Puzzle 1
Using all of the letters A to Z, each once only, complete these common words. There are currently 3 different answers, can you find them all?
--ust
-ue-n
o-o-e
--and
-a-er
id-o-
-uc-y
e-i-t
hea--
tou--
-rie-
-ry-t
h-m-n
Puzzle Copyright © Kevin Stone
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Hint
The first word begins with the letter J.
Answers
--ust = joust (JO)
-ue-n = queen (QE)
o-o-e = ozone (ZN)
--and = brand (BR)
-a-er = water (WT)
id-o- = idiom (IM)
-uc-y = lucky (LK)
e-i-t = exist (XS)
hea-- = heavy (VY)
tou-- = tough (GH)
-rie- = fried (FD)
-ry-t = crypt (CP)
h-m-n = human (UA)
Alternative Answers
Joust, queen, ozone, brand, wafer, idiom, lucky, exist, heavy, tough, tried, crypt, human.
Joust, queen, ozone, bland, wafer, idiot, mucky, exist, heavy, tough, drier, crypt, human.
Can you find another?
Puzzle 2
Gymnastics Gala - Logic Puzzles
During a recent gymnastics gala, four athletes competed in three events, the vault, the beam, and the floor.
Chris came third on the floor.
The athlete who came second in the vault won the beam event.
The athlete who came second on the beam won the floor.
Emery beat Alex in every event.
The athlete who won the vault came second on the floor.
Billie came last in the vault.
Can you determine who finished where in each event?
Puzzle Copyright © Kevin Stone
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Hint
Could Alex have come first or second in any event?
Answer
Vault Beam Floor
#1 Emery Chris Billie
#2 Chris Billie Emery
#3 Alex Emery Chris
#4 Billie Alex Alex
Reasoning
By (1), Chris came third on the floor. By (6), Billie came last in the vault.
Vault Beam Floor
#1
#2
#3 Chris
#4 Billie
By (2), the athlete who came second in the vault won the beam event, let's mark these as X.
By (3), the athlete who came second on the beam won the floor, let's mark these as Y.
By (5), the athlete who won the vault came second on the floor, let's mark these as Z.
Vault Beam Floor
#1 Z X Y
#2 X Y Z
#3 Chris
#4 Billie
By (4), Emery beat Alex in every event, therefore Alex can't have won any event, nor come second (because of X, Y, and Z). So, Alex came third on the vault, and last on the floor.
Vault Beam Floor
#1 Z X Y
#2 X Y Z
#3 Alex Chris
#4 Billie Alex
Looking at the vault, Z can't be Chris because Z came second on the floor (but we know Chris came last). So, Z must be Emery, and X is Chris.
Vault Beam Floor
#1 Emery Chris Y
#2 Chris Y Emery
#3 Alex Chris
#4 Billie Alex
Looking at the floor, Y is Billie.
Vault Beam Floor
#1 Emery Chris Billie
#2 Chris Billie Emery
#3 Alex Chris
#4 Billie Alex
By (4), Emery beat Alex on the beam.
Vault Beam Floor
#1 Emery Chris Billie
#2 Chris Billie Emery
#3 Alex Emery Chris
#4 Billie Alex Alex
Puzzle 3
Gardener's Spring Fayre - Logic Puzzles
At the recent spring fayre, four keen gardeners were displaying their fine roses.
In total there were four colours and each rose colour appeared twice.
From the clues below, can you tell who had which colour roses?
Gardener Green had a yellow rose.
Gardener Yellow did not have a red one.
Gardener Red had a blue rose, but not a green one.
Gardener Blue did not have a yellow one.
One person with a red rose also had a green one.
One person with a yellow rose also had a blue one.
One of the persons with a green rose had no red.
Neither of the persons with a yellow rose had a green one.
No person had two roses of the same colour.
No two persons had the same two colour roses, and their names provide no clues.
Puzzle Copyright © Kevin Stone
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Hint
Start with a grid with all of the colours each gardener could have for each flower, and then eliminate some based on the clues given.
Answer
Name
Rose1
Rose2
Gardener Blue
red
green
Gardener Green
yellow
red
Gardener Red
blue
yellow
Gardener Yellow
blue
green
Reasoning
We can create a grid of the possible flowers each person could have. Initially, they each have the possibility of BGRY (blue, green, red, yellow) and as we work our way through the clues we can eliminate certain options.
Person F1 F2
Blue BGRY BGRY
Green BGRY BGRY
Red BGRY BGRY
Yellow BGRY BGRY
From (1), (2), (3) and (4) we can reduce the options to:
Blue BGR BGR
Green Y BGRY
Red B BRY
Yellow BGY BGY
By (9) this becomes:
Blue BGR BGR
Green Y BGR
Red B RY
Yellow BGY BGY
By (8) we get:
Blue BGR BGR
Green Y BR
Red B RY
Yellow BGY BGY
Only Blue can satisfy (5), and Yellow is the only person who can have the other G.
Blue R G
Green Y BR
Red B RY
Yellow G BY
By (8), Yellow can't be Y.
Blue R G
Green Y BR
Red B RY
Yellow G B
Since there are only two of each colour, we can reduce further:
Blue R G
Green Y R
Red B Y
Yellow G B
Puzzle 4
Below, 10 nine-letter words have been broken into chunks of three letters.
The chunks have been moved around, no chunk is used twice, and all of the chunks are used.
Can you determine what the 10 words are?
hol rec ant imp fer con
man dif ice pol acq htn
ort myt new air sag duc
dec lig kch ogy ing tan
uit ent gle tal tor ent
Puzzle Copyright © Kevin Stone
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Hint
The first letters of the words are: A, C, D, D, I, L, M, N, P, R.
Answers
acq + uit + tal = acquittal
con + duc + tor = conductor
dec + kch + air = deckchair
dif + fer + ent = different
imp + ort + ant = important
lig + htn + ing = lightning
myt + hol + ogy = mythology
new + sag + ent = newsagent
pol + ice + man = policeman
rec + tan + gle = rectangle
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