## Is this question even solvable? It is too hard I feel

##### This topic has expert replies
Senior | Next Rank: 100 Posts
Posts: 44
Joined: 17 Apr 2015
Thanked: 6 times
Followed by:2 members
GMAT Score:740

### Is this question even solvable? It is too hard I feel

by nchaswal » Tue May 31, 2016 8:23 am How can this question be even solved without doing intricate mathematics?
It is GMAT. So what?

### GMAT/MBA Expert

Elite Legendary Member
Posts: 10392
Joined: 23 Jun 2013
Location: Palo Alto, CA
Thanked: 2867 times
Followed by:508 members
GMAT Score:800
by [email protected] » Tue May 31, 2016 9:14 am
Hi nchaswal,

This question is essentially a giant multi-step "estimation" question, BUT you have to use the answer choices to your advantage and the work that you do to answer the first question will actually HELP you to answer the second. Here's how:

Using the given formulas for circumference and surface area:

C = 2(pi)(R)
SA = 4(pi)(R^2)

The first sphere has circumference = 5.5m
It's radius is...
5.5 = 2(pi)(R)
5.5/(2pi) = R ....don't do anything more to this....

It's surface area is....
SA = 4pi(5.5/2pi)^2
SA = 4pi[5.5^2/4pi^2)
SA = 5.5^2/pi

5.5 is between 5 and 6, so 5.5^2 is between 25 and 36
We need a rough estimate for....
(25 to 36)/pi

If we say pi = 3 (Note: we ALL know that this isn't super-accurate, but it works in this question. You'll see why this is helpful in a moment...)
(25 to 36)/pi = between 8 and 12
Let's say it's about 10....

With a Surface Area of 10 meters^3 and a cost of $92 per meter^3, we have about.... 10(92) =$920

• Page 1 of 1