If we only assume Statement 1, then by examining these two cases, we show that we do not know whether the median changes. 

Suppose the original set is 

If we only know that one of the elements added is 50, then it is possible that the median does not change - this happens if the other element added is 70:

has median 60.

It is also possible that the median does change - this happens if the other element added is 50.

has median 55.

A similar argument can be used to demonstrate that Statement 2 is insufficient.

However, suppose we know both. 60 is the sixth-highest element of the old data set, and, since one element greater than 60 and one less than 60 are added, 60 is the seventh-highest element of the new thirteen-element set. We therefore know the median did not change.

Both statements can be shown to be equivalent to the continued inequality 

by way of the Multiplication Property of Inequality. We will demonstrate this as follows:

Multiply each expression in

(Statement 1)

by :

.

Multiply each expression in

(Statement 2) 

by :

The median of five numbers (an odd number) is the number in the middle when they are arranged in ascending order. Therefore, it can be established from either statement alone that of the five, the median is .

Assume Statement 1 alone. By the Addition Property of Inequality, if 

,

then  can be added to each quantity to give an equivalent inequality:

.

By extension, the continued inequality in Statement 1 is equivalent to 

.

A similar argument holds for Statement 2. By the Multiplication Property of Inequality, if 

, then

and . 

By extension, the continued inequality in Statement 2 is also equivalent to 

.

The median of five numbers (an odd number) is the number in the middle when they are arranged in ascending order. Therefore, it can be established from either statement alone that of the five, the median is .

Assume Statement 1 alone. The median of five numbers (an odd number) is the number in the middle when they are arranged in ascending order. By Statement 1, this number is , so the question is answered.

Statement 2 alone, however, gives that the mean is . It is possible that the mean and the median can be one and the same or two different numbers.

Case 1: 

The mean is 

making this consistent with Statement 2.

The median is the middle element, .

Case 2: 

again, making this consistent with Statement 2.

The median is the middle element, .

In a data set of 19 elements, 19 being odd, the median is the element that occurs in the  position when they are arranged in ascending (or descending) order. Neither statement alone tells us what that middle element is. But put together, they tell us what the ninth and eleventh elements would be when arranged in ascending (or descending) order. Since both are 72, the element between them - the tenth element, and, thus, the median - must be 72. 

The answer is that both statements together are sufficient to answer the question, but neither statement alone is sufficient to answer the question.

From the initial information, we can determine that  of the balls are green.

The first statement alone tells you that half of the blue marbles and half of the red marbles are large, so half of the green marbles must be large as well. But this alone does not tell you how many marbles there are total.

The second statement alone tells you how many large marbles are red or green, but you do not have any way of figuring out how many small marbles or total marbles are. 

But if you put the statements together, you know the following:

Half of the red marbles and half of the green marbles comprise the thirty-six large marbles:

There are 120 marbles.

The answer is that both statements together are sufficent to answer the question, but neither statement alone is sufficient.

Both  and  satisfy the conditions of Statement 1; the first fraction is in lowest terms, the second is not.

Both  and  satisfy the conditions of Statement 2; the first fraction is in lowest terms, the second is not.

Neither statement alone is enough to answer the question. Both together, however, are enough to prove the fraction is not in lowest terms; a fraction which satisfies both statements has both its numerator and denominator divisible by 2.

The two statements together are insufficient.  and  each satisfy the conditions of both statements, but the first fraction is in lowest terms and the second is not.

The two statements together provide insufficient information.

  and   are two examples of fractions that satisfy both conditions; note that the first is represented in decimal form by a terminating decimal, and the second, by a repeating decimal.

From the question, we know that only one of the fractions in the answer choices is larger than . The correct answer must therefore be the largest of the answer choices. This observation is important, because it means we can eliminate some of the answer choices without performing any calculations. For example, is smaller than , so eliminate from the possibilities. is smaller than both and , so eliminate it as well. Similarly, is smaller than , so eliminate too.

This leaves us with only two answer choices: and

First consider . We can rewrite this as , which is clearly smaller than . Eliminate .

By process of elimination, we've thus shown that is the correct answer choice.