Assume both statements to be true. Examine these two scenarios.

Case 1: The metal rod is 56 inches long.

If four inches are removed from each end - that is, if eight inches total are removed - the rod will have length  inches, which is equal to  feet.

If twenty inches are removed, the rod will have length  inches, which is equal to one yard. 

Case 2: The metal rod is 92 inches long.

If four inches are removed from each end - that is, if eight inches total are removed - the rod will have length  inches, which is equal to  feet.

If twenty inches are removed, the rod will have length  inches, which is equal to  yards.

Both lengths satisfy the conditions given in both statements, so, together, the statements provide insufficient information.

Assume Statement 1 alone. Since one meter is equal to about 3.3 feet, we can divide the minimum and maximum lengths in feet by 3.3 to get these lengths in meters:

Minimum length: 7 feet

 meters

Maximum length: 8 feet

 meters

The minimum and maximum areas of the sheet in square meters are the squares of these:

Since the area of the sheet must be a whole number of square meters, the only possble area is 5 square meters.

Assume Statement 2 alone. Since one meter is equal to about 3.3 feet, we square this to get the conversion factor from square meters to square feet:

One square meter is equal to  square feet.

Divide the minimum and maximum areas in square feet by 10.89 to convert them to square meters: 

Minimum area: 40 square feet

 square meters:

Maximum area: 60 square feet

 square meters:

The area in square meters is a whole number between 3.7 and 5.5, so the area in square meters is either 4 or 5 . Without additional information, however, we cannot narrow it down further.

From Statement 1 alone, it can be determined that there are eight ounces more that a whole number of pounds; equivalently, the number of ounces in can is 8 greater than a multiple of 16, such as 24, 40, 66, etc. However, we cannot narrow it down further.

From Statement 2 alone, we can narrow the possibilities down to eleven: 50, 51, up to 60 ounces. We cannot narrow it down further.

Assume both statements to be true. We can start with 8 and keep adding 16 to find a number that 8 greater than a multiple of 16 and that is between 50 and 60:

The only possible weight of the coffee is 56 ounces.

Assume Statement 1 alone. Since the length of one side is at minimum 4 feet, or equivalently,  yards, its area is at least the square of this, or  square yards. Since the length of one side is at most 6 feet, or, equivalently,  yards, its area is at most the square of this, or 4 square yards. Since the area must be equal to a whole number of yards, we can only narrow the area down to 2, 3, or 4 square yards.

Assume Statement 2 alone. Since the length of one side is at minimum 44 inches, or equivalently,  yards, its area is at least the square of this, or  square yards. Since the length of one side is at most 54 inches, or, equivalently,  yards, its area is at most the square of this, or   square yards. Since the area must be equal to a whole number of yards, the only possible area for the sheet is 2 square yards.

Assume both statements are true. Since, as known from Statement 2, the weight of the bag of oranges is given by a whole number of pounds, then the weight in ounces must be a multiple of 16. From Statement 1, this number is between 110 and 130 inclusive. However, there are two multiples of 16 - 112 and 128 - that fall in this range. Therefore, the question cannot be answered for certain.

Assume Statement 1 alone. One yard is equal to 36 inches, so one square yard is equal to  square inches. The maximum area of 3,000 square inches is equal to  square yards; since the area is a whole number of square yards less than this, we can narrow the area down to 1 or 2 square yards. Without further information, we cannot narrow this down further.

Assume Statement 2 alone. One yard is equal to 3 feet, so one square yard is equal to  square feet. The maximum area of 15 square feet is equivalent to  square yards; since the area is a whole number of square yards less than this, the only possible area is one square yard. 

If Statement 1 alone is assumed, since the length of the rod is given by a whole number of feet, then the length in inches must be a multiple of 12 - 12 inches, 24 inches, etc. However, no other information can be found.

If Statement 2 alone is assumed, there are eleven possible lengths of the rod - 40 inches, 41 inches, and so on up to 50 inches.

Now assume both statements. The length of the rod in inches must be a multiple of 12 between 40 and 50; this narrows it down to one possibility, 48 inches. 

Assume Statement 1 alone. Since one meter is equal to 100 centimeters, the minimum and maximum lengths in meters are  meters and  meters. The squares of these lengths provide the minimum and maximum areas of the sheet:

From the main body of the problem, the area in square meters must be a whole number, so its area must be seven meters.

Assume Statement 2 alone. One meter is equal to 1,000 millimeters, so the minumum and maximum lengths are   meters and  meters. These are the same boundaries derived from Statement 1, so again, the area can be determined to be seven meters.

To determine if two lines are perpendicular, only the slope needs to be considered. The slopes of perpendicular lines are the negative reciprocals of each other. Knowing a single point on the line is not sufficient, as an infinite number of lines can pass through and individual point.

Statement 1 alone establishes nothing about the angle  makes with , as it is not part of the triangle. Statement 2 alone only establishes that .

Assume both statements are true. Then  is an altitude of an equilateral triangle, making it - and  - perpendicular with the base  - and .