Ratios and Proportions in Fish Tanks

Question

I was recently studying for a test and got this problem correct. However, I don't think I would be able to answer a similar type of question correct again.

The question is the following:

A marine aquarium has a small tank and a large tank, each containing only red and blue fish. In each tank, the ratio of red fish to blue fish is 3 to 4. The ratio of fish in the large tank to fish in the small tank is 46 to 5. What is the ratio of blue fish in the small tank to red fish in the large tank?

Thank you for taking the time!

Answer 1

HINT

Let indicate with

• $x,y$ the number of red and blue fish in the small tank then $4x=3y$
• $z,w$ the number of red and blue fish in the large tank then $4z=3w$

then we know that

$$\frac{z+w}{x+y}=\frac{46}{5}$$

and need to find $y/z$, then note that

$$\frac{z+4z/3}{3y/4+y}=\frac{46}{5}\implies \frac{z}{y}\frac{\frac73}{\frac74}=\frac{z}{y}\frac43=\frac{46}{5}\implies \frac{y}{z}=\frac{4\cdot 5}{3\cdot 46}=\frac{20}{138}=\frac{10}{69}$$

Answer 2

Let the number of red fish in the small tank be $r$, the number of blue fish in the small tank be $b$, the number of red fish in the large tank be $R$ and the number of blue fish in the big tank be $B$.

We are told:$$ \begin{align} 4r &= 3b\tag 1\\ 4R&=3B\tag 2\\ 5(R+B)&=46(r+b)\tag 3 \end{align}$$ and we are asked to solve for ${B}/{r}.$

All you need to do is to express $b$ in terms of $r$ from (1), express $R$ in terms of $B$ from (2), and substitute the values in (3). That will give you an equation involving only $B$ and $r$.

Answer 3

Small Tank : Red fish = 3x, Blue fish =4x

Big Tank: Red fish = 3y, Blue fish =4y

$ y=46\lambda $, $x=5\lambda $

Thus

Blue fish in the small tank= $$4x=20\lambda $$

Red fish in the large tank =

$$ 3y=138\lambda $$ Thus the ration of blue fish in the small tank to red fish in the large tank is $$\frac {20\lambda}{138\lambda}= 20/138.$$