The results of a recent Chess tournament between five close rivals is described below:
Dale finished before Alex. Emery finished after Bailey. Alex finished before Chris. Emery finished after Dale. Bailey finished before Alex. Dale finished after Bailey. Chris finished before Emery.
Who finished where?
Hint
Try rewriting each statement so that they are all 'finished before' statements instead.
Answer
Bailey won …
… then Dale, Alex, Chris, and Emery.
Reasoning
By placing the person who finished higher to the left, we have:
Dale finished before Alex (Dale > Alex)Emery finished after Bailey (Bailey > Emery)Alex finished before Chris (Alex > Chris)Emery finished after Dale (Dale > Emery)Bailey finished before Alex (Bailey > Alex)Dale finished after Bailey (Bailey > Dale)Chris finished before Emery (Chris > Emery)
From (1) and (3):
Dale > Alex > Chris
Adding (7):
Dale > Alex > Chris > Emery
Adding (6):
Bailey > Dale > Alex > Chris > Emery
Which is the final placing for all the players, and can be double-checked with the remaining clues.
?
Puzzle 18
What comes next in this sequence:
Dave · Edward · Duncan · Neil · Liam · ?
Choose from:
Each name begins with the last letter of the preceding name.
??
Puzzle 19
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?
cer ent ead rat uti spr
ful oun pro ann ope ock
nce een oat est liv ion
nou sev ion ast hou akf
bre bea ens dim bed seb
Hints
The first letters of the words are: P, B, D, L, O, A, B, H, S, B.
Answers
pro + nou + nce = pronounce
bea + uti + ful = beautiful
dim + ens + ion = dimension
liv + est + ock = livestock
ope + rat + ion = operation
ann + oun + cer = announcer
bed + spr + ead = bedspread
hou + seb + oat = houseboat
sev + ent + een = seventeen
bre + akf + ast = breakfast
??
Puzzle 20
How many people must be at a party before you are likely to have two having the same birthday (but not necessarily the same year)?
Reasoning
By likely, we mean greater than 50% chance.
With one person there is a 0 percent chance that you'll have two people with the same birthday.
With two people the probability that they won't share a birthday is 364 ÷ 365. The probability that they will share a birthday is therefore 1 − (364 ÷ 365).
With three people the probability that they won't share a birthday is the same as for two people, times 363 ÷ 365. So the probability that three people will share a birthday is 1 − (364 ÷ 365) x (363 ÷ 365).
Notice that with each additional person added, the probability that they shares a birthday with one of the previous persons goes up, because there are fewer "free" days remaining.
We keep adding people until the %age is greater than 50%.
