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Problems 10 and 14 compute a surface area and the volume of a frustum respectively. Aha problems involve finding unknown quantities (referred to as Aha) if the sum of the quantity and part(s) of it are given. You have found it correctly." The 14th problem of the Moscow Mathematical calculates the volume of a frustum.


Seven of the twenty-five problems are geometry problems and range from computing areas of triangles, to finding the surface area of a hemisphere (problem 10) and finding the volume of a frustum (a truncated pyramid). The 10th problem of the Moscow Mathematical Papyrus asks for a calculation of the surface area of a hemisphere (Struve, Gillings) or possibly the area of a semi-cylinder (Peet). It is a well-known mathematical papyrus along with the Rhind Mathematical Papyrus.The Moscow Mathematical Papyrus is older than the Rhind Mathematical Papyrus, while the latter is the larger of the two.


Problems 1, 19, and 25 of the Moscow Papyrus are Aha problems. Problem 11 asks if someone brings in 100 logs measuring 5 by 5, then how many logs measuring 4 by 4 does this correspond to? The volume is found to be 56 cubic units, which is correct. You are to add the 16 and the 8 and the 4; result 28.

For instance problem 19 asks one to calculate a quantity taken 1 and ½ times and added to 4 to make 10. Problem 23 finds the output of a shoemaker given that he has to cut and decorate sandals. Find the remainder of this 8 after subtracting 2/3 + 1/6 + 1/18. The text of the example runs like this: "If you are told: a truncated pyramid of 6 for the vertical height by 4 on the base by 2 on the top: You are to square the 4; result 16.



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