Multiplying fractions is very easy. All you do is multiply all the numerators by each other and all the denominators by each other. You do not have to do anything that has to do with fancy common denominators like you do for adding and subtracting. For a question like this, it is often easiest just to cancel factors before you start your final multiplication. First, note:

\(\displaystyle \frac{44}{18}*\frac{3}{11}*\frac{5}{8}=\frac{44*3*5}{18*11*8}\)

Now, cancel the \(\displaystyle 11\) from the \(\displaystyle 44\):

\(\displaystyle \frac{4*3*5}{18*8}\)

Next, the \(\displaystyle 4\) in the numerator cancels with the \(\displaystyle 8\) in the denominator:

\(\displaystyle \frac{3*5}{18*2}\)

Finally, the \(\displaystyle 3\) in the numerator cancels with the \(\displaystyle 18\) in the denominator:

\(\displaystyle \frac{5}{6*2}=\frac{5}{12}\)

Multiplying fractions is very easy. All you do is multiply all the numerators by each other and all the denominators by each other. You do not have to do anything that has to do with fancy common denominators like you do for adding and subtracting. For a question like this, it is often easiest just to cancel factors before you start your final multiplication. First, note:

\(\displaystyle \frac{15}{2}*\frac{12}{45}*\frac{3}{4}=\frac{15*12*3}{2*45*4}\)

Now, cancel the \(\displaystyle 4\) in the denominator with the \(\displaystyle 12\) in the numerator:

\(\displaystyle \frac{15*3*3}{2*45}\)

Next, the \(\displaystyle 3*3\) in the numerator cancels with the \(\displaystyle 45\) in the denominator:

\(\displaystyle \frac{15}{2*5}\)

Finally, cancel the \(\displaystyle 5\) in the denominator with the \(\displaystyle 15\) in the numerator:

\(\displaystyle \frac{3}{2}\)

Begin by distributing the group on the left side of the equation. Remember that it is easy to multiply fractions. You only need to multiply the denominators and numerators. There are no "fancy" steps in between.

Therefore,

\(\displaystyle \frac{1}{3}(4x+\frac{12}{5})=2\)

is the same as:

\(\displaystyle (\frac{1}{3}*4x+\frac{1}{3}*\frac{12}{5})=2\)

You can cancel part of the second fraction out, so you get:

\(\displaystyle \frac{4x}{3}+\frac{4}{5}=2\)

Now, subtract \(\displaystyle \frac{4}{5}\) from both sides:

\(\displaystyle \frac{4x}{3}=2-\frac{4}{5}\)

Simplifying the right side of the equation, you get...

\(\displaystyle \frac{4x}{3}=\frac{10}{5}-\frac{4}{5}\)

\(\displaystyle \frac{4x}{3}=\frac{6}{5}\)

Now, multiply both sides by \(\displaystyle \frac{3}{4}\):

\(\displaystyle \frac{3}{4}*\frac{4x}{3}=\frac{6}{5}*\frac{3}{4}\)

Simplify:

\(\displaystyle x=\frac{18}{20} = \frac{9}{10}\)

Let's make the two quantities look the same. 

Quantity A: 120 miles / 5 gallons =  24 miles / gallon

Quantity B: 25 miles / gallon

Quantity B is greater.

In order to solve quantitative comparison problems, you must first deduce whether or not the problem is actually solvable. Since this consists of finding the solution to an \(\displaystyle x\)-coordinate on a line where nothing too complicated occurs, it will be possible.  

Thus, your next step is to solve the problem.

Since \(\displaystyle y = \frac{3}{4}x-2\) and \(\displaystyle y = \frac{-1}{3}\), you can plug in the \(\displaystyle y\)-value and solve for \(\displaystyle x\):

\(\displaystyle y = \frac{3}{4}x-2\)

Plug in y:

\(\displaystyle \frac{-1}{3} = \frac{3}{4}x-2\)

Add 2 to both sides:

\(\displaystyle \frac{5}{3} = \frac{3}{4}x\)

Divide by 3/4.  To divide, first take the reciprocal of 3/4 (aka, flip it) to get 4/3, then multiply that by 5/3:

\(\displaystyle \frac{20}{9} = x\)

Make the improper fraction a mixed number:

\(\displaystyle 2\tfrac{2}{9}\)

 Now that you have what x equals, you can compare it to Quantity B. 

Since \(\displaystyle 2\tfrac{2}{9}\) is bigger than 2, the answer is that Quantity A is greater

Remember that when you divide by a fraction, you multiply by the reciprocal of that fraction.  Therefore, this division really is:

\(\displaystyle \frac{1}{5}\cdot \frac{15}{4}\)

At this point, it is merely a matter of simplification and finishing the multiplication:

\(\displaystyle \frac{1}{5}\cdot\frac{15}{4}=\frac{15}{20}=\frac{3}{4}\)

To begin with, most students find it easy to remember that...

\(\displaystyle \frac{7}{\frac{4}{5}}=\frac{\frac{7}{1}}{\frac{4}{5}}\)

From this, you can apply the rule of division of fractions.  That is, multiply by the reciprocal:

Therefore,

\(\displaystyle \frac{\frac{7}{1}}{\frac{4}{5}} = \frac{7}{1} \cdot \frac{5}{4}=\frac{35}{4}\)

Since nothing needs to be reduced, this is your answer.