Puzzle 197
Can you find a word that begins with the letter I …
… and by the addition of the letter A becomes another word that is pronounced the same?
Puzzle Copyright © Kevin Stone
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Hint
The final word starts with AI.
Answer
Isle becomes aisle.
Puzzle 198
My first is in bridge, but not in ridge.
My second is in awake, and in mistake.
My third is in danger, but not in ranger.
My fourth is in strange, and in orange.
My fifth is in spline, and in nine.
My last is in river, and in diver.
My whole likes the darkness.
What am I?
Puzzle Copyright © Kevin Stone
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Hint
The first letter can only be B.
Answer
A badger.
Reasoning
My first is in bridge, but not in ridge = B
My second is in awake, and in mistake = A K E
My third is in danger, but not in ranger = D
My fourth is in strange, and in orange = R A N G E
My fifth is in spline, and in nine = I N E
My last is in river, and in diver = R I V E
Puzzle 199
Can you find a word which can go in front of each of these words to make a new word?
colour
cress
fall
proof
shed
way
Puzzle Copyright © Kevin Stone
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Hint
The word begins with the letter W.
Answer
Water.
Watercolour, watercress, waterfall, waterproof, watershed, waterway.
Puzzle 200
As the auditor for my local theme park, I noticed that on Saturday there were 4,296 children and 2,143 adults and the takings were £98,718.
However, on Sunday, there were 5,146 children and 2,807 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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