You turn 2 switches "on" and leave 1 switch "off" and wait about a minute. Then enter the room, but just before you enter, turn one switch from "on" to "off". Once in the room, feel the lightbulb - if it is warm, but off, it has to be the last switch you turned off. If it is on, it has to be the switch left on. If it is cold and is off, it has to be the switch you left in the off position.
One thing Julie could do is freeze a tray of ice-cubes, and turn the tray of ice upside down in her freezer. When she comes home, she should check the tray. If the ice cubes are still in the tray, the food is safe to eat. If the trays are empty, it's time to clean out the freezer. She will have to make a judgment call if the ice-cubes are only slightly thawed.
At the start, 1inch of the yellow pencil gets smeared with wet paint. As the blue pencil is moved downward, a second inch of the blue pencils smears a second inch of the yellow pencil.
Each pair of down and up movesof the blue pencil smears 1 more inch of each pencil. 5 pairs of moves will smear 5 inches. This together with the initial inch, makes 6 inches for each pencil.
Fill the 5 oz. and 11 oz. Containers from the 24 oz. container. This leaves 8 oz. in the 24 oz. bottle. Next empty the 11 oz. bottle by pouring the contents into the 13 oz. bottle. Fill the 13 oz. bottle from the 5 oz. container (with 2 oz.) and put the remaining 3 oz. in the 11 oz. bottle. This leaves the 5 oz. container empty. Now pour 5 oz. from the 13 oz. bottle into the 5 oz. bottle leaving 8 oz. in the 13 oz. bottle. Finally pour the 5 oz. bottle contents into the 11 oz. bottle giving 8 oz. in this container.
Most people seem to think that the thing to do is weight six coins against six coins, but if you think about it, this would yield you no information concerning the whereabouts of the only different coin. As we already know that one side will be heavier than the other. So that the following plan can be followed, let us number the coins from 1 to 12. For the first weighing let us put on the left pan coins 1,2,3,4 and on the right pan coins 5,6,7,8. There are two possibilities. Either they balance, or they don't. If they balance, then the different coin is in the group 9,10,11,12. So for our second weighing we would put 1,2 in the left pan and 9,10 on the right. If these balance then the different coin is either 11 or 12. Weigh coin 1 against 11. If they balance, the different coin is number 12. If they do not balance, then 11 is the different coin. If 1,2 vs 9,10 do not balance, then the different coin is either 9 or 10. Again, weigh 1 against 9. If they balance, the different coin is number 10, otherwise it is number 9. That was the easy part. What if the first weighing 1,2,3,4 vs 5,6,7,8 does not balance? Then any one of these coins could be the different coin. Now, in order to proceed, we must keep track of which side is heavy for each of the following weighings. Suppose that 5,6,7,8 is the heavy side. We now weigh 1,5,6 against 2,7,8. If they balance, then the different coin is either 3 or 4. Weigh 4 against 9, a known good coin. If they balance then the different coin is 3, otherwise it is 4. Now, if 1,5,6 vs 2,7,8 does not balance, and 2,7,8 is the heavy side, then either 7 or 8 is a different, heavy coin, or 1 is a different, light coin. For the third weighing, weigh 7 against 8. Whichever side is heavy is the different coin. If they balance, then 1 is the different coin. Should the weighing of 1,5, 6 vs 2,7,8 show 1,5,6 to be the heavy side, then either 5 or 6 is a different heavy coin or 2 is a light different coin. Weigh 5 against 6. The heavier one is the different coin. If they balance, then 2 is a different light coin.
The answer is: a dice. An explanation: "It's always 1 to 6": the numbers on the faces of the dice, "it's always 15 to 20": the sum of the exposed faces when the dice comes to rest after being thrown, "it's always 5": the number of exposed faces when the dice is at rest, "but it's never 21": the sum of the exposed faces is never 21 when the dice is at rest, "unless it's flying": the sum of all exposed faces when the dice is flying is 21 (1 + 2 + 3 + 4 + 5 + 6).
The difference between the real time and the time of the mirror image is two hours and ten minutes (two and a half hours, minus the twenty minutes of cycling). Therefore, the original time on the clock at home that morning could only have been five minutes past seven: The difference between these clocks is exactly 2 hours and ten minutes (note that also five minutes past one can be mirrored in a similar way, but this is not in the morning!). Conclusion: The boy reaches school at five minutes past seven plus twenty minutes of cycling, which is twenty-five minutes past seven!...
He works in the engine room of a liner! To get to work, he walks along the decks from his cabin facing forwards, and down the ladders between decks facing backwards. However, when he finishes, he only needs to face forwards to climb the ladders again and walk along the deck back to his cabin.
Mary is not a nurse. The way to solve this riddle, is to consider statements 4, 5, and 6 and create a chart of all possible true and false answers. Next, fill in the chart according to statements 1 through 3. You will discover that there is only one line where only one of statements one, two and three are true. Thus, it is determined that: Statement 4 and 5 are false and statement 6 is true.