Hard Puzzles 

Workings

Hint
Think of an alternative alphabet.

Answer
N.

From epsiloN, last letters of the Greek alphabet, alpha, beta, gamma, delta, epsilon, zeta, eta, theta, iota, kappa, lambda, mu.

Workings

Hint
It doesn't actually matter how many men or women there were.

Answer
17,760 dollars.

If there were M men, then there were (3552 - M) women.

So one ninth of the men each received 45 dollars and one twelve of the women each received 60 dollars. So the total received was:
= 45 x 1 x M  +  60 x  1 x (3552 - M)
       —              ——
       9              12

Simplify the two fractions by dividing the 45 by 9, and the 60 by 12 respectively.

= 5 x M + 5 x (3552 - M)

= 5M + 17760 - 5M

= 17760

Workings

Hint

Answer
 

Workings

Hint
This is quite a tricky puzzle, and knowledge of algebra would certainly help.

Answer
The children tickets were £14, and the adult tickets were £18.

Reasoning
There are a number of methods for solving this problem, including:

Using a spreadsheet. Using a computer program. Using the intersection of lines on a graph. Using an online equation solver. Solving simultaneous equations using algebra. Solving simultaneous equations using inverse matrices.
Here is my solution using simultaneous equations and algebra.

First construct two algebraic equations, where 'c' is the number of children, and 'a' is the number of adults:

[1]   4296c + 2143a = 98718
[2]   5146c + 2807a = 122570

To make the number in front of 'c' the same on both, we multiply [1] by 5146 and [2] by 4296 to give:

[3]   22107216c + 11027878a = 508002828
[4]   22107216c + 12058872a = 526560720

Now we can do [4] – [3] to give:

   1030994a = 18557892

Divide throughout by 1030994 so that:

   a = 18

Substituting a = 18 in [1] will give:

   4296c + 2143 x 18 = 98718
   4296c + 38574     = 98718
   4296c             = 60144
       c             = 14

Double-Checking
c = 14 and a = 18

4,296 x 14 + 2,143 x 18 = 98,718

and

5,146 x 14 + 2,807 x 18 = 122,570