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Optimization / Maximum Number of... Problem question

by infiniti007 » Sat Feb 28, 2015 4:11 pm
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
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cats and dogs

by GMATGuruNY » Sat Feb 28, 2015 4:49 pm
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.
Last edited by GMATGuruNY on Fri Aug 10, 2018 8:12 am, edited 1 time in total.
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by [email protected] » Sat Feb 28, 2015 5:17 pm
Hi infiniti007,

Mitch's approach is spot-on. These types of questions are relatively rare on the Official GMAT (you might see 1, but you might not see any). One of the "design elements" of these types of questions is that more than one answer is achievable given the restrictions in the prompt, so you have to be careful that you answer the question that is ASKED.

The number of cats must be a multiple of 5
The number of dogs must be a multiple of 4
The total number is 55

As Mitch showed, 15 cats and 40 dogs fits this description.

There's another way to get to 55 pets though: 35 cats and 20 dogs.

In the first scenario, we have (1/5)(15) + (1/4)(40) = 3 + 10 = 13 pets
In the second scenario, we have (1/5)(35) + (1/4)(20) = 7 + 5 = 12 pets

Since the prompt asks for the GREATEST possible number of pets, 13 is clearly the answer. But the prompt could have asked for the least possible or the difference between the greatest and the least. In those situations, slightly different work would be required.

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by infiniti007 » Sat Feb 28, 2015 7:24 pm
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!
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by infiniti007 » Sun Mar 01, 2015 11:26 am
Does anyone have or know of additional practice problems like this type?
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by DavidG@VeritasPrep » Sun Mar 01, 2015 11:42 am
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
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by DavidG@VeritasPrep » Sun Mar 01, 2015 11:45 am
OA: C
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by [email protected] » Sun Mar 01, 2015 12:47 pm
HI infiniti007,

Yes, the approach that you've outlined is correct!

Before you spend too much time working on this concept, you have to remember how rare it is (relative to the other categories that WILL appear multiple times throughout the Quant section). Unless you're already scoring at a high level in the Quant, spending any volume of study time on this one concept is not a good use of your time.

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by vinni.k » Tue Feb 14, 2017 5:14 am
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:
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by [email protected] » Tue Feb 14, 2017 10:45 am
Hi vinni.k,

Yes - if the question asked for the MINIMUM number of pets that could have been adopted, then your solution would be correct.

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by Matt@VeritasPrep » Wed Feb 15, 2017 6:49 pm
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!
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by Matt@VeritasPrep » Wed Feb 15, 2017 6:52 pm
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.
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by Matt@VeritasPrep » Wed Feb 15, 2017 7:01 pm
We also have a great blog on this topic from our fearless leader, Brian Galvin:

www.veritasprep.com/blog/2014/05/gmat-t ... -problems/
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by Jay@ManhattanReview » Thu Feb 16, 2017 5:24 am
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
Many great approaches. Here's my take...:)

We have #of cats + #of dogs = 55.

We have new guests: 1/5th of cats and 1/4th of dogs.

The aim is to maximize the value of (#of cats/5 + #of dogs/4).

Point to note: For the same number, its 1/4th would be greater that its 1/5th.

Since # of new dogs are 1/4th of the earlier number, we must maximize the number of dogs or minimize the number of cats.

Another point to note that #of cats must be a multiple of 5 and #of dogs must be a multiple of 4.

We have #of cats + #of dogs = 55.

The minimum qualified value for #of cats = 5 (divisible by 5). This gives #of dogs = 55-5=50. But 50 is not divisible by 4, so this is not possible.

Let's try with #of cats = 10 (divisible by 5). This gives #of dogs = 55-10=45. But 45 is not divisible by 4, so this is not possible.

Let's try with #of cats = 15 (divisible by 5). This gives #of dogs = 55-15=40. Bingo! 40 is divisible by 4, so this is the solution.

The greatest possible number of pets could be = [spoiler]15/5 + 40/4 = 3 + 10 = 13[/spoiler].

The correct answer: C

Hope this helps!

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by Jay@ManhattanReview » Thu Feb 16, 2017 5:35 am
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
Hi infiniti007,

Just realized that you asked for an algebraic solution.

So here it is.

Say #of cats = x and #of dogs = y.

Thus, x + y = 55 ----(1).

We have to maximize (x/5 + y/4).

From eqn (1), we get x = 55 - y

By plugging in the value of x = 55 - y in (x/5 + y/4), we get,

x/5 + y/4 = (55-y)/5 + y/4

=> 11-y/5 +y/4 = (220-4y+5y)/20

=> (220+y)/20

=> 11 + y/20

We have to maximize y/20. Since y/20 must be an integer and y < 55, the maximum value of y would be 40.

Thus, the maximum number of pets could be = 11 + 40/20 = 11 + 2 = 13.

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