decimal
points here. So if I push this to the right
two, I push this to the right two, and this is my
new decimal point. So 5 goes into 5-- well,
5 goes into 5 one time, so it's 100. Right? So 5 divided by 0.05 is 100. So we get 10 to the m minus k,
is equal to 100-- what's 100? That's the same thing
as 10 squared. Times 10 to the 7th. So 10 to the m minus k is equal
to 10 to the ninth. So m minus k must
be equal to 9. And that is choice A. See you in the next video.