First, add up the total number of ducks, geese, and chickens.

\(\displaystyle 15+12+20=47\)

Now, write the fraction of these animals that are geese and the fraction of these animals that are chickens.

\(\displaystyle \text{Geese}=\frac{15}{47}\)

\(\displaystyle \text{Chickens}=\frac{20}{47}\)

Now, since we want the ratio of geese to chickens, we write the fractions as thus:

\(\displaystyle \frac{\text{Geese}}{\text{Chickens}}=\frac{\frac{15}{47}}{\frac{20}{47}}\)

Divide and simplify the resulting fraction to find the ratio.

\(\displaystyle \frac{\frac{15}{47}}{\frac{20}{47}}=\frac{15}{47}\div\frac{20}{47}=\frac{15}{47}\times\frac{47}{20}=\frac{15}{1}\times\frac{1}{20}=\frac{15}{20}=\frac{3}{4}=3:4\)

There is more than one way to solve this problem. You can either figure out how much the chicken costs per pound and multiply that cost by three pounds, or you can set up a proportion and solve for the cost of three pounds of chicken that way.

1) Solving the Problem Using Cost per Pound

First, find how much the chicken costs per pound.

\(\displaystyle \frac{33.75}{15}=2.25\)

Since chicken costs \(\displaystyle \$2.25\) per pound, multiply this by the number of pounds we need to get the cost.

\(\displaystyle \$2.25\times3=\$6.75\)

\(\displaystyle 3\) pounds of chicken cost \(\displaystyle \$6.75\).

2) Solving the Problem Using a Proportion

You can set up a proportion to figure out how much \(\displaystyle 3\) pounds of chicken costs:

\(\displaystyle \frac{3\:lbs}{15\:lbs}=\frac{x}{\$33.75}\)

Cross multiply:

\(\displaystyle 3\cdot 33.75=15x\)

Solve for \(\displaystyle x\), the cost of \(\displaystyle 3\) pounds of chicken:

\(\displaystyle 101.25=15x\)

\(\displaystyle \frac{101.25}{15}=x\)

\(\displaystyle x=6.75\)

Let \(\displaystyle 2x\) be the number of freshmen, \(\displaystyle 3x\) be the number of sophomores, and \(\displaystyle 5x\) be the number of juniors.

Now, since we have \(\displaystyle 60\) students,

\(\displaystyle 2x+3x+5x=60\)

\(\displaystyle 10x=60\)

\(\displaystyle x=6\)

Since we want to find the number of juniors, we need to find the value of \(\displaystyle 5x\).

\(\displaystyle 5x=5(6)=30\)

Let \(\displaystyle 5x\) be the number of mammals,\(\displaystyle 6x\) be the number of reptiles, and \(\displaystyle 9x\) be the number of birds.

Since the zoo has \(\displaystyle 200\) animals,

\(\displaystyle 5x+6x+9x=200\)

\(\displaystyle 20x=200\)

\(\displaystyle x=10\)

Because we want the number of birds, we need to find the value of \(\displaystyle 9x\).

\(\displaystyle 9x=9(10)=90\)

Let \(\displaystyle 2x\) be the number of freshmen, \(\displaystyle 3x\) be the number of sophomores, \(\displaystyle 8x\) be the number of juniors, and \(\displaystyle 12x\) be the number of seniors.

Because the high school has \(\displaystyle 2000\) students,

\(\displaystyle 2x+3x+8x+12x=2000\)

\(\displaystyle 25x=2000\)

\(\displaystyle x=80\)

Since we want to find out how many juniors there are, we need the value of \(\displaystyle 8x\).

\(\displaystyle 8x=80\times8=640\)

Set up the following proportion, with \(\displaystyle x\) being the number of freshmen.

\(\displaystyle \frac{4}{5}=\frac{x}{150}\)

Now, cross-multiply and solve for \(\displaystyle x\).

\(\displaystyle 5x=600\)

\(\displaystyle x=120\)

Let \(\displaystyle 9x\) be the number of crabs and \(\displaystyle 5x\) be the number of seagulls.

Since there are \(\displaystyle 182\) crabs and seagulls on the beach,

\(\displaystyle 9x+5x=182\)

\(\displaystyle 14x=182\)

\(\displaystyle x=13\)

Because the question asks for the number of crabs, we need to find the value of \(\displaystyle 9x\).

\(\displaystyle 9x=13\times9=117\)

Write the numbers of boys and girls as a fraction, then simplify.

\(\displaystyle \frac{45}{36}=\frac{5}{4}\)

\(\displaystyle \frac{5}{4}\) can also be written as \(\displaystyle 5:4\)

Write the number of seniors and numbers of freshmen as a fraction:

\(\displaystyle \frac{30}{36}=\frac{5}{6}\)

That fraction is equivalent to \(\displaystyle 5\text{ to }6\).

Let \(\displaystyle x, 3x, \text{ and }6x\) be the values of the angles.

Since all the angles in a triangle need to add up to \(\displaystyle 180\),

\(\displaystyle x+3x+6x=180\)

\(\displaystyle 10x=180\)

\(\displaystyle x=18\)

Because we want the value of the largest angle, we need to find the value of \(\displaystyle 6x\).

\(\displaystyle 6x=18(6)=108\)