Dear
yousufa,
I'm happy to help with these.
These are two bodacious questions, so I am going to treat each as a separate entry. First, the one about the architect and the spherical stones.
This is a question screaming out for approximations. The answers are widely spaced, and the question even says "
select the value that is closest to ...", which is practically an engraved invitation to approximate. See this blog:
https://magoosh.com/gmat/2012/the-power- ... mat-quant/
First of all, one tricky thing --- the formula for surface area is given in terms of radius, but they give us circumferences. We ourselves need to know
c = 2(pi)r
Rearrange this to
r = c/[2(pi)]
Substitute this into the surface area formula:
SA = 4(pi)(r^2) = 4(pi)[(c^2)/[4((pi)^2)] = (c^2)/(pi)
Now, multiply that surface area times $92 for the cost
Cost = (92)*(c^2)/(pi)
Right away, I am going to approximate --- pi = something a little over three, so 92 (a little over 90) divided by something a little over three is approximately 90/3 = 30, so we can approximate
Cost = 30*(c^2)
For c = 5.5
Avoid the temptation to use the calculator. The calculator is a waste of time. What is 5.5 squared? Well, 5^2 = 25, and 6^2 = 36. We are just approximating, so let's say (c^2) is about halfway between 25 and 36, around 30. (Good to pick a nice round number in estimating.)
Cost = 30*30 = 900
There's the answer for the first column.
For c = 7.85.
Well, 7^2 = 49 and 8^2 = 64, and 7.85 is much closer to 8 than to 7, so approximate (c^2) is closer to 64 than to 49, so around 60 (just to pick a nice round number).
Cost = 30*60 = 1800
There's the answer for the second column.
Does this approach make sense?
Mike
