The arithmetic mean of  and  is equal to .

Statement 1 alone gives us this value directly. 

From Statement 2 alone, the value can be determined by dividing both sides by 4:

The median of three or five numbers (both odd numbers) is the number in the middle when they are arranged in ascending order. Statement 1 alone does not answer the question of which one is the median; both of the following orderings are consistent with that statement:

 - median 

 - median .

For similar reasons, Statement 2 alone does not answer the question.

Now, assume both statements to be true. From Statement 1, exactly one of  and  is less than , and the other is greater. From Statement 2, exactly one of  and  is less than , and the other is greater. Therefore, two of the five values are less than  and two are greater, making  the middle element, or median.

Assume Statement 1 alone. The sequences

and

are both arithmetic, each term being the previous term plus the same number (in the first case, this common difference is 40; in the second, it is 20). The fourth term is 120 in both cases, The second and third terms of the first sequence have arithmetic mean ; the second and third terms of the second sequence have arithmetic mean .  Therefore, the mean of those two terms cannot be determined for certain. A similar argument holds for Statement 2 alone being insufficient.

Now assume both statements. Let  be the common difference of the sequences mentioned in Statement 2. By Statement 2, 0 is the first term, so the sequence will be

By the first statement, the fourth term is 120, so

The second terms is and the third term is , and their arithmetic mean is .

An arithmetic sequence is one in which each term is formed by adding the same number to its preceding term - the common difference.

Let  be the first term, and  be the common difference. The first five terms are

The arithmetic mean of two numbers is half the sum of the numbers. The arithmetic mean of the first and third terms is

,

which is the second term. Statement 2 alone gives this number as 100.

Now assume Statement 1 alone. Consider these two sequences, both of which can be seen to be arithmetic with fifth term 130:

The arithmetic mean of the first and third terms differ, as can be seen by looking at the second terms; in the first sequence, it is 127, and in the second, it is 100. That makes Statement 2 inconclusive.

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.