This problem merely requires careful working out of each part. Begin by simplifying the first fraction:

\(\displaystyle \frac{3+\frac{3}{7}}{\frac{2}{3}-4}\)

The numerator will be:

\(\displaystyle 3+\frac{3}{7} = \frac{21}{7}+\frac{3}{7} = \frac{24}{7}\)

The denominator will be:

\(\displaystyle \frac{2}{3}-4 = \frac{2}{3}-\frac{12}{3} = -\frac{10}{3}\)

Thus, we have the following fraction:

\(\displaystyle \frac{\frac{24}{7}}{-\frac{10}{3}}\)

Remember that you must multiply the numerator by the reciprocal of the denominator:

\(\displaystyle \frac{24}{7} * -\frac{3}{10} = -\frac{72}{70}\)

Now, work on the second fraction:

\(\displaystyle \frac{2 + \frac{1}{2}}{3}\)

This fraction is much easier. After simplifying the numerator, you get:

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

This is the same as:

\(\displaystyle \frac{5}{2} \cdot \frac{1}{3} = \frac{5}{6}\)

Thus, we come to our original expression! It is:

 \(\displaystyle -\frac{72}{70} - \frac{5}{6}\)

The common denominator of these fractions is \(\displaystyle 210\). Thus, you have:

\(\displaystyle \frac{-72 * 3 - 5 * 35}{210} = \frac{-216 -175}{210} = -\frac{391}{210}\)

As it is, quantities A and B have different denominators, so making a comparison can be tricky. Making a common denominator will allow for comparison of just the numerators, so that would make a good first step:

Quantity A:

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

Quantity B:

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

Disregarding equal denominators, since they'll always have a positive value, compare the numerators. If \(\displaystyle x^2+4x\) is subtracted from quantity A and from quantity B, we're left with \(\displaystyle 4\) for the former and \(\displaystyle 3\) for the latter.

Quantity A is greater.

One way to approach this question is to try to reduce the complexity of each quantity. By subtracting the value of Quantity A, \(\displaystyle x^2+5x+ 4\) from both A and B, we can make a new comparison:

Quantity A': \(\displaystyle x^2+5x+ 4-(x^2+5x+ 4 )=0\)

Quantity B': \(\displaystyle x^2+10x+21-(x^2+5x+4)=5x+17\)

Now, it's easy to see that for different values of \(\displaystyle x\) Quantity B may be greater, lesser, or equal to Quantity A.

Since \(\displaystyle x\) is not restricted in its possible values, the relationship cannot be determined.

To find out how many rhododendrons Rita will need to prune in an hour, we must first find out how many she needs to prune.

If Rhoda can prune \(\displaystyle 42\) rhododendrons in \(\displaystyle 3\) hours, then she can prune \(\displaystyle 84\) in \(\displaystyle 6\) hours, and if Rhonda can prune \(\displaystyle 40\) in \(\displaystyle 2\) hours, she can prune \(\displaystyle 120\) in \(\displaystyle 6\) hours.

If this is not readily apparent, it can be found by finding out how many each prunes in one hour, then multiplying by \(\displaystyle 6\).

Rhoda:\(\displaystyle 6(\frac{42}{3})=6(\frac{14}{1})=84\)

Rhonda: \(\displaystyle 6(\frac{40}{2})=6(\frac{20}{1})=120\)

Between Rhoda and Rhonda, \(\displaystyle 204\) of the \(\displaystyle 300\) rhododendrons can be pruned, leaving \(\displaystyle 96\) for Rita.

Since she has \(\displaystyle 6\) hours for the task, her rate of pruning can be found to be:

\(\displaystyle \frac{96}{6}\frac{rhododendrons}{hours}=16\frac{rhododendrons}{hour}\)