Riddle #474

From, left to right: #1 Mrs Jennifer - blue marbles #2 Mrs Sharon - red marbles #3 Mr Cruz - white marbles #4 Mr Danny - green marbles If we separate and label the clues, and label the houses #1, #2, #3, #4 from left to right we can see that: a. Mrs Jennifer's house is somewhere to the left of the green marbles one. b. The third one along is white marbles. c. Mrs Sharon owns a red marbles house d. Mr Cruz does not live at either end. e. Mr Cruz lives somewhere to the right of the blue marbles house. f. Mr Danny lives in the fourth house g. The first house is not made from red marbles. By (g) #1 isn't made from red marbles, and by (b) nor is #3. By (f) Mr Danny lives in #4 therefore by (c) #2 must be red marbles, and Mrs Sharon lives there. Therefore by (d) Mr Cruz must live in #3, which, by (b) is the white marbles house. By (a) #4 must be green marbles (otherwise Mrs Jennifer couldn't be to its left) and by (f) Mr Danny lives there. Which leaves Mrs Jennifer, living in #1, the blue marbles house.

Devour, rumour and contour.

First of all, the brute-force approach does not work. If the Camel starts by picking up the 1000 bananas and try to reach point B, then he will eat up all the 1000 bananas on the way and there will be no bananas left for him to return to point A. So we have to take an approach that the Camel drops the bananas in between and then returns to point A to pick up bananas again. Since there are 3000 bananas and the Camel can only carry 1000 bananas, he will have to make 3 trips to carry them all to any point in between. When bananas are reduced to 2000 then the Camel can shift them to another point in 2 trips and when the number of bananas left are <= 1000, then he should not return and only move forward. In the first part, P1, to shift the bananas by 1Km, the Camel will have to Move forward with 1000 bananas – Will eat up 1 banana in the way forward Leave 998 banana after 1 km and return with 1 banana – will eat up 1 banana in the way back Pick up the next 1000 bananas and move forward – Will eat up 1 banana in the way forward Leave 998 banana after 1 km and return with 1 banana – will eat up 1 banana in the way back Will carry the last 1000 bananas from point a and move forward – will eat up 1 banana Note: After point 5 the Camel does not need to return to point A again. So to shift 3000 bananas by 1km, the Camel will eat up 5 bananas. After moving to 200 km the Camel would have eaten up 1000 bananas and is now left with 2000 bananas. Now in the Part P2, the Camel needs to do the following to shift the Bananas by 1km. Move forward with 1000 bananas – Will eat up 1 banana in the way forward Leave 998 banana after 1 km and return with 1 banana – will eat up this 1 banana in the way back Pick up the next 1000 bananas and move forward – Will eat up 1 banana in the way forward Note: After point 3 the Camel does not need to return to the starting point of P2. So to shift 2000 bananas by 1km, the Camel will eat up 3 bananas. After moving to 333 km the camel would have eaten up 1000 bananas and is now left with the last 1000 bananas. The Camel will actually be able to cover 333.33 km, I have ignored the decimal part because it will not make a difference in this example. Hence the length of part P2 is 333 Km. Now, for the last part, P3, the Camel only has to move forward. He has already covered 533 (200+333) out of 1000 km in Parts P1 & P2. Now he has to cover only 467 km and he has 1000 bananas. He will eat up 467 bananas on the way forward, and at point B the Camel will be left with only 533 Bananas.

Assume there are N people at the party. Note that the least number of people that someone could shake hands with is 0, and the most someone could shake hands with is N-1 (which would mean that they shook hands with every other person). Now, if everyone at the party really were to have shaken hands with a different number of people, then that means somone must have shaken hands with 0 people, someone must have shaken hands with 1 person, and so on, all the way up to someone who must have shaken hands with N-1 people. This is the only possible scenario, since there are N people at the party and N different numbers of possible people to shake hands with (all the numbers between 0 and N-1 inclusive). But this situation isn't possible, because there can't be both a person who shook hands with 0 people (call him Person 0) and a person who shook hands with N-1 people (call him Person N-1). This is because Person 0 shook hands with nobody (and thus didn't shake hands with Person N-1), but Person N-1 shook hands with everybody (and thus did shake hands with Person 0). This is clearly a contradiction, and thus two of the people at the party must have shaken hands with the same number of people. Pretend there were only 2 guests at the party. Then try 3, and 4, and so on. This should help you think about the problem. Search: Pigeonhole principle

The match.

None, the boat rises with the tide.

A coffin.

Flip the first switch and keep it flipped for five minutes. Then unflip it, and flip the second switch. Go into the room. If the lightbulb is off but warm, the first switch controls it. If the light is on, the second switch controls it. If the light is off and cool, the third switch controls it.

N for November.

pi * z * z * a

Just once, because after you subtract anything from it, it's not 25 anymore.