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 2
Which four vegetables have been merged together?
cab swe rpr eed oaa etn
Hint
The fourth player is the key to this tricky question.
Answer
9 points.
Respectively the scores were 7, 14, 20, 30, 23, 9.
Reasoning
If we label the six players A, B, C, D, E, and F, we know that:
[1] A + B + C + D + E + F = 103
and from the clues:
A = B ÷ 2 B = C − 6 C = D x 2 ÷ 3 E = D − A F = E − 14
Note that it could be E = A − D or E = D − A, but using A − D we'd end up with a negative value for E later, which isn't allowed (so we'd have to try again with D − A anyway).
As we have no information for D, it's best to find all of the other letters in terms of D. These steps are left as an exercise (e.g. use C in the equation for B), but the result is:
A = ( D − 9) ÷ 3 B = (2D − 18) ÷ 3 C = (2D ) ÷ 3 D = (3D ) ÷ 3 E = (2D + 9) ÷ 3 F = (2D − 33) ÷ 3
Writing it as D = 3D ÷ 3 makes things slightly clearer in the next step.
In the illustration we have a sketch of Sir Edwyn de Tudor going to rescue his love, who was held captive by a neighbouring wicked baron.
Sir Edwyn calculated that if he rode at fifteen miles an hour he would arrive at the castle an hour too soon, while if he rode at ten miles an hour he would get there just an hour too late.
Now, it was of the first importance that he should arrive at the exact time appointed, in order that the rescue that he had planned should be a success, and the time of the tryst was five o'clock, when the captive would be taking afternoon tea.
The puzzle is to discover exactly how far Sir Edwyn de Tudor had to ride.
Sir Edwyn De Tudor, Amusements In Mathematics, Henry Ernest Dudeney.
If Sir Edwyn left at noon and rode 15 miles an hour, he would arrive at four o'clock, which is an hour too soon. If he rode 10 miles an hour, he would arrive at six o'clock, which is an hour too late. But if he went at 12 miles an hour, he would reach the castle of the wicked baron exactly at five o'clock, which is the time appointed.
The text above is the answer given in the book, and below is a method of finding the answer.
If we call the distance to the castle, D and use the fact that Time = Distance ÷ Speed, we have:
Travelling at 15 mph:
Time1 = D ÷ 15 (an hour too soon)
Travelling at 10 mph:
Time2 = D ÷ 10 (an hour too late)
The time gap between these two times is 2 hours, therefore
