A Sphere is Inscribed in a Cube with an Edge of 10. What is GMAT Problem Solving

Question: A sphere is inscribed in a cube with an edge of 10. What is the shortest possible distance from one of the vertices of the cube to the surface of the sphere?

(A) 10(√3−1)
(B) 5
(C) 10(√2−1)
(D) 5(√3−1)
(E) 5(√2−1)

“A sphere is inscribed in a cube with an edge of 10. What is”- is a topic of the GMAT Quantitative reasoning section of GMAT. This question has been taken from the book “GMAT Official Guide”. To solve GMAT Problem Solving questions a student must have knowledge about a good amount of qualitative skills. The GMAT Quant topic in the problem-solving part requires calculative mathematical problems that should be solved with proper mathematical knowledge.

Solution and Explanation:
Approach Solution 1:
As the sphere is inscribed in the cube then the edges of the cube equal to the diameter of a sphere --> Diameter = 10.

Next, diagonal of a cube equals to
Diagonal= √10^2 + 10^2 + 10^2 = 10√3.

Now half of (Diagonal minus Diameter) is a gap between the vertex of a cube and the surface of the sphere, which will be the shortest distance:
x= Diagonal−Diameter/2 = 10∗√3−10/2 = 5(√3−1)

Correct Answer: D

Approach Solution 2:
Subtract the diameter of the sphere from the length of the diagonal and divide by two.

Diameter of the sphere = 10, since the sphere is inscribed inside the cube, it touches the faces and thus the diameter of the sphere = edge of the cube

length−of−the−diagonal = √3∗10, since diagonal of the cube is √3∗(edge−of−the−cube)

(√3∗10−10)/2

10∗(√3−1)/2

5∗(√3−1)

Correct Answer: D

Approach Solution 3:
The shortest distance would be the (diagonal of the cube - diameter of the sphere)/2

Diagonal = 10√3

Diameter = 10

The shortest distance = 10√3-10/2 = 10(√3-1)/2 = 5(√3-1)

Correct Answer: D

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