Puzzle 17
Chess Contest - Logic Puzzles
At the local games evening, four friends were competing in the chess and croquet competitions.
Frankie beat Glen in chess, Dale came third, and the 16-year-old won.
Frankie came second in croquet, the 15-year-old won, Dale beat the 18-year-old, and the 19-year-old came third.
Emery is 3 years younger than Glen.
The person who came last in chess came third in croquet, and only one person got the same position in both games.
Can you find the ages of the people and their positions in the two games?
Puzzle Copyright © Kevin Stone
workings
hint
answer
print
Share link – www.brainbashers.com/puzzle/ziwd
Hint
How old could Glen be?
Answer
Name Age Chess Croquet
Dale 15 3 1
Emery 16 1 4
Frankie 18 2 2
Glen 19 4 3
Reasoning
The possible ages are: 15, 16, 18, 19.
Breaking the clues down, and numbering them:
Frankie beat Glen in chess.
Dale came third (chess).
The 16-year-old won (chess).
Frankie came second in croquet.
The 15-year-old won (croquet).
Dale beat the 18-year-old (croquet).
The 19-year-old came third (croquet).
Emery is 3 years younger than Glen.
The person who came last in chess came third in croquet.
Only one person got the same position in both games.
By (4), Frankie came 2nd in croquet.
Name Age Chess Croquet
Dale
Emery
Frankie 2
Glen
By (2), Dale came 3rd in chess.
Name Age Chess Croquet
Dale 3
Emery
Frankie 2
Glen
By (8), Emery is 15 or 16, and Glen is 18 or 19.
By (9), the person who came 4th in chess, came 3rd in croquet, and by (7), they are 19. This can't be Dale (3rd in chess), or Frankie (2nd in croquet), or Emery (aged 15 or 16), so it must be Glen.
Name Age Chess Croquet
Dale 3
Emery
Frankie 2
Glen 19 4 3
By (6), Dale can't have been 4th in croquet, so must be 1st. Leaving Emery in 4th.
Name Age Chess Croquet
Dale 3 1
Emery 4
Frankie 2
Glen 19 4 3
By (10), only one person got the same position in both games, this can't be Emery 4th in both (as Glen is already 4th in chess), so must be Frank, 2nd in both. Leaving Emery 1st in chess.
Name Age Chess Croquet
Dale 3 1
Emery 1 4
Frankie 2 2
Glen 19 4 3
By (3) and (5) we know the ages of the winners in each competition, and Frankie is therefore 18.
Name Age Chess Croquet
Dale 15 3 1
Emery 16 1 4
Frankie 18 2 2
Glen 19 4 3
Puzzle 18
Can you find anagrams of each of the words below?
Each of the words has more than one anagram, but the initial letters of the correct anagrams form another anagram related to the word QUESTION.
bruise
warned
please
listen
veined
trance
Puzzle Copyright © Kevin Stone
workings
hint
answers
print
Share link – www.brainbashers.com/puzzle/zwoh
Hint
Find anagrams of the words in this order: PLEASE, TRANCE, LISTEN, WARNED, VEINED, BRUISE.
Answers
ANSWER.
Reasoning
The initial letters are RWASEN, which is an anagram of ANSWER.
bruise becomes rubies
warned becomes wander
please becomes asleep
listen becomes silent
veined becomes envied
trance becomes nectar
Puzzle 19
Pre-Championship Rehearsal - Logic Puzzles
At last month's rehearsal, four top athletes competed in two 400-metre qualifiers.
As the results were expected to be mislaid, various notes were taken to ensure the accuracy of the overall placings:
No one finished both races in the same position.
In both races, Sam beat the runner whose surname was Watson.
Billie Taylor came third in the second race, and Pat came last in the first race.
In the second race, the athlete whose surname was Arnold won, and the athlete whose surname was Bowler came last.
In the first race, Billie beat Jamie, but Jamie beat Sam.
Can you determine who finished where in each of the races?
Puzzle Copyright © Kevin Stone
workings
hint
answer
print
Share link – www.brainbashers.com/puzzle/zgbd
Hint
Look at Clue 3 first, and then Clue 5.
Answer
Race1 Race2
————————————————————————
#1 Billie Sam
Taylor Arnold
————————————————————————
#2 Jamie Pat
Bowler Watson
————————————————————————
#3 Sam Billie
Arnold Taylor
————————————————————————
#4 Pat Jamie
Watson Bowler
Reasoning
The four names were: Billie, Jamie, Pat, Sam
The four contestants were: Arnold, Bowler, Taylor, Watson.
By (3), Billie Taylor came third in the second race, and Pat came last in the first race, giving:
Race1 Race2
————————————————————————
#1
————————————————————————
#2
————————————————————————
#3 Billie
Taylor
————————————————————————
#4 Pat
By (5), in the first race Billie beat Jamie, and Jamie beat Sam, giving:
Race1 Race2
————————————————————————
#1 Billie
Taylor
————————————————————————
#2 Jamie
————————————————————————
#3 Sam Billie
Taylor
————————————————————————
#4 Pat
By (2), Sam beat the runner with the surname Watson in the first race, which means that Pat's surname is Watson, giving:
Race1 Race2
————————————————————————
#1 Billie
Taylor
————————————————————————
#2 Jamie
————————————————————————
#3 Sam Billie
Taylor
————————————————————————
#4 Pat
Watson
By (1), Pat can't have also finished in last place in race 2, and by (2), Sam beat the Pat Watson, so Pat must have finished in second place in race 2. And Sam came first, therefore Jamie came last, giving:
Race1 Race2
————————————————————————
#1 Billie Sam
Taylor
————————————————————————
#2 Jamie Pat
Watson
————————————————————————
#3 Sam Billie
Taylor
————————————————————————
#4 Pat Jamie
Watson
By (4), in the second race the athlete whose surname was Arnold won, and the athlete whose surname was Bowler came last, so we know Sam and Jamie's surnames, and we can complete the grid:
Race1 Race2
————————————————————————
#1 Billie Sam
Taylor Arnold
————————————————————————
#2 Jamie Pat
Bowler Watson
————————————————————————
#3 Sam Billie
Arnold Taylor
————————————————————————
#4 Pat Jamie
Watson Bowler
Puzzle 20
Here we have a rectangular room, measuring 30 feet by 12 feet, and 12 feet high.
There is a spider in the middle of one of the end walls, 1 foot from the ceiling (A).
There is a fly in the middle of the opposite wall, 1 foot from the floor (B).
What is the shortest distance that the spider must crawl in order to reach the fly?
The Spider and the Fly – The Canterbury Puzzles, Henry Ernest Dudeney.
workings
hint
answer
print
Share link – www.brainbashers.com/puzzle/zwql
Hint
Going down and across the floor isn't the only route.
Answer
40 feet.
Explanation Diagram
If you imagine the room to be a cardboard box, you can 'unfold' the room in various ways, and each route gives a different answer.
We can use Pythagoras' theorem (a2 + b2 = c2) to calculate the distances:
distance2 = horizontal2 + vertical2
distance = √(horizontal2 + vertical2)
Route #1
distance = 1 + 30 + 11 = 42 feet.
Route #2
horizontal = 6 + 30 + 6 = 42 feet.
vertical = 10 feet.
distance = √(422 + 102) ≈ 43.174 feet.
Route #3
horizontal = 1 + 30 + 6 = 37 feet.
vertical = 6 + 11 = 17 feet.
distance = √(372 + 172) ≈ 43.178 feet.
Route #4
horizontal = 1 + 30 + 1 = 32 feet.
vertical = 6 + 12 + 6 = 24 feet.
distance = √(322 + 242) = 40 feet.
< previous next >
Note: BrainBashers has a Dark Mode option – I recommend not using your browser's dark mode or extensions for BrainBashers