# GMAT: MathÂ 11

## Video transcript

We're on problem 61. If 1 kilometer is approximately
0.6 miles, which of the following best
approximates the number of kilometers in 2 miles? So we want to know 2 miles. So we could set up a ratio. We could say, 0.6 miles are to 2
miles as 1 kilometer is to x kilometers. And you could cross multiply. You could say 0.6
is equal to 2. And that x is equal to 2 divided
by 0.6, which is the same thing as 20 divided by 6. Right? 20 divided by 6 is
equal to 3 2/6. Right? 3 1/3. 3 1/3 kilometers. And that makes sense, right? A mile is longer than
a kilometer. So 2 miles is about
3 1/3 kilometers. And that's not one of the
choices, but if we wrote this as an improper fraction, 3 times
3 is 9, plus 1 is 10. So that equals 10/3 kilometers,
which is choice A. Let me do a brighter color. I don't know if you
can even see this. 62. Lucy invested $10,000 in a
new mutual fund account exactly 3 years ago. The value of the account
increased by 10% during the first year, increased by 5%
during the second year, and decreased by 10% during
the third year. What is the value of
the account today? So you start with $10,000. When you grow by 10%, that's
the same thing as multiplying by 1.1. Right? It's like 1 plus 0.1
times whatever. So that's multiplying by 1.1. And then you grow
by another 5%. That's like multiplying
by 1.05. And then when you decrease
by 10%, that's like multiplying by 0.9. So we need to figure
out what this is. So this is a little bit
of math for us to do. So 1.1 times 10,000. That's at least fairly
straightforward. That's 11,000. Right? That's how much we have
after year one. 11,000 and we have to multiply
that times 1.05. And then we lose 10%. So 11,000 times 1.05. Well, just to make sure I get
right, just so I don't make a careless mistake, let's
actually do it. 11,000 times 1.05. So 5 times 11,000 is 55,000. And then we should
add two 0's. The 0 doesn't matter,
so we add two 0's. And then 1 times 11,000
is 11,000. You add them up. You get 0, 0, 0, 5, 5, 1, 1. And we have two digits behind
the decimal point. So we got 11,550. Which is right. It would have been weird
if we got 100,000 or if we got 1,000. We only had a 5% change. So then we have 11,550. And now it's going to
decrease by video. 10%. We could multiply it by 0.9. I don't know, it might be easier
just to take out 10%. Well, let's multiply
it by 0.9. 9 times 0 is 0. 9 times 0 is 0. 9 times 5 is 45. 9 times 1 is 9, plus 4 is 13. 9 times 1 is 9, plus 1 is 10. And we have one digit behind
the decimal point. So it's right there. So we're left with $10,350,
and that is choice A. Next problem. 63. And they give us this table. They say a computer chip
manufacturer expects the ratio of the number of defective chips
to the total number of chips in all future shipments
to equal the corresponding ratio for shipments
as combined, as shown in the table. What is the expected number
in a shipment. OK, so let me write down what
they wrote, so that we see the same thing. So shipment. And they labeled them
S1, S2, S3, and S4. And they need to ask us the
number of defective chips in the shipment. I know you can't read
my handwriting. 2, 5, 6, and 4. And then they say the total
number of chips in each shipment is 5,000, 12,000,
18,000, and 16,000. Now let's reread the question. Maybe it'll make a little
bit more sense. A computer chip manufacturer
expects the ratio of the number of defective chips to the
total number of chips -- so that's defective to total --
in all future shipments to equal the corresponding ratio
for shipments S1 through S4 combined, as shown in
the table above. Let me make sure I read that. To equal the corresponding ratio
for shipments S1, S2, S3, and S4 combined, as shown
in the table above. What are the expected number
of defective chips in a shipment of 60,000? They're saying, essentially,
if we take these shipments combined, what percentage
were defective? And that's the same number we
would expect to be defective in a shipment of 60,000. So let's figure out what
percentage are defective. So how many total
were defective? 2 plus 5 is 7, 7 plus 6 is
13, 13 plus 4 is 17. And how many total chips
did we ship? 17,000. 17 plus 18 is 35. 35 plus 16 is 41, is equal
to 51,000 chips. So we could view it
a couple of ways. It's 17 for every 51,000
were defective. But actually, if you haven't
realized it already, 51 is divisible by 17. That's one that a lot
of people miss. 17 times 3. So then if we divide both of
these numbers by 17, we can say that 1 chip for every 3,000
is defective, if we take shipments S1 through
S4 combined. 1 chip in every 3,000. And so if we have that same
ratio for 60,000 chips, well, how many groups of 3,000
are there in 60,000? Well, there are 20
of them, right? next We could say 1 in 3,000
is equal to x over 60,000. And you could eyeball
this to some degree. Or you could just say, what
can I multiply both the numerator and the denominator
by to get 60,000. Well, if I multiply the
numerator and the denominator by 20, I get 20/60,000. You could cross multiply. There's a bunch of ways
you could do this. But the bottom line is
you would expect 20 defective chips. And that is choice B. Problem 64. I'll do it in this color. Let's see, they tell us that A
is equal to a set of numbers. I don't know what the
context is yet. 2, 3, 4, and 5. And they tell us that B is
equal to another set of numbers, 4, 5, 6, 7, and 8. 2 integers will be randomly
selected from the sets above, 1 integer from set A and
1 integer from set B. What is the probability
that the sum of the 2 integers will equal 9? So let's think about all of the
combinations that equal 9. Right? If I pick 2 from this, I have
to pick 7 from here, right? If I pick 3 from this set, I
have to pick 6 from here. If I pick 4 from this set, I
have to pick 5 from here. And if I pick 5 from this set,
I have to pick 4 from here. Right? So no matter what, I'm going to
pick one of these numbers from set 1. It's going to be 2,
3, or 4 or 5. And depending which of these I
pick, I have to pick the exact right number from set B. So think of it this way. I don't know what I've picked
in set A, right? I don't know. Let's say we've picked
one of them. We don't know which of
these we've picked. And we know that exactly one
of these is going to be the right number to add
to whatever we've picked to equal 9. And we have to pick that
exactly right one. And we're going to be picking it
out of a selection of 1, 2, 3, 4, 5 numbers. So we have a 1 in 5 chance of
having them add up to 9. And 1 in 5 is the same thing
as 0.2, which is choice B. I hope that makes sense. You could say, look, I don't
care what I pick from set 1. I could pick any of these, and
in order for it to add up to 9, no matter what I pick from
here, there's a 1/5 chance that I picked the right
corresponding number that adds up to 9 there. Because each of these
have only one. If I pick 3, I have
to pick 5, right? So if I pick 2, I
have to pick 7. If I pick 3, I have to pick 5. If I pick 3, I have to pick 6. If I pick 4, I have to pick 5. And if I pick 5, I have
to pick 4, right? And so you could say, well,
the chances of each of these are 1/5. Actually I'm not going
to go down that road. All you have to say is that just
overly complicates it. All you have to say is,
look, I'm going to pick one of these. And once I have one of those
in my hand, there's a 1/5 chance that I picked the exact
one that adds up to whatever I just picked to equal 9. And the answer is B. And now I'm out of time. See you in the