The average (arithmetic mean) age of the buildings on a certain city

This is a ‘Could be’ type of question. So, the best approach would be to take simple values to make a statement true, ONCE. Any statement that is true once, ‘Could be’ true.

From the question stem, we know that we have four buildings, each of age 2 years. To find the average age of the buildings, we can use the equation,

Average age of the buildings = \(\frac{Sum of ages of all buildings}{Number of buildings}\).

We know that the average age of the buildings is greater than 40 years and also that none of the buidlings are more than 80 years old. So, 80 is the upper bound for us.
We are trying to find a possible number of buildings on one city block.

Let’s try and see if we can make statement I true. Statement I says that there could be 8 buildings on the block.
We already know that 4 of these 8 buildings are 2 years old. So, the sum of ages of these buildings is 8 years. If we take the other 4 buildings to be 80 years old, we get 320 years as the sum of ages of these buildings. Therefore, the average age = \(\frac{328}{8}\) = 41 years. This satisfies the condition that the average age should be more than 40 years.

So statement I could be true. Any answer option not containing statement I can be ruled out. Options B and C go out.
Possible answer options are A, D and E.

Instead of trying statement II, let’s try statement III. This way, we will be able to eliminate more options, regardless of how it turns out.
Of 40 buildings, 4 buildings are 2 years old, summing up to 8 years. If we take all the other 36 buildings as 80 years old, we get their sum to be 2880 years. Therefore, the average age = \(\frac{2888}{40}\) = 72.2 years. So, statement III could be true.
Therefore, options A and D can be ruled out. The correct answer option is E.

As mentioned in some of my posts on ‘Could be’ questions, a good idea is to prove a statement true once and just be done with it. You do not have to worry about the statement turning out to be false in some other cases, because that’s not what the question is asking you to evaluate.

Hope this helps!

assume no. of buildings to be 8 then sum of the ages (average age is assumed 40.0001) = 320. if 4 buildings are 2 years old then the average of rest of the 4 must be (320-4*2)/4 = 312/4 = 78. which is less than 80.

As the number of buildings increases the average of the rest of the buildings will fall. IMO option E

Solution:

We are given that 4 buildings were built two years ago (that is, each of them has an age of 2) and the average age of all buildings is greater than 40 years old and no buildings are more than 80 years old. Now, let’s check each given Roman numeral choice. That idea is to see if we can show the remaining buildings (i.e., those that were not built two years ago) could have an average age of no more than 80 years old. (Note: In each of the inequalities below, x is the average age of the remaining buildings.)

I. 8 (Since 4 have an age of 2 each, there are 4 remaining buildings.)

[4(2) + 4x]/8 > 40

8 + 4x > 320

4x > 312

x > 78

We see that x is greater than 78; it could still be less than or equal to 80, such as 79 or 80. Therefore, 8 could be the number of buildings.

II. 11 (Since 4 have an age of 2 each, there are 7 remaining buildings.)

[4(2) + 7x]/11 > 40

8 + 7x > 440

7x > 432

x > 61.7

We see that x is greater than 61.7; it could definitely be less than or equal to 80. Therefore, 11 could be the number of buildings.

As we can see, the higher the number of buildings, the lower the average age of the remaining buildings. Therefore, we can safely conclude that 40 could also be the number of buildings.

Answer: E

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.