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## Integrated math 3

### Course: Integrated math 3 > Unit 13

Lesson 5: Modeling with rational functions- Analyzing structure word problem: pet store (1 of 2)
- Analyzing structure word problem: pet store (2 of 2)
- Rational equations word problem: combined rates
- Rational equations word problem: combined rates (example 2)
- Rational equations word problem: eliminating solutions
- Reasoning about unknown variables
- Reasoning about unknown variables: divisibility
- Structure in rational expression

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# Analyzing structure word problem: pet store (1 of 2)

CCSS.Math: , , ,

Sal solves a word problem about the unknown number of bears, cats, and dogs in a pet store. This is part 1 where Sal uses a visual reasoning. Created by Sal Khan.

## Want to join the conversation?

- Could you not also input the lowest possible numbers like 1 bear over 3 cats 2 dogs and 1 bear ( 1b / 3c+2d+1b) to figure out an answer depending on what yo wanted to find. so 1 bear would be 1 over 6 or would the 1 bear on the numerator cancel out the bear on the denominator?(25 votes)
- no because you cant cancel the bear out because their is addition going on

example: 1/ 3+2+1 USES ADDITION

1/ 3*2*1 USES MULTIPLICATION-you can use it then(14 votes)

- I don't think that it makes any logical sense. Think of it as a common sense problem. Can someone explain?(9 votes)
- Here is how I thought about it. 1/3 is the same as 1/1+1+1 and we can replace 1 with a variable that represents an arbitrary number of units. In this case if we went with b like the problem in the video, we would get a ratio of b/b+b+b in order to be equivalent to 1b/3b, or 1/3 when simplified.

Though since d>b and c>d, we know that the denominator, c+d+b, has to be larger than 3b and when you increase the denominator while keeping the numerator the same you end up with a smaller fraction - for example 1/3 compared to 1/6, which is the largest possible ratio that b/c+d+b could be if we went with the least possible number of animals (1b, 2d, 3c).(18 votes)

- that's only true if a, b, and d were all consecutive integers (odd or even).

because a>d>b could be 5>2>1 which is 2/ 8 which is 1/4 not 1/3

?(5 votes)- Well, you have answered your question yourself here, the question is not an equality statement but rather a question of what is bigger or smaller to another thing. In this case, c>d>b, you suggest the combination 5>2>1 the equation is b/c+d+b which would be 1/1+2+5 = 1/8, and 1/8 is definitely smaller than 1/3. Hope that helps :)(17 votes)

- Where can I find a video that shows how to plot linear inequalities for coordinates?? Thanks(5 votes)
- https://www.khanacademy.org/math/algebra/linear-equations-and-inequalitie/graphing-linear-inequalities/v/graphing-inequalities

I would check out all the videos in this section, it should help you a lot.(4 votes)

- Can I say that,

since c > d > b, then c+d >> b, (">> = much greater")

and since we have c+d+b in the denominator this will be

c+d+b >> c+d , and therefore b/c+d+b < 1b/3b .

Since the c+d+b in the denominator, there for the value will be smaller .

Thank you(3 votes)- Yes, your reasoning is correct. Since c > d > b, we can conclude that c + d >> b. Therefore, c + d + b is much greater than c + d, which implies that b / (c + d + b) is less than b / (c + d). Since b / (c + d) is equivalent to b / 3b + x + y, we can conclude that b / (c + d + b) is less than b / (3b + x + y). And since the denominator 3b + x + y is greater than 3, we can conclude that b / (c + d + b) is less than 1/3.(1 vote)

- How can we tell how much greater the #dogs is to the # bears?(2 votes)
- You can't, at least not from the information given in this video. All we know are that there are more dogs than bears, not how many more.(3 votes)

- We have dogs(D) and cats(C). There are more dogs than cats and there are at least one cat.

Compare the expressions C/C+D and 0.4. Which statement is correct? =, > or <

The correct answer say there are not enough informations.

But i want to know how to express the thought that its not C/C+D = 2/5? are they the same ratio?

My thought was that the two expressions convey the same meaning and i realize that the quantity of the variables might be different.(2 votes)- I usually assess these types of problems by trying different values for the variables.

Your problem tells you there is at least 1 cat. So, C = 1, 2, 3, 4, etc.

And, there are more dogs than cats.

The only way to get c/(c+d) = 2/5 would be for c=2 and d=3 (or multiples of these)

But, you could do c=1; d=2. In this case: c/(c+d) = 1/3 (smaller than 2/5)

And, we could also do c=4; d=5. In this case: c/(c+d) = 4/9 = 0.444... with 4 repeating. So, it is now larger than 2/5.

Hopefully you can see from these examples that depending on the values of C and D, all the possibilities are correct. This is why there is not enough info.

Hope this helps.(3 votes)

- if c>d>b, then c equals b plus something, b+x, and d equals b plus something, b+y. Then our expression turns into: b/(b+x)+(b+y)+b which is the same as b/3b+x+y deviding by b we get: 1/(3+(x/b)+(y/b)), and since the denominator is 3 plus something and all variables are positive, we conclude (3+(x/b)+(y/b)) is bigger than 3, and therefore 1/(3+(x/b)+(y/b)) is smaller than 1/3.(3 votes)
- That is correct. Using the given information that c > d > b, we can express c and d in terms of b, as c = b + x and d = b + y, where x and y are positive numbers. Substituting these expressions into the given expression, we get b/(b+x) + (b+y) + b = b/3b+x+y.

Dividing both sides by b, we get 1/(3+(x/b)+(y/b)). Since x/b and y/b are positive, (x/b)+(y/b) is positive, and so 3+(x/b)+(y/b) is greater than 3. Therefore, 1/(3+(x/b)+(y/b)) is less than 1/3.

Hence, we have shown that the given expression is less than 1/3, given the assumption that c > d > b.(1 vote)

- Who else noticed the name of the title? Dogs cats, and bears? I think Sal meant Dogs, cats, and bears.(3 votes)
- 1:22If you just wanted to simplify the fraction of animals that are bears, and you cancelled out the b in the top and bottom, would the numerator become a 1, or would it be a 0, and cease to be a fraction?(1 vote)
- You can't cancel the b in (b) / (b + c + d) because you are doing addition.

This will only work if you had (b) / (b)(c)(d)(1 vote)

## Video transcript

Let's say that someone's running
a fairly unusual pet store that only has three types
of animals in it. It has cats, and we
represent the number of cats with the letter c. It has dogs, and we
represent the number of dogs by the letter d. And it has bears. That's what makes it unusual. And we're going to represent
the number of bears with the letter b. Now, this person also tells us
that at this unusual pet store, the number of cats is greater
than the number of dogs, which is greater than the
number of bears. Now based just on
this information, they then ask us which
expression is greater. b divided by c plus
d plus b or 1/3. And I encourage you to pause
the video now and come up with your own rationale behind
which of these is greater. Or maybe neither is greater. Maybe they're equal. Or maybe you just don't
have enough information from what this pet store owner
told us to actually figure out. So pause the video now. So let's reason through what
this expression is actually representing. This is b, the number of
bears, over the number of cats, plus the number of dogs,
plus the number of bears. So there's the number of
bears over the sum of all the animals. So this really is the
fraction that are bears. Now so this question
really reduces to is the fraction that are
bears greater than a third, less than a third, equal to
a third, or can we not tell? And there's a bunch of different
ways to do this problem. And I'll try to expose
you to many ways in this video and the next. So let's do a visual one
doing different cases. Let's do the scenario where
the fraction that are bears are greater than one third. So let's visualize this. So this little diagram
right over here. Let's say this is all of
the pets in the pet store. And I've divided it into
thirds right over here. If the fraction that are bears
is greater than one third, then it might look
something like this. So this is a third
under the dotted lines. If I want to be greater than
1/3, I'll go a little bit more than greater than 1/3. So we go greater than
1/3 right over there. But if the fraction of
bears is greater than 1/3, well the fraction of dogs has
to be even greater than that. So it has to be at least
as much as this blue area. So it's going to be
even greater than that. And the fraction
that are cats has to be even greater than that. And I didn't even make
this one that great. It has to be more
than this blue area. And the fraction
that are cats has to be even greater than that. And you see, you can't
have three things that are all greater than
1/3 adding up to a whole. You can't have all three parts
of it being greater than 1/3. So this scenario breaks down. Another way you could
have thought about it is if this expression right over
here is greater than a third, then c plus d over this
has to be less than 2/3. But that means one
of them, c or d, would have to be definitely
less than a third in order for that to work out. So at least thinking
about this way, you know that this thing
cannot be greater than a third. Let's think about whether
it can be equal a third. Let's imagine if this
expression, the fraction that are bears, were
equal to a third. So let's make our diagram again. So in this situation, the
bears are exactly 1/3. So I'll just color in this
third right over here. So exactly 1/3 are bears. But we know that
the number of dogs is greater than the
number of bears. So the fraction
that are dogs has to be greater than the
fraction that are bears. So the fraction
that are dogs would have to be greater than 1/3. But if the fraction that are
dogs are greater than 1/3, then all we have left is
something less than 1/3 for the cats. That's actually the
largest fraction of the animals are cats. So we know that this also
is not a possibility. So what we're
really left with is that this has to
be less than 1/3. And let's make sure
that's reasonable. So, let's paste another
example right over here. So if the bears
are less than 1/3-- let me make it a little
bit more dramatic, so that you can make it clear. So this is less than a third. let's say that the
purple, the dogs, let's say they're
roughly exactly 1/3. Then you could make
a scenario where you have the largest
proportion of cats. So this would be the
cats right over here. That's not the same
color as the cats. Let me do the same color
as that pink color. So this right over
here is our cats. So this is complete believable. c over c plus d
plus b does indeed look bigger than d over c plus
d plus b, which does indeed look bigger than b
over c plus d plus b. This is completely believable
to be less than 1/3. So based on just this visual
argument that we just made, you could say hey, look. This right over here
is the larger quantity. In the next video,
I'll give what I would call more of an
analytical argument, where I won't draw diagrams
to show that this has to be less than 1/3.