Total area of hall \(\displaystyle =60ft*50ft=3000ft^{2}\)

At \(\displaystyle 1\) person per \(\displaystyle 5\) square feet, \(\displaystyle 3000ft^{2}\div5ft^{2}\) per person  = \(\displaystyle 600\) people

Let \(\displaystyle x\) be the number of \(\displaystyle 10\)-person tables, and \(\displaystyle y\) be the number of \(\displaystyle 20\)-person tables. Since there are \(\displaystyle 40\) tables in the cafeteria, \(\displaystyle x+y=40\). \(\displaystyle 10x\) represents the number of people sitting at \(\displaystyle 10\)-person tables, and \(\displaystyle 20y\) represents the number of people sitting at \(\displaystyle 20\)-person tables. Since the cafeteria can seat \(\displaystyle 600\) people, \(\displaystyle 10x+20y=600\). Now we have \(\displaystyle 2\) equations and \(\displaystyle 2\) unknowns, and can solve the system. To do this, multiply the first equation by \(\displaystyle 10\) and subtract it from the second equation. This yields \(\displaystyle 0x+10y=200\); solving for \(\displaystyle y\) tells us there are \(\displaystyle 20\) tables that seat \(\displaystyle 20\) people. Since \(\displaystyle x+y=40\), \(\displaystyle x=20\), so there are \(\displaystyle 20\) tables that seat \(\displaystyle 10\) people. The ratio of \(\displaystyle x:y\) is therefore \(\displaystyle 1:1\).

Calculate the number of feet walked in an hour.  Then calculate what fraction of an hour \(\displaystyle 40\) minutes is.

\(\displaystyle 5,280*4=21,120\) feet walked in an hour

\(\displaystyle 60\) minutes in an hour, so \(\displaystyle 40\) minutes = \(\displaystyle \frac{2}{3}\) hour \(\displaystyle \left ( \frac{40}{60} \right )\)

\(\displaystyle 21,120*\left ( \frac{2}{3} \right )=14,080\) feet walked in \(\displaystyle 40\) minutes

First, we need to figure out how many red, blue, and green marbles are in the bag before any are removed. Let \(\displaystyle 5x\) represent the number of red marbles. Because the marbles are in a ratio of \(\displaystyle 5:2:1\), then if there are \(\displaystyle 5x\) red marbles, there are \(\displaystyle 2x\) blue, and \(\displaystyle 1x\) green marbles. If we add up all of the marbles, we will get the total number of marbles, which is \(\displaystyle 240\).

\(\displaystyle 5x+2x+1x=240\)

\(\displaystyle 8x=240\)

\(\displaystyle x=30\)

Because the number of red marbles is \(\displaystyle 5x\), there are \(\displaystyle 5*30\), or \(\displaystyle 150\) red marbles. There are \(\displaystyle 2*30\), or \(\displaystyle 60\) blue marbles, and there are \(\displaystyle 1*30\), or \(\displaystyle 30\) green marbles.

So, the bag originally contains \(\displaystyle 150\) red, \(\displaystyle 60\) blue, and \(\displaystyle 30\) green marbles. We are then told that one-third of the red marbles is removed. Because one-third of \(\displaystyle 150\) is \(\displaystyle 50\), there would be \(\displaystyle 100\) red marbles remaining. Next, two-thirds of the green marbles are removed. Because \(\displaystyle \frac{2}{3}*30=20\), there would be \(\displaystyle 10\) green marbles left after \(\displaystyle 20\) are removed.

To summarize, after the marbles are removed, there are \(\displaystyle 100\) red, 60 blue, and \(\displaystyle 10\) green marbles. The question asks us for the fraction of blue marbles in the bag after the marbles are removed. This means there would be \(\displaystyle 60\) blue marbles out of the \(\displaystyle 170\) left in the bag. The fraction of blue marbles would therefore be \(\displaystyle \frac{60}{170}\), which simplifies to \(\displaystyle \frac{6}{17}\).

The answer is \(\displaystyle \frac{6}{17}\).

\(\displaystyle 3\) filters can support a total of \(\displaystyle 120\) creatures/plants. The fish he wants need \(\displaystyle 2\) plants for every \(\displaystyle 5\) fish. The plant needs \(\displaystyle 4\) cleaning fish per \(\displaystyle 3\) plants. Thus for every \(\displaystyle 15\) of the fish he wants, he needs \(\displaystyle 6\) plants and \(\displaystyle 8\) cleaning fish.

This gives us a total of \(\displaystyle 29\) creatures. We can complete this number \(\displaystyle 4\) times, but then we are left with \(\displaystyle 4\) spots open that the filters can support.

This is where the trick arises. We can actually add one more fish in.  Since \(\displaystyle 1\) plant supports up to \(\displaystyle 2.5\) fish \(\displaystyle \left ( 2:5 \right )\), and 2 cleaning fish support up to \(\displaystyle 1.5\) plants, we can add \(\displaystyle 1\) fish, \(\displaystyle 1\) plant, and \(\displaystyle 2\) cleaning fish to get a total of \(\displaystyle 120\) creatures. If we attempt to add \(\displaystyle 2\) fish, then we must also add the \(\displaystyle 1\) plant, but then we don't have enough space left to add the \(\displaystyle 2\) cleaning fish necessary to support the remaining plant.

Thus, Tom can buy at most \(\displaystyle 61\) of the fish he originally wanted to get.

In order to maintain a proportion, each value in the ratio must be multiplied by the same value:

Before and after the snakes arrive, the number of lizards stays constant.

Before new snakes — Snakes : Lizards = \(\displaystyle 3x:5x\)

After new snakes — Snakes : Lizards = \(\displaystyle 4x:5x\)

Before the new snakes arrive, there are \(\displaystyle 3x\)snakes. After the \(\displaystyle 15\) snakes are added, there are \(\displaystyle 4x\) snakes. Therefore, \(\displaystyle 3x+15=4x\). Solving for \(\displaystyle x\) gives \(\displaystyle x=15\).

There are \(\displaystyle 5x\) lizards, or \(\displaystyle 5*15=75\) lizards.

Let \(\displaystyle 10x\) be the number of store employees, \(\displaystyle 3x\) the number of store managers, and \(\displaystyle x\) the number of corporate managers.

\(\displaystyle 10x+3x+x=126\)

\(\displaystyle 14x=126\)

\(\displaystyle x=9\)

\(\displaystyle 10x=10*9=90\), so the number of store employees is \(\displaystyle 90\).

\(\displaystyle 3x=3*9=27\), so the number of store managers is \(\displaystyle 27\).

\(\displaystyle x=9\), so the number of corporate managers is \(\displaystyle 9\).

Therefore, the number of employees who are either store managers or corporate managers is \(\displaystyle 27+9=36\).

We can use dimensional analysis to solve this problem.  We will create ratios from the conversions given.

\(\displaystyle 30\, jagged*\frac{2\, smooth}{6\, jagged}*\frac{1\, shiny}{3\, smooth}=3.333\, shiny\)

Since Joaquin cannot trade for part of a shiny rock, the most he can get is \(\displaystyle 3\) shiny rocks.

First, we should write a fraction for each ratio given:

\(\displaystyle \frac{2\, z}{5\, b}\, and\, \frac{7\, m}{3\, b}\)

Next, we will multiply these fractions by each other in such a way that will leave us with a fraction that has only Z and M, since we want a ratio of these two flowers only.

\(\displaystyle \frac{7\, m}{3\, b}\, *\, \frac{5\, b}{2\, z}=\frac{35\, m}{6\, z}\)

So the final answer is \(\displaystyle 35:6\)

The number of hours required to mow the lawn remains constant and can be found by taking the original \(\displaystyle 8\) workers times the \(\displaystyle 6\) hours they worked, totaling \(\displaystyle 48\) hours. We then split the total required hours between the \(\displaystyle 5\) works that remain, and each of them have to work \(\displaystyle 9\) and \(\displaystyle \frac{3}{5}\) hours: \(\displaystyle \frac{48}{5} = 9\frac{3}{5}\) .