Problem Solving

I came across two problems which don't seem like GMAT problems. Especially the second one which is about Right Circular Cone. Don't have the OA on either. Hoping to get detailed explanations. Thanks

if you think these ones are not GMAT like you should ponder over them?
I agree the first one has breadth and the second question has wastage terminology hardly met in any formal math literature.

Yet check another post here in PS, it seems GRE question was posted by a fellow GMATter ...

A sphere would hold maximum volume for a given surface area.

Given SA = 2(18+12+24) = 108

SA of sphere = 4Pi r^2 = 108

r ~ 2.9

Volume = 4/3 Pi r^3 = 105.5

Old Volume = 6*3*4 = 72

Change = 105-72 = 33

not sure if a sheet can be made into a sphere..

@shankar-why would a sphere have maximum volume for a given SA.Can you explain please.
For second question,
Volumes of the two figures would be same.
So,
1/3 * pi * r ^2 * h = 2/3 * pi * r ^ 3

h = 2r.Not sure though if it's correct.

if a semi circular sheet having radius 4 units is made into a right circular cone then slant height of the cone is EQUAL to the radius of the semi circular sheet and the circumference of the cone base is EQUAL to the semi circular circumference(minus the length of the straight line AOB (refer the diagram))

The circumference of the cone base = 2*pi*radius of cone base = pi*4
radius of the cone base = 2 units

Then height, radius(of the cone) and the slant height make a right angled traingle.
Slant height = 4 units
Radius of the cone = r = 2 units
height = 2√3

p.s: please read 4m in the attachment as 4 units

Its a property of a sphere

Of all the solids having a given volume, the sphere is the one with the smallest surface area; of all solids having a given surface area, the sphere is the one having the greatest volume.

ankush123251 wrote:@shankar-why would a sphere have maximum volume for a given SA.Can you explain please.

Oh ok.I can relate that to for a given perimeter,square has maximum area and for given area it has minimum perimeter.