BRAIN TWISTER

1.man carries chicken, man leaves chicken and comes back, man gets fox, man leaves fox and gets chicken, man leaves chicken and gets corn, man leaves fox and corn to get chicken, man gets chicken.
2.The old man temporarily added his camel to the 17, making a total of 18 camels. 
1/2 = 9
1/3 = 6
1/9 = 2
for a total of 17. He then takes his camel back and rides away.
3.turn switch 1 on for a while .. about 5 mins or more .. then turn it off. turn switch 2 on, go inspect: the hot bulb is controlled by switch 1, the lit bulb is controlled by switch 2, the unlit and cold bulb is controlled by switch 3
4.None. A hole only has air in it, no dirt. Sorry that was an easy one!
5.When they meet, they're both at the same spot, so they are both the same distance from NYC
6.They are female boxers.
7.The dates are in BC
8.A must not see two white hats on B & C, or he would know his own must be black. So his answer establishes that at least one of the hats on B & C is black. If B saw that C was wearing a white hat, then B would know his own was black. Since B could not say that truthfully, then C must be wearing black and he answers so correctly.
9.The Reason the census taker could not figure out the childrens ages is because, even with knowing the number on the house next door there were still two possibilities. The only way that the product could be 36 and still leave two possibilities is if the sum equals 13. These possibilities being 9+2+2 and 6+6+1. When the home owner stated that her "Oldest"child is sleeping she was giving ths census taker the fact that there is an "oldest." The childrens ages are 9,2 and 2.
10.The man is on a bridge when he spots the police car. He's more than halfway across, so the quickest way off the bridge is to run forward
11.The "easy" (ha!) way to solve this problem is to work it backwards. A diagram would help, but that would make my explanation more difficult, so activate your imaginations. Consider the greatest distance our truck can travel before all fuel is used: it's 180 km. (1 unit fuel (litre) = 1 unit distance (km)) The optimum solution will have the truck using its last dribble of fuel as it arrives at its destination. Therefore, the last leg will be 180 km. This means that the last (Z) fuel dump must contain sufficient fuel from previous trips plus what's onboard at the last trip to equal 180. In order to get 180 units of fuel to Z, at least 2 other trips are required from some prior dump location. It is safe to assume that the truck made one round trip starting with a full load and returned empty, and that it also made a final one-way trip starting with another full load of 180. (And we don't need to be warp-space scientists to figure out that at least one (hmmm..., why's that, Lem?) additional trip was necessary. In order to cache 180 total units of fuel at this site, the truck used an additional 180 units for travel in three trips, each of which was 60 km. (180/3) This requires that the quantity of fuel at the previous Dump Location = 360. (I'd like to think that there are many mental light bulbs popping on out there... "Oh. Sure. That means...") Moving right along (in a reverse direction, of course,) the previous dump must have had enough fuel cached plus enough for the travel to z-1. Using the same method, this requires five trips minimum. We're limited by that 180 fuel capacity, remember, so the further back we go, the shorter will be the distances and the more numerous will be the trips between fuel dumps. Hence, the distance to Dump z-2 is 180/5 = 36 km. See the pattern yet? Working backwards, we've traveled 180*1 + 60*3 + 36*5.. The method of solution was the real puzzle, not the actual answer.
12.They were Siamese twins!
13.Zero crows are left. The crows all flew away when they heard the farmer shooting his gun.
14.The man is on a bridge when he spots the police car. He's more than halfway across, so the quickest way off the bridge is to run forward.
15.(1) The one you did after the one you did before this one IS this one. In other words: (2)"....the puzzle you solved after you solved the puzzle you solved before you solved this one" IS this one. (3) Hence the question may be rephrased this way: "If the puzzle you solved before this one was harder than THIS ONE, was the puzzle you solved before this one harder than THIS ONE?" (4) The answer to this question is obviously YES.
16.You ask either of the twins "If I asked your twin which path to take to get to the hospital, which way would he tell me to go?" And then I would go in the opposite path of the one I'm told. Because: 1) If I'm talking to the truth telling twin, then he will tell me what his twin would say, which would be a lie. So the direction indicated would be wrong. 2) If I'm talking to the lying twin, then he would lie about what his twin would say, which would be the truth. So the lying twin would still indicate the wrong path.
17.Since the counterfeit note was used in every transaction, they are all invalid. Therefore, everybody stands in the same position to his/her creditor as before the banker picked up the counterfeit note.
18.20
19.MANE, MEAN, AMEN, NAME
20.NO! It would be very unwise of Jack to accept the bet. To find the chances that the 3 coins will fall alike or not alike, consider all the possible ways that the 3 coins can fall, as follows: 
1:HHH
2:HHT
3:HTH
4:HTT
5:THH
6:THT
7:TTH
8:TTT
Each of the 8 possibilities is equally likely to occur. Note that only 2 of them show all the coins alike. this means that the chances of all 3 coins being alike are 2 out of 8, or one quarter. There are 6 ways that the coins can fall without being all alike. Therefore the chances that this will happen are three-quarters.
21.Once a girl removed 2 balls from her box, she narrowed the possible combinations in her box to 2. If she was able to deduce the color of the third ball, it must have been because the label on her box showed one of the two possible combinations, forcing the actual contents to be the other combination- remember, all the boxes were incorrectly labeled. The first girl's box contained either BBB or WBB, and it's label must have read either WBB or BBB for her to have guessed the color of the third ball. Similarly, the second girl's box must have contained either WBB or WBW, with it's label reading either WBW or WBB. The third girl's box must have contained either WWW or WBW- but her box couldn't be labled with either of these, otherwise she, too, would have been able to deduce the color of her third ball. The only distribution satisfying all these conditions is: 
GIRL 1 2 3 4
LABEL WBB WBW BBB
ACTUAL BBB WBB ?
The fourth girl instantly realized that her box must have been labeled WWW; since it could not actually contained WWW, the third girl' box must have contained that combination, leaving her (fourth girl's) box to contain WBW.
22.IT IS MUCH TOO HOT TO DO THESE THINGS RIGHT NOW!
23.5
24.20
25. 
    
	Seat#1 - Mrs. Dunlop
	Seat#2 - Mrs. Barker
	Seat#3 - Mr. Collins
	Seat#4 - Mr. Andrews
	Seat#5 - Mrs. Collins
	Seat#6 - Mr. Dunlop
	Seat#7 - Mrs. Andrews
	Seat#8 - Mr. Barker
26.The supervisor gave the repair person the number broken in like terms (5 booths out of the first 8 are broken). If number 8 were working the supervisor would have said that 5 booths out of the first 7 are broken. When giving the number broken, the supervisor gave it based on the total number broken. Therefore number 8 is the only one the repair person can be sure is broken.
27.The second option should be recommended.
28.40 years old & 10 years old
29.The missing digit is 8. If you spell the numbers out (ie. one, two, three, etc..) the next digit in the sequence begins with the last letter of the previous digit. Therefore we have Five, eight, three, eight, two, one
30.General, Colonel, Lt. Colonel, Major, CAPTAIN. Answer is 'C'
31.3
32.MISTAKEN
33.use the periodic table of elements 
Packers Krush Chicago
34.The banker and accountant are female. The refuse collector is the only male out of the three and has to be John.
35.July = 10 .... The answer is: each syllable = 5
36.Is the statement that you are reliable equivalent to the statement that Atlantis is still around? OR "Do you believe that the statement that you are a Fairy is equivalent to the statement that Atlantis is still around?" OR something similar.
37.It does make a difference.
A submerged body displaces its volume, a floating body displaces its weight. Since a penny is denser than water, dropping it into the canoe will raise the water level higher.
38.The next letter in the sequence O T T F F S S is E
The letters are the initial letters of One, Two, Three ... but you know the rest!
39.Einstein's Riddle
House Color Yellow Blue Red Green White
Nationality Norweigian Dane Brit German Swede
Beverage Water Tea Milk Coffee Beer
Brand of Cigar Dunhill Blends Pall Mall Prince Bluemasters
Kind of Pet Cats Horses Birds Fish Dogs
40.Answer 41
41.Answer LIFE
42.Answer POST
43.Answer 71 (& = 2, # = 19, @ = 31)
44.Answer LOST, LOTS and SLOT
45.Answer COCONUT = 23p, FIG = 13p, LETTUCE = 17p (Add the alphabetical value of the first and last letter)
46.Answer: 97
47.Answer: B (The bottom row of a keyboard)
48. 
     TALL  ( Boy )  HOOD
     COVE  ( Red )  RAFT
     FOND  ( Ant )  HEM
      ROB  ( Ins )  TILL
    DOUGH  ( Nut )  HATCH
     FORE  ( See )  PAGE
Answer 'Brains'
49.Answer: 'Six'
50.Answer: 'Each word contains a herb' 
             arCHIVE  DILLy-dally
                   ?
               mesSAGE  BAYonet
51.Answer '10.5 inches per hour'
Let's say that it craws D inches. Then it takes D/21 hours to crawl to the patch and D/7 hours to crawl back.
The total time is D/21 + D/7 = 4D/21 hours to travel 2D inches, or 2/21 hours to travel one inch. Therefore the average speed is 21/2 = 10.5 inches/hour.
52.Answer: The letters are the first and last letters of: 'o'
Mercury, Venus, Earth, Mars, Jupiter, Saturn, Uranus, Neptune and Pluto
53.Answer: '140'
The alphabetical positions of all the letters are added to give the amount
54.Answer: South
The series: south, east, north, south, west, east spirals clockwise from the top left-hand corner !
55.Answer: 6 pounds and 44 pence!
56.Answer: The letters are the first three of a precious stone and the corresponding letter is the initial of its color: 
AMEthyst = Purple,    RUBy = Red,    SAPphire = Blue,     EMErald = Green
57.Answer: The words are Radar, Oddly, Sales and Tryst.
58.Answer: '6' -- Any small square of four numbers totals 20 !
59.Answer: The words become Dahlia, Raisin, Orchid and Fuchsia. So Raisin is the odd one out !
60.Answer: The maximum possible is 63 !
61.Answer: Bedrock, Bedpost, Bedspread, Bedtime, Bedevil and Bedclothes !
62.Answer: 18.71 using divide, plus, multiply and minus. -15.29 using divide, minus, multiply and plus ! 
	3 / 7 + 5 x 4 - 3 = 18.71
	3 / 7 - 5 x 4 + 3 = -15.29
63.Answer: Blaze, then Blade, then Glade, then Grade and finally Grape !
64.Answer: 46p. A satsuma costs 7p, a pear costs 14p and an apple costs 16p !
65.Answer: 5:15 pm !
66.Answer: Yes, with 0.63 gallons left over !
67.Answer: 36. A vowel is worth two and a consonant worth six. These are totaled in each place name !
68.Answer: Isaac Newton, Albert Einstein and Louis Pasteur !
69.Answer: 13 !. Add 2, then 4, then 6, then 8 !
70.Answer: Oscar Wilde !
71.Answer: 100 miles. Each vowel is worth 300 and each consonant is worth -100. These are totalled in each city name to give the distance !
72.Solution: TWENTY NINE.
73.Solution: One is in Eastern Oregon (Mountain time); the other in Western Florida (Central time), and the phone call takes place on daylight-savings-time changeover day at 1:30am.
74.Solution: 
	dgooddodgooddo
	odoogggdodgogg
	ogogdoodgooddd
	dgdoooggoogdgo
	ogdgogdgoggogd
	dddgddodoogdoo
	odgoggdooggood
75.Solution: Herb (American English only), Job, Nice, Polish
76.Solution: "That that is, is; that that is not, is not; is not that it? It is."
77.Solution: "Herein" The words it contains are: "he," "her," "here," "ere," "re," "rein," "I," and "in."
78.Solution: A towel.
79.Solution: An ear of corn.
80.Solution: A rug.
81.Solution: A cold.
82.Solution: A nose.
83.Solution: An embarrassed skunk. Or A zebra painted red. Or A sunburned penguin. Or A newspaper.
84.Solution: A stamp.
85.Solution: A sponge.
86.Solution: Fire.
87.Solution: A coffin.
88.Solution: A shoe.
89.Solution: A tissue. Or A wave.
90.Solution: A river.
91.Solution: Silence.
92.Solution: Suicide.
93.Solution: An onion.
94.Solution: A shadow.
95.Solution: A tree.
96.Solution: A splinter.
97.Solution: A hole.
98.Solution: Tomorrow.
99.Solution: Breath.
100.Solution: Smoke.
101.Solution: An eye.
102.Solution: Stars. Or Dreams.
103.Solution: A dictionary.
104.Solution: The letter 'e'.
105.Solution: An egg.
106.Solution: A pig.
107.Solution: The wind.
108.Solution: One's heart.
109.Solution: A snail.
110.Solution: An icicle.
111.Solution: A ship.
112.Solution: Time.
113.Solution: A candle.
114.Solution: An iceberg.
115.Solution: A cipher.
116.Solution: A stove, fire, and smoke.
117.Solution: Counterfeit money.
118.Solution: The sun.
119.Solution: Nothing.
120.Solution: The number PI, the digits of which are illustrated by the length of each word in the riddle.
121.Solution: Roasting meat on a spit.
122.Solution: A glove.
123.Solution: A stable.
124.Solution: A cloud.
125.Solution: A human does. As a baby, it crawls; in old age, uses a cane.
126.Solution: The word "wholesome."
127.Solution: An electric fan.
128.Solution: Push the cork into the bottle and shake the coin out.
129.Solution: Put the valuable object into the box, secure it with one of your locks, and send the box to your friend. Your friend should then attach one of his own locks and return it. When you receive it again, remove your lock and send it back. Now your friend can unlock his own lock and retrieve the object.
130.Solution: After you draw one of the papers, swallow it. The jailer will be forced to check the remaining paper to determine what the one you drew said. The jailer will of course see a paper with "DEATH" written on it, assume you drew the one with "LIFE" written on it, and set you free.
131.Solution: You can make 120 cubes (10 full trays) in the time it takes to freeze two trays. First, fill four of the trays with water and turn the other three upside down and use them to space the four apart. That gives you 48 cubes. Next, empty the four trays and put two ice cubes in diagonally opposed corners of each of six of the trays. Fill the remaining holes -- and the entire seventh tray -- with water. Using the ice cubes to hold the trays apart, stack all seven (the seventh tray should go on top), and freeze them. You'll get an additional 72 cubes. You can get 72 cubes for every batch except the first, for which spacer ice cubes are not yet available.
132.Solution: A square manhole cover can fall into the hole on the diagonal; a round manhole cover cannot be dropped into the hole.
133.Solution: The first surgeon operates with the first glove (glove A) inside the second glove (glove B). The second surgeon operates using just glove B. The third surgeon operates using glove A, turned inside out, inside glove B.
134.Solution #1: Almost all of it. First tie the ropes together at the ends. then climb one of the ropes. Tie a loop in the rope as close to the ceiling as possible. Hang from the loop, then cut the rope just below it (but don't drop it, or you'll be stuck on the ceiling). Run the rope through the loop and tie it to your waist. Swing over to the other rope and pull the rope that's going through the loop tight. Cut the other rope as close as possible to the ceiling, holding tight to the end of it. Once the rope is loose, you will swing down below the loop. Let yourself down to the ground by letting the rope out. At the bottom, untie yourself and pull the rope completely through the loop. At this point you'll have all of the rope except what was used to tie the loop.
Solution #2: All the rope, provided that the knife is such that: it can support your weight when half the blade is stuck in ceiling, and it can bend at the blade-handle junction without breaking. Climb to top of the first rope, bringing other rope along. At the top, stretch the second rope toward you and circle it around your waist. Cut the first rope at the ceiling, hang on, and swing down on the second rope. Climb back up to the ceiling. Pull up the other end of the second rope, and tie it to the end of the first rope. Loop the tied ropes around the handle of the knife. In one (quick) motion, cut the second rope at the ceiling while pushing the knife into the ceiling. Bend the handle of the knife enough so that the rope won't slip out. Go down, hanging on the double rope with both hands. At the floor, jerk hard to release the knife and the rest of the rope.
Solution #3: Climb up the first rope, carrying up the end of the second with you. Cut the first rope, flush with the ceiling, swinging down on the second. Climb up the second and cut that flush with the ceiling too, but only part way, until it can just barely hold the rest of your weight. Slowly climb down until you're just short of the bottom. Tie a small loop here and climb up about two feet. Secure the loop about your feet and fall to the ground (roughly three feet). Just before your feet hit the ground the force of the fall will snap the rope at the top.
135.Solution: A nickel and a half dollar. Only one is not a nickel.
136.Solution: Seven.
137.Solution: The sun is shining; there's no rain.
138.Solution: Whatever color yours are. You're the bus driver.
139.Solution: Two.
140.Solution: Short.
141.Solution: There isn't any smoke. It's an electric train.
142.Solution: All of them.
143.Solution: They were two of triplets.
144.Solution: The only place each window could have a southern exposure is on the north pole. So the bear must have been a polar bear. The answer, therefore, is white.
145.Solution: Two. Nobody said it had to be a matching pair.
146.Solution: It's daytime; the sun is out.
147.Solution: He's born in room number 1964 of a hospital and dies in room number 1984.
148.Solution #1: The other end of the rope isn't tied to anything.
Solution #2: The haystack is only 10 feet away from where the other end of the rope is tied, and the horse is 30 feet away on the other side.
149.Solution: A ton.
150.Solution: Post Office.
151.Solution: One. Everybody else was coming from St. Ives.
152.Solution: Eight thousand. When books sit on shelves, the first page of the book is the rightmost page, and the last page is the leftmost page. So you can't count the pages in the first and last volumes.
153.Solution: A minute and a half.
154.Solution #1: Ten letters.
Solution #2: Nineteen letters long.
Solution #3: Twenty five letters in length.
Solution #4: Two words.
Solution #5: Three words long.
Solution #6: Four words in length.
Solution #7: Four syllables.
Solution #8: Five syllables long.
Solution #9: Six syllables in length.
Solution #10: The answer is one sentence long.
Solution #11: How long.
155.Solution: Dead.
156.Solution: Six. (Three for each team.)
157.Solution: One.
158.Solution: The match.
159.Solution: An hour. You take one right away, take the next a half-hour later, and the last one a half-hour after that.
160.Solution: I don't. In 46 B.C., they wouldn't have known how many years before Christ it was.
161.Solution: Not if he isn't dead yet.
162.Solution: 70. (30 divided by 2 is 15, but 30 divided by 1/2 is 60.)
163.Solution: None. Noah was the one with the ark.
164.Solution: Meat.
165.Solution: Of course. They just don't celebrate it.
166.Solution: No. If he has a widow, he's dead.
167.Solution: The umpire and the catcher.
168.Solution: When the numbers are expressed in roman numerals, this works out: If you take I from XIX, you are left with XX.
169.Solution: John sat down in the chair. Jack ran around it twice, then said, "I'll be back in a week to run the third time around!"
170.Solution: No more than his own weight.
171.Solution: Just two -- one on each side.
172.Solution: Incorrectly.
173.Solution: The sun never rises and sets at the same time of day. At the poles, the sun only rises in the morning and only sets in the evening. Even at the poles, it can't do both simultaneously.
174.Solution: "Language." Ignore the first two sentences, which are irrelevant to the question. The third sentence says, "There are three words in 'the English language.'" The first is "the"; the second is "English"; and the third is "language." Language is something we use every day.
Unfortunately, this puzzle is frequently asked of others by those who do not know the answer; as such, the wording of the puzzle is inadvertently altered in such a way that there is no correct answer. There are a couple other archaic words in English that end in "gry"; however, none of these are correct answers, because none of them are things we use every day. (Besides, then the puzzle would be a trivia question rather than a riddle.)
175.Solution: None. Roosters do not lay eggs.
176.Solution: You would impact between the 2nd and 3rd second after the cable broke, so you would want to jump 2 seconds after the break. However, considering the calculations involved, you would probably end up as a heap of screaming bloody mess at the bottom of the elevator shaft before you figured the answer out. Besides, there's no telling if jumping really works.
177.Solution: About half way. After that, he starts running out of the woods.
178.Solution: From now until hell freezes over. No telescope yet invented has that kind of magnification. Besides, Alpha Centuri is a star and your friend couldn't get within miles of it without being vaporized.
179.Solution: Pretty good! The dice are obviously loaded.
180.Solution: As many as you can carry, if its not U.S. currency.
181.Solution:
10 is the score for the first roll.
15 is the score for the second roll.
9 is the score for the third roll.
20 is the score for the fourth roll.
12 is the score for the fourth roll.
182.Solution: The only lockers that remain open are perfect squares (1, 4, 9, 16, etc) because they are the only numbers divisible by an odd number of whole numbers; every factor other than the number's square root is paired up with another. Thus, these lockers will be "changed" an odd number of times, which means they will be left open. All the other numbers are divisible by an even number of factors and will consequently end up closed. So the number of open lockers is the number of perfect squares less than or equal to one thousand. These numbers are one squared, two squared, three squared, four squared, and so on, up to thirty-one squared. (Thirty-two squared is greater than one thousand, and therefore out of range.) So the answer is thirty-one.
183.Solution: Use the first two cuts to cut an 'X' in the top of the cake. Now you have four pieces. Make the third cut horizontal, which will divide the four pieces into eight. Think of a two by two by two Rubik's cube. There's four pieces on the top tier and four more just underneath it.
184.
185.
186.Solution: Not at all. He'll earn $5,368,709.12 on the thirtieth day alone.
187.Solution: Let d be the distance to the store, T be the time it gets to get there, t be the time it takes to get back, and A be the average speed (which is what we want to find out). As we know from elementary mathematics, distance equals rate times time: 
  d = 20T
  T = d/20
  d = 30t
  t = d/30 

Now that we have expressions for T and t, we can come up with an equation that describes the round trip:

  2d = A(T + t)
  2d = A(d/20 + d/30)
  2d = A(3d/60 + 2d/60)
  2d = A(5d/60)
  A = 120d/5d
  A = 24
So the average speed is 24 mph. If this seems strange to you, consider that more time is spent traveling at 20 mph than time spent at 30 mph, so the "20 mph" figure should count more toward the average.
188.Solution: Let d be the distance to the store, T be the time it gets to get there, t be the time it takes to get back, and R be the speed you travel on the return trip (which is what we want to find out). As we know from elementary mathematics, distance equals rate times time: 
  d = 20T
  T = d/20
  d = Rt
  t = d/R 

Now that we have expressions for T and t, we can come up with an equation that describes the round trip:

  2d = 40(T + t)
  2d = 40(d/20 + d/R)
  2d = 40d(1/20 + 1/R)
  1 = 20(R/20R + 20/20R)
  20R = 20(R+20)
  R = R + 20
Here we have worked our way into a paradox. The reason, simply, is that you have to travel back at an infinite speed to make your average speed 40 mph. This may seem strange, but consider that, the faster your return trip, the quicker you make it, and consequently, this faster speed has a lesser impact on the average speed. If you traveled the return trip instantaneously, this would be equivalent to traveling double the distance in the same amount of time as the one-way trip. So if the rate of speed of the return trip is infinite, you do indeed get an average speed of 40 mph.
189.Solution: There is only one solution, discounting mirror image solutions and rotations:

190.Solution: Since the trains are 100 miles apart, and the trains are traveling toward each other at 40 and 60 mph, the trains will collide in one hour. The bird will have been flying for an hour at 90 miles per hour at that point, so the bird will have traveled 90 miles.
191.Solution: The problem is with the division that takes place between the fourth and fifth equations. Since a = b, a - b is zero, and you can't divide by zero.
192.Solution: Actually there's nothing wrong with it. Ten does equal 9.99999..., as this proof clearly shows.
193.Solution: If the product of his three children's ages is 72, there are the following possibilities:
1 * 1 * 72 = 72
1 * 2 * 36 = 72
1 * 3 * 24 = 72
1 * 4 * 18 = 72
1 * 6 * 12 = 72
1 * 8 * 9 = 72
2 * 2 * 18 = 72
2 * 3 * 12 = 72
2 * 4 * 9 = 72
2 * 6 * 6 = 72
3 * 3 * 8 = 72
3 * 4 * 6 = 72
Isaac later gives Albert the sum of their ages, but we don't know what number he says. We do, however, know that Albert can't figure it out from that information. So, we take the possibilities listed above and add them up:
1 + 1 + 72 = 74
1 + 2 + 36 = 39
1 + 3 + 24 = 28
1 + 4 + 18 = 23
1 + 6 + 12 = 19
1 + 8 + 9 = 18
2 + 2 + 18 = 22
2 + 3 + 12 = 17
2 + 4 + 9 = 15
2 + 6 + 6 = 14
3 + 3 + 8 = 14
3 + 4 + 6 = 13
The only way Albert wouldn't be able to figure out Isaac children's ages by knowing the sum is if the sum was 14, because there are two possibilities. So either the children's ages are 2, 6, and 6, or 3, 3, and 8. But Isaac points out that he has a youngest child. So the ages must be 2, 6, and 6.
194.Solution: The hour hand is exactly on a minute mark five times per hour -- on the hour, twelve minutes past the hour, twenty four minutes past, thirty six minutes past, and forty eight minutes past. Let X be the number of hours, and Y be the number of minutes past the hour. When the hour hand is on a minute mark, the position of the hour hand is 5X + Y/12, and the position of the minute hand is Y. On the first occasion, Y = 5X + Y/12 + 6. This is equivalent to 60X = 11Y - 72. Since Y can only take one of the values in the set { 0, 12, 24, 36, 48 }, it can be determined that the only legal values for the equation are X = 1 and Y = 12. So the time is 1:12. Similarly, the second occasion's equation is 60X = 11Y - 84. The only legal values here are X = 3 and Y = 24. So the time is 3:24. Between 1:12 and 3:24, two hours and twelve minutes have elapsed
195.Solution: Let c be the number of chickens, and r be the number of rabbits.
r + c = 72
4r + 2c = 200
To solve the equations, we multiply the first by two, then subtract the second.
2r + 2c = 144
2r = 56
r = 28
c = 44
So there are 44 chickens and 28 rabbits in the cage.
196.Solution: Let g be the total units of grass in the field before the sheep are turned out. Let s be the number of units of grass each sheep eats per day. We must determine the constant values of g and r from the information given in the problem. The total number of grass units eaten equals the number of days times the amount of grass the sheep eat per day. From this, we can construct two equations: g + 20r = 20(10s)
g + 10r = 10(15s)
Reducing the equations, we obtain:
g + 20r = 200s
g + 10r = 150s
Subtract one equation from the other:
10r = 50s
So r equals 5s. Substituting back, we discover g equals 100s. Now we construct an equation for the case of having 25 sheep turned out in the field, where x is the number of days it takes the sheep to eat all the grass:
g + xr = x(25s)
100s + x(5s) = x(25s)
20sx = 100s
x = 5
So 25 sheep would consume all the grass in the field in 5 days.
197.Solution: Let a be the number of red squeegees you buy. Let b be the number of yellow squeegees you buy. Let c be the number of blue squeegees you buy. We know two equations:
a + b + c = 100
6a + 3b + 0.1c = 100
By multiplying the first equation by 6 and then 3, then subtracting these two equations from the second, we can come up with two more equations:
6a + 6b + 6c = 600
3a + 3b + 3c = 300
3b = 500 - 5.9c
3a = 2.9c - 200
Normally, two equations isn't enough to solve for three variables. But we know that a and b are nonnegative integers. So, if 3b >= 0, then 500 - 5.9c >= 0. This means c <= 84.75. Also, if 3a >= 0, then 2.9c - 200 > 0. This means c >= 68.97.However, since buying blue squeegees is the only way to spend a fraction of a dollar, the number of blue squeegees we buy must cost an even dollar amount. The only two numbers of blue squeegees we can buy between 68.97 and 84.75 that satisfy this condition are 70 and 80. If we substitute 80 for c in the last two equations listed above, we can solve for a, which equals 10.67, and b, which equals 9.33. But the values of a and b must be integers, so we know this is not the solution. If we substitute 70 for c in the last two equations listed above, we can solve for a, which equals 1, and b, which equals 29. So one red squeegee must be bought for $6.00, 29 yellow squeegees must be bought for $87.00, and 70 blue squeegees must be bought for $7.00.
198.Solution #1: Let x be the age of the man at the time of his death. His age consists of the number of years he spent as a child plus the number of years he spent as a teenager plus the number of years between his coming of age and his marriage plus five plus his son's age (half his own) plus four. In mathematical terms, this is:
x = x/6 + x/12 + x/7 + 5 + x/2 + 4
x = 14x/84 + 7x/84 + 12x/84 + 42x/84 + 9
x = 75x/84 + 9
84x/84 - 75x/84 = 9
9x/84 = 9
x/84 = 1
x = 84
So the man lived to be 84 years old.

Solution #2: The man lived 1/12 of his life as a teenager. There are seven teen years. 7 x 12 = 84.
199.Solution #1: Let s be the total distance of the journey to the hotel. Let v be walking speed. So 2v is the bike's speed, and 8v is the train's speed. Let I be the time it took for Isaac to complete the journey, and A be the time it took for Albert to complete the journey. Since distance equals rate times time, we have two equations, one for I and one for A:
I = (s/2)/8v + (s/2)/v = s/16v + s/2v
A = s/2v
Note that I exceeds A by s/16v. Albert will reach the hotel first.

Solution #2: The problem may be solved more easily with simple logic. If the bicycle is twice as fast as walking, the time it takes to bike the whole way is equal to the time it takes to walk half the way. So if the train's speed is anything shy of infinite, biking will still be faster.
200.Solution: You can buy any number of nuggets that is evenly divisible by three except for three just by using combinations of boxes of 6 and boxes of 9. (Use at most one box of 9, then multiples of boxes of 6.) If the number is not divisible by three, use a box of 20. If, after a box of 20 is purchased, the remaining number is divisible by three, you're all set. Otherwise, use a second box of 20. The remaining number will necessarily be divisible by 3, and you're all set. So the largest number that cannot be purchased would be one that requires two boxes of 20 before the remainder is reduced to a number divisible by three. Since three is the only number evenly divisible by three that cannot be purchased, the largest impossible number is 3 + 20 + 20 = 43.
201.Solution: Let x equal the number of marbles in a small bag, and y represent the number of marbles in a large bag. We know that the following equation holds:
7x + 18y = 233
Ordinarily, it is not possible to solve for two unknowns with a single equation; however, we also know that x and y are positive integers and that x is less than y. We can narrow the search by solving for the highest and lowest possible values for y:
Assume y = x:
7y + 18y = 233
25y = 233
y = 9.32
Since y must be an integer, y >= 10.
Assume y = 1:
7 + 18y = 233
18y = 226
y = 12.555
Since y must be an integer, y <= 12.
So the only possible values for y are 10, 11, and 12. Substituting each of these values into the original equation, we can find the orresponding values of x. Respectively, they are 7.57, 5, and 2.42. Since x must be an integer, the correct solution is x = 5 and y = 11. So the small bags of marbles contained five each, and the large bags contained eleven each.
202.Solution: For starters, the answer is not 1/2. When the coin is drawn, there are four possibilities, each of which is equally likely: 
      Coin Drawn                     Side Shown                        Other Side
    --------------------------------------------------------------------------------
      Double-Headed Coin               Heads                              Heads
      Double-Headed Coin               Heads (the other heads)            Heads
      Ordinary Coin                    Heads                              Tails
      Ordinary Coin                    Tails                              Heads
The problem tells us that the last possibility did not occur. Therefore, there are three remaining possibilities, each of which is equally likely. Of the three, two of the possibilities will show heads on the other side; only one will show tails on the other side. So the probability that the other side of the coin is heads is two thirds.
203.Solution: It should only take nine minutes to cook the egg. If you want to try to figure out how it is done in this short amount of time before seeing the answer, stop reading now. To start, flip both hourglasses over and put the egg in the water. When the four-minute hourglass runs out, flip it back over immediately. When the seven minute hourglass runs out, flip that back over immediately too. One minute later, the four-minute hourglass will run out again. At this point, flip the seven minute hourglass back over. The seven minute hourglass had only been running for a minute, so when it is flipped over again it will only run for a minute more before running out. When it does, exactly nine minutes will have passed, and the egg is done.
204.Solution: First figure out how much three of each time of stamp costs (63 cents). Now you need to figure out which two stamps to get a fourth of that will make the total cost some number of cents divisible by ten. You can't get enough to make 80 cents with two stamps, so the total cost must be 70 cents and the cost of the two extra stamps must be 7 cents. So the two extra stamps are a two-cent stamp and a five-cent stamp.
205.Solution: Let's break the question down into steps. We know that a boy and a half can eat a hot dog and a half in a minute and a half. So how many hot dogs could six boys eat in a minute and a half? We have the same amount of time, but four times as many boys, so the answer is four times as many hot dogs -- six, to be precise. But now let's consider what six boys could eat in six minutes. We now have four times as much time, so the answer is four times as many hot dogs -- specifically, 24.
206.Solution: The original number of coins must be a number such that you can subtract one and multiply by four fifths and get an integer. These numbers are 6, 11, 16, 21, 26, and so on.
But the pile remaining after the first pirate has taken his gold must also have this property. So the possibilities for the original number are 16, 36, 56, 76, 96, and so on.
The pile remaining after the second pirate has taken his gold must also have this property. So the possibilities for the original number are 76, 156, 236, 316, 396, and so on.
The pile remaining after the third pirate has taken his gold must also have this property. So the possibilities for the original number are 316, 636, 956, 1276, 1596, and so on.
The pile remaining after the fourth pirate has taken his gold must also have this property. The smallest possibility for this is 1276.
This number is the number of gold pieces in the chest the fourth pirate left behind (for the fifth pirate to divide). The fourth pirate hid a quarter of this number, plus one extra, just before the fifth pirate got there. So the third pirate left behind 1276 + 1276/4 + 1 = 1596 gold pieces.
The third pirate hid a quarter of this number, plus one extra, just before the fourth pirate got there. So the second pirate left behind 1596 + 1596/4 + 1 = 1996 gold pieces.
The second pirate hid a quarter of this number, plus one extra, just before the third pirate got there. So the first pirate left behind 1996 + 1996/4 + 1 = 2496 gold pieces.
The first pirate hid a quarter of this number, plus one extra. So the original number of coins must have been 2496 + 2496/4 + 1 = 3121 gold pieces.
207.Solution: 50 factorial includes, as factors, 10, 20, 30, 40, and 50. Therefore, the value of 50 factorial must end in at least five zeroes. The number given only ends in three zeroes. The correct value of 50 factorial is close to this, however. It's 30414093201713378043612608166064768844377641568960512000000000000.
208.Solution: 4 and 12.
209.Solution: A half-dollar, a quarter, and four dimes.
210.Solution: After ten sales of five apples, all the three-for-fifty-cents apples are sold; the remainder is still sold at five for a dollar when they should be sold at two for fifty cents.
211.Solution: No. Six dozen dozen dozen is 6 x 12 x 12 x 12. A half dozen dozen dozen is just 6 x 12 x 12.
212.Solution: Alligator breeder #1 had 49 alligators, and alligator breeder #2 had 35.
213.Solution: 99 99/99
214.Solution: Let S be the number of sheep and C be the number of chickens. So:
2S = C
5S + 3C = 99
We can rephrase the first equation thusly:
6S - 3C = 0
And then we can add this to the second equation, which yields:
11S = 99
By solving for S, we find that S equals 9. By substituting back in one of the original equations, we find that C equals 18. So there are nine sheep and eighteen chickens.
215.Solution: The circumference of a circle is 2 x pi x r, where r is the radius of the circle. If you want a rope that is one yard above the ground, this radius is larger by one yard. Let R be this new radius. So R = r + 1.
Let x be the amount of extra rope required by the eccentric's rival. So:
x = (2 x pi x (r + 1)) - (2 x pi x r) x = (2 x pi x r) + (2 x pi) - (2 x pi x r) x = 2 x pi
So x is about 6.2832 yards. Note that this answer does not depend on the radius of the circle. If the eccentric and his rival were attempting to tie up a baseball rather than the earth, the amount of additional required rope would be the same amount.
216.Solution: The head and tail are each three inches long; the rest is nine.
217.Solution #1: A half dollar, 39 dimes, and 60 pennies.

Solution #2: A dollar, a half dollar, 28 dimes, and 70 pennies.
218.Solution: 62.5 chocolates.
219. Answer: A Wheelbarrow.
220. Answer: A Mountain.
221. Answer: A Windmill.
222. Answer: One is hard up, while the other is soft down.
223. Answer: An Arrow.
224. Answer: Your Temper.
225. Answer: Smoke.
226. Answer: The letters I & S.
227. Answer: A Ball
228. Answer: I am my fathers son, so that mans father must be me. So that man must be my son.
229. Answer: A Bell
230. Answer: Thursday.
231. Answer: Too wise you are, Too wise you be; I see you are, Too wise for me.
232. Answer: Bob accidentally put his hands over the horse's eyes. If a horse can't see he will automatically stop.
233. Answer: The person slept on a waterbed. His killer used the scissors to cut the bed open and drown him.
234. Answer: Halfway, any farther and he would be running out
235. Answer: Wild Man Dave said: "You will boil me in water." The fairies were faced with a dilemma. If they boil him in water, that would make his statement true, which means he should have been fried in oil. They can only fry him in oil if he makes a true statement, but if they do, it would make his final statement false. The fairies had no way our of their situation so they were forced to set Wild Man Dave free.
236. Answer: King Tut was born in 20 B.C. There were 120 years between the birth of King Eros and the death of King Tut, but since their ages amounted to only 100 years, there must have been 20 years when neither existed. This would be a period between the death of King Eros, 40 B.C., and the birth of King Tut, 20 B.C.
237. Answer: At that moment, the time and day could be written as: 12:34, 5/6/78
238. Answer: Because they can't...Penguins live in Antarctica.
239. Answer: If he was born before 1783, then Massachusetts would still be a British colony.
240. Answer: The vowels. A, E, I, O and U.
241. Answer: Your mother.
242. Answer: If you only move half the distance, then you will always have half the distance remaining no matter how small the number
243. Answer: Inkstand
244. Answer: A Coconut.
245. Answer: Nothing!
246. Answer: The Letter E.
247. Answer: His plan is to dig the tunnel and pile up the dirt to climb up to the window to Escape
248. Answer: The horse operates a mill and travels in a circular clockwise direction. The two outside legs will travel a greater distance than the two inside legs.
249. Answer: A Clock
250. Answer: A Last Name
251. Answer: Sawdust
252. Answer: Sabrina had $50 and Samantha had $30
253. Answer: A Piano
254. Answer: Your name
255. Answer: One Thousand
256. Answer: YOUR FINGERNAILS
257. Answer: The Future
258. Answer: CHARCOAL
259. Answer: Footsteps.
260. Answer: A Turtle.
261. Answer: Salt.
262. Answer: A Violin.
263. Answer: The Office of the President of the United States.
264. Answer: deck of cards
265. Answer: Your hands
266. Answer: The letter R.
267. Answer: A Shadow.
268. Answer: Death.
269. Answer: Your own eyes.
270. Answer: Haste.
271. Answer: A Bar of Soap.
272. Answer: Memories.
273. Answer: A Dandelion.
274. Answer: A Doll.
275. Answer: Your fears.
276. Answer: The name of the ship.
277. Answer: A Telephone Book.
278. Answer: A Clock.
279. Answer: On a Map.
280. Answer: A Strawberry.
281. Answer: A Thimble.
282. Answer: A Book.
283. Answer: I'm their daughter.
284. Answer: A Rainbow
285. Answer: A Tree.
286. Answer: A See-Saw.
287. Answer: A Candle
288. Answer: A Butterfly
289. Answer: A needle and thread.
290. Answer: His Job. ( or work )
291. Answer: A Waterfall
292. Answer: Time
293. Answer: An Umbrella.
294. Answer: The Wind.
295. Answer: Cain (who killed Abel, leaving only himself, Adam and Eve).
296. Answer: Fleas
297. Answer: Milk.
298. Answer: BAG. All of the words in group A can begin with the word SAND
299. Answer: Tomorrow
300. Answer: The found a cabin of an airplane that had crashed there with 3 bodies in it.
301. Answer: To the right. It's always the same direction
302. Answer: No. This is a situation where you lose even if you win. Assuming the other person is being wise, they would take your $2 and say, "I lose", and give you $1 in return. You win the bet, but you're out $1
303. Answer: None. A pear tree does not bear plums.
304. Answer: Sunshine
305. Answer: It's cheaper to take two friends at the same time. In this case, you would only be buying three tickets, whereas if you take the same friend twice you are buying four tickets.
306. Answer: Nine Years.
307. Answer: It's the shortest sentence in the English language that users every letter of the alphabet.
308. Answer: 37 days. At the end of day one, the worm would be at the one foot mark. At the end of the 35th day the worm would be at the 35 foot mark. On the 36th day the worm travels from 35 feet to 39 feet but slips back to 36 feet at the end of the 36th day. On the 37th day, the worm travels up four feet from the 36th foot mark and is consequently out of the hole.
309. Answer: Because you need a camera...wooden legs don't take very good pictures
310. Answer: Butthead's solution is to rotate the three burgers. If the three burgers are 1,2 and 3 with sides A and B for each, the first five minutes sides 1A and 2A are cooked. The second burger is put aside and for the next five minutes, 1B and 3A are cooked. The first burger is now done and in the final five minutes, 2B and 3B are cooked.
311. Answer: The same reason seven dollars is more than six. Because there is one more.
312. Answer: She was a clergy woman.
313. Answer: She was born in the southern hemisphere.
314. Answer: Which is heavier, a pound of feathers or a pound of gold? A pound of feathers. Some would say a pound is a pound, but the fact is: Gold is a precious metal and is therefore weighed in the Troy system of measurement. This means that a pound of gold weighs only 12 oz and a pound of feathers weighs 16 oz.
315. Answer: It should be written as $13,212
316. Answer: Nine rungs will be above the water. As the tide rises, so will the ship and the ladder.
317. Answer: 1 - $50 bill, 1 - $5 bill, 4 - $2 bills
318. Answer: The Answer: is that one girl took the basket. She took the last apple while it was in the basket.
319. Answer: A window.
320. Answer: Sally's mother slid a newspaper under a door and made Sally stand on one side of the door and her brother on the other.
321. Answer: Two hours. Alicia can complete 1/3 of the job in one hour and Mark can complete 1/6 of the job in one hour; therefore, together they can complete 1/3 plus 1/6 or 1/2 of the job in one hour. Consequently, the entire job can be completed in just two hours. Or, you can figure it by saying (a x b)/(a b)=time spent.
322. Answer: I am the pupil of an eye.
323. Answer: They were the same man. Grover Cleveland served two terms as president of the United States, but the terms were not consecutive.
324. Answer: None. You will not find unlisted phone numbers in a phone directory.
325. Answer: He put a hole in the barrel to make it weigh less.
326. Answer: Australia was always the biggest island in the world, even before it was discovered.
327. Answer: Oil
328. Answer: The two men were partners playing doubles.
329. Answer: I'm Knowledge
330. Answer: Darkness
331. Answer: A Telephone
332. Answer: He was "walking" and not driving his cab
333. Answer: A SHOE
334. Answer: BLACK BOARD OR CHALK BOARD
335. Answer: TOSS IT IN THE AIR
336. Answer:
ARE YOU SLEEPING
ARE YOU DEAD
ARE YOU ME
WHAT IS THE OPPISITE OF YES