SUPER EASY MATH PROBLEM!!! EASY 10 Points!!!?

Question

A baker made a cake in the shape of a cube to serve at a
banquet for a group of mathematicians. Unfortunately, he slipped and the cake went sailing into a vat of chocolate frosting. The baker fished the cake out of the vat and was relieved to find that it had retained its shape, a perfect cube, while becoming completely coated with frosting. He then cut the cake into smaller cube-shaped pieces of equal size and served one to each mathematician. There were no pieces left over. The cake was especially enjoyed by the guests whose pieces had a lot of frosting. Many were disappointed that their pieces had no frosting, and one complained to the baker that the number of pieces without frosting was exactly 8 times the number of pieces with frosting on 3 sides.
A How many mathematicians attended the banquet?

B How many of the pieces of cake had:
1 No frosting
2 Frosting on one side only
3 Frosting on exactly 2 sides
4 Frosting on 3 sides
5 Frosting on more than 3 sides?

Answer 1

The number of pieces without frosting is pcs0fr = 8 * #pcs3fr (number of pieces with frosting on three sides, on a cube we have 8 corners which are the pieces that do have frosting on three sides which answers question B4).
Therefore, pcs0fr = 8 * 8 pcs3fr = 64 (which answers question B1)
Yet the pieces with no frosting create an inner cube inside the big cube with frosting; therefore, the number of pieces on the edge of the small cube is
s = 64^(1/3) = 4 pieces on each side.
Moreover, to recreate the bigger cube we need to add one cubic piece with frosting on both sides or each small cube's row,
so the side of the bigger cube is made of
S = s + 2 = 4 + 2 = 6 pieces.
The number of mathematicians is equal to the number of pieces of the bigger cube
#pcs = S ^ 3 = 6 ^ 3 = 216 mathematicians (answering question A).
B2) on each side of the bigger cube we have a number of pieces = S * S = 6 * 6 = 36 pieces,
yet we substitute four corners that have frosting on three sides, and substitute four pieces from each side that have frosting on two sides to get the number of pieces without frosting on one side
X = 36 - 4 - (4 * 4) = 32 - 16 = 16
But a cube has six sides, so the total number of pieces with frosting on one side is
pcs1fr = 6 * X = 6 * 16 = 96 pieces
B3) On every side we have S = 6 pieces, two of them are corner pieces with frosting on three sides, and we have twelve sides in a cube; therefore, the number of pieces with only two sides of frosting is:
pcs2fr = 12 * (S - 2) = 12 * (6 - 2) = 12 * 4 = 48 pcs
B5) A piece with the greatest number outer sides is a corner piece, and a corner piece has frosting only on three sides; therefore, there is no piece with frosting on more than three sides. (also if we add the numbers from B1 + B2 + B3 + B4 = A; therefore there are no more pieces to fit in this category which is a more mathematician prove than simply stating the obvious).

In sum:
A) 216 mathematicians
B1) 64 pieces
B2) 96 pieces
B3) 48 pieces
B4) 8 pieces
B5) 0 pieces

Answer 2

A.) Since each edge of the cube is x pieces on an edge, then there are x³ pieces, and since there were no pieces left over, then there were also x³ mathematicians at the banquet.

Imagine a cube x units on an edge cut into x³ equal pieces. If we hull the center out, leaving a 1 piece thick wall on all 6 facets, then the width and height of each wall is (x - 2) pieces. The area of each wall then is (x - 2)² pieces. To find the volume which has been hulled out, we simply multiply (x - 2)² by the length of the piece hulled out. Since each wall is 1 piece thick, then the length of the piece hulled out is also (x - 2) pieces long. Then the volume of the chunk or cube hulled out is (x - 2)(x - 2)² = (x - 2)³ pieces. Since each of these pieces is on the interior of the cake, they have no frosting on them. So (x - 2)³ pieces have no frosting on them. One mathematician remarked that the number of pieces without frosting equaled 8 times the number with frosting on 3 sides. And because there are 8 corner pieces, which would each have frosting on three sides, then that means (x - 2)³ = 8 (8) = 64 was the number of pieces without frosting and the volume of the piece hulled out. The cube root of 64 is 4. That means (x - 2) = 4, so x = 4 + 2 = 6 is the length of each edge, and x³ = (6)³ = 216 is the total number of pieces, which must also be the number of mathematicians who attended the banquet, since there is a one-to-one correspondence. To verify this, we simply add the number of interior pieces which we have hulled out, the number of pieces on each face, the number of pieces on each edge, and the number of corner pieces:

# Interior pieces (x - 2)³ = (4)³ = 64
# Pieces on each face: 6 (x - 2)² = 6 (4)² = 6 (16) = 96
# Edge pieces = 12 (x - 2) = 12 (4) = 48
# Corner pieces = 8

T = (x - 2)³ + 6 (x - 2)² + 12 (x - 2) + 8
T = 64 + 96 + 48 + 8
T = 216 = (6)³.

So there were 216 mathematicians at the banquet.

B.) 64 pieces had no frosting on them, because they were interior pieces.

96 face pieces had one side covered with frosting.

48 edge pieces had two sides covered with frosting.

8 corner pieces had frosting on three sides

This last question is tricky. There actually was 1 piece which had fronting on all 4 sides. That was the large cube before it was cut up. Therefore, the answer could technically be 1 or 0 depending on the size of the piece one is talking about.

Answer 3

the tax is 6% and the only precise fee of the e book is 8.40 3 so what you're searching for is the unique value. 8.40 3 is one hundred% of the fee so subtract 6% from 8.40 3 so 8.40 3 * 0.ninety 4 = 7.ninety 4 is the unique value earlier tax it somewhat is the main outstanding answer only try it in opposite 7.ninety 4 * a million.06 = 8.40 3