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There are 3 kinds of animals: Cats, dogs, bears. # of cats is $c$, # of dogs is $d$ and # of bears is $b$. It's said that $$c>d>b.$$ The question is:

Which expression is greater $\dfrac{b}{c+d+b}$ or $\dfrac{1}{3}$?

The thing that I didn't understand is what $\dfrac{b}{c+d+b}$ actually means. And how can I solve for it?

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    You can't solve for it, there isn't enough information. But...maybe you can show that $c+d+b>3b$? – 2017-02-13
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    @JazzyMatrix Please don't edit out information; it's important to the question that $b, c, d$ represent certain quantities and are *not* just numbers -- especially since OP specifically asks for an interpretation of the fraction. – 2017-02-13

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Since $$c\gt d\gt b$$ we have $$c+d+b\gt b+b+b=3b$$ As $c,d,b\ge 0$, $$\frac{1}{c+d+b}\lt\frac{1}{3b}$$ $$\frac{b}{c+d+b}\lt\frac{b}{3b}$$ $$\frac{b}{c+d+b}\lt\frac{1}{3}$$

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    @Steve No, and I have edited my answer to make it clearer – 2017-02-13
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    why c+d+b>b+b+b? – 2017-02-13
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    @Steve Because $c\gt b$ and $d\gt b$. So replacing them by $b$ makes the expression smaller. – 2017-02-13
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    It seems I'm so bad at logic as I'm still struggling to understand( – 2017-02-13
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To properly show the inequality, we would need a bit of algebra. That's covered in the other answer. I want to focus on why you should believe this is true.

If there are $b$ bears, $c$ cats, and $d$ dogs, then the ratio $r = \dfrac{b}{b + c + d} = \dfrac{\text{number of bears}}{\text{number of animals}}$ is just asking for the fraction of animals that are bears.

Since we have the least number of bears, at most $\frac{1}{3}$ of the animals can be bears; $r \le \frac{1}{3}$. But since we have strictly fewer bears than any other animal, we must have $r < \frac{1}{3}$.

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    So is $r = \dfrac{b}{b + c + d}$ number of bears? – 2017-02-13
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    No, $b$ is the number of bears, and $b + c + d$ is the total number of animals. So that's the fraction of all animals that are bears. – 2017-02-13
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    I got it, can you answer one more question: why isn't the number of bears compared to one third? I mean, here the fraction of animals that are bears is compared to a third. What would be if "b" itself was compared? – 2017-02-15
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    $b$ is a a whole number of bears; 1, 2, 3, etc -- surely we more than $1/3$ of a bear :) It just so happens that fewer than $1/3$ *of the animals* are bears, based on the given information. – 2017-02-15
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I prefer to keep things in fractional form, and to try to get the same numerator to compare.

$$\frac{b}{c+d+b} = \frac{b}{c+d+b}\cdot\frac{\frac{1}{b}}{\frac{1}{b}}= \frac{\frac{b}{b}}{\frac{c+d+b}{b}}=\frac{1}{\frac{c}{b}+\frac{d}{b}+1}\lt\frac{1}{3}$$