Free math worksheets

- Did Paul add or remove any cubes when he rearranged them? - What does the number of unit cubes tell you about the volume? - Would rearranging \(18\) identical building blocks change the total amount of space occupied by the blocks?

1. The original shape contains \(18\) unit cubes. Each cube has a volume of \(1\,\text{cm}^3\), so its total volume is \(18\times1\,\text{cm}^3=18\,\text{cm}^3\). 2. Paul uses the same \(18\) cubes to make the rectangular prism. He does not add or remove any cubes. 3. Therefore, both shapes have a volume of \(18\,\text{cm}^3\). Rearranging the cubes changes the shape, but not the volume.

No. Both shapes use the same \(18\) unit cubes, so each has a volume of \(18\,\text{cm}^3\).

- Which object is smallest, and which is largest? - Think about how cubic millimeters compare with cubic centimeters. - Recall the relationship between liters and cubic decimeters. - Which units are commonly used for household containers and for large structures?

1. A grain of rice is the smallest object, so \(30\,\text{mm}^3\) is reasonable. 2. A USB flash drive is small but much larger than a grain of rice, so \(8\,\text{cm}^3\) is reasonable. 3. A typical juice carton holds \(1\,\text{L}=1000\,\text{cm}^3\). 4. A full cleaning bucket can reasonably hold about \(10\,\text{L}\). 5. A large grain silo is the largest object, so \(400\,\text{m}^3\) is reasonable.

- grain of rice \(\rightarrow30\,\text{mm}^3\) - USB flash drive \(\rightarrow8\,\text{cm}^3\) - carton of orange juice \(\rightarrow1000\,\text{cm}^3\) - full cleaning bucket \(\rightarrow10\,\text{L}\) - large grain silo \(\rightarrow400\,\text{m}^3\)

- A stack can contain more than one cube even though it has only one top face. - Count the cubes in every stack. - Distinguish a square face from a cubic unit of volume.

1. The front row has three stacks of height \(1\), and the back row has three stacks of height \(2\). 2. Add all cubes: \(1+1+1+2+2+2=9\). The volume is \(9\) cubic units. 3. Sam counted one top face for each of the \(6\) occupied stacks. A top face is a square face, not a unit of three-dimensional volume. The three back stacks each contain a second cube below their top face, so counting top faces misses cubes.

a) \(9\) cubic units b) Sam counted visible top faces instead of all unit cubes. The building contains \(9\) unit cubes, so its volume is \(9\) cubic units.

- Count the cubes in each building independently. - Compare the totals before deciding whether one set of cubes could make the other building. - Think about what changes when cubes are rearranged and what stays unchanged.

1. Building A has four stacks of height \(2\), so its volume is \(4\times2=8\) cubic units. 2. Building B has one stack of height \(4\) and four stacks of height \(1\), so its volume is \(4+1+1+1+1=8\) cubic units. 3. Because both buildings use \(8\) unit cubes, the cubes from A can be rearranged to make B without changing the total volume. 4. A solid can change shape while its volume stays the same when the same unit cubes are rearranged.

a) Building A: \(8\) cubic units; Building B: \(8\) cubic units b) Yes. Both buildings use \(8\) unit cubes. c) Rearranging the same unit cubes can change a solid’s shape without changing its volume.

- Add all the stack heights in the plan. - Use \(V=s\times s\times s\) for one cube. - Multiply the volume of one cube by the total number of cubes.

1. Add the stack heights: \(2+1+2+2+1+2=10\). The structure contains \(10\) cubes. 2. One cube has volume \(5\times5\times5=125\,\text{cm}^3\). 3. The total volume is \(10\times125=1250\,\text{cm}^3\).

1. \(10\) cubes 2. \(125\,\text{cm}^3\) 3. \(1250\,\text{cm}^3\)

- Count each structure separately, stack by stack. - Remember that a stack's height includes every cube below its top cube. - Compare each total with \(15\).

1. Structure A contains \(5 + 5 + 5 + 5 = 20\) cubes. 2. Structure B contains \(2 + 3 + 6 + 4 = 15\) cubes. 3. Structure C contains \(4 + 3 + 4 + 3 = 14\) cubes. 4. Therefore, Structure B contains exactly \(15\) unit cubes.

Structure B

- Add the stack heights in each row. - Then add the three row totals. - Use the symmetry to check that the front and back row totals are equal.

1. The front row contains \(1 + 2 + 1 = 4\) cubes. 2. The middle row contains \(2 + 3 + 2 = 7\) cubes. 3. The back row contains \(1 + 2 + 1 = 4\) cubes. 4. The total is \(4 + 7 + 4 = 15\) cubes.

\(15\) unit cubes

- Count each structure row by row or stack by stack. - Include the cubes below the top cube in every tall stack. - Record all three totals before comparing them.

1. Structure A contains \(1 + 1 + 3 + 1 + 1 = 7\) cubes. 2. Structure B contains \(1 + 1 + 2 + 2 + 2 = 8\) cubes. 3. Structure C contains \(1 + 3 + 3 = 7\) cubes. 4. Structure B has the greatest total, with \(8\) unit cubes.

Structure B, with \(8\) unit cubes

- Use the symmetry to group stacks with the same height. - Count the height-\(3\) stacks, then the height-\(2\) stacks, and finally the center stack.

1. The four corner stacks each have height \(3\), for \(4\times3=12\) cubes. 2. The four middle-edge stacks each have height \(2\), for \(4\times2=8\) cubes. 3. The center stack has height \(1\). Altogether, the building contains \(12+8+1=21\) cubes.

The building contains \(21\) unit cubes.

- Determine each box's width, depth, and height from the outlines. - Multiply the three dimensions. - Compare the results.

1. Box 1 measures \(4\times3\times5\), so it holds \(60\) unit cubes. 2. Box 2 measures \(2\times6\times5\), so it also holds \(60\) unit cubes. 3. The boxes have equal volumes.

Box 1 holds \(60\) unit cubes, and Box 2 holds \(60\) unit cubes.

- Count the cubes in each stack, including the cubes below the top cube. - Work one row at a time so you do not skip a stack. - Compare the totals after you have counted both structures.

1. Structure A has \(1 + 0 + 1 = 2\) cubes in the front row and \(2 + 1 + 2 = 5\) cubes in the back row. Its total is \(2 + 5 = 7\) cubes. 2. Structure B has \(1 + 1 + 0 = 2\) cubes in the front row and \(3 + 1 + 1 = 5\) cubes in the back row. Its total is \(2 + 5 = 7\) cubes. 3. Both structures contain the same number of cubes.

Yes. Structure A has \(7\) cubes, and Structure B also has \(7\) cubes.

- Identify the height of the center stack. - Add the stack heights from one end to the center and then back down.

1. The stack heights from left to right are \(1, 2, 3, 4, 3, 2,\) and \(1\). 2. Add them: \(1 + 2 + 3 + 4 + 3 + 2 + 1 = 16\).

\(16\) unit cubes

- Count the structure one row at a time from front to back. - A stack of height \(2\) contains the top cube and the cube directly below it.

1. The front row contains \(1 + 1 + 0 = 2\) cubes. 2. The middle row contains \(1 + 1 + 1 = 3\) cubes. 3. The back row contains \(2 + 2 + 1 = 5\) cubes. 4. Altogether, the structure contains \(2 + 3 + 5 = 10\) cubes.

The structure contains \(10\) unit cubes.

- Count each structure row by row or stack by stack. - Include every cube below the top cube in a tall stack. - Subtract the smaller total from the larger total.

1. Structure A contains \(1 + 1 = 2\) cubes in the front row and \(2 + 1 = 3\) cubes in the back row, for a total of \(2 + 3 = 5\) cubes. 2. Structure B contains \(0 + 1 + 0 = 1\) cube in the front row and \(2 + 2 + 1 = 5\) cubes in the back row, for a total of \(1 + 5 = 6\) cubes. 3. Since \(6 - 5 = 1\), Structure B contains one more cube.

1) Structure B 2) \(1\) unit cube

- Add all the numbers in the plan to find the total number of cubes. - From the front, taller stacks hide shorter stacks directly behind them. - Find the greatest number in the middle column of the plan.

1. Add the stack heights: \(1+2+1+2+4+2+1+2+1=16\). The building contains \(16\) unit cubes. 2. For the front view, find the greatest height in each column of the plan. The middle column has heights \(2\), \(4\), and \(2\), so its visible height is \(4\).

a) \(16\) unit cubes b) \(4\) cubes high

- Each number in the plan gives the height of one stack. - First find the total number of cubes needed for the building. - Subtract that number from the starting supply.

1. Add the occupied stack heights by rows: \(1+2+3+2=8\), \(2+1+2+1=6\), and \(1+1=2\). 2. The building uses \(8+6+2=16\) cubes. 3. Jonah has \(20-16=4\) cubes left.

Jonah has \(4\) cubes left.

- Count each structure carefully, including cubes below the top cubes. - Find the total for each structure. - Subtract the smaller total from the larger total.

1. Structure A contains \(1 + 1 + 0 = 2\) cubes in the front row and \(2 + 1 + 1 = 4\) cubes in the back row, for a total of \(2 + 4 = 6\) cubes. 2. Structure B contains \(1 + 1 + 1 = 3\) cubes in the front row and \(2 + 2 + 1 = 5\) cubes in the back row, for a total of \(3 + 5 = 8\) cubes. 3. Maria must remove \(8 - 6 = 2\) cubes from Structure B.

Maria must remove \(2\) unit cubes.

- Total cube count will not distinguish A, B, and C; each represents \(9\) cubes. - Compare the four stack-height entries position by position with S. - Adding one cube changes exactly one stack height by \(+1\).

1. Plan S has stack heights \(1,2\) in the front row and \(2,3\) in the back row, for \(8\) cubes total. 2. Adding one cube must increase exactly one cell by \(1\) and leave every other cell unchanged. 3. In A, only the back-right stack changes, from \(3\) to \(4\), so A can result from adding one cube. 4. In B, the front-right and back-right stacks increase while the back-left stack decreases. In C, the front-right stack decreases and the back-right stack increases by \(2\). Neither can result from adding exactly one cube without moving others. 5. All three options total \(9\) cubes, so total count cannot choose among them.

Option A. It is the only plan in which exactly one corresponding stack is one cube taller and every other stack is unchanged.

- Look first at the number of positions in the base. - Check whether every stack has the same height. - Connect cubes per layer with the number of layers.

1. The base has \(3\times2=6\) cube positions, so one horizontal layer contains \(6\) cubes. 2. Every stack is \(2\) cubes high, so there are \(2\) identical layers. 3. The prism contains \(6\times2=12\) unit cubes, so its volume is \(12\) cubic units.

a) \(6\) cubes b) \(2\) layers c) \(12\) cubic units

- Count occupied base positions, not just the rectangular outline. - Does every position in the rectangular base actually contain a cube? - A cubic unit counts only when that unit-sized space is filled once.

1. Five base positions contain one unit cube, but one base position is empty. 2. Therefore, the structure contains \(5\) unit cubes, not \(6\). 3. A volume of \(6\) cubic units would require all \(6\) unit-cube spaces to be filled with no gap. The empty position is a gap, so Rowan's count is too large.

No. The volume is \(5\) cubic units. One of the \(6\) base positions is empty, so the unit cubes do not fill all \(6\) cubic-unit spaces.

- How many decimeters are in one meter? - Picture the cubes arranged in layers. How many cubes are in one layer, and how many layers are there?

1. Since \(1\,\text{m}=10\,\text{dm}\), exactly \(10\) small cubes fit in one row along each edge. 2. One layer contains \(10\times10=100\) small cubes. The large cube is \(10\,\text{dm}\) high, so it contains \(10\) layers. Therefore, the total number of small cubes is \(10\times10\times10=1000\).

a) \(10\) cubes b) \(1000\) cubes, because \(10\times10\times10=1000\).

- Determine how many cubes fit along each dimension. - For the larger cubes, check whether each box dimension is a multiple of \(2\,\text{cm}\). - Think about the \(5\,\text{cm}\) by \(4\,\text{cm}\) base one layer at a time.

1. For the \(1\,\text{cm}\) cubes, \(5\) fit along the length, \(4\) along the width, and \(3\) along the height. Therefore, \(5\times4\times3=60\) cubes fit. 2. For the \(2\,\text{cm}\) cubes, \(2\) fit along the \(5\,\text{cm}\) length, \(2\) fit along the \(4\,\text{cm}\) width, and \(1\) fits along the \(3\,\text{cm}\) height. 3. Therefore, \(2\times2\times1=4\) larger cubes fit completely inside. 4. The box dimensions \(5\,\text{cm}\) and \(3\,\text{cm}\) are not multiples of \(2\,\text{cm}\). The remaining strips of space are too narrow for another whole cube.

a) \(60\) cubes b) \(4\) cubes c) Some space remains empty because \(5\,\text{cm}\) and \(3\,\text{cm}\) are not divisible by \(2\,\text{cm}\) with no remainder.

- First find how many cubes are in one complete base layer. - Then determine how many equal layers make \(18\) cubes. - Think of the prism as a stack of equal floors.

1. One layer contains \(3\times3=9\) cubes. 2. Divide the total number of cubes by the number in each layer: \(18\div9=2\). 3. The prism is \(2\) cubes high.

The rectangular prism is \(2\) cubes high.

- Use the length and width to find the number of cubes in the bottom layer. - Compare the prism’s height with the edge length of one cube. - Multiply the number of cubes in one layer by the number of layers. - Imagine taking the prism apart and placing every cube in one straight row.

1. The bottom layer contains \(5\times3=15\) cubes. 2. Because each cube is \(1\,\text{cm}\) high, a height of \(2\,\text{cm}\) gives \(2\) layers. 3. The entire prism contains \(15\times2=30\) cubes. 4. A row of \(30\) cubes with edge length \(1\,\text{cm}\) is \(30\times1\,\text{cm}=30\,\text{cm}\) long.

a) \(15\) cubes b) \(2\) layers c) \(30\) cubes d) \(30\,\text{cm}\)

- Determine how many times the bottom layer fits into the total number of cubes. - Find two whole numbers whose product is \(15\). - Picture how the cubes could be arranged on the bottom of the box.

1. Each layer contains \(15\) cubes, so the number of layers is \(45\div15=3\). 2. The length and width of the bottom layer must be whole-number factors of \(15\). One possible pair is \(5\,\text{cm}\) and \(3\,\text{cm}\), since \(5\times3=15\).

a) \(3\) layers b) One possible answer is \(5\,\text{cm}\) by \(3\,\text{cm}\).

- Read the stack heights in the diagram one position at a time. - Count how many positions contain a first, second, and third cube. - Relate the total number of \(1\,\text{cm}^3\) cubes to the figure's volume.

1. Read the stack heights from the diagram. The bottom layer fills all \(9\) positions of the \(3\times3\) footprint, so it contains \(9\) cubes. 2. Eight stacks reach at least the second layer, so the middle layer contains \(8\) cubes. 3. Four corner stacks reach the third layer, so the top layer contains \(4\) cubes. 4. Altogether, the figure contains \(9+8+4=21\) cubes. 5. Each cube has volume \(1\,\text{cm}^3\), so \(V=21\times1\,\text{cm}^3=21\,\text{cm}^3\).

The total volume is \(21\,\text{cm}^3\).

- Compare the stack heights one row at a time. - Keep track of which locations are blank. - After choosing the matching plan, add its stack heights to find the total number of cubes.

1. Read the structure row by row. The front row has two blank locations followed by a stack of height \(1\). 2. The middle row has a blank location followed by stacks of heights \(1\) and \(2\). 3. The back row has stacks of heights \(1\), \(2\), and \(3\). 4. Only plan C shows all three rows in those positions. 5. Add the stack heights in plan C: \(1 + 1 + 2 + 1 + 2 + 3 = 10\).

a) Plan C b) \(10\) unit cubes

- Count each stack in the diagram, including cubes below the top cube. - How many unit cubes fill a \(3 \times 3 \times 3\) rectangular prism? - Subtract the number already present from the completed prism's total.

1. Count the cubes by rows. The front row contains \(1 + 1 + 0 = 2\) cubes, the middle row contains \(2 + 2 + 1 = 5\) cubes, and the back row contains \(3 + 3 + 2 = 8\) cubes. The structure contains \(2 + 5 + 8 = 15\) cubes. 2. A complete \(3 \times 3 \times 3\) rectangular prism contains \(3 \times 3 \times 3 = 27\) unit cubes. 3. The number of cubes to add is \(27 - 15 = 12\).

a) \(15\) unit cubes b) \(12\) unit cubes

- How many unit cubes fit in a complete \(3 \times 3 \times 3\) cube? - Count how many cubes are already shown. - Work row by row so that no cubes are missed. - Subtract the number present from the total capacity.

1. A completed \(3 \times 3 \times 3\) cube contains \(3\times3\times3=27\) unit cubes. 2. Count the cubes already present by rows: the back row has \(3+2+2=7\), the middle row has \(2+1+0=3\), and the front row has \(1+0+0=1\). Altogether, \(7+3+1=11\) cubes are present. 3. The number still needed is \(27-11=16\).

\(16\) unit cubes are still needed.

- For each view, use the tallest stack along each line of sight. - Keep the columns or rows in order. - To find the volume, add the heights of all the stacks.

1. From the front, use the tallest stack in each left-to-right column. The heights are \(3\), \(2\), and \(1\), so the front view is \((3, 2, 1)\). 2. From the right, use the tallest stack in each front-to-back row. The heights are \(3\), \(2\), and \(1\), so the right view is \((3, 2, 1)\). Therefore, Option A is correct. 3. Add the heights of all nine stacks: \(3+1+0+2+2+1+0+1+1=11\). The building contains \(11\) unit cubes, so its volume is \(11\) cubic units.

a) Option A b) \(11\) cubic units

- Use the outline to determine each box's length, width, and height. - Count the cubes already shown in each box. - Subtract the existing cubes from the full volume.

1. Box A measures \(3\times3\times4\), so its total volume is \(36\) unit cubes. 2. Box A contains \(4+1+1+4+4=14\) cubes, so \(36-14=22\) cubes are missing. 3. Box B measures \(5\times2\times3\), so its total volume is \(30\) unit cubes. 4. Box B contains \(3+1+1+1+1+3=10\) cubes, so \(30-10=20\) cubes are missing. 5. Since \(22>20\), Box A needs more cubes.

a) Box A has a total volume of \(36\) unit cubes. Box B has a total volume of \(30\) unit cubes. b) Box A needs more cubes; \(22\) cubes are missing.

- Count the stacks carefully, including cubes hidden under other cubes. - How do you find the volume of a cube from its edge length? - Once you know the volume of one cube, how can you find the volume of the entire building?

1. Count the cubes by rows of stacks. The back row has \(3+2+1=6\) cubes, the middle row has \(2+1=3\) cubes, and the front row has \(1\) cube. The building contains \(6+3+1=10\) cubes. 2. One small cube has volume \(3\times3\times3=27\,\text{cm}^3\). 3. The total volume is \(10\times27\,\text{cm}^3=270\,\text{cm}^3\).

1. \(10\) cubes 2. \(27\,\text{cm}^3\) 3. \(270\,\text{cm}^3\)

- Use the finished prism's three dimensions. - Add all the stack heights in the plan. - Subtract the cubes already present from the total needed.

1. The finished prism's volume is \(4\times3\times3=36\,\text{cm}^3\), so it needs \(36\) unit cubes. 2. Add the stack heights shown in the plan: \(2+1+0+0+3+3+2+1+3+2+1+0=18\). 3. The number still needed is \(36-18=18\).

a) The finished prism's volume is \(36\,\text{cm}^3\). b) \(18\) more unit cubes are needed.

- Count the cubes by adding the stack heights. - Find the volume of one small cube. - Use \(1\,\text{dm}^3=1000\,\text{cm}^3\) for the conversion.

1. Add the stack heights: \(1+0+1+1+0+1+2+2+2=10\). The structure contains \(10\) cubes. 2. One cube has volume \(5\times5\times5=125\,\text{cm}^3\). 3. The total volume is \(10\times125=1250\,\text{cm}^3\). 4. Since \(1\,\text{dm}^3=1000\,\text{cm}^3\), \(1250\div1000=1.25\,\text{dm}^3\).

a) The structure contains \(10\) cubes. b) The total volume is \(1250\,\text{cm}^3\). c) The total volume is \(1.25\,\text{dm}^3\).

- Find the tallest stack in each column and each row. - A side view turns the building's rows into visible columns. - Add every stack height to find the volume.

1. The front-view column heights are \((3, 2, 1)\), so the back view is \((1, 2, 3)\). 2. The tallest stacks in the three front-to-back rows have heights \((3, 2, 2)\). From the left, that order is reversed, giving \((2, 2, 3)\). Therefore, the view is from the left. 3. Add the heights of all nine stacks: \(3+1+1+1+2+1+2+0+1=12\). The volume is \(12\) cubic units.

a) From the left b) \(12\) cubic units

- Find the tallest stack and its location. - Match every occupied position and stack height. - Add the plan's stack heights to find the volume.

1. Read the stack heights by location. The back-left stack has height \(3\), the stack directly to its right has height \(1\), the center stack has height \(2\), and the front-right stack has height \(1\). 2. Plan A shows exactly those positions and heights. Plans B and C place one or more stacks differently, so Plan A is correct. 3. Add the stack heights: \(3+1+2+1=7\). The building has volume \(7\) cubic units.

a) Plan A b) \(7\) cubic units

- A and C have the same occupied cells, so compare where the height-\(3\) stack sits within that footprint. - Rotate the entire grid, including the stack heights, not just the outline. - Use volume as a check, but remember that all three totals are equal.

1. Plans A and C have the same L-shaped footprint, so footprint alone cannot decide which plan matches B after rotation. 2. Rotating Plan A by \(90^\circ\) gives Plan B: the height-\(3\) stack moves with the corner of the L-shaped footprint, and the four height-\(1\) stacks move to the corresponding rotated cells. 3. Plan C keeps the same footprint as A but places the height-\(3\) stack at the end of an arm instead of at the corner, so no rotation makes it match A or B. 4. Each plan represents \(7\) cubes. Equal volume and even the same footprint can therefore occur with different stack-height arrangements.

a) Plans A and B b) A: \(7\) cubic units; B: \(7\) cubic units; C: \(7\) cubic units c) Equal volume does not determine the stack arrangement; A and C even share the same footprint but are not rotations of the same building.

- Compare total cube counts and numbers of occupied positions first. Are those enough to distinguish the options? - Then compare the height at each front-row and back-row position. - Remember that the first grid row is the front row; on a plan it is drawn at the bottom.

1. Building A has front-row stack heights \(1,2\) followed by an empty position, and back-row stack heights \(3,4\) followed by an empty position. This matches Plan 1. 2. Building B has stack heights \(1\) and \(2\) at the front corners and \(3\) and \(4\) at the back corners, with the middle column empty. This matches Plan 2. 3. Plan 3 also totals \(10\) cubes in \(4\) occupied positions, but its stack heights are arranged differently from both buildings. 4. Therefore, total volume and occupied-position count cannot determine the match; the height at each specific top-view position must agree.

a) Building A matches Plan 1. Building B matches Plan 2. Plan 3 is not used. b) The stack height at each top-view position must match; equal total cube count and equal numbers of occupied positions are not sufficient.

- Begin with the front row and compare it with the bottom table row. - Each plan entry gives the height of one stack. - After completing the plan, add all six stack heights.

1. In the front row, the stack heights from left to right are \((2, 1, 2)\). Therefore, \(b = 2\). 2. In the back row, the stack heights are \((3, 2, 3)\). Therefore, \(a = 2\). 3. Add all six stack heights: \(2 + 1 + 2 + 3 + 2 + 3 = 13\).

1) \(a = 2\) and \(b = 2\) 2) \(13\) unit cubes

- Read the front row from the bottom row of Plan P. - Compare the stack heights in both structures. - After choosing the match, add its stack heights.

1. Plan P has front-row stack heights \((2, 1)\) and back-row stack heights \((1, 1)\). 2. Structure A has those heights in the same locations. Structure B switches the two front-row heights. 3. Add the stack heights in structure A: \(2 + 1 + 1 + 1 = 5\).

a) Structure A b) \(5\) unit cubes

- For the front view, use the tallest stack in each left-to-right column. - Add every stack height to find the volume. - Look for cubes hidden behind cubes at the same height.

1. In the left column, the tallest stack is \(2\) cubes high, so \(2\) square faces are visible. In the right column, the tallest stack is \(3\) cubes high, so \(3\) square faces are visible. Altogether, \(2+3=5\) square faces are visible. 2. Add all stack heights: \(2+1+0+3=6\). The volume is \(6\) cubic units. 3. One cube is directly behind another cube at the same height, so its front face is hidden. That is why the front view shows \(5\) squares even though the building contains \(6\) cubes.

a) \(5\) square faces b) \(6\) cubic units c) One cube's front face is hidden behind another cube.

- Use the tallest stack in each left-to-right column for the front view. - Add all stack heights to find each volume. - Matching views and volumes do not always mean the buildings are identical.

1. Building A has visible front heights \((3, 1)\): the left column is \(3\) cubes high and the right column is \(1\) cube high. Building B has the same visible front heights, so the front views match. 2. Building A has volume \(3+1+1+1=6\) cubic units. Building B has volume \(3+1+2+0=6\) cubic units. 3. The buildings are not identical. They have the same front view and volume, but the stack heights behind the front row are arranged differently.

a) Yes. Both front views have column heights \((3, 1)\). b) Each building has volume \(6\) cubic units. c) No. Their hidden stack arrangements are different.

- For the front view, find the tallest stack in each column. - For the right-side view, find the tallest stack in each row. - Add every stack height to find the volume.

1. From the front, the visible column heights are \(3\), \(1\), and \(2\). Therefore, \(3+1+2=6\) square faces are visible. 2. From the right, the visible row heights are \(1\), \(2\), and \(3\). Therefore, \(1+2+3=6\) square faces are visible. 3. Add all stack heights: \(1+0+1+2+0+1+3+1+2=11\). The volume is \(11\) cubic units.

a) \(6\) square faces b) \(6\) square faces c) \(11\) cubic units

- Compare each stack with the corresponding plan entry. - Add the corrected stack heights to find the volume. - Compare the correct total with the total from the incorrect plan.

1. The front row of building G has heights \((1, 2, 1)\), which matches the front row of plan P. The back row of the building has heights \((2, 2, 2)\). 2. The middle entry of the plan's back row is \(3\), but the corresponding stack is only \(2\) cubes high. The \(3\) should be replaced with \(2\). 3. The actual volume is \(1+2+1+2+2+2=10\) cubic units. 4. The incorrect plan totals \(11\) cubic units, so it gives a volume that is \(1\) cubic unit too large.

a) The \(3\) in the middle of the back row is wrong. It should be \(2\). b) \(10\) cubic units c) \(1\) cubic unit too large

- Count the cubes in the front and back rows separately. - Each unit cube contributes one cubic unit of volume. - Compare the building's volume with the volume of the containing rectangular prism.

1. The front row contains \(3\) cubes. The back row has three stacks of height \(2\), so it contains \(3 \times 2=6\) cubes. The building contains \(3+6=9\) cubes. 2. Because each cube has volume \(1\) cubic unit, the building's volume is \(9\) cubic units. 3. The containing rectangular prism has volume \(3 \times 2 \times 2=12\) cubic units. The number of missing cubes is \(12-9=3\).

a) \(9\) cubes b) \(9\) cubic units c) \(3\) unit cubes

- Count the cubes in one layer, then double the result. - Compare a full layer with the empty square in the center.

1. A full \(4 \times 4\) layer contains \(4\times4=16\) cubes. 2. The empty \(2 \times 2\) center removes \(2\times2=4\) cubes, so one frame layer contains \(16-4=12\) cubes. 3. The frame is \(2\) cubes high, so it contains \(12\times2=24\) cubes.

The frame contains \(24\) unit cubes.

- For each building, record the height of every stack. - Add the stack heights for each building separately.

1. Building A contains \(1+(2+1)+(3+2+1)=10\) cubes. 2. Building B contains \((2+1)+(2+2+1)+(3+2+2)=15\) cubes. 3. Building C contains \((2+2)+(4+2+2)=12\) cubes. 4. Therefore, Building B has volume \(15\) cubic units.

Building B has volume \(15\) cubic units.

- Find the number of cubes in the full block. - Find the number of cubes removed for the shaft. - Subtract the removed cubes from the original total.

1. The full cube contains \(5\times5\times5=125\) unit cubes. 2. The shaft contains \(3\times3\times5=45\) unit cubes. 3. The number remaining is \(125-45=80\).

\(80\) unit cubes remain.

- For the bottom layer, count one cube at every occupied position. - For the total, count every cube in each stack. - Check whether the position at the far right of the back row is occupied.

1. The bottom layer has one cube at each occupied position. There are \(2\) occupied positions in the front row and \(3\) in the back row, so the bottom layer contains \(2 + 3 = 5\) cubes. 2. The front row contains \(2 + 1 + 0 = 3\) cubes, and the back row contains \(2 + 2 + 1 = 5\) cubes. The entire structure contains \(3 + 5 = 8\) cubes.

The bottom layer contains \(5\) unit cubes. The entire structure contains \(8\) unit cubes.

- Record the height of the stack at each occupied position. - List the back row first and the front row last. - Add all the stack heights to find the total number of cubes.

1. The back row has stack heights \((2, 2, 2)\), the middle row has heights \((1, \text{empty}, 1)\), and the front row has heights \((1, 1, 1)\). 2. Add the stack heights: \(2+2+2+1+1+1+1+1=11\) cubes.

a) Back row: \((2, 2, 2)\); middle row: \((1, \text{empty}, 1)\); front row: \((1, 1, 1)\) b) The building contains \(11\) unit cubes.

- Match the front row of the 3D building to a row in the plan. - Use the compass orientation to name that side. - Add all plan entries to find the volume.

1. The front row of the 3D building has stack heights \((2, 3, 1)\). 2. In Plan P, this pattern is the top row, which is the north row. Because the north row appears at the front of the 3D view, the building is viewed from the north. 3. Add all stack heights in the plan: \(1+1+1+1+2+1+2+3+1=13\). The volume is \(13\) cubic units.

a) From the north b) \(13\) cubic units

- Match the front row of the building with the bottom row of each plan. - Add all stack heights in the correct plan. - Find the number of unit cubes in a solid \(3 \times 3 \times 3\) cube.

1. Plan A matches every stack height and position. 2. Add the stack heights: \(3+2+7 \times 1=12\) cubes. 3. A solid \(3 \times 3 \times 3\) cube contains \(27\) unit cubes. The number missing is \(27-12=15\).

a) Plan A b) \(12\) cubes c) \(15\) cubes

- Count the cubes row by row, including cubes in each stack. - Find the volume of one cube with an edge length of \(10\,\text{cm}\). - How many cubic centimeters equal one liter?

1. The front row contains \(3\) cubes. The back row has two corner stacks with \(3\) cubes each, for \(3+3=6\) cubes. The building contains \(3+6=9\) cubes. 2. One cube has volume \(10\times10\times10=1000\,\text{cm}^3\). 3. Since \(1000\,\text{cm}^3=1\,\text{L}\), each cube has a volume of \(1\,\text{L}\). 4. The total volume is \(9\times1\,\text{L}=9\,\text{L}\).

The total volume is \(9\,\text{L}\).

- Add all the stack heights in the diagram. - Find the volume of one cube before finding the structure's volume. - For part b, use the stated numbers of cubes along the prism's three dimensions.

1. Add the stack heights: \(1+0+0+1+0+0+1+1+0+2+2+1=9\). The structure contains \(9\) cubes. 2. One cube has volume \(2\times2\times2=8\,\text{cm}^3\), so the structure has volume \(9\times8=72\,\text{cm}^3\). 3. The completed rectangular prism contains \(4\times3\times2=24\) cubes. 4. The number of additional cubes needed is \(24-9=15\).

a) The volume is \(72\,\text{cm}^3\). b) \(15\) more cubes are needed.

- Count the stacks row by row from front to back. - A stack's height equals the number of cubes at that plan location. - Include zeros for locations with no stack.

1. Add the stack heights: \(1 + 2 + 1 + 2 + 3 + 3 + 3 = 15\). 2. The front row is \((1, 2, 1)\), the middle row is \((0, 2, 0)\), and the back row is \((3, 3, 3)\).

a) \(15\) unit cubes b) Front: \((1, 2, 1)\) Middle: \((0, 2, 0)\) Back: \((3, 3, 3)\)

- Match each building stack with its plan cell. - The front row of the building is the bottom row of the table. - Compare the current total with four stacks of height \(4\).

1. The front-left stack is \(3\) cubes high, so \(x=3\). 2. The back-right stack is \(2\) cubes high, so \(y=2\). 3. The building contains \(3+1+2+2=8\) cubes. Four stacks of height \(4\) contain \(4 \times 4=16\) cubes. Therefore, \(16-8=8\) cubes must be added.

a) \(x=3\) and \(y=2\) b) \(8\) cubes

- Track where each stack moves during the rotation. - Keep each stack height unchanged. - Add all stack heights to find the volume.

1. Before the rotation, the front row is \((2, 1)\) and the back row is \((3, 0)\). 2. After a \(90^\circ\) clockwise rotation, the front row is \((1, 0)\) and the back row is \((2, 3)\). Plan B shows this arrangement. 3. The building contains \(2+1+3+0=6\) unit cubes, so its volume is \(6\) cubic units both before and after the rotation. 4. A rotation changes only the positions of the stacks; it does not add or remove cubes.

a) Plan B b) \(6\) cubic units before and \(6\) cubic units after c) The same cubes are only rearranged by the rotation.

- Break the staircase into sections with the same depth. - Find the capacity of one full layer of the box. - Remember to include cubes hidden below the upper cubes.

1. The four steps are each \(4\) cubes deep. They contain \(4\times1=4\), \(4\times2=8\), \(4\times3=12\), and \(4\times4=16\) cubes. The building contains \(4+8+12+16=40\) cubes. 2. The \(4 \times 4 \times 4\) box holds \(4\times4\times4=64\) cubes. 3. The number still needed is \(64-40=24\).

1. \(40\) unit cubes 2. \(64\) unit cubes 3. \(24\) unit cubes

- Count the occupied positions around the empty center and determine each stack's height. - Use the three box dimensions to find the total cube capacity. - For the fraction, compare the number of occupied cubes with the box's total capacity.

1. The outer ring consists of \(8\) stacks with \(2\) cubes in each stack, so the structure contains \(8\times2=16\) cubes. 2. The box holds \(3\times3\times2=18\) unit cubes. Therefore, \(18-16=2\) cubes are needed to fill it. 3. The structure occupies \(\frac{16}{18}\) of the box's capacity. Simplifying by dividing numerator and denominator by \(2\) gives \(\frac{8}{9}\).

1) \(16\) unit cubes 2) \(2\) unit cubes 3) \(\frac{8}{9}\)

- Add every number in the plan to find the total number of cubes. - For the front view, compare the stacks that line up behind one another. - In each visible column, only the tallest aligned stack determines the height. - For the left-side view, compare the stacks in each row from front to back.

1. Add all the stack heights: \(1+2+0+3+1+2=9\). The building contains \(9\) unit cubes. 2. For the front view, use the greatest height in each column: left \(\max(1,3)=3\), middle \(\max(2,1)=2\), and right \(\max(0,2)=2\). 3. For the left-side view, use the greatest height in each row: front \(\max(1,2,0)=2\) and back \(\max(3,1,2)=3\).

a) \(9\) unit cubes b) Left: \(3\); middle: \(2\); right: \(2\) c) Front: \(2\); back: \(3\)

- For the front view, find the tallest stack in each column. - For the left-side view, find the tallest stack in each row. - Add every stack height to find the total number of cubes.

1. For the front view, take the greatest stack height in each column: \(\max(2,2,2)=2\), \(\max(2,1,0)=2\), and \(\max(0,0,2)=2\). The front-view heights are \((2, 2, 2)\). 2. For the left-side view, take the greatest height in each row. The back, middle, and front rows each have greatest height \(2\), so the left-side-view heights are \((2, 2, 2)\). 3. Add the stack heights: \((2+2+0)+(2+1+0)+(2+0+2)=4+3+4=11\).

a) \((2, 2, 2)\) b) \((2, 2, 2)\) c) \(11\) unit cubes

- For each plan, find the tallest stack in each column. - Add all plan entries to find each volume. - Compare the positions of stacks, not only the totals.

1. In Plan A, the greatest height in the left column is \(3\), and the greatest height in the right column is \(2\). Its front view is \((3, 2)\). 2. In Plan B, the greatest heights are \(2\) and \(3\). Its front view is \((2, 3)\). Therefore, Plan A matches the given front view. 3. Plan A has volume \(3+1+2+2=8\) cubic units. Plan B has volume \(1+3+2+2=8\) cubic units. 4. The plans have equal volumes but place their stacks differently, so their front views are different.

a) Plan A b) Each plan represents \(8\) cubic units. c) Equal volume does not determine where the stacks are located.

- Each number in the top-view plan is a stack height. - Add all of the known stack heights first. - Compare that known total with the structure's total of \(15\) cubes.

1. Add the known stack heights in the plan: \(3+2+1+2+1+2+2+0=13\). 2. The structure contains \(15\) cubes altogether, so the missing stack height is \(15-13=2\).

\(x=2\)

- For each view, identify which stacks line up behind one another. - The tallest stack in each line determines the visible column height. - To find the volume, add the heights of all stacks in the plan.

1. Looking from the front, the greatest heights in the four plan columns are \(3\), \(2\), \(3\), and \(2\). Looking from the back reverses their order, so the back-view heights are \((2, 3, 2, 3)\). 2. For the right-side view, find the greatest stack height in each row from front to back. The row maxima are \(3\), \(2\), and \(3\), so the right-side heights are \((3, 2, 3)\). 3. Add all stack heights in the plan: \((3+1+2)+(1+2+1)+(1+3)=6+4+4=14\). Therefore, the volume is \(14\) cubic units.

a) \((2, 3, 2, 3)\) b) \((3, 2, 3)\) c) \(14\) cubic units

- A complete \(2 \times 2 \times 2\) cube has stack height \(2\) in every plan cell. - Identify the plan cells that are one cube short and look at the shape those missing positions make. - Both candidates have the same number of cubes, so compare their shapes rather than their counts.

1. In a complete \(2 \times 2 \times 2\) cube, every plan cell would have stack height \(2\). In S, three cells have height \(1\), so the missing top-layer cubes occupy three cells in an L shape. 2. Piece A is an L-shaped arrangement of three connected cubes. After a rotation, it can occupy exactly those three missing top-layer positions. 3. Piece B is a straight row of three connected cubes. Rotating a straight three-cube piece cannot make it occupy an L-shaped set of positions. 4. Both pieces contain three cubes, so cube count alone cannot decide; the spatial arrangement of the cubes must match the complement of S.

Piece A. Both pieces contain three cubes, but only the L-shaped piece can match the three missing top-layer positions.

- At each base position, the two stack heights must add to \(2\). - A position with height \(2\) in arrangement A needs height \(0\) in the matching arrangement. - Compare the required heights with arrangements B, C, and D.

1. At each position in the \(2\times2\) base, the two stack heights must add to \(2\). 2. Arrangement A has stack heights \((2, 1)\) in one row and \((1, 0)\) in the other row. 3. The matching arrangement must therefore have heights \((0, 1)\) and \((1, 2)\). 4. Arrangement B has exactly these stack heights.

Arrangement B fits arrangement A exactly.

- Count all the cubes in the complete structure G. - Count all the cubes in blue part B. - Subtract the blue-part total from the complete-structure total.

1. The complete structure G contains \(2 + 2 + 3 + 2 = 9\) unit cubes. 2. Blue part B contains \(1 + 1 + 2 + 0 = 4\) unit cubes. 3. The other part contains \(9 - 4 = 5\) unit cubes.

The other part contains \(5\) unit cubes.

- Determine how many cubes fit along each inside dimension. - How many cubes are in one layer, and how many complete layers fit? - What happens when the box height is not a multiple of the cube edge length? - Can part of a cube be packed as an additional whole cube?

1. In a standard box, \(50\div5=10\) cubes fit along the length, \(30\div5=6\) fit along the width, and \(20\div5=4\) fit along the height. 2. Therefore, \(10\times6\times4=240\) cubes fit in the standard box. 3. In the shorter box, only \(3\) complete cubes fit along the height because \(3\times5=15\) and \(4\times5=20>18\). 4. Therefore, \(10\times6\times3=180\) cubes fit in the shorter box. 5. Dividing the two volumes would treat all empty space as though it could be combined into space for additional cubes. The unused \(3\,\text{cm}\) of height is not tall enough for another complete cube.

a) \(240\) cubes b) \(180\) cubes. Only whole cubes can be packed, and the remaining \(3\,\text{cm}\) of height cannot hold another cube.

- First express all dimensions in the same unit. - How many cubes fit along each side of the crate? - How many cubes cover the bottom in one layer, and how many layers fit? - Which metric volume unit is equal to one liter?

1. Convert the crate dimensions to centimeters: \(1.2\,\text{m}=120\,\text{cm}\), \(8\,\text{dm}=80\,\text{cm}\), and the height is \(60\,\text{cm}\). 2. Along the three dimensions, \(120\div20=6\), \(80\div20=4\), and \(60\div20=3\) cubes fit. 3. One layer has \(6\times4=24\) cubes. With \(3\) layers, the crate holds \(24\times3=72\) cubes. 4. In decimeters, the crate measures \(12\,\text{dm}\times8\,\text{dm}\times6\,\text{dm}\). Its volume is \(12\times8\times6=576\,\text{dm}^3\). 5. Since \(1\,\text{dm}^3=1\,\text{L}\), the volume is \(576\,\text{L}\).

a) \(72\) cubes b) Each layer has \(6\times4=24\) cubes, and \(3\) layers give \(24\times3=72\) cubes. c) \(576\,\text{L}\)

- Use the tallest stack in each row for the side view. - Add all stack heights to find each volume. - A side view can hide cubes behind taller stacks.

1. For Building A, the tallest stack in each row is \(2\), so the left-side view shows \(2+2+2=6\) square faces. 2. For Building B, the tallest stack heights by row are \(2\), \(3\), and \(2\), so the left-side view shows \(2+3+2=7\) square faces. Therefore, Building B has more visible squares. 3. Building A has volume \(1+2+1+1+2+1+1+2+1=12\) cubic units. Building B has volume \(2+1+1+0+3+0+1+1+2=11\) cubic units. 4. Building B has more squares visible from the left, but Building A has the greater volume. A side view records only the tallest stack in each row, not every cube.

a) Building B: \(7\) visible square faces compared with \(6\) for Building A b) Building A: \(12\) cubic units; Building B: \(11\) cubic units c) No. Building B has more visible squares, but Building A has greater volume.

- Find the total number of cubes in the completed large cube. - Divide the current building into horizontal layers and count each layer. - Look for a pattern in the layer counts. - Subtract the number present from the completed total.

1. A complete \(4 \times 4 \times 4\) cube contains \(4\times4\times4=64\) unit cubes. 2. Count the building by horizontal layers from top to bottom: \(1,3,6,10\). It contains \(1+3+6+10=20\) cubes. 3. The number still needed is \(64-20=44\).

\(44\) unit cubes must be added.

- First determine how many unit cubes fill the entire glass case. - Count the blue building by stacks or by layers. - Compare the building's volume with the case's total volume to form a fraction. - How many cubes represent exactly half of the case?

1. The glass case can hold \(4\times4\times4=64\) unit cubes. 2. The outer frame has \(12\) stacks that are each \(2\) cubes high, for \(12\times2=24\) cubes. The inner square has \(4\) stacks that are each \(4\) cubes high, for \(4\times4=16\) cubes. The building contains \(24+16=40\) cubes. 3. The filled fraction is \(\frac{40}{64}=\frac{5}{8}\). 4. Half of the case is \(64\div2=32\) cubes. Therefore, \(40-32=8\) cubes must be removed.

a) \(40\) unit cubes b) \(\frac{5}{8}\) c) \(8\) cubes

- Mentally rotate each building from above. - Compare where the one-cube stack is attached to the two taller stacks. - Count the cubes in every building before deciding what volume can show.

1. In Building 1, two adjacent stacks of height \(2\) form the corner of an L-shape, and a one-cube stack is attached at the third position. Building 2 has the same relative arrangement after a \(90^\circ\) rotation. 2. In Building 3, the one-cube stack is attached at the other end of the L-shape, so it cannot be obtained from Building 1 or 2 by a rotation. 3. Each building has volume \(2+2+1=5\) cubic units. 4. Equal volume tells only the number of cubes. Buildings can use the same number of cubes in different arrangements, as Building 3 shows.

a) Buildings 1 and 2 b) Each building has volume \(5\) cubic units. c) Equal volume does not determine the arrangement of the cubes.

- Find the total number of cubes in the building. - Two identical stacked pieces must contain the same number of cubes. - Locate the empty position in the front or back row. - The bottom plan row represents the front of the building.

1. The building has five stacks of height \(2\) and one empty position, so it contains \(5 \times 2=10\) cubes. 2. Because the two pieces are identical and stacked directly, each piece contains one cube at each occupied position. Each piece therefore contains \(5\) cubes. 3. The empty position is in the middle of the front row. Plan A shows exactly that arrangement. Plan B has only \(4\) cubes, and Plan C places the empty position in the back row.

Plan A