Algebra Trick Question Students in a class were asked about the number of brothers and sisters they had. 17 said they had at least one brother. 18 said they had at least one sister. 6 said they had at least one brother AND one sister. 5 said they had no brothers or sisters. What was the least possible number of students in the class?


Well I actually have 2 answers to this question, which both are probably wrong. My first guess is 11, since it is clearly stated in the question that 5 had no siblings and 6 had atleast one brother and one sister.

My second guess is 1, since there has to be atleast a student in the class.
Out of the 17 who had at least one brother, 11 had no sisters.  Combined with the 18 who had at least one sister, this gives us 29 students.  Adding in the 5 who had no brothers or sisters, we get 34.  So there is exactly 34 students.

$$\text{no sister/brother $+$ no sisters some brothers $+$ sisters, with or w/out brothers}=34$$
It says that there are 18 with at least one sister, and 5 without brothers/sisters. So there have to be at least 23. Now we have to look if something implies that this has to be higher. Because these eighteen said that they had at least one sister, those 17 with at least one brother can all be combined with this, so we get a minimum of 23.