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solid geometry - Volume of a cone, cube, and cylinder

For COMPETITION
Number of Total Problems: 8.
FOR PRINT ::: (Book)

Problem Num : 1

From : NCTM

Type: Complex

Section:solid geometry

Theme:Length
Adjustment# :

Difficulty: 2

Category Volume of a cone, cube, and cylinder

Analysis

Solution/Answer

Problem Num : 2

From : NCTM

Type: Understanding

Section:solid geometry

Theme:Length
Adjustment# : 0

Difficulty: 1

Category Volume of a cone, cube, and cylinder

Analysis

Solution/Answer
Answer:

Problem Num : 3

From : AMC10

Type:

Section:solid geometry

Theme:
Adjustment# : 0

Difficulty: 1

'
A company sells peanut butter in cylindrical jars. Marketing research suggests that using wider jars will increase sales. If the diameter of the jars is increased by $25\%$ without altering the volume, by what percent must the height be decreased?
$mathrm{(A) } 10 qquad mathrm{(B) } 25 qquad mathrm{(C) } 36 qquad mathrm{(D) } 50 qquad mathrm{(E) } 60$
'

Category Volume of a cone, cube, and cylinder

Analysis

Solution/Answer
When the diameter is increased by $25\%$ , it is increased by $frac54$ , so the area of the base is increased by $left(frac54 ight)^2=frac{25}{16}$ .
To keep the volume the same, the height must be $frac{1}{frac{25}{16}}=frac{16}{25}$ of the original height, which is a $36\%$ reduction $Rightarrowmathrm{(C)}$ .

Answer:

Problem Num : 4

From : AMC10

Type:

Section:solid geometry

Theme:
Adjustment# : 0

Difficulty: 1

'
A wooden cube $n$ units on a side is painted red on all six faces and then cut into $n^3$ unit cubes. Exactly one-fourth of the total number of faces of the unit cubes are red. What is $n$ ?
$mathrm{(A) } 3qquad mathrm{(B) } 4qquad mathrm{(C) } 5qquad mathrm{(D) } 6qquad mathrm{(E) } 7$
'

Category Volume of a cone, cube, and cylinder

Analysis

Solution/Answer
Since there are $n^2$ little faces on each face of the big wooden cube, there are $6n^2$ little faces painted red.
Since each unit cube has $6$ faces, there are $6n^3$ little faces total.
Since one-fourth of the little faces are painted red,
$frac{6n^2}{6n^3}=frac{1}{4}$
$frac{1}{n}=frac{1}{4}$
$n=4Longrightarrow mathrm{(B)}$

Answer:

Problem Num : 5

From : AMC10

Type:

Section:solid geometry

Theme:
Adjustment# : 0

Difficulty: 1

'
Centers of adjacent faces of a unit cube are joined to form a regular octahedron. What is the volume of this octahedron?
$mathrm{(A) } frac{1}{8}qquadmathrm{(B) } frac{1}{6}qquadmathrm{(C) } frac{1}{4}qquadmathrm{(D) } frac{1}{3...$
'

Category Volume of a cone, cube, and cylinder

Analysis

Solution/Answer
We can break the octahedron into two square pyramids by cutting it along a plane perpendicular to one of its internal diagonals. The cube has edges of length 1 so all edges of the regular octahedron have length $frac{sqrt{2}}{2}$ . Then the square base of the pyramid has area $left(frac{1}{2}sqrt{2} ight)^2 = frac{1}{2}$ . We also know that the height of the pyramid is half the height of the cube, so it is $frac{1}{2}$ . The volume of a pyramid with base area $B$ and height $h$ is $A=frac{1}{3}Bh$ so each of the pyramids has volume $frac{1}{3}left(frac{1}{2} ight)left(frac{1}{2} ight) = frac{1}{12}$ . The whole octahedron is twice this volume, so $frac{1}{12} cdot 2 = frac{1}{6} Longrightarrow mathrm{(B)}$ .

Answer:

Problem Num : 6

From : AMC10

Type:

Section:solid geometry

Theme:
Adjustment# : 0

Difficulty: 1

'
A cube with side length $1$ is sliced by a plane that passes through two diagonally opposite vertices $A$ and $C$ and the midpoints $B$ and $D$ of two opposite edges not containing $A$ or $C$ , as shown. What is the area of quadrilateral $ABCD$ ?

$mathrm{(A)} frac{sqrt{6}}{2}qquadmathrm{(B)} frac{5}{4}qquadmathrm{(C)} sqrt{2}qquadmathrm{(D)} frac{5}{8}qq...$
'

Category Volume of a cone, cube, and cylinder

Analysis

Solution/Answer
Since $AB = AD = CB = CD = sqrt{.5^2+1^2}$ , it follows that $ABCD$ is a rhombus. The area of the rhombus can be computed by the formula $A = frac 12 d_1d_2$ , where $d_1,,d_2$ are the diagonals of the rhombus (or of a kite in general). $BD$ has the same length as a face diagonal, or $sqrt{1^2 + 1^2} = sqrt{2}$ . $AC$ is a space diagonal, with length $sqrt{1^2+1^2+1^2} = sqrt{3}$ . Thus $A = frac 12 imes sqrt{2} imes sqrt{3} = frac{sqrt{6}}{2} mathrm{(A)}$ .

Answer:

Problem Num : 7

From : AMC10

Type:

Section:solid geometry

Theme:
Adjustment# : 0

Difficulty: 1

'
One dimension of a cube is increased by $1$ , another is decreased by $1$ , and the third is left unchanged. The volume of the new rectangular solid is $5$ less than that of the cube. What was the volume of the cube?
$extbf{(A)} 8 qquad extbf{(B)} 27 qquad extbf{(C)} 64 qquad extbf{(D)} 125 qquad extbf{(E)} 216$
'

Category Volume of a cone, cube, and cylinder

Analysis

Solution/Answer
Let the original cube have edge length $a$ . Then its volume is $a^3$ . The new box has dimensions $a-1$ , $a$ , and $a+1$ , hence its volume is $(a-1)a(a+1) = a^3-a$ . The difference between the two volumes is $a$ . As we are given that the difference is $5$ , we have $a=5$ , and the volume of the original cube was $5^3 = 125Rightarrowoxed{ ext{(D)}}$ .

Answer:

Problem Num : 8

From : AMC10

Type:

Section:solid geometry

Theme:
Adjustment# : 0

Difficulty: 1

'
A solid cube has side length 3 inches. A 2-inch by 2-inch square hole is cut into the center of each face. The edges of each cut are parallel to the edges of the cube, and each hole goes all the way through the cube. What is the volume, in cubic inches, of the remaining solid?
$extbf{(A)} 7 qquad extbf{(B)} 8 qquad extbf{(C)} 10 qquad extbf{(D)} 12 qquad extbf{(E)} 15$
'

Category Volume of a cone, cube, and cylinder

Analysis

Solution/Answer
Solution 1

Imagine making the cuts one at a time. The first cut removes a box $2 imes 2 imes 3$ . The second cut removes two boxes, each of dimensions $2 imes 2 imes 0.5$ , and the third cut does the same as the second cut, on the last two faces. Hence the total volume of all cuts is $12 + 4 + 4 = 20$ .
Therefore the volume of the rest of the cube is $3^3 - 20 = 27 - 20 = oxed{7 extbf{(A)}}$ .

Solution 2

We can use Principle of Inclusion-Exclusion to find the final volume of the cube.
There are 3 "cuts" through the cube that go from one end to the other. Each of these "cuts" has $2 imes 2 imes 3=12$ cubic inches. However, we can not just sum their volumes, as the central $2 imes 2 imes 2$ cube is included in each of these three cuts. To get the correct result, we can take the sum of the volumes of the three cuts, and subtract the volume of the central cube twice.
Hence the total volume of the cuts is $3(2 imes 2 imes 3) - 2(2 imes 2 imes 2) = 36 - 16 = 20$ .
Therefore the volume of the rest of the cube is $3^3 - 20 = 27 - 20 = oxed{7 extbf{(A)}}$ .

Solution 3

We can visualize the final figure and see a cubic frame. We can find the volume of the figure by adding up the volumes of the edges and corners.
Each edge can be seen as a $2 imes 0.5 imes 0.5$ box, and each corner can be seen as a $0.5 imes 0.5 imes 0.5$ box.
$12cdot{frac{1}{2}} + 8cdot{frac{1}{8}} = 6+1 = oxed{7 extbf{(A)}}$ .

Answer:

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