Can you help me with this problem?

yumi2012 wrote:Thanks in advance

Dear Yumi,
I'm happy to help.

Here's a text version of the question:
An architect is planning to incorporate several stone spheres of different sizes into the landscaping of a public park, and workers who will be applying a finish to the exterior of the spheres need to know the surface area of each sphere. The finishing process costs $92 per square meter. The surface area of a sphere is equal to 4(pi)r^2, where r is the radius of the sphere.
In the table, select the value of that closest to the cost of finishing a sphere with a 5.5-meter circumference as well as the cost of finishing a sphere with a 7.85-meter circumference.

First of all, notice circumference has the formula
c = 2(pi)r
Square this
c^2 = 4((pi)^2)(r^2)
But surface area (SA) = 4(pi)r^2
So
c^2 = 4((pi)^2)(r^2) = (pi)*[4(pi)r^2] = (pi)*SA
or
SA = (c^2)/(pi)
That's a hugely handy shortcut.

Now, we will briefly use the calculator to square the two circumferences ----
(5.5)^2 = 30.25
(7.85)^2 = 61.6225

Now, I am going to estimate like mad. For the first sphere,
Estimate (5.5)^2 as just 30, and estimate (pi) as 3. Then, SA = 10 sq meters
Estimate the cost as $90/ sq m, so total cost = 90*10 = $900, answer (A).

The second sphere has approximately double the (c^2), so it will be double the cost of the first ----- $1800, answer (C).

Does all this make sense?
Mike

An Architect is planning to incorporate several stone spheres of difference sizes into the lanscaping of a public park and workers who will be applying a finish to the exterior of the spheres need to know the surface area of each spere. The finishing process costs $92 per square meter. The surface area is equal to 4(pi)r^2, where r is the radius of the sphere.

Select the value that is the closest to the cost of finishing a sphere with a 5.50 meter circumference and the cost of finishing a sphere with a 7.85 meter circumference.

A) $900
B) $1,200
C) $1,800
D) $2,800
E) $3,200
F) $4,500

To apply the surface area formula, 4(pi)r^2, we need to calculate the radius of each sphere.

We know that circumference = 2(pi)(radius)

Sphere #1: 5.50 meter circumference
This means that 2(pi)(radius) = 5.5
So, radius = 5.5/(2pi)

Aside: in a moment, you'll see why I didn't evaluate 5.5/(2pi)

Now, we'll apply the surface area formula
Surface area = 4(pi)r^2
= 4(pi)[5.5/(2pi)]^2
= 4(pi)[30/(4pi^2)] ...approximately

Aside: 5.5^2 = 30.25
There's a nice way to make this calculation in your head.
Here's a free video on how to do so: https://www.gmatprepnow.com/module/gmat- ... ts?id=1024

Okay, moving along
Surface area = 4(pi)[30/(4pi^2)]....approximately
= 30/pi ....approximately
= a number a little less than 10 (square meters)

Since each square meter of finishing costs $92, the total cost is a little less than $920
Only answer choice A works


Sphere #2: 7.85 meter circumference
This means that 2(pi)(radius) = 7.85
So, radius = 7.85/(2pi)

Now, we'll apply the surface area formula
Surface area = 4(pi)r^
= 4(pi)[7.85/(2pi)]^2
= 4(pi)[60/(4pi^2)] ...approximately
= 60/pi ....approximately
= a number a little less than 20 (square meters)

Each square meter of finishing costs $92, so the total cost = (92)(20), which is approximately $1800
Only answer choice C is close

Cheers,
Brent