Puzzle 201
Last week, I travelled from Birmingham to Glasgow (via bicycle).
On the first day, I travelled one quarter of the distance.
On day two, I travelled one half of the remaining distance.
On day three, I travelled three quarters of the remaining distance.
Yesterday I travelled one third of the remaining distance.
I now have 21 miles left to travel.
How far is it from Birmingham to Glasgow in total?
Puzzle Copyright © Kevin Stone
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Hint
How many miles were left at the start of yesterday?
Answer
336 miles.
Reasoning
To work these numbers out, it is perhaps easier to start at the end and work backwards.
Yesterday I travelled 1/3 of the distance to leave 21 miles, therefore there were 31.5 miles (=21 ÷ 2/3) at the start of yesterday.
On day 3 I travelled 3/4 of the distance to leave 31.5 miles, therefore there were 126 miles (=31.5 ÷ 1/4) at the start of day 3.
On day 2 I travelled 1/2 of the distance to leave 126 miles, therefore there were 252 miles (=126 ÷ 1/2) at the start of day 2.
On the first day, I travelled 1/4 of the distance to leave 252 miles, therefore there were 336 miles (=252 ÷ 3/4) at the start of the first day.
Double-Checking
On the first day, I travelled one quarter of the distance (84, leaving 252).
On day two, I travelled one half of the remaining distance (126, leaving 126).
On day three, I travelled three quarters of the remaining distance (94.5, leaving 31.5).
Yesterday I travelled one third of the remaining distance (10.5).
I now have 21 miles left to travel.
Puzzle 202
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
Puzzle 203
Can you find the country hidden in the following paragraph?
Aliens landed in downtown Chicago last night. Most locals stepped outside to see the spaceship's massive wingspan. Amazingly, seven people failed to see the sight before them, as they took shelter from the bright light that shone from above.
Puzzle Copyright © Kevin Stone
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Hint
The country spans mode than one word.
Answer
Panama.
Aliens landed in downtown Chicago last night. Most locals stepped outside to see the spaceship's massive wingsPAN. AMAzingly, seven people failed to see the sight before them, as they took shelter from the bright light that shone from above.
Puzzle 204
Which is heaviest...
a pound of feathers
a pound of wood
a pound of steel
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Hint
Imagine a set of scales.
Answer
They are all the same!
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