In order to solve this problem, it is best to work backwards by "plugging in" the answer choices to see which one yields a correct answer. 

If 29 is the largest of the three odd consecutive numbers, then that means that the numbers being added together would be 25, 27, and 29. 

Given that \(\displaystyle 25+27+29=81\), \(\displaystyle 29\) is the correct answer. 

In this sequence, every subsequent number is equal to one third of the preceding number:

\(\displaystyle 108, 36, 12, 3, x\)

\(\displaystyle 108\div3=36\)

\(\displaystyle 36\div3=12\)

\(\displaystyle 12\div3=4\)

Given that \(\displaystyle 4\div3 =\frac{4}{3}\), that is the correct answer. 

For a set to be a subset of \(\displaystyle K\), all of its elements must also be elements of \(\displaystyle K\) - that is, all of its elements must be multiples of 5. An integer is a multple of 5 if and only if its last digit is 5 or 0, so all we have to do is examine the last digit of each number in all four sets. Every number in every set ends in 5 or 0, so every number in every set is a multiple of 5. This makes all four sets subsets of \(\displaystyle K\).

The elements of the set \(\displaystyle C \cap D\) - that is, the intersection of \(\displaystyle C\) and \(\displaystyle D\) - are exactly those in both sets. We can test each of the six elements in \(\displaystyle D\) for inclusion in set \(\displaystyle C\) by dividing each by 3 and noting which divisions yield no remainder:

\(\displaystyle D = \left \{ 683, 705, 759, 832, 852, 944\right \}\)

\(\displaystyle 683 \div 3 = 227 \textrm{ R }2\)

\(\displaystyle 705 \div 3 = 235\)

\(\displaystyle 759 \div 3 = 253\)

\(\displaystyle 832 \div 3 = 277 \textrm{ R }1\)

\(\displaystyle 852 \div 3 = 284\)

\(\displaystyle 944 \div 3 = 314 \textrm{ R }2\)

\(\displaystyle C\) and \(\displaystyle D\) have three elements in common, so \(\displaystyle C \cap D\) has that many elements.

For a set to be a subset of \(\displaystyle C\), all of its elements must be elements of \(\displaystyle C\) - that is, all of its elements must be multiples of 4. A set can therefore be proved to not be a subset of \(\displaystyle C\) by identifying one element not a multiple of 4.

We can do that with all four given sets:

\(\displaystyle \left \{ 24, 36, 42, 84, 88\right \}\): \(\displaystyle 42 \div 4 = 10 \textrm{ R }2\)

\(\displaystyle \left \{ 12, 40, 66, 92, 100\right \}\): \(\displaystyle 66 \div 4 = 16 \textrm{ R }2\)

\(\displaystyle \left \{ 18, 28, 44, 68, 76\right \}\): \(\displaystyle 18 \div 4 = 4 \textrm{ R }2\)

\(\displaystyle \left \{ 32, 52, 64, 74, 92\right \}\): \(\displaystyle 74\div 4 = 18 \textrm{ R }2\)

The correct response is therefore "None of the other responses are correct."

By dividing the numerator of each fraction by its denominator, each fraction can be rewritten as its decimal equivalent:

\(\displaystyle \frac{3}{5} = 3 \div 5 = 0.6\)

\(\displaystyle \frac{5}{7} = 5\div 7 = 0.714285...\)

\(\displaystyle \frac{7}{9}= 7 \div 9 = 0.777...\)

\(\displaystyle \frac{9}{11} =9 \div 11 = 0.8181...\)

Of the four, \(\displaystyle \frac{5}{7}\) and \(\displaystyle \frac{7}{9}\) fall between 0.6 and 0.8 exclusive. The correct response is "two"

For a set to be a subset of \(\displaystyle L\), all of its elements must also be elements of \(\displaystyle L\) - that is, all of its elements must be multiples of 4. An integer is a multple of 4 if and only the number formed by its last two digits is also a multiple of 4, so all we have to do is examine the last two digits of each number in all four sets. 

Of all of the numbers in the four sets listed, only 8,878 has this characteristic:

\(\displaystyle 78 \div 4 = 19 \textrm{ R }2\)

8,878 is not a multiple of 4, so among the sets from which to choose,

\(\displaystyle \left \{ 4876, 3296, 8878, 9004, 7612\right \}\)

is the only set that is not a subset of \(\displaystyle L\).

If every number that appears in set \(\displaystyle P\) also appears in set \(\displaystyle Q\), that means that set \(\displaystyle Q\) must be broader than set \(\displaystyle P\). 

Any number that is a multiple of 16 will also be a multiple of 8 (characteristic of set \(\displaystyle Q\)); therefore, if set \(\displaystyle P\) consists of multiples of 16, set \(\displaystyle Q\) will include all those numbers. 

Therefore, Set \(\displaystyle P\) can consist of multiples of 16. 

In this set, each subsequent fraction is half the size of the preceding fraction; (the denominator is doubled for each successive fraction, but the numerator stays the same). Given that the last fraction in the set is \(\displaystyle \frac{1}{16}\), it follows that the subsequent fraction will be \(\displaystyle \frac{1}{32}\). 

A prime number only has two factors, one and itself.  \(\displaystyle 13\) and \(\displaystyle 23\) are the only values given that have that property.  But the range of the set is between \(\displaystyle 1\) and \(\displaystyle 20\) which means that the answer cannot be \(\displaystyle 23\) since it is greater than \(\displaystyle 20\).  That means the final answer is \(\displaystyle 13\).