Puzzle 21
What letter is missing from this sequence:
A A A A {?} A A A A A A
Clue: the answer is not A!
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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.
Puzzle 22
A rich aristocrat decided that he would give every man 45 dollars and every woman 60 dollars.
Only one ninth of the men and only one twelfth of the women collected their dues.
Can you tell me how much money the aristocrat spent if there were 3552 people in total?
Puzzle Copyright © Kevin Stone
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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
Puzzle 23
Below is a very special grid, around each shaded number are 8 white squares. However, each white square should have a number from 1 to 7. Once filled in, these 8 numbers will sum to the shaded number. In addition, once completed correctly, no row nor column will contain a duplicate number within a white square. For example, the top row may be 5 6 4 2 3 1 7, etc. This is a very difficult puzzle, and many people resort to using a computer to help them.
Puzzle Copyright © Kevin Stone
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Answer
Puzzle 24
As the auditor for my local theme park, I noticed that on Saturday there were 4296 children and 2143 adults and the takings were £98,718.
However, on Sunday, there were 5146 children and 2807 adults and the takings were £122,570.
How much were the children's tickets and adult's tickets?
Puzzle Copyright © Kevin Stone
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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
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