SAT Question

Test Preparation
1

10 students have dogs and 15 have cats. If a total of 19 students have only one of these animals, how many students have both cats and dogs?

How do I solve this?

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3

3

((10+15)-19)/2 = 3

4
  1. Venn diagram.
5

@Readyforthewind how did you come up with that

6

It’s 3.

Solution 1: Draw a Venn diagram. Let D = # students who have only dogs, C = # students who have only cats, and X = # students who own a dog and a cat. Then

D+X = 10 (10 students have dogs)
C+X = 15 (15 students have cats)
C+D = 19 (19 students have only one type of animal)

We want to find X. To do this, add the first two equations to get C+D+2X = 25. C+D = 19, so 19 + 2X = 25 --> X = 3.

Solution 2: You could use Inclusion-Exclusion (for any finite sets A, B, |A∪B| = |A| + |B| - |A∩B|). We can let A = {students who have dogs}, B = {students who have cats}. Let X be the number of students who have both types of pet. Then |A∪B| = 19+X, |A| = 10, |B| = 15, |A∩B| = X. Substituting,

19+X = 10 + 15 - X
X = 3