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Current time:0:00Total duration:11:13

We're on problem 55. At a certain picnic each of the
guests was served either a single scoop or a double
scoop of ice cream. How many of the guests
were served a double scoop of ice cream? So let's say that d is the
number of guests that were served double scoop and
that's what we're trying to figure out. So what are their statements? Statement number one says at the
picnic 60% of the guests were served a double
scoop of ice cream. So 60% percent of the total were
served a double scoop. Let's see if we can write
that algebraically. So 60% of the guests. So the total number of guests
are the number of guests that were served double scoops plus
the number of guests that were served single scoops. Then we have all of
the guests, right? This is this is equal to the
total number of guests. And then they tell us that
60% of them were served double scoop. So the number that were
served double scoop is 60% of the total. So that statement number one is
just an algebraic equation. And this alone though doesn't
help me solve this equation. Maybe in the next statement
they'll tell us how many total guests there are. What this is. And that'll help me
but I don't know. Let's see, statement number two,
they say a total of 120 scoops of ice cream were
served to all the guess at the picnic. So once again, they're telling
us that there are 120 scoops of ice cream, they're
not telling us that there are 120 guests. So let's see, what are the
total number of scoops. So there's the number of singles
groups and each of those are-- the number of guests
that got singles scoops are s and then the number of
scoops is just 1 times that because they each
got one scoop. And then you have the
number of guests that got double scoops. But then the number of scoops
they got is going to be 2 times that. So this is the total numbers
scoops, right? The number of people that got
single scoops times 1 plus the number that got double scoops
times 2 and that is going to be equal to the total
number of scoops. And this alone isn't going to
help us solve how many double scoops were served, it depends
on the singles scooped and all of that. But if you look at both of these
statements we have two linear equations and
two unknowns. So at this point you should
immediately say, unless somehow these are the same
equation or somehow these are two parallel lines, these should
intersect and allow you to solve this equation. So immediately-- especially if
you don't want to waste time-- you should say that both
statements together are sufficient. And if you want me to prove this
to you, let me see if I can solve this. So I'll do it in a
different color. So let me do some
substitution. So here we get s is equal
to 120 minus 2d. And we can substitute
this back here. So you get 0.6 times d plus s,
which is 120 minus 2d, is equal to d. And so you get 0.6 times 120
minus d is equal to d. And what's 0.6 times 120? That's 72, right? 6 times 12 is 72. 72 minus 0.6d is equal to d. So you get 72 is equal to 1.6d
and then 72 divided by 1.6 is equal to 45. So there, we solved the problem.
d is equal to 45. 45 double scoops were served. And then we could substitute
back into the equation to figure how many single scoops or
how many total guests there were, but this is all that
we had to figure out. And we were able to using both
statements combined. To the next problem. 56. What is the value of xy? OK. Statement number one: they
tell us that y is equal to x plus 1. And that's not going to tell me
what xy is, I mean there's a bunch of x's and a bunch of
different y's depending on what I choose. Two: they tell us y is equal
to x squared plus 1. Once again, this by itself-- it
completely depends on which xy's I pick, right? I can pick any combination
and get different values. You could try that
out yourself. But I have two equations
and two unknowns. It's a little tricky because one
of them's non-linear, but let's see if we can figure
it out with substitution. If y equals this and y equals
that, we could set these two equal to each other and see
if we can solve for x. So we get x plus 1, that's the
first one, should equal this one, because they're
both equal to y. x squared plus 1. Let's see, we can subtract 1
from both sides and you get x is equal to x squared. And let's see, you could say--
well, you could divide-- if we assume that if x does not
equal 0-- so there's two solutions here, right? x could be equal to 0. That's completely legitimate. They never told us that
x isn't equal to 0. That would satisfy
that equation. And then what other
x would satisfy? x could be equal to 0 or
x could be equal to 1. Either of those are completely
legitimate answers here, right? Fair enough. If x is 0 this is true,
0 equals 0. If x is 1, 1 is equal
to 1 squared. x can't be negative 1. Negative 1 is not equal
to negative 1 squared. So these are the two
solutions for x. So right now it seems like
it's a little shady, but remember, we're not trying to
solve for x or solve for y, we're trying to solve for xy. So maybe when we solve for y
something, I don't know, interesting could happen. Let's solve for y. If you assume x is equal
to 0 what's y? y is equal to 0 plus 1. y is
equal to 1 in this case. That's one solution. And if you put x is equal
to 0 here you also get y is equal 1. So this is one set
of solutions. And in this reality xy is going
to be equal to what? xy is going to be equal
to 0, right? 0 times 1 is 0. Now if x is equal
to 1, what is y? Well, if x is equal to 1,
y is equal to 1 plus 1, y is equal to 2. Well, then xy is equal to 2. So using both of the statements,
I still don't know whether xy is equal to 0 or 2. Someone would have to tell me
that x is not equal to 0, x is equal to 0, and only then can I
really solve this equation. So the answer to this, even
though we have two equations and two unknowns, because one
of them was a quadratic and had two solutions, we actually
don't know what xy is. So the answer is that the
statements together are not sufficient or E. That's interesting because they
actually make you do a little bit more work than just
eyeballing it and saying, oh, I have two equations
to unknowns. You've got to be a little
careful when the equations are non-linear. At least I think that's right. Alright, question 57. They want to know what
1/x plus 1/y-- some police sirens outside. I live in Palo Alto, there's
not normally a high crime scene here but you never know. Anyway, so 1/x plus 1/y. So statement number one is
x plus y is equal to 14. Actually let me try to simplify
this a little bit. If I were to try to add these
two, what could I do? Well, when you add two fractions
you have to find a common denominator. An easy common denominator
for x and y is just xy. So 1/x is what over xy? Well, it's y/xy plus-- right,
this is the same thing as 1/x-- plus, and if I had
xy here, what's 1/y? Well, that's x/xy. So that can be rewritten
as y plus x over xy. And now when you rewrite this
statement like this, then statement one looks a
little interesting. They gave us the numerator at
least. They haven't told us what xy is but they told us that
y plus x, or x plus y is equal to 14. So statement one, it's kind of
part of the puzzle but it's not completely helping us. Let's see what statement
two tells us. I suspect, if this is solveable,
they'll have to tell us what xy-- yep, they
tell us what xy is. They tell us that xy
is equal to 24. So then we're done, right? This simplifies to this
and they're telling us what this is. x plus y is 14. So it equals 14/24-- not that
we have to really solve for it-- but that is
equal to 7/12. But as soon as we got--
you say, oh, that's that, that's that. You're done. You say, both statements
together are sufficient. So C. 58. If d denotes a decimal is d
greater than or equal to 0.5? They put a 0 here so you make
sure you notice the decimal. d, greater than or
equal to 0.5. Statement number one tells us
that when d is rounded to the nearest 10th the
result is 0.5. So let's see, it could be 0.52
and when you round it to the nearest 10th it becomes 0.5. But on the other hand, it could
be 0.46 and when you're round it to the nearest
10th you also get 0.5. So since statement one says it
could be either one of these, it really doesn't help us to
know whether we're greater than or equal to 0.5. This one is greater;
this one isn't. So statement one by itself
doesn't help. Statement two: when d is rounded
to the nearest integer the result is 1. Now this is interesting. If something is rounded to the
nearest integer it has to be greater than 0.5, right? The smallest number that when
you round it to the nearest integer equal to 1
is 0.5, right? So it has to be 0.5 or it has to
be 0.51, because 0.499999, when you're round it to the
nearest integer, is going to be equal 0. That's the only way it can
get rounded to 1, if you're 0.5 or above. So actually statement two alone
is all you need to know that d is greater than
or equal to 0.5. So that is choice B. Statement two alone
is sufficient. Alright, I've realized
I've run out of time. See you in the next video.