Puzzle 185
Pre-Championship Rehearsal - Logic Puzzles
At last month's rehearsal, four top athletes competed in two qualifying 400-metre races.
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
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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 186
Below are thirteen 5-letter words, each of which has had two of its letters removed.
In total, these 26 letters are A to Z.
The remaining letters in each word are in the correct order.
There are no words that are spelled differently based on location (favour / favor, etc) and there are no plurals.
Can you find the original words?
bag
ded
dia
uoa
ide
oel
ums
pyy
gou
nal
cra
umb
dty
Puzzle Copyright © Kevin Stone
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Hint
These are the first letters of the words: B, D, D, Q, I, V, B, P, G, F, C, J, D.
Answer
bag + de = badge
ded + az = dazed
dia + ry = diary
uoa + qt = quota
ide + nx = index
oel + vw = vowel
ums + bp = bumps
pyy + gm = pygmy
gou + hl = ghoul
nal + fi = final
cra + ck = crack
umb + jo = jumbo
dty + us = dusty
Puzzle 187
Which country has been hidden in the paragraph below:
The local habitat around a railway track can be very interesting. For example, supporting the track is a sleeper, under which you can find the lesser-spotted great weevil.
Puzzle Copyright © Kevin Stone
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Hint
The country has four letters.
Answer
Peru.
The local habitat around a railway track can be very interesting. For example, supporting the track is a sleeper, under which you can find the lesser-spotted great weevil.
Puzzle 188
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. From Amusements In Mathematics by Henry Ernest Dudeney (1917).
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Hint
The gap between the two options is 2 hours.
Answer
The distance must have been sixty miles.
If Sir Edwyn left at noon and rode 15 miles an hour, he would arrive at four o'clock – an hour too soon. If he rode 10 miles an hour, he would arrive at six o'clock – 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 – 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
Time2 - Time1 = 2
D ÷ 10 - D ÷ 15 = 2
Multiply throughout by 30:
3D - 2D = 60
D = 60 miles.
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