MATHCOUNTS® Problem of the Week Archive Trick‐or‐Treat! – October 29, 2012 Problems & Solutions Originally posted: 10/25/2010 Maribel found a perfectly spherical pumpkin and decided to carve it to create a Jack‐
O‐Lantern. She hollowed out the pumpkin leaving only the shell which was 3/4 inch thick. Maribel cut and removed pieces of the pumpkin shell to create the Jack‐O‐
Lantern’s eyes, nose and mouth. For each eye she removed one right triangular prism with legs measuring 2 inches and 3 inches. For the nose and mouth she removed 10 prisms, each with bases that were equilateral triangles with side length 1 inch. (a) How many cubic inches of pumpkin shell did Maribel remove to make the Jack‐O‐Lantern’s two eyes? Express your answer as a common fraction. (b) How many cubic inches of pumpkin shell did Maribel remove to make the Jack‐O‐Lantern’s nose and mouth? Express your answer as a common fraction in simplest radical form. To determine the volume we need to multiply the area of each triangle by the thickness of the pumpkin
shell. (a) Since each eye is a right triangular prism, A = 1/2(2)(3) = 3 in2. Therefore, for each eye, V = 3(3/4) =
9/4 in3. So the total number of cubic inches of pumpkin shell removed for both eyes is 9/4(2) = 18/4 = 9/2 in3.
(b) Since the nose and mouth are made from prisms with equilateral triangles, we first need to find the
height of each triangle to determine the area. The altitude of an equilateral triangle is the longer leg of a
30-60-90 right triangle, with the shorter leg equal to half the length of the side, or 1/2 inch. Therefore, the
altitude is 1/2( 3 ) =
triangular prisms, V =
3 /2. So for each equilateral triangle A = 1/2(1)( 3 /2) =
3 /4. For each of these
3 /4(3/4) = 3 3 /16. Since there are 10 of these triangular prisms, the total number
of cubic inches of pumpkin shell removed for the nose and mouth is 3 3 /16(10) = 30 3 /16 = 15 3 /8 in3.
Mr. Thomas returned from his trip to the orchard with a basket of apples. His children went bobbing for apples, for which they used 16 of the apples. That left Mr. Thomas with only 2/3 the number of apples he brought back from the orchard. Mr. Thomas decided to use 3/4 of the remaining apples to make 4 apple pies. Mr. Thomas had just enough apples left to make 2 caramel apples for each of his children. How many children does Mr. Thomas have? Let a represent the total number of apples Mr. Thomas brought back from the orchard. The 16 apples used
for bobbing for apples represent 1/3 of the total number of apples, so 1/3a = 16  a = 48. Subtract 16 from
48 and we see that leaves 32 apples. The apple pies are made from 3/4(32) = 24 apples, leaving 32 – 24 = 8
apples with which Mr. Thomas makes caramel apples. We are told that each of his children receives 2
caramel apples, so Mr. Thomas has 8 ÷ 2 = 4 children.
On Friday, a house is haunted by a combined total of 51 ghosts, goblins and ghouls, and there were 1/2 as many ghosts as there were goblins. On Saturday, 2/3 of the ghouls each became a ghost. On Sunday, 11 of the ghosts each became a goblin, and the ratio of ghouls to goblins was then 1/3. At that time how many ghosts were there? Let A be the number of ghosts on Friday, B be the number of goblins on Friday and C be the number of
ghouls on Friday. We are told that A + B + C = 51. We are also told that the number of ghosts is half the
number of goblins. That means A = (1/2)B. By Sunday the number of goblins has increased by 11 which is
B + 11, and there are 1/3 as many ghouls or 1/3C. Now the ratio of ghouls to goblins on Sunday is 1/3, so
1/3 = (1/3C)/(B + 11). Cross multiplying we see that C = B + 11. Now substituting for A and C in the original
equation A + B + C = 51 we have ((1/2)B) + B + (B + 11) = 51  (5/2)B + 11 = 51  (5/2)B = 40  B = 16. So on
Sunday there are 16 goblins. That means the number of ghosts A + (2/3)C – 11 can also 0be written
(1/2)B + (2/3)(B + 11) – 11. Since B = 16, we have (1/2)(16) + (2/3)(16 + 11) – 11 = 8 + (2/3)(27) – 11 =
8 + 18 – 11 = 15 ghosts.