Easy math puzzles
Poll
 | 21 votes (45.65%) | ||
| | 14 votes (30.43%) | ||
| | 6 votes (13.04%) | ||
| | 3 votes (6.52%) | ||
| | 12 votes (26.08%) | ||
| | 3 votes (6.52%) | ||
| | 6 votes (13.04%) | ||
| | 5 votes (10.86%) | ||
| | 12 votes (26.08%) | ||
| | 10 votes (21.73%) |
46 members have voted
January 15th, 2025 at 5:03:36 PM
permalink
New Question! I bolded the differences from the previous one.
The king wants to marry off his daughter. Three suitors apply -- a logician, politician and a lawyer. The king may ask two of them a yes/no question, for which the suitor must know the answer. The logician always tells the truth. The politician always lies. The lawyer says either randomly. The three men know each other. The king seeks only to avoid choosing the lawyer for his daughter. What question should he ask of the two suitors?
The king wants to marry off his daughter. Three suitors apply -- a logician, politician and a lawyer. The king may ask two of them a yes/no question, for which the suitor must know the answer. The logician always tells the truth. The politician always lies. The lawyer says either randomly. The three men know each other. The king seeks only to avoid choosing the lawyer for his daughter. What question should he ask of the two suitors?
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
January 16th, 2025 at 5:17:10 PM
permalink
I guess the puzzle above was too tough. I agree it is a hard one.
Here is something easier.
Find the area of the red square.
Here is something easier.
Find the area of the red square.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
January 16th, 2025 at 6:36:42 PM
permalink
I assume the numbers are the areas of the smaller squares, and that those are all squares
You don't need the yellow square on the left. In fact, a harder problem might have been to calculate that square's area.
The lengths of the sides of the squares on the right are: green, sqrt(3); yellow, 2 sqrt(3); blue, 3 sqrt(3)
The three white triangles formed by the red square and the three other squares are similar.
Since the white triangle next to the yellow square has height 2 sqrt(3) and base length (3 sqrt(3) - 2 sqrt(3)) = sqrt(3), its hypotenuse is sqrt(15).
Since the white triangles are similar, and the height of the top one = 1/2 the height of the middle one, the hypotenuse of the top one = 1/2 the hypotenuse of the middle one = sqrt(15) / 2.
Also, since the height of the bottom one = 3/2 the height of the middle one, the hypotenuse of the middle one = 3/2 the hypotenuse of the middle one = 3/2 sqrt(15).
The length of the side of the red square = the sum of the lengths of the hypotenuses of the three white triangles = 3 sqrt(15), so the area is the square of that, or 135.
You don't need the yellow square on the left. In fact, a harder problem might have been to calculate that square's area.
The lengths of the sides of the squares on the right are: green, sqrt(3); yellow, 2 sqrt(3); blue, 3 sqrt(3)
The three white triangles formed by the red square and the three other squares are similar.
Since the white triangle next to the yellow square has height 2 sqrt(3) and base length (3 sqrt(3) - 2 sqrt(3)) = sqrt(3), its hypotenuse is sqrt(15).
Since the white triangles are similar, and the height of the top one = 1/2 the height of the middle one, the hypotenuse of the top one = 1/2 the hypotenuse of the middle one = sqrt(15) / 2.
Also, since the height of the bottom one = 3/2 the height of the middle one, the hypotenuse of the middle one = 3/2 the hypotenuse of the middle one = 3/2 sqrt(15).
The length of the side of the red square = the sum of the lengths of the hypotenuses of the three white triangles = 3 sqrt(15), so the area is the square of that, or 135.
January 16th, 2025 at 9:13:59 PM
permalink
Assumption - all coloured boxes are square.
Solution:
Let the length of the side of the red square = x.
The length of the sides of the green, yellow and blue boxes are the sq. root of their respective areas.
Green Box: sq. root 3
Yellow Box: sq. root 12 = 2* sq. root 3
Blue Box: sq. root 27 = 3* sq. root 3
By deduction the base lengths of the three white triangles are:
Green Box: (sq. root 3)/2
Yellow Box: sq. root 3
Blue Box: 1.5*(sq. root 3)
The sum of the lengths of the hypotenuses of the three triangles equals the length of the side of the red square.
x = sq. root[(6*sq. root 3)² + (3*sq. root 3)²]
= sq. root 135
The area of the red square is 135 sq. units.
Casino Enemy No.1
January 17th, 2025 at 4:44:27 AM
permalink
Dave and Don -- I agree.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
January 19th, 2025 at 11:34:53 AM
permalink
Folks,
Not exactly a math puzzle, but here goes!
In a standard game of Monopoly:
1. How can you go bankrupt by bankrupting another player?
2. How can you legally avoid going bankrupt in that situation?
Here is a link to a PDF with the Monopoly rules:
https://www.hasbro.com/common/instruct/monins.pdf
Dog Hand
P.S. My favorite Monopoly token is, naturally, the dog.
Not exactly a math puzzle, but here goes!
In a standard game of Monopoly:
1. How can you go bankrupt by bankrupting another player?
2. How can you legally avoid going bankrupt in that situation?
Here is a link to a PDF with the Monopoly rules:
https://www.hasbro.com/common/instruct/monins.pdf
Dog Hand
P.S. My favorite Monopoly token is, naturally, the dog.
January 19th, 2025 at 4:27:14 PM
permalink
I love Monopoly, but I don't quite understand what you're asking.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
January 20th, 2025 at 5:10:43 PM
permalink
Quote: WizardI love Monopoly, but I don't quite understand what you're asking.
link to original post
Wiz,
I am asking under what circumstances when you bankrupt a player, you yourself also go bankrupt immediately.
Dog Hand
January 21st, 2025 at 1:30:51 AM
permalink
I can't remember all the cards you can draw but there used to be one where you got $10/£10 from every player. So It's possible you can bankrupt a player from your move.
However I think the puzzle uses the rule that when you go bankrupt you hand everything over, including mortgaged properties, and the receiver now has to pay 10% or unmortgage those, and thus also goes bankrupt. This can probably only happen with Player B owning Old Kent Road and Player A having mortgaged Mayfair, and neither player has money. Player A must have drawn the "Return to Old Kent Road" card (otherwise would have £200). Player A hasn't passed Go to get there but has to pay £2. Player B receives mortgaged Mayfair but can't raise the £40, even by mortgaging Old Kent Road; so also goes bankrupt.
As Player B, your correct tactic should have been to mortgage Old Kent Road at the end of the last move.
However I think the puzzle uses the rule that when you go bankrupt you hand everything over, including mortgaged properties, and the receiver now has to pay 10% or unmortgage those, and thus also goes bankrupt. This can probably only happen with Player B owning Old Kent Road and Player A having mortgaged Mayfair, and neither player has money. Player A must have drawn the "Return to Old Kent Road" card (otherwise would have £200). Player A hasn't passed Go to get there but has to pay £2. Player B receives mortgaged Mayfair but can't raise the £40, even by mortgaging Old Kent Road; so also goes bankrupt.
As Player B, your correct tactic should have been to mortgage Old Kent Road at the end of the last move.
January 21st, 2025 at 2:49:26 AM
permalink
Quote: charliepatrickI can't remember all the cards you can draw but there used to be one where you got $10/£10 from every player. So It's possible you can bankrupt a player from your move.
However I think the puzzle uses the rule that when you go bankrupt you hand everything over, including mortgaged properties, and the receiver now has to pay 10% or unmortgage those, and thus also goes bankrupt. This can probably only happen with Player B owning Old Kent Road and Player A having mortgaged Mayfair, and neither player has money. Player A must have drawn the "Return to Old Kent Road" card (otherwise would have £200). Player A hasn't passed Go to get there but has to pay £2. Player B receives mortgaged Mayfair but can't raise the £40, even by mortgaging Old Kent Road; so also goes bankrupt.
As Player B, your correct tactic should have been to mortgage Old Kent Road at the end of the last move.
link to original post
Ding! Full credit for first question, but only partial credit for the second.
Dog Hand
January 21st, 2025 at 3:05:54 AM
permalink
Thanks - I had forgotten about the second question.Quote: DogHand...Ding! Full credit for first question, but only partial credit for the second...
You could always fail to ask for the rent, although the rules are ambiguous if you own the set (the other property would already have been mortgaged in this case), or you could just mortgage Old Kent Road "at any time", say just after the dice had been rolled/card picked and shown, but before any movement.
