Q 68. Cubes and Dice
A cubical vessel of side 1 m is filled completely with water. How many millilitres of water is contained in it (neglect thickness of the vessel)?
- Volume of the cube = Side × Side × Side = 1 × 1 × 1 = 1 cubic meter Now, 1 cubic meter = 1000000 mililiters.
Q 10. Cubes and Dice
A solid cube is painted yellow, blue and black such that opposite faces are of same colour. The cube is then cut into 36 cubes of two different sizes such that 32 cubes are small and the other four cubes are big. None of the faces of the bigger cubes is painted blue. How many cubes have only one face painted?
• By the condition described in the question the cube can be divided as shown below.
• For the sake of simplicity, let the side of original cube be 12 units.
• So the new structure is as shown in image.
• The required cubes are marked with R. So a total of 4 and 4 = 8 cubes will have only one face painted.
Q 72. Cubes and Dice
Each face of a cube can be painted in black or white colours. In how many different ways can the cube be painted?
• 0 sides White : 1 way (all sides Black)
• 1 side White : 1 way (all other sides Black)
• 2 sides White : 2 ways (1 way with adjacent sides White, and 1 way with opposite sides White)
• 3 sides White : 2 ways (1 way where three White sides have same corner, and 1 way where opposite sides are White and one center side is White)
• 4 sides White : 2 ways (same as 2 sides Black)
• 5 sides White : 1 way (same as 1 side Black)
• 6 sides White : 1 way (same as 0 sides Black)
• So total=1+1+2+2+2+1+1 = 10 ways.
Q 4. Cubes and Dice
How many diagonals can be drawn by joining the vertices of an octagon?
To find the number of diagonals in a polygon, we can use the formula:
\[ \text{Number of diagonals} = \frac{n(n-3)}{2} \]
where \( n \) is the number of sides (or vertices) of the polygon.
For an octagon, \( n = 8 \).
Step-by-step Calculation:
● Step 1: Substitute \( n = 8 \) into the formula:
\[ \text{Number of diagonals} = \frac{8(8-3)}{2} \]
● Step 2: Simplify the expression inside the parentheses:
\[ 8 - 3 = 5 \]
● Step 3: Multiply the results:
\[ 8 \times 5 = 40 \]
● Step 4: Divide by 2 to find the number of diagonals:
\[ \frac{40}{2} = 20 \]
Therefore, the number of diagonals that can be drawn by joining the vertices of an octagon is 20.
Q 35. Cubes and Dice
A solid cube of 3 cm side, painted on all its faces, is cut up into small cubes of 1 cm side. How many of the small cubes will have exactly two painted faces?
To solve this problem, we need to understand how the cube is divided and how the painting affects the smaller cubes.
1. Initial Setup:
○ We have a solid cube with a side length of 3 cm.
○ This cube is painted on all its faces.
2. Cutting the Cube:
○ The cube is cut into smaller cubes, each with a side length of 1 cm.
○ Since the original cube is 3 cm on each side, it will be divided into \(3 \times 3 \times 3 = 27\) smaller cubes.
3. Analyzing the Smaller Cubes:
● Corner Cubes: These are the cubes at the corners of the original cube. Each corner cube will have 3 painted faces. There are 8 corner cubes in total.
● Edge Cubes: These are the cubes along the edges of the original cube, excluding the corners. Each edge of the original cube has 3 cubes, but the corner cubes are not included in this count. Therefore, each edge has 1 cube with exactly 2 painted faces. Since the original cube has 12 edges, there are \(12 \times 1 = 12\) such cubes.
● Face Cubes: These are the cubes on the faces of the original cube, excluding the edges. Each face of the original cube has a \(3 \times 3\) grid of cubes, but the edge cubes are not included in this count. Therefore, each face has 1 cube with exactly 1 painted face. Since the original cube has 6 faces, there are \(6 \times 1 = 6\) such cubes.
● Inner Cubes: These are the cubes inside the original cube, not touching any face. There is only 1 such cube, and it has no painted faces.
4. Conclusion:
○ The number of smaller cubes with exactly two painted faces is 12, which corresponds to the edge cubes.
Therefore, the correct answer is Option A: 12.
Q 68. Cubes and Dice
The outer surface of a 4 cm x 4 cm x 4 cm cube is painted completely in red. It is sliced parallel to the faces to yield sixty four 1 cm x 1 cm x 1 cm small cubes. How many small cubes do not have painted faces?
To solve this problem, we need to determine how many of the smaller 1 cm x 1 cm x 1 cm cubes do not have any painted faces after the larger cube is sliced.
Step-by-step Explanation:
1. Understanding the Cube Structure:
○ The original cube is a 4 cm x 4 cm x 4 cm cube.
○ It is sliced into 64 smaller cubes, each measuring 1 cm x 1 cm x 1 cm.
2. Identifying the Inner Cubes:
○ The outer layer of the cube is painted, which means all cubes on the surface have at least one face painted.
○ To find the cubes that do not have any painted faces, we need to focus on the inner cubes that are not exposed to the surface.
3. Calculating the Inner Layer:
○ The inner cubes form a smaller cube inside the original cube.
○ This inner cube is formed by removing the outer layer of 1 cm thickness from each face of the original cube.
○ The dimensions of the inner cube are (4 - 2) cm x (4 - 2) cm x (4 - 2) cm = 2 cm x 2 cm x 2 cm.
4. Counting the Inner Cubes:
○ The inner cube is composed of 2 x 2 x 2 = 8 smaller cubes.
Therefore, there are 8 small cubes that do not have any painted faces. These are the cubes located entirely within the inner 2 cm x 2 cm x 2 cm cube.
Hence, the correct answer is Option A: 8.
Q 49. Cubes and Dice
A piece of tin is in the form of a rectangle having length 12 cm and width 8 cm. This is used to construct a closed cube. The side of the cube is:
- Surface area of constructed cube = Area of rectangle
- Hence, 6a2 = 12*8
- So, a = 4 cm. Hence, the answer is (c).
Q 69. Cubes and Dice
A cube has all its faces painted with different colours. It is cut into smaller cubes of equal sizes such that the side of the small cube is one-fourth the big cube. The number of small cubes with only one of the sides painted is:
To solve this problem, we need to understand the structure of the cube and how it is divided:
1. Initial Setup:
○ We have a large cube with each face painted a different color.
○ The large cube is divided into smaller cubes, where the side of each small cube is one-fourth the side of the large cube.
2. Division of the Cube:
○ Since the side of each small cube is one-fourth of the large cube, the large cube is divided into \(4 \times 4 \times 4 = 64\) smaller cubes.
3. Identifying Small Cubes with One Painted Face:
○ Small cubes with only one face painted are located on the faces of the large cube but not on the edges or corners.
○ Each face of the large cube is a \(4 \times 4\) grid of small cubes.
○ The cubes on the edges and corners of each face will have more than one face painted.
4. Counting the Small Cubes with One Painted Face:
○ On each face of the large cube, the inner \(2 \times 2\) grid of small cubes will have only one face painted.
○ This inner grid consists of \(2 \times 2 = 4\) small cubes per face.
5. Total Calculation:
○ Since the large cube has 6 faces, and each face contributes 4 small cubes with only one face painted, the total number of such small cubes is:
\[
6 \times 4 = 24
\]
Therefore, the number of small cubes with only one of the sides painted is 24.
Q 54. Cubes and Dice
Each of the six different faces of a cube has been coated with a different colour i.e., V, I, B, G, Y and O. Following information is given:
1. Colours Y, O and B are on adjacent faces.
2. Colours I, G and Y are on adjacent faces.
3. Colours B, G and Y are on adjacent faces.
4. Colours O, V and B are on adjacent faces.
Which is the colour of the face opposite to the face coloured with O?
To solve this problem, we need to determine the arrangement of colors on the cube's faces based on the given conditions. Let's analyze the information step by step:
● Condition 1: Colours Y, O, and B are on adjacent faces.
● Condition 2: Colours I, G, and Y are on adjacent faces.
● Condition 3: Colours B, G, and Y are on adjacent faces.
● Condition 4: Colours O, V, and B are on adjacent faces.
From these conditions, we can deduce the following:
● Y is a common color in conditions 1, 2, and 3, indicating that Y is adjacent to O, B, I, and G.
● B is a common color in conditions 1, 3, and 4, indicating that B is adjacent to Y, G, and O.
● O is adjacent to Y and B (from condition 1) and V (from condition 4).
Since Y is adjacent to O, B, I, and G, it cannot be opposite to any of these colors. Therefore, Y must be opposite to V.
Now, let's determine the face opposite to O:
○ O is adjacent to Y, B, and V.
○ Since Y is opposite to V, O cannot be opposite to V.
○ Therefore, the face opposite to O must be one of the remaining colors, which are I or G.
Since G is adjacent to B and Y (from condition 3), and O is adjacent to B and Y (from condition 1), G cannot be adjacent to O. Thus, G must be opposite to O.
Therefore, the color of the face opposite to the face colored with O is G.
Q 51. Cubes and Dice
A cube has six numbers marked 1, 2, 3, 4, 5 and 6 on its faces. Three views of the cube are shown below: What possible numbers can exist on the two faces marked A and B, respectively on the cube ?
Looking at the first two figures we see that 2 is opposite to 6 and 3 is opposite to 4. The third figure tells us that 5 is opposite to 1. Thus, the numbers A and B, which are adjacent to 5 could be any of the following four, 6 and 4, 4 and 2, 2 and 3, and 3 and 6. Out of these only 2 and 3 is given as a option, so (a) is the right answer.
Q 47. Cubes and Dice
Six squares are coloured, front and back, red (R), blue (B), yellow (Y), green (G), white (W) and orange (O) and are hinged together as shown in the figure given below. If they are folded to form a cube, what would be the face opposite the white face?
To solve this problem, we need to determine the arrangement of the squares when they are folded into a cube. The key is to understand the layout of the squares and how they relate to each other when folded.
Let's assume the squares are labeled as follows based on their colors:
○ Square 1: Red (R)
○ Square 2: Blue (B)
○ Square 3: Yellow (Y)
○ Square 4: Green (G)
○ Square 5: White (W)
○ Square 6: Orange (O)
When these squares are hinged together and folded into a cube, each square will form one face of the cube. The challenge is to determine which face is opposite the white face (W).
Step-by-step Analysis:
● Identify Adjacent Faces:
When folded into a cube, each face will have four adjacent faces. The face opposite to a given face will not share an edge with it.
● Determine the Opposite Face:
If we consider the layout of the squares before folding, we can visualize how they will come together to form a cube. Typically, in such puzzles, the opposite faces are determined by the layout pattern.
● Visualize the Folding:
Imagine folding the squares into a cube. The white face (W) will have one face directly opposite to it. By visualizing or sketching the folding process, we can determine which face ends up opposite the white face.
● Conclusion:
After visualizing the folding process, we find that the blue face (B) is opposite the white face (W) when the squares are folded into a cube.
Therefore, the face opposite the white face is Blue (B), which corresponds to Option C.
Q 49. Cubes and Dice
Three views of a cube following a particular motion are given below: What is the letter opposite to A?
To determine which letter is opposite to 'A' on the cube, we need to analyze the given views of the cube and deduce the relative positions of the letters.
● View 1: Let's assume the visible faces show letters A, B, and P.
● View 2: Assume the visible faces show letters A, M, and H.
● View 3: Assume the visible faces show letters B, P, and M.
Step-by-step Analysis:
○ From View 1 and View 2, we see that 'A' is adjacent to both 'B' and 'M'.
○ From View 1 and View 3, we see that 'B' is adjacent to both 'A' and 'P'.
○ From View 2 and View 3, we see that 'M' is adjacent to both 'A' and 'P'.
Since 'A' is adjacent to 'B', 'M', and 'P', the only letter left that can be opposite to 'A' is 'H'.
Therefore, the letter opposite to 'A' is H.