Problem Solving

Would appreciate any help on approach and solving the following problem. Is there a way to set it up algebraically?

On Monday, an animal shelter housed 55 cats and dogs and by Friday, exactly 1/5 of the cats and 1/4 of the dogs had been adopted. If no new cats or dogs were brought to the shelter during this period, what is the greatest possible number of pets that could have been adopted from the animal shelter between Monday and Friday?

A) 11
B) 12
C) 13
D) 14
E) 20

infiniti007 wrote:Would appreciate any help on approach and solving the following problem. Is there a way to set it up algebraically?

On Monday, an animal shelter housed 55 cats and dogs and by Friday, exactly 1/5 of the cats and 1/4 of the dogs had been adopted. If no new cats or dogs were brought to the shelter during this period, what is the greatest possible number of pets that could have been adopted from the animal shelter between Monday and Friday?

A) 11
B) 12
C) 13
D) 14
E) 20

Goal:
To MAXIMIZE the number of pets adopted.
Since the fraction of dogs adopted (1/4) is greater than the fraction of cats adopted (1/5), the number of pets adopted will be maximized if we MAXIMIZE THE NUMBER OF DOGS.
:
Since 1/4 of the dogs are adopted, the total number of dogs must be a MULTIPLE OF 4.
Since 1/5 of the cats are adopted, the total number of cats must be a MULTIPLE OF 5.

Options:
Dogs = 52, cats = 55-52 = 3.
Dogs = 48, cats = 55-58 = 7.
Dogs = 44, cats = 55-44 = 11.
Dogs = 40, cats = 55-40 = 15.

Only the option in red yields a multiple of 5 for the number of cats.
Implication:
Greatest number of dogs that could be adopted = (1/4)(40) = 10.
Greatest number of cats that could be adopted = (1/5)(15) = 3.
Thus:
Greater number of pets that could be adopted = 10+3 = 13.

The correct answer is C.

Thanks Mitch and Rich, I really appreciate the explanations.

To Rich's point about the question possibly asking for the LEAST possible # of pets that could have been adopted - would the following approach work in general?

Goal:
To MINIMIZE the number of pets adopted
Since the fraction of dogs adopted (1/4) is greater than the fraction of cats adopted (1/5), the number of pets adopted will be minimized if we MINIMIZE THE NUMBER DOGS or, alternatively: MAXIMIZE THE NUMBER OF CATS.

Options:
Cats = 50, dogs = 5
Cats = 45, dogs = 10
Cats = 40, dogs = 15
Cats = 35, dogs = 20

Only the option in red yields a multiple of 5 for the number of cats and a multiple of 4 for the number of dogs.

So, the number of dogs that could be adopted = (1/4)(20) = 5
The number of cats that could be adopted = (1/5)(35) = 7

Thus:
The least number of pets that could be adopted = 5 + 7 = 12.

Please let me know if this would be the right way to approach optimizing for the minimal number of something.

Also - as a side question - if this were hypothetically a Data Sufficiency question, asking simply: What were the number of pets that could have been adopted from the shelter between Monday and Friday?

And,

Statement 1 simply saying: cats + dogs = 55

Statement 2 simply saying: (1/5) of the cats and (1/4) of the dogs had been adopted between Monday and Friday

I'm thinking that the answer would be: E (Statements 1 and 2 TOGETHER are not sufficient) since we could have more than one solution when combining both statements. Is this accurate?

Thank you again!

Does anyone have or know of additional practice problems like this type?

Ask and you shall receive:

In a home library consisting of 108 books, some hardcover and some softcover, exactly 2/3 of the hard cover and exactly 1/4 of the softcover books are nonfiction. What is the greatest possible number of nonfiction books in this home library?

A) 18
B) 40
C) 67
D) 72
E) 96

OA: C

Nice question,

What if the question asked - what is the least possible number of pets that could have been adopted from the animal shelter ?

Then we must minimize the number of dogs. I think in that case we have to do the following

we have (1/5)(35) + (1/4)(20) = 7 + 5 = 12 pets

Am i correct ? :roll:

GMATGuruNY wrote: Goal:
To MAXIMIZE the number of pets adopted.

I don't think I've ever been happier with the start of a solution!

infiniti007 wrote:Does anyone have or know of additional practice problems like this type?

Here's a good one (not written by me):

A dental licensure exam requires a 75 percent minimum score in order to pass each section. Did Jennifer pass the 30-question third section?

(1) Jennifer recorded eight more correct answers on the second half of the third section than she did on the first half of the third section.

(2) Jennifer answered one more question correctly on the third section than she did on the 28-question second section, which she passed.

We also have a great blog on this topic from our fearless leader, Brian Galvin:

www.veritasprep.com/blog/2014/05/gmat-t ... -problems/