at a school 15 students were absent on monday, 12 absent on tuesday, and 9 absent on wednesday. if 22 students were absent at least once during these three days, what is the maximum number of students who could have been absent on all three days.
A:5
B:6
C:7
D:8
E:9
please show how you worked this out
2006-07-10 15:06:32 · 12 answers · asked by hello 2 in Science & Mathematics ➔ Mathematics
7 -- did it in my head but
7 x 3 = 21.
15+12+9=36
36-21=15
15+7=22
2006-07-10 15:11:35 · answer #1 · answered by Anonymous · 3⤊ 0⤋
It is 7.
15 - 7 = 8
12 - 7 = 5
9 - 7 = 2
7 + 8 + 5 + 2 = 22
It cannot be 9.
If 9 were absent all three days , then that means
15 - 9 = 6 were also absent on Monday
12 - 9 = 3 were also absent on Tuesday
9 - 9 = 0 were also absent on Wednesday
9 + 6 + 3 = 18 students were absent at least one day.
This is false because you were given that 22 students were absetn
You could also try this with 8, and you will get a number less than 22.
8 + (15-8) + (12-8) + (9-8)
= 8 + 7 + 4 + 1
= 20
2006-07-10 15:33:15 · answer #2 · answered by MsMath 7 · 0⤊ 0⤋
There are a total of 36 absences (15 + 12 + 9). If there were 9 people absent on all 3 days that would account for 27 absences, leaving only 9 more slots... 9 + 9 = 18, but that isn't enough for the 22 students.
If there were 8, then it would be 24 absences, leaving 12 slots. But again 8 + 12 = 20, so that isn't correct.
So the answer is obviously 7 (for 21 absences) leaving 15 more students that were absent once on either Monday or Tuesday.
You could also do this via algebra:
Let n be the number of students absent all 3 days.
36 - (3n) + n = 22
36 - 2n = 22
14 = 2n
n = 7
The answer is:
C. 7
2006-07-10 15:34:08 · answer #3 · answered by Puzzling 7 · 0⤊ 0⤋
In order to get a maximum number of students absent all three days we'll need to assume that every other absent student would only be absent on one of the days. Call the number of students absent all three days X. So every one of the 22 absentees would fall into one of four groups: either they're absent only on Monday, only on Tuesday, only on Wednesday, or all three days. The first group would be (15-X) students, as there were 15 absentees on Monday and we'll subtract out those absent all three days. Similar logic gets us (12-X) students absent only on Tuesday and (9-X) students absent only on Wednesday. Then we can set up an equation adding up all four groups and setting it equal to the total 22, then solve for X.
--- (15-X)+(12-X)+(9-X)+X = 22
--- (15+12+9)-X-X+(X-X) = 22
--- 36 - 2X + 0 = 22
--- 36 = 22 + 2X
--- 14 = 2X
--- X = 7
If you need to you can then check this answer...
--- (15-7)+(12-7)+(9-7) +7
--- 8+5+2+7
This equals 22, which is what we were expecting.
2006-07-10 15:51:37 · answer #4 · answered by Kyrix 6 · 0⤊ 0⤋
C:7 is the correct answer. I worked this out backwards by trying the answers. We'll say the same 7 students were out for each of the three days. We'll assume that every other student out on a given day is a different student that was not out any of the other two days. So on the first day 8 other students beyond 7 were out. On the second 5 were out. On the third 2 were out. If you add 7+8+5+2 you get 22, the number of students supposed to be out. If you try answers D and E you get too small a number(less than 22). The other choices are possible but they would not be the maximum under the given circumstances.
2006-07-10 15:37:44 · answer #5 · answered by jvcc06 3 · 0⤊ 0⤋
Is this a real math question? I'm pretty sure the answer is 9 because it is the most that could have been absent on wednesday. since there are no other stipulations on the number out of 22 that could have been absent on any given day, the maximum that could have been out on all three days has to be the maximum on the lowest absentee day. its just logic, not really math, if I am right
2006-07-10 15:13:10 · answer #6 · answered by tok913 3 · 0⤊ 0⤋
There are 36 total absences (15 + 12 + 9)
There are 22 unique absent students
The difference gives us 14 'bonus absences' (absences in excess of one)
So we let each student have one day as part of the 'unique' constraint, leaving us 2 days to distribute the 14 bonuses over, leaving a maximum of 7 bonus per day.
Which is less than nine, the other constraint, so we're happy.
I vote for seven.
2006-07-10 16:51:40 · answer #7 · answered by samsyn 3 · 0⤊ 0⤋
Simple: Only 9 students were absent on Wednesday so the maximum number of studens who could have been absent on all three days is 9.
2006-07-10 16:02:55 · answer #8 · answered by Anonymous · 0⤊ 0⤋
C:7
2006-07-10 16:28:45 · answer #9 · answered by Anonymous · 0⤊ 0⤋
I believe it is 7
2006-07-10 15:38:07 · answer #10 · answered by Shorti 2 · 0⤊ 0⤋