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Two animal shelters, Cedar Shelter and Harbor Shelter, each house only 17 Mar 2026, 19:00
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85% (hard)

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45% (02:24) correct 55% (02:32) wrong based on 92 sessions

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Two animal shelters, Cedar Shelter and Harbor Shelter, each house only cats and dogs. Cedar Shelter has more cats than dogs while Harbor Shelter has more dogs than cats. Does Cedar Shelter have more animals in total than Harbor Shelter?

(1) The number of cats at Harbor Shelter is 2 more than twice the number of dogs at Cedar Shelter.

(2) The number of cats at Cedar Shelter is 2 more than twice the number of dogs at Harbor Shelter.

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Two animal shelters, Cedar Shelter and Harbor Shelter, each house only 26 Mar 2026, 01:00
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GMAT Club Official Solution:

Two animal shelters, Cedar Shelter and Harbor Shelter, each house only cats and dogs. Cedar Shelter has more cats than dogs while Harbor Shelter has more dogs than cats. Does Cedar Shelter have more animals in total than Harbor Shelter?

(1) The number of cats at Harbor Shelter is 2 more than twice the number of dogs at Cedar Shelter.

Let x be the number of dogs at Cedar. Then Harbor has 2x + 2 cats, but Harbor’s number of dogs is still unknown other than being greater than its cats. Cedar’s number of cats is also unknown other than being greater than x. So Cedar’s total could be smaller than Harbor’s or larger than Harbor’s. Not sufficient.

(2) The number of cats at Cedar Shelter is 2 more than twice the number of dogs at Harbor Shelter.

Harbor Shelter has more dogs than cats, so if it has y dogs, then it has fewer than 2y animals in total. (2) says Cedar Shelter has 2y + 2 cats. Since Cedar Shelter has at least 2y + 2 animals in total, Cedar must have more animals than Harbor. Sufficient.

Answer: B.
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Two animal shelters, Cedar Shelter and Harbor Shelter, each house only 18 Mar 2026, 01:54
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Statement 1:
When there is direct inequality the greater portion has no limits at all so it can be anything.
INSUFFICIENT.

Statement 2:
This is interesting!
We have a direct comparison between the two top ratios of both shelters! And it also says one of it is more that double!
Meaning the smaller value of one shelter even if you take its closest value, the total for it cant exceed or even equal the one of the larger value!

Meaning:
Take 3>2 for harbor
then using the formula you have minimum total for cedar as 8 which is clearly more than 3+2.
SUFFICIENT.

Answer: Option B
Bunuel
Two animal shelters, Cedar Shelter and Harbor Shelter, each house only cats and dogs. Cedar Shelter has more cats than dogs while Harbor Shelter has more dogs than cats. Does Cedar Shelter have more animals in total than Harbor Shelter?

(1) The number of cats at Harbor Shelter is 2 more than twice the number of dogs at Cedar Shelter.

(2) The number of cats at Cedar Shelter is 2 more than twice the number of dogs at Harbor Shelter.

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Two animal shelters, Cedar Shelter and Harbor Shelter, each house only 18 Mar 2026, 03:49
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Case 1: insufficient
Case 2: lets say the no of dog in harbor is d, maximum cat in harbor can be d-1, total 2d-1.
no of cats in cedar is 2d+2, minimum dog is lets say 0, still 2d+2>2d-1, so its cedar with higher no of animals
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Two animal shelters, Cedar Shelter and Harbor Shelter, each house only 18 Mar 2026, 23:49
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Good variable setup is what makes this clean. Let me walk through it.

Let:
c1 = cats in Cedar, d1 = dogs in Cedar
c2 = cats in Harbor, d2 = dogs in Harbor

We're told c1 > d1 (Cedar: more cats) and d2 > c2 (Harbor: more dogs). Question: is c1 + d1 > c2 + d2?

1. Statement (1): c2 = 2d1 + 2.
This tells us Harbor's cat count in terms of Cedar's dog count. But c1, d2 are free to be anything.
Try d1=1, c1=2 (Cedar total=3), c2=4, d2=5 (Harbor total=9). Cedar < Harbor.
Try d1=10, c1=100 (Cedar total=110), c2=22, d2=23 (Harbor total=45). Cedar > Harbor.
Two opposite answers — Statement 1 alone is not sufficient.

2. Statement (2): c1 = 2d2 + 2.
Cedar's cat count is pinned to Harbor's dog count. Now something interesting happens.

Cedar total = c1 + d1 = (2d2 + 2) + d1.
Since both shelters house cats AND dogs, d1 >= 1.
So Cedar total >= 2d2 + 3.

Harbor total = c2 + d2. Since d2 > c2, we know c2 <= d2 - 1.
So Harbor total <= (d2 - 1) + d2 = 2d2 - 1.

Cedar total >= 2d2 + 3 and Harbor total <= 2d2 - 1.
Cedar always has more animals. The answer to the question is always YES. Statement 2 alone is sufficient.

Answer: B.

The trap here is that Statement 1 feels like it's giving a useful relationship, so many people rush to "C." But Statement 1 connects Harbor's cats to Cedar's dogs while leaving Cedar's total completely free to swing either way. Statement 2 directly links Cedar's largest category (cats, since c1 > d1) to Harbor's largest category (dogs, since d2 > c2), and that constraint is tight enough to lock in the comparison.
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