The 25 students in a sixth-grade class each reported whether or not they have at least one dog, at least one cat, or at least one parakeet. The results are shown in the Venn diagram above, with two of the numbers represented by x and y. If all of the students have at least one of these pets, what is the value of x + y?

If x + y were 6, the total number of students with pets would be less than 25. This would imply that the counts for at least one dog, one cat, or one parakeet are insufficient to account for all 25 students, contradicting the problem's assertion that every student has at least one pet.

Choosing 7 for x + y would similarly lead to an incorrect total. The remaining counts from the Venn diagram would not provide enough students to sum to 25, suggesting that there are still students unaccounted for, which is not possible as all students reported having pets.

If we consider x + y to be 10, it would again cause discrepancies in the total pet count. The numbers would push the total students reported to be greater than 25, leading to an impossible scenario where more students have pets than are present in the class.

An x + y value of 14 would make it exceedingly high relative to the total of 25 students. This would leave too few students remaining to balance out the total in the Venn diagram, indicating that not all students could have at least one pet.