Parts of the seating arrangement in a theatre is shown in the diagram below - Leaving Cert Mathematics - Question 7 - 2018

We know that the last row has 50 seats. Setting the equation from part (a):

$28 + (n - 1) = 50$

Solving for n:

$n - 1 = 50 - 28$ $n - 1 = 22$ $n = 23$

Thus, there are 23 rows of seats in total in the theatre.

The total number of seats can be calculated using the formula for the sum of an arithmetic series. The number of seats in the nth row is:

S_n = rac{n}{2} imes ( ext{first term} + ext{last term})

Here, the first term is 28, the last term for 23 rows is 50:

S_{23} = rac{23}{2} imes (28 + 50) = rac{23}{2} imes 78 = 23 imes 39 = 897

So, the total number of seats in the theatre is 897.

Given that there are 600 people in the first 7 rows, we need to calculate how many seats are filled:

• Seats in the first 7 rows are:
  • Row 1: 28
  • Row 2: 29
  • Row 3: 30
  • Row 4: 31
  • Row 5: 32
  • Row 6: 33
  • Row 7: 34

Total for the first 7 rows:

$28 + 29 + 30 + 31 + 32 + 33 + 34 = 217$

Now, we also need to find n:

Using the equation derived previously:

\rightarrow n(n - 1) = 1200$$ Solving, we have: $$n^2 - n - 1200 = 0$$ Using the quadratic formula, we find: $$n = rac{1 + ext{sqrt}(1 + 4800)}{2} = 34$$ Thus, 600 - 217 = 383 people need to be seated in row 8, which implies that: $$ ext{Row 8 has } 35 ext{ seats.}$$

To find the total income from ticket sales, we use the number of tickets sold:

• Adult tickets sold: 276
• Children's tickets sold: 212

Calculating the total:

$276 imes 25 + 212 imes 12 = 6900 + 2544 = €9,444$

Let x represent the number of children's tickets. Given that the ratio of adult tickets (3x) to children's tickets is known:

4x = 752\ x = 188\

Thus:

• Children's tickets: 188
• Adult tickets: 3x = 564

Let the cost of a children's ticket be y:

Adult tickets cost 2.5y, so:

188 imes 2.5y + 564y = 17578\

Solving for y gives:

\rightarrow y = £11$$