In a certain class, 1/3 of the students are honors students, and 1/4 of the students play varsity sports. If 12 students play varsity sports AND are honor students, what is the least possible number of students in the class?
2007-04-28 16:51:45 · 4 answers · asked by kapitalist 2 in Science & Mathematics ➔ Mathematics
48.
You have 12 students who do both so the least number of students is these are the ONLY ones who are honor students is 12 / (1/3) = 36 and the least number of students is these are the ONLY ones who are varsity sports students is 12 / (¼) = 48. So there are at least 48 students and, if this is divisible by three (the 1/3 factor for the honor students), then it is the least number of students fitting the criteria. If not, then more are required. Well, 48 * (1/3) = 16 so it works out. 48 is therefore the least possible number of students in the class.
2007-04-28 16:57:06 · answer #1 · answered by roynburton 5 · 0⤊ 1⤋
1/3 is larger than 1/4th, so there are more honors students than varsity sports students.
Let's assume that every sports student is an honors student. 1/3-1/4 = 1/12, so 1/12 of the students are only honors students. 2/3 of the students are in neither group. 2/3 + 1/12 = 9/12 = 3/4. So 1/4 of the students are in both groups. That means the group is 48.
The more students who play sports but are not honors, the larger the class gets.
2007-04-28 17:02:37 · answer #2 · answered by TychaBrahe 7 · 1⤊ 0⤋
1/3= 12/x
so you do the porporition
so you take
12x3
which =36/1
which = 36
do the same for 1/4
so the least possible is 48 because it is the highest number :]
2007-04-28 17:00:35 · answer #3 · answered by Anonymous · 0⤊ 0⤋
I concur with Susie that the respond is 0. I drew a Venn Diagram and extra (A union B union C), (A union B), (A union C), A purely, B purely, and C purely. the end result replaced into 23. hence (B union C) is 0.
2016-10-04 01:55:46 · answer #4 · answered by gizzi 4 · 0⤊ 0⤋