Of the 60 families in a certain neighborhood, 38 have a cat.

[BTGmoderatorDC]

Of the 60 families in a certain neighborhood, 38 have a cat. How many of the families in this neighborhood have a dog?

(1) 28 of the families in this neighborhood have a cat but not a dog
(2) The number of families in the neighborhood who have a dog and a cat is the same as the number of families who have neither a cat nor a dog.

OA B

Source: Veritas Prep

[Jay@ManhattanReview]

Of the 60 families in a certain neighborhood, 38 have a cat. How many of the families in this neighborhood have a dog?

(1) 28 of the families in this neighborhood have a cat but not a dog
(2) The number of families in the neighborhood who have a dog and a cat is the same as the number of families who have neither a cat nor a dog.

OA B

Source: Veritas Prep

Say,

the number of families that have only dogs = d;
the number of families that have only cats = c;
the number of families that have both cats and dogs = b;
the number of families that have neither cat nor dog = n

From the given information, we have d + c + b + n = 60 and b + c = 38; thus, d + n = 60 - 38 = 22

We have to get the value of d + b.

Let's take each statement one by one.

(1) 28 of the families in this neighborhood have a cat but not a dog.

=> c = 28 => b = 38 - 28 = 10. Cannot get the value of d + b. Insufficient.

(2) The number of families in the neighborhood who have a dog and a cat is the same as the number of families who have neither a cat nor a dog.

=> b = n

We know that b + c = 38 => n + c = 38 since b = n. Thus, d + b = (d + c + b + n) - (n + c) = 60 - 38 = 22. Sufficient.

The correct answer: B

Hope this helps!

-Jay
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