A cube has each edge 2 cm and a cuboid is 1 cm long, 2 cm wide and 3 cm high. The paint in a certain container is sufficient to paint an area equal to 54 cm. Which one of the following is correct? 1. both cube and cuboid can be painted 2.only cube can be painted 3.only cuboid can be painted 4.neither cube nor cuboid can be painted

Options

A. Both cube and cuboid can be painted
B. Only cube can be painted
C. Only cuboid can be painted
D. Neither cube nor cuboid can be painted

Answer

To solve the problem, we need to calculate the surface area of both the cube and the cuboid, and then compare the total area that needs to be painted with the area that the paint can cover.

### Step 1: Calculate the surface area of the cube
The formula for the surface area of a cube is given by:
\[ \text{Surface Area} = 6 \times (\text{side})^2 \]
Given that each edge of the cube is 2 cm:
\[ \text{Surface Area of the Cube} = 6 \times (2 \, \text{cm})^2 = 6 \times 4 \, \text{cm}^2 = 24 \, \text{cm}^2 \]

### Step 2: Calculate the surface area of the cuboid
The formula for the surface area of a cuboid is given by:
\[ \text{Surface Area} = 2(lb + bh + hl) \]
Where \( l \) is the length, \( b \) is the breadth, and \( h \) is the height. Given:
- Length \( l = 1 \, \text{cm} \)
- Breadth \( b = 2 \, \text{cm} \)
- Height \( h = 3 \, \text{cm} \)

Now substituting the values:
\[ \text{Surface Area of the Cuboid} = 2(1 \times 2 + 2 \times 3 + 3 \times 1) = 2(2 + 6 + 3) = 2 \times 11 = 22 \, \text{cm}^2 \]

### Step 3: Calculate the total surface area to be painted
Now, we need to find the total surface area of both the cube and the cuboid:
\[ \text{Total Surface Area} = \text{Surface Area of the Cube} + \text{Surface Area of the Cuboid} \]
\[ \text{Total Surface Area} = 24 \, \text{cm}^2 + 22 \, \text{cm}^2 = 46 \, \text{cm}^2 \]

### Step 4: Compare with the paint area
The paint in the container can cover an area of 54 cm². Now we compare the total surface area to be painted with the paint area:
\[ \text{Total Surface Area} = 46 \, \text{cm}^2 < 54 \, \text{cm}^2 \]

### Conclusion
Since the total surface area of both the cube and the cuboid (46 cm²) is less than the area that the paint can cover (54 cm²), we can conclude that both the cube and the cuboid can be painted.

### Final Answer
1. Both cube and cuboid can be painted.