DS: "Each of the offices in a certain building.. "

[guest]

"Each of the offices in a certain building has a floor area of 200, 300, or 350 square feet. How many offices are on the first floor of the building?"

(1) There is a total of 9,500 square feet of office space on the first floor of the building

(2) Ten of the offices on the first floor have floor areas of 350 square feet each.


I got the answer by plugging and chuging different values for 200*x 300*y and 350*z to determine if there were different possibilities, but it took about 5 minutes. anyone know to do this sort of problem with a formula? any shortcuts?

Thanks


(answer is E)

[RonPurewal]

you can follow your gut to the conclusion that answer 1 is insufficient - i.e., that there are a ridiculously huge number of ways of making the sum total of 9500 out of a collection of 200's, 300's, and 350's. Also, #2 is clearly insufficient by itself, because there could be any number of offices (as long as it's more than ten).

taking the two together, just subtract the 3500 square feet taken up by the ten offices in question, and you have 6000 square feet left. therefore, the question is now something like this:
'is there more than one way to make 6000 out of a combination of 200's and 300's?'
(the previous sentence has been edited from its prior version, which mistakenly still considered the possibility of more 350's)

the answer to that question is definitely yes, so it's (e). if you don't see why right away, notice that you could change the number of offices, while not changing the number of square feet, just by replacing two 300-sqft offices with three 200-sqft offices (or vice versa).

[guest]

thanks for the help. That's more or less how I solved the problem, except I broke the square footage down to: "how many offices that are either 2, 3 or 3.5 square feet can fit in a total of 60 square feet (95 square feet - 35 square feet)". I find it easier to work with smaller numbers.

[RonPurewal]

[Guest]

In this post 'Is there more than one way to make 6000 out of a combination of 200's, 300's, and 350's?' , Why should we consider any more 350's when they have specifically mentioned that there are only 10 such offices.. I know it does not affect the answer over here but I have faced this situation where I was'nt sure whether to consider any extras... Thnx

[RonPurewal]

In this post 'Is there more than one way to make 6000 out of a combination of 200's, 300's, and 350's?' , Why should we consider any more 350's when they have specifically mentioned that there are only 10 such offices.. I know it does not affect the answer over here but I have faced this situation where I was'nt sure whether to consider any extras... Thnx

you have a good point. the post in question has been edited.
thanks.

[pratik.munjal]

Here's how I re-did the question (got it wrong in the first attempt)

200x + 300y + 350z (from the question)

Statement 1: 200x + 300y + 350z=9500. There could be many values for x, y or z. Not sufficient.

Statement 2: 10 offices x 350 sq feet per office. That means a remainder of 6000 sq feet. Again, many numbers possible for x and y. (Note that I am still referring to my own equation here). Many values means the absence of a unique solution. Not sufficient.

Combining the two statements:

I still can't determine x and y. I can guess values, and many values will satisfy the equation.

Hence, E.

[jnelson0612]

Yes, agreed, pratik.