The median of five numbers (an odd number) is the number in the middle when they are arranged in ascending order. 

These two orderings are both consistent with the ordering given in Statement 1:

 - median 

 - median .

Therefore, Statement 1 alone provides insufficient information to answer the question. For a similar reason, so does Statement 2. 

Assume both statements to be true. Then  is greater than both  and  and less than both  and . That makes  the middle element, and, thus, the median.

The area of a rectangle is the product of its length and width; the area of a triangle is half the product of its height and the length of its base. Therefore, from Statement 1, we get that 

.

From Statement 2, we get that

, or, equivalently,

, or

In other words, the two statements are equivalent, so one of two things happens - either statement alone is sufficient, or both together are insufficient. We show that the latter is the case:

Case 1: 

The mean of the two is .

Case 2: 

The mean of the two is .

Therefore, knowing the area of the rectangle with these dimensions is not helpful to determining their arithmetic mean. This makes Statement 1, and, equivalently, both statements together, unhelpful.

The average or mean is found by taking all of the scores and dividing by the total number of scores. Remember, we must first find the lowest score and not include that in the calculation. Therefore, we get: 

 when rounded. 

Knowing the mean of the set is enough to calculate :

Knowing the median is enough to deduce . Since there are ten elements - an even number - the median is the arithmetic mean of the fifth and sixth highest elements. If , those two elements are 47 and 49, making the median 48. If , those two elements are both 49, making the median 49. This forces  to be 48, making the median 48.5.

Therefore, either statement alone is sufficient to answer the question.

If we are given only that 15 occurs four times in the data set, it is possible that another number can occur up to six times; similarly, if we are given only that 16 occurs three times, it is possible that another number can occur up to seven times. Either way, the mode - the most frequently occurring data value - cannot be determined.

However, if we know both facts, then no other data value can occur more than three times, so 15 must be the mode.

Therefore, the answer is that both statements are sufficient, but not one alone.

Suppose  but  is unknown. The above set, with known elements ordered, is

The median is the arithmetic mean of the two middle elements, when the elements are ordered.

Regardless of the value of  , the two middle elements must both be 84, making the median 84. 

Now, suppose  but  is unknown. The above set, with known elements ordered, is

The median cannot be determined with certainty. For example, if , as stated before, the median is 84. But if , the middle elements are 84 and 87, making the median 85.5.

The answer is that Statement 1 alone is sufficent to answer the question, but Statement 2 alone is not.

If we know that , then the set is known to have one 24, five 26's, three 27's, two 28's, and one 29. The only way the set can have two modes is for  and ; this makes 27 occur five times, just as frequently as 26. 

If we know , however, the set  is known to have one 24, four 26's, four 27's, two 28's, and one 29. There are two ways for the set to have two modes (26 and 27): for  and , or  for  and . 

The answer is that Statement 1 alone is sufficient to answer the question, but not Statement 2.

The median of a data set with an even number of elements is the arithmetic mean of the two elements that fall in the middle when the elements are arranged in ascending order. These two elements are both 6, so 6 is the median.

The mode of a data set is the element that occurs most frequently. Since 6 appears thre times, 7 appears two times, and all other elements appear once each, the mode is 6.

Therefore, the median and the mode are equal.

The data set has four 6's and no more than two of any other element - and there cannot be more than four of any other element regardless of the value of  - so  6 must be one of the modes. For the set to be bimodal, there must be four of another element. Since  occurs twice, it must be set to a number known to occur exactly two other times. There are, however, two choices, 5 and 7, so Statement 1 is insufficient.

Statement 2 is sufficient, as seen below:

If , as given in Statement 1, the set can have one mode (the other four numbers are different from each other and from  and ), two modes (for example,  ) or three modes (for example,  ) . Therefore, Statement 1 alone is not enough. 

If none of , , , or  are equal to each other, as given in Statement 2, the set can have one mode (for example, , the other numbers are different), two modes (,  and  are different), or no modes (all six different numbers).

If both statements are true, however, there are two possibilities - , with the other four elements being different, or , with one number being the same and the other three different. Either way, the set is known to have one mode.