Mathematical and Logical Puzzles
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Send me good mathematical/logical
puzzles and answers ; will post with complements to you.
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Q1 |
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1. A man is condemned to death.
He is given to say a last statement; if it is truth he is hanged to death and if it is
false he is poisoned. He uttered his last sentence from which he escaped. What is the last
statement ?
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Q2 |
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2. In a 3 way junction there is
a house in which there are 2 brothers. Elder one always tell the TRUTH and the younger one
alway LIE. Here a man comes from one road and wanted to go for a particular city (say city
A) and he don't know which way to take out of the two. He knows that in that house there
are 2 brothers who always made TRUE and FALSE statements but he donot know who is who. At
one time there is one one brother at home. The traveller asks only one question from one
of the brothers who is there at that moment and chooses the correct way for his
destination. What is that question and whats the road he chooses ?
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Q3 |
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3. A,B,C are aligned according
to their height. A the shortest. C the tallest. A, B,C are faced left. They cannot see
back. C can see A&Bs' head. B can see only A's head and A cannot see anybody's head.
After C there is D who holds a gun. ie they stand like A B C D(with the gun)
D has 3 white hats and 2 black hats
ABC are blind fold and hats are put onto
their head. After that,
Now C can see A&B's hat colors,
B can
see A's hat color
Now respectively C,B,As' chance to predict the color of the hat that they are wearing (one
at a time starting with C). If someone made a mistake he will be shot to death(by D).
A,B,C are intelligent and try to escape from the death.
They all can hear previous guy's answer and if the previous guy is shot
to death the sound of gun fire.
This is what happend.
First C uttered the answer, but shot to death.
Then B uttered the answer, but again shot to death.
Finnally A uttered the answer and it is the correct answer
So he escaped from the death.
What is the answer and explain
how A thought(actually you) that it is the color of
A's hat.
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Q4 |
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4. You are given two threads which takes 1hr to burn from one end to the
other. Using these 2 threads and a matchmox(or lighter) how will you measure 45 minutes
You are not allowed to break these threads.
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Q5 |
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5. A warden of a prison comes
back to the prison from a party very drunk. Inhis state, he accidentally turns the
key to open all 100 of the jail cells.He realizes what he's done, so he goes back and
turns the key on every othercell (thus, now cell 1 is open, cell 2 is closed, cell three
is open, etc.).He realizes that he hasn't closed all of them, so he goes back and turns
thekey on every third cell (thus, now cell 3 is closed, and cell 6, which heclosed when he
turned every other key, is now open). Still drunk, he goesback and turns the key on
every 4th, then every 5th, 6th, 7th, etc. until heturns the key on every 100th cell (thus,
only on cell #100). After all this,which cells remain open?
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Q6 |
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6. There are 4 people that need
to cross a bridge. Only two can go across at a time and one has to come back with the
flashlight each time. They take 1 minute, 2 minutes, 4 minutes and 5 minutes. They only
walk as fast as the slowest one. They are able to all cross the bridge in 12 minutes. How?
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Q7 |
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very good one
7. Pretend you have two strings which burn at random
rates but which will alwaysbe completely burned in exactly one hour. Thus, after 1/2
hour the stringswill NOT necessarily be half burned length-wise, only time wise. How
do youuse these strings to time exactly 45 minutes without using any
othertime-keeping
device?
It is inaccurate to assume that each 4th of
string A will burn in exactly 15 minutes. Three of the fourths could take one
minute, and the other could takea full hour (thus, the entire string burns in one hour, as
said).
The string, remember, burns at random rates
in random places. If you were tocut it into four pieces and light each SEPARATELY so
that one would beginburning when another stopped, the sum of the time would be one hour.
That isall you can assume. For example, if you lit all of them together,
three ofthe fourths could burn in exactly one minute, and the last fourth could take57
minutes. Thus the entire string burns in one hour. That is the onlyassumption
you can make. You must look a little beyond just the string.
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Q8 |
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8. There are 3 settlers and 3
indians on one side of the creek. How can you get them all over to the other side by using
a boat that holds two(they can only get there by boat). You can never have more indians
than settlers on one side at a time because the indianswill kill the settler.
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Q9 |
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9. You ask a person to pick a number. Say they pick the number 20, 20 has six lettres in it, so therefore 20 is 6, then you see that 6 has three lettre's in it, so 6 is 3, 3 has five lettres so, 3 is 5, 5 has four lettre's so 5 is 4, and four is cosmic. this pattern works for any number, at all. so quickly, take the number 13 (thirteen), 13 is 8 (eight), 8 is 5 (five), 5 is four and four is cosmic.
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Q10 |
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10. if a baot is exactly 10ft long, and there is a ladder hanging from the top
and it goes 6ft down the boat, and the tide is coming in and rises 1ft perhour how long will it take before the ladder touches the water?
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Q11 |
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11. Three men, members of a safari, are captured by cannibals in the jungle. The men are given one chance to escape with their lives. The men are lined up and bound to stakes such that one man can see the backs of the other two, the middle man can see the back of the front man, and the front man can't see anybody. The men are shown five hats, three of which are black and two of which are white. Then the men are blindfolded, and one of the five hats is placed on each man's head. The remaining two hats are hidden away. The blindfolds are removed. The men are told that if just one of the men can guess what hat he's wearing, they may all go free. Time passes. Finally, the front man, who can't see anyone, correctly guesses the color of his hat. What color was it, and how did he guess correctly?
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Q12 |
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12. Ten men stand in a line facing same direction.
Random Red or White hat is placed on their head(one hat per person).
One cannot see his own color(of the hat) but can see everybody else's in front of him.
The first man(Man1) cannot see anybody else's whereas the last man(Man10) can see 9hats.
They know ten men got 10 hats but we do not know how many Reds or Whites were there.
The game organizer starts asking men from the last in line(Man10, Man9,... Man2, Man1)
the color of individual's own hat.
When a man gives correct answer he is awarded a one point.
Individual's answers is heard by everybody else in the line. Their correctness exposed at the end.
What is the strategy the group should take to maximize their group score.
Hint : They can get 9 points.
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Q13 |
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13. X number of cards are placed on table A. Out of X, Y number of them are face down.
You have to move cards (one at a time) from table A to table B.
After the move both tables should have equal number of cards face down.
You are blind folded and sit in the middle of tableA and tableB.
(means you can touch the cards but you cannot see whether they are face up or down)
You are given the values of X and Y.
How do you accomplish this ?
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Q14 |
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14.
AB=BC=CD=DA=PQ=QR=RS=SP=12
AP=BQ=CR=DS=30
DM=MC=PN=NQ=6
MX=NY=1
Consider this as a room where X is switch and Y is bulb.
What is the shortest length of wire drawn from X to Y(along the
walls/floor/ceiling)
(Wire cannot drawn through space)
Answer is less than : 42
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A1 |
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"I should be
poisoned" think a bit and analyse. Still if you don't get keep on reading.
Explanation : If "I should be
poisoned" is TRUE(by sentense) you have to poison him. But according to the rule of
the riddle "He should be hanged". Because of the contradiction no one can hang
him or poison him. This solves by CONTRADICTION.
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A2 |
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2. "If my brother
is here instead of you what would he say for the question "What is the correct way to
city A" ? " Then he chooses the other road whatever
the brother say.
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A3 |
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A cannot be black, because if A was black, C or B
would live.
If C sees A is black:
if C sees B is also black
then he knows he is white, and lives.
Cannot happen.
else C sees B is white
B knows this because if he were black
and A were black, C would live.
No matter how C answers, B has to be white, if A
is black, or C would live.
Therefore A cannot be black, and has to be white for
both C and B to die.
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A4 |
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4. burn both ends of one thread
and one end of the other. After the first thread is burnt completely burn the other
end of the second thread. The total time taken by threads to burn will be 45 minutes.
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A5 |
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5. All the cells that are
perfect squares (1,4,9,16,25,36,49,64,81,100)remain open because they have an odd number
of multiples so the key will be turned an odd number of times on them, making them open at
the end. Allothers have an even number of multiples and will stay closed.
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A6 |
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6. 1&2 : 2mins, 1back :
1min, 4&5 : 5mins, 2back : 2mins, 1&2 : 2mins = 12mins
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A7 |
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7. The first thing you must
understand is that the time which it takes the strings to burn is proportional to the
number of flames present on the string. If you light both ends of one of the strings,
it will take HALF of an hour to burn, by definition. Thus, to time 45 minutes, here's
what you'd do. Light string A at both ends while you simultaneously light string Bon
one end. When string A is completely burned up, 30 minutes will have
passed. At
the moment that this happens, you will have 30 minutes left on string B. Right as
string A burns out, light the OTHER end of string B, which will now only burn for 15 more
minutes because you have two flames on a30-minute string. When it's gone, 45
minutes total have passed.
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A8 |
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email the answer to get it
verified
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A9 |
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email the answer to get it
verified
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A10 |
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10.Let I = Indians and S = Settlers
One Side of Creek Boat Other side of Creek
I I I S S S
I I S S I S ->
I I S S <- S I
S S S I I -> I
S S S <- I I I
I S S S -> I I
I S <- S I I S
I I S S -> I S
I I
<- I S S S
I I I -> S S S
I
<- I S S S I
I I
-> S S S I
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A11 |
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A11. If black is 1 and white is 0
All possible combinations :
a. 000
b. 001
c. 010
d. 011
e. 100
f. 101
g. 110
h. 111
a. does not happen ; if e fist buy can answer
if c or g, second guy could have answered
In all other cases 3rd guy wears black.
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A12 |
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A12. If one sees odd number of (say) Reds he calls RED. Then the one in
front
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A13 |
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A13. Pick a card from tableA , flip the side and place on tableB
You do this until you reach Y.
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A14 |
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A14.
YZ=30+1+1, XZ=12+6+6
XY2=ZY2+XZ2
XY = 40
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