A Solid Metal Cylinder of Height Equal to Double the Base Diameter GMAT Problem Solving

Question: A solid metal cylinder of height equal to double the base diameter is melted and from this molten mass three identical spheres are made. What is the percentage change in the total surface area? (Area of a sphere = \(\frac{4}{3}\pi r^{3}\). The surface area of a sphere = \(\frac{4}{3}\pi r^{2}\))

1. 70%
2. 50%
3. 26%
4. 20%
5. 15%

Answer:
Solution with Explanation:
Approach Solution (1):

Let radius of cylinder = R
And, radius of each sphere = r
Height of cylinder, H = 2 * diameter of cylinder = 2 * 2R = 4R
Volume of cylinder = \(\pi R^2H = \pi*R^2*4R = 4\pi * R^3\)
Total surface area of cylinder = \(2\pi *R+2\pi * RH = 10\pi * R^2\)
Since three spheres are made out of the cylinder,
3 * Volume (each sphere) = Volume (cylinder)
\(3*\frac{4}{3}\pi*r^3 = 4\pi*R^2\)
Simplifying we get: R = r
Surface area of each sphere = \(4\pi * r^2 = 4\pi * R^2\)
Percentage difference in surface area = \(\frac {Area(all spheres) - Area(cylinder)}{Area(cylinder)} = 100\)% = \(\frac{12\pi * R^2 - 10\pi * R^2}{10\pi R^2} * 100\)% = \(\frac{2}{10}*100\)% = 20%

Correct Option: D

Approach Solution (2):

(1) Find the volume of the cylinder and break it down into 3 spheres (there is a misprint: the volume of 1 sphere = \(\frac{4}{3}\pi r^{2}\)
Cylinder has radius = r
Cylinder has height = 2 * diameter = 2 (2r) = 4r
Volume of cylinder = \(\pi r^rh\)
V = \(\pi r^2(4r)\)
V = \(\pi 4(r)^3\)

Now break this volume, melt it down, and create 3 spheres
Each sphere will have volume = \(\frac{1}{3} * \pi * 4 * r^3\)
Which is the general formula for the volume or the sphere given radius = r
The surface area of a sphere is = 4 * (Area of 4 circles with radius r) = \(4* \pi *r^2\)
We have 3 of these spheres, so multiplying this result by 3 to add up all 3 surface areas of the spheres: \(12* \pi *r^3\)
(2) Calculate the surface area of the original right cylinder
The surface area of a right cylinder = (2 Areas of identical base and top circles) + (Circumference of one base circle) * (Height of cylinder)
SA = \(2(\pi)(r)^2 + 2(\pi)(r)(4r)\)
SA = \(10\pi(r^2)\)
The new surface area of the 3 spheres = \(12*\pi*r^2\)
An increase from original 10 to New 12 corresponds to a 20% increase in the surface area

Correct Option: D

Approach Solution (3):

Volume of cylinder = \(\pi R^2H\)
Height of cylinder, H = 2 * diameter of cylinder = 2 * 2R = 4R
Therefore, Volume of cylinder = \(\pi R^2H = 4\pi * R^3\)
This solid metal cylinder is melted into 3 identical spheres
3 * Volume (each sphere) = Volume (cylinder)
\(3*\frac{4}{3}\pi*r^3 = 4\pi * R^3\)
Total surface area of cylinder = \(2\pi*R + 2\pi*RH = 10\pi * R^2\)
Surface area of each sphere = \(4\pi*r^2 = 4\pi * R^2\)
Percentage difference in surface area = \(\frac {Area(all spheres) - Area(cylinder)}{Area(cylinder)} = 100\)% = \(\frac{12\pi * R^2 - 10\pi * R^2}{10\pi R^2} * 100\)% = \(\frac{2}{10}*100\)% = 20%

Correct Option: D

“A solid metal cylinder of height equal to double the base diameter is melted and from this molten mass three identical spheres are made. What is the percentage change in the total surface area? (Area of a sphere =\(\frac{4}{3}\pi r^{3}\). The surface area of a sphere =\(\frac{4}{3}\pi r^{2}\))”- is a topic of the GMAT Quantitative reasoning section of GMAT. This question has been taken from the book “GMAT Official Guide Quantitative Review”. 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.

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