Question: A solid metallic cube is melted to form five solid cubes whose volumes are in the ratio 1 : 1 : 8: 27: 27. The percentage by which the sum of the surface areas of these five cubes exceeds the surface area of the original cube is nearest to:

1. 10
2. 20
3. 30
4. 50
5. 60

Correct Answer: (D)

Solution and Explanation:
Approach Solution 1:

The problem statement informs that:

Given:

• A solid metallic cube is melted to form five solid cubes.
• The volumes of each cube are in the ratio of 1 : 1 : 8 : 27: 27.

Find out:

• The percentage by which the sum of the surface areas of these five cubes exceeds the surface area of the original cube.

The ratio of volumes of 5 smaller cubes and the original big cube= 1 : 1 : 8 : 27 : 27 : 64
Therefore, the ratio of sides = 1 : 1 : 2 : 3 : 3 : 4 (since as per the formula, the volume of a cube = Side^3)
Therefore, the ratio of areas = 1 : 1 : 4 : 9 : 9 : 16
The sum of the areas of the 5 smaller cubes is 24 parts while that of the big cube is 16 parts.
Hence, the sum is [(24-16)/16]*100 = 50% greater.

Approach Solution 2:

The problem statement states that:

Given:

• A solid metallic cube is melted to form five solid cubes.
• The volumes of each cube are in the ratio of 1 : 1 : 8 : 27: 27.

Find out:

• The percentage by which the sum of the surface areas of these five cubes exceeds the surface area of the original cube.

Let the length of each side of the original cube is “a” cm.
Therefore, the surface area of the original cube = 6 a^2 sq. cm ….(1)
The volume of the original cube = a^3 cubic cm

The ratio of the volumes of the cube is 1 : 1 : 8 : 27: 27.
Therefore, the sum of the ratios = (1+1+8+27+27) = 64

Hence, the volumes of the 5 smaller cubes will be:
a^3 / 64, a^3 / 64, 8a^3 / 64, 27a^3 / 64 and 27 a^3 / 64 cubic cm respectively.

The sides of the smaller cubes can be determined by doing the cube roots of their volumes.
Therefore, the sides of the smaller cubes will be:
a / 4, a / 4, 2a / 4, 3 a / 4 and 3a / 4 cm respectively

The surface areas of these individual smaller cubes will be:
6 a^2 / 16, 6 a^2 / 16, 6 a^2 / 4, 54 a^2 / 16 and 54 a^2 / 16 sq. cm respectively.

Therefore the sum of the surface areas of the 5 smaller cubes
= (6a^2 + 6a^2 + 24 a^2 + 54 a^2 + 54 a^2) / 16
= 144 a^2 / 16 = 9 a^2 sq. cm…. (2)

Therefore, by comparing (1) and (2), the percentage increase in surface area
=(9a^2 - 6a^2 ) * 100 / 6a^2
= 3a^2 * 100 / 6a ^2
= 50 %

Approach Solution 3:

The problem statement suggests that:

Given:

• A solid metallic cube is melted to form five solid cubes.
• The volumes of each cube are in the ratio of 1 : 1 : 8 : 27: 27.

Find out:

• The percentage increase in the sum of the surface areas of these five cubes from the surface area of the original cube.

Let’s assume the volumes of the five cubes be x, x, 8x, 27x and 27x.
Let’s assume the sides of the cubes be a, a, 2a, 3a and 3a. (where x = a^3 )
Let’s assume the side of the original cube be A.

Therefore, A^3 = x + x + 8x + 27x + 27x
= 64x
=(4a)^3

Therefore, A = 4a

Original surface area = 96a^2 (since the surface area of a cube = 6* side^2)
New surface area of the cube = 6(a^2 + a^2 + 4a^2 + 9a^2 + 9a^2)= 144a^2

Therefore, percentage increase = \(\frac{144a^2-96a^2}{96a^2} * 100\) = \(\frac{48a^2}{96a^2} * 100\) = 50%

“A solid metallic cube is melted to form five solid cubes whose volumes”- is a topic of the GMAT Quantitative reasoning section of the GMAT exam. The GMAT Problem Solving questions test the abilities of candidates in interpreting quantitative problems. The candidates can practice varieties types of questions from GMAT Quant practice papers to improve their calculative skills.

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