Question 178126: An auditorium with 20 rows of seats has 10 seats in the first row and 1 more seat in each successive row. students taking exams are allowed to sit in any row but not next to another student in that row. what is the maximum number of students that can be seated for an exam?
Found 3 solutions by Mathtut, gonzo, solver91311:
Answer by Mathtut(3670)   (Show Source):
You can put this solution on YOUR website! each even row's max will be value of the row divided by 2
each odd row's max will be value of the row divided by 2 +1
:
there are 10 even rows and 10 odd rows.
:
if you take the max of the 20th row which is odd plus the max of the 1st row which is even you will arrive at 20 which is the same as the number of rows we have.....hmmmmmmmmm!!!!....if you continue to pair in this manner, 19th row with the 2nd row ,18 with the 3rd each time you will arrive at 20 students max:
therefore the max value of students is
:
20(10)=200
:
kind of reminds me of Gauss's sum of a series equation
Answer by gonzo(654) (Show Source):
You can put this solution on YOUR website! all the even rows contain 1/2 the number of students.
there are 10 even rows, each containing half the number of students as seats.
you start with 10 seats which means the first even row contains 5 students.
this winds up being an arithmetic series where n = 10.
you don't start off with 1, however.
you want to start with 5.
this means that you will need to subtract the sum of the first 4 numbers.
4 + 10 = 14
you take the sum of 1 through 14 and you subtract the sum of 1 through 4 and you will get the sum of 5 through 14 as your answer.
s = n*(n+1)/2
when n = 14, you get:
s = 14*15/2 = 7*15 = 105
when n = 4, s = 4*5/2 = 10
105-10 = 95
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you have 10 odd rows.
each odd row contains 1/2 the number of students truncated to the nearest integer + 1.
as an example:
11 students = 11/2 = 5.5 students = 5 + 1 = 6
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the first odd row contains 6.
each succeeding odd row contains 1 more.
the odd rows are also an arithmetic series.
you are going to start with 6 so you would need to subtract the first 5.
5 + 10 = 15 total
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when n = 15, you get:
s = 15*16/2 = 15*8 = 120
when n = 5, you get:
s = 5*6/2 = 15
120 - 15 = 105
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the sum of the number of students in the even and odd rows should be
95 + 105 = 200
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your answer is:
maximum number of students that can be seated is 200.
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here's an example of an even row with 10 seats.
x is the student, . is the space.
x.x.x.x.x.
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here's an example of an odd row with 11 seats.
x.x.x.x.x.x
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here's what your total seating arrangement will look like, starting with 10 seats in the first row and adding one seat as you go down.
01-x.x.x.x.x. = 5
02-x.x.x.x.x.x = 6
03-x.x.x.x.x.x. = 6
04-x.x.x.x.x.x.x = 7
05-x.x.x.x.x.x.x. = 7
06-x.x.x.x.x.x.x.x = 8
07-x.x.x.x.x.x.x.x. = 8
08-x.x.x.x.x.x.x.x.x = 9
09-x.x.x.x.x.x.x.x.x. = 9
10-x.x.x.x.x.x.x.x.x.x = 10
11-x.x.x.x.x.x.x.x.x.x. = 10
12-x.x.x.x.x.x.x.x.x.x.x = 11
13-x.x.x.x.x.x.x.x.x.x.x. = 11
14-x.x.x.x.x.x.x.x.x.x.x.x = 12
15-x.x.x.x.x.x.x.x.x.x.x.x. = 12
16-x.x.x.x.x.x.x.x.x.x.x.x.x = 13
17-x.x.x.x.x.x.x.x.x.x.x.x.x. = 13
18-x.x.x.x.x.x.x.x.x.x.x.x.x.x = 14
19-x.x.x.x.x.x.x.x.x.x.x.x.x.x. = 14
20-x.x.x.x.x.x.x.x.x.x.x.x.x.x.x = 15
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the odd numbered rows have an even number of seats.
the even numbered rows have an odd number of seats.
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sum of number is 1*5 + 2*(2 thoruh 14) + 1*15
sum is:
5
12
14
16
18
20
22
24
26
28
15
total is: 200
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this agrees with the arithmetic sequence logic which i suspect is where they were driving you.
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Answer by solver91311(24713)  (Show Source):
You can put this solution on YOUR website!
The first row has 10 seats, the second row, 11, and so on; so the 20th row must have 29 seats.
In each of the rows with an even number of seats, the maximum number of students is simply half of that number of seats, so the first row can seat 5, the third row can seat 6, and so on. So let's first deal with the number of students that can be seated in the rows with an even number of seats. First calculate the number of seats in these rows:
The sum of any series of integers is given by:
n}{2})
where a is the first number in the series, l is the last number in the series, and n is the number of elements in the series.
For the rows with an even number of seats,  ,  , and, since there are an even number of rows, namely 20, exactly half of them or 10 must be rows with an even number of seats so  .
10}{2} = \frac {380}{2} = 190)
But only half of these seats can be occupied, so the rows with an even number of seats can accomodate  students.
Now for the rows with an odd number of seats. Depending on whether the first seat is occupied or not, the 2nd row, containing 11 seats can accomodate either 6 (first seat occupied) or 5 (first seat not occupied) students. Since the question is asking for maximum capacity, we need to consider the situation where the first seat is occupied for each of these rows.
However, just adding up the number of seats in these rows and then dividing by 2 is insufficient because, in effect, we would be dividing the odd number of seats in each row by 2, giving us a half person for each of these rows, making our overall count  students short. Therefore:
10}{2} = \frac {400}{2} = 200)
Now we divide this by 2 and add the extra 5 that this method undercounts:
 + 5 = 105)
Add this to the 95 that we obtained for the even rows to get the final tally:
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