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 rubieswarned becomes wanderplease becomes asleeplisten becomes silentveined becomes enviedtrance becomes nectar
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Puzzle 6
Place the following words into the grid:
MERCURY VENUS EARTH MARS SATURN URANUS NEPTUNE PLUTO
Jupiter is NOT in the list.
Here is a snippet of section D of the curious multiple-choice entrance exam into the exclusive BrainBashers puzzle club.
Q1. Which is the first question where c) is the correct answer?
a) Q3
b) Q4
c) Q1
d) Q2
Q2. Which is the first question where a) is the correct answer?
a) Q4
b) Q2
c) Q3
d) Q1
Q3. Which is the first question where d) is the correct answer?
a) Q1
b) Q2
c) Q4
d) Q3
Q4. Which is the first question where b) is the correct answer?
Reasoning
Let's look at Q1 first, and check each possible option:
If Q1's answer was a):
… we'll leave checking this for now, because …
If Q1's answer was b):
… we'll leave checking this for now, because …
If Q1's answer was c):
it would reference itself, and we'd learn nothing new, and we'd then have to check Q2.
So, let's start with Q2 first instead of Q1.
If Q2's answer was a):
an immediate contradiction, as Q4 can't be the first answer with a) as we've assumed Q2's answer is a).
If Q2's answer was b):
an immediate contradiction, as Q2 can't be the first answer with a) as we've assumed Q2's answer is b).
If Q2's answer was c):
Q3's answer would be a)
Q1's answer would be d)
confirming Q2's answer as c)
leaving Q4's only possible answer as b)
no contradiction.
If Q2's answer was d):
Q1's answer would be a) (*)
Q3's answer would be c)
Q4's answer would be d)
Q1's answer would be b)
but this contradicts the section marked (*).
Therefore, the only possible answer with no contradiction is when Q2's answer is c).
We can then follow Q2's answer, which leads to the answers to Q1 to Q3. Q4 must then reference itself.
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Puzzle 8
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 – 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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