Question

A 9 cm solid metallic cube and a solid metallic cuboid having dimensions 5 cm, 13 cm, 31 cm are melted and recast into a single cube. What is the total surface area (in cm2) of the new cube?

• A. 1362
• B. 865
• C. 2744
• D. 1176

Hint:

Familiarize yourself with cubes of numbers up to 20.
Knowing that \( 14^3 = 2744 \) instantly gives you the side length of the new cube as $14\text{ cm}$.
Also, note that the surface area must be a multiple of 6 (from \( 6a^2 \)).
Checking the options:
\( 1362 \div 6 = 227 \) (not a perfect square)
$865$ is not divisible by 6
$2744$ is not divisible by 6
\( 1176 \div 6 = 196 = 14^2 \) (perfect match!)
This option analysis can solve the problem in seconds without even calculating the volume!

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
This problem is based on the concept of melting and recasting 3D geometric shapes.
When solid objects are melted and recast into a new shape, the total volume remains conserved.
We need to compute the combined volume of the original cube and the cuboid.
This combined volume will equal the volume of the newly formed single cube.
From this volume, we can find the side length of the new cube, and subsequently calculate its total surface area.

Step 2: Key Formula or Approach:

• Volume of a cube with side $s$ is: \( V_{\text{cube}} = s^3 \).
  
• Volume of a cuboid with dimensions $l$, $b$, and $h$ is: \( V_{\text{cuboid}} = l \times b \times h \).
  
• Conserved total volume: \( V_{\text{new cube}} = V_{\text{original cube}} + V_{\text{cuboid}} \).
  
• Let the side length of the new cube be $a$. Thus, \( a^3 = V_{\text{new cube}} \implies a = \sqrt[3]{V_{\text{new cube}}} \).
  
• Total Surface Area of the new cube is: \( \text{TSA} = 6a^2 \).
  

Step 3: Detailed Explanation:

• First, calculate the volume of the original solid metallic cube.
  
• The side length of the original cube is $s = 9\text{ cm}$.
  \[ V_{\text{original cube}} = 9^3 = 9 \times 9 \times 9 = 729\text{ cm}^3 \]
  
• Next, calculate the volume of the original solid metallic cuboid.
  
• The dimensions of the cuboid are length $l = 5\text{ cm}$, width $b = 13\text{ cm}$, and height $h = 31\text{ cm}$.
  \[ V_{\text{cuboid}} = 5 \times 13 \times 31 = 65 \times 31 = 2015\text{ cm}^3 \]
  
• Since both metallic bodies are melted and recast together, the total volume of the new cube is:
  \[ V_{\text{new cube}} = V_{\text{original cube}} + V_{\text{cuboid}} \]
  \[ V_{\text{new cube}} = 729 + 2015 = 2744\text{ cm}^3 \]
  
• Let $a$ be the side of the newly formed cube.
  \[ a^3 = 2744 \]
  
• To find $a$, take the cube root of $2744$:
  \[ a = \sqrt[3]{2744} \]
  
• We know that \( 10^3 = 1000 \) and \( 20^3 = 8000 \). Since $2744$ ends in 4, its cube root must end in 4. Let us test $14$:
  \[ 14 \times 14 \times 14 = 196 \times 14 = 2744 \]
  
• This confirms that the side length of the new cube is $a = 14\text{ cm}$.
  
• Finally, calculate the total surface area ($\text{TSA}$) of the new cube using the formula $6a^2$:
  \[ \text{TSA} = 6 \times a^2 \]
  \[ \text{TSA} = 6 \times 14^2 = 6 \times 196 \]
  \[ \text{TSA} = 1176\text{ cm}^2 \]
  
• Thus, the total surface area of the new cube is $1176\text{ cm}^2$.
  

Step 4: Final Answer:
The total surface area of the recast cube is $1176\text{ cm}^2$, which corresponds to Option (D).