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Five pumpkins are weighed two at a time in all ten sets of two. The weights are recorded as 16, 18,19, 20, 21, 22, 23, 24, 26, and 27 pounds. All individual weights are also integers. How much does each pumpkin weigh?

Two glasses, one nearly filled with liquid A, one with liquid B. Take a teaspoon of A and put it in B, stir and put a teaspoon of the mixture back in A. Which now is the purest??

So, there are 5 pumpkins - lets call them a, b, c, d and e, and assume they are in ascending weight order.And they're weighed in pairs:a+b, a+c, a+d, a+e, b+c, b+d, b+e, c+d, c+e and d+eThis means that each pumpkin was weighed 4 times, and so the total of all the weighings/4 is the total weight of all 5 pumpkins16+18+19+20+21+22+23+24+26+27 = 216 ... 216/4 = 54lbsNow, if a and b are the smallest, then a+b = 16lbsIf d and e are the largest, then d+e = 27 lbsSo 54lb - (a+b)+(d+e) = c54 - (16+27) = 11So pumpkin c weighs 11lb.And from there, it should be relatively easy to work out all the other weights:a+b = 16a+c = 18 (and therefore a= 7lb, and b=9lb)a+d = 19 (and therefore d= 12lb)...d+e = 27 (and therefore e=15lb)So:a= 7lbb= 9lbc= 11lbd= 12lbe= 15lb

The Three SuspectsKyle, Neal, and Grant were rounded up by their mother yesterday, because one of them was suspected of having grabbed a few too many cookies from the cookie jar. The three brothers made the following statements under very intensive questioning:• Kyle: I'm innocent.• Neal: I'm innocent.• Grant: Neal is the guilty one.

Three switches. If the lamps are of a type which gets warm...............One on.One off.One on for a minute or so.Could even do 4.. One on for several miniutes.. One on for 30 secs.. Bit like when detectives want to know if a car has been driven very recently.. The engine and exhaust will be hot.

Are you allowed to pull out the bathplug?Did he lock up the visitor?If he did, he would soon have his institution full of visitors

Here is another east oneA man is trapped in a room. The room has only two possible exits: two doors. Through the first door there is a room constructed from magnifying glass. The blazing hot sun instantly fries anything or anyone that enters. Through the second door there is a fire-breathing dragon. How does the man escape?

Another more based on astronomyYou awake inside a small transparent capsule sitting on the surface of Venus. From a small speaker you hear a voice that says, "We will leave you here either for a day or a year. If you choose to stay a day, we will give you $1 million. If you choose to stay a year, we will give you $2 million. Either way, you will have sufficient food and water. We will make sure the temperature is a constant 70 degrees Fahrenheit. We will also supply cable TV."What is your choice? (Don't let money decide your answer).

Lightarrow,Lightarrow,I like you approach to this not so simple little problemHard SolutionIt is intuitive that at all times the dogs will form the corners of a square, ever decreasing in size and rotating about the center. Let C be the center of the square at point (0,0). Let the dogs be at initial points (.5,.5), (.5,-.5), (-.5,-.5), and (-.5,.5). Let dog 1 bet at (.5,-.5) who is chasing dog 2 at (.5,.5). Consider the line from C to dog 1. Next consider the line from dog 1 to dog 2. The angle between these lines is 135 degrees. Theorem: Let C be a fixed point, D a point on a curve, r the distance from C to D, x the polar angle formed by r, and y the angle formed by the angle between CD and the tangent line to the curve at point D. Then tan(y)=r/(dr/dx).

So the line of sight from the center to the initial point of dog 1 would be be at a 270 degree angle. The line of sight from dog 1 to dog 2 would be a vertical line of 90 degrees. The angle between these lines is 270 degrees. The tangent of 270 is -1. So we have r/(dr/dx) = -1. Thus r(x)=c*e-x. At the moment the dogs start walking x equals 0. At this moment the distance from the center to any of the dogs is 2-1/2. So r(x)=2-1/2*e-x. The arc length is given by the formula: Integral from 0 to infinity of ((f(x))2+(f'(x))2)1/2

= Integral from 0 to infinity of ((2-1/2*e-x)2 + (-2-1/2*e-x)2)1/2 = Integral from 0 to infinity of (2-1*e-2x + 2-1*e-2x)1/2 = Integral from 0 to infinity of e-x = -e-x from 0 to infinity = =0 - (-e0) = 0 -(-1) = 1 The path the dogs take is called a logarithmic spiral. It is an interesting paradox that the dogs will make an infinite number of circles, yet the total distance is constant. Every revolution the size of the square will to e-2*pi/21/2 = 0.03055663 times it's size before the revolution. Easy SolutionDavid Wilson suggested the more simple solution which follows... Suppose dog A is pursuing dog B who is pursuing dog C. During the entire pursuit, the dogs remain at the corners of a square, and angle ABC is a constant 90 degrees. That is, B's path towards C is always perpendicular to A's path towards B, and B has zero velocity in the direction along A's path. Since B neither approaches nor recedes from A during the walk, A simply covers the intial separation of 1.

Antonio Molins went further to state that as dog A is heading direcctly towards dog B, dog B is moving towards from dog A at rate equal to cosine of any of the interior angles of the shape made by the dogs. He also provided the following graphics. For example consider a pentagon. Remember that for a n-gon of n sides each angle will be 180*(1-(2/n)). So a pentagon will have angles of 108°. Let's assume each dog walks at a rate of 1 unit per minute, where the initial shape has sides of 1 unit. As pointed out by David Wilson we can forget about the spiral paths but instead think of the dogs walking in straight lines. In the case of the penagon for every one unit dog A draw towards dog B's initial position, dog B will draw cos(180°)=-0.309

units towards dog A. The question is, how long will it take for them to meet? Let the answer to that question be t. In t seconds dog A will cover t units, and dog B will cover .309*t units. The total distance traveled by A equals the distance traveled by B plus 1 (the initial separation). So solving for t: t=0.309+1 --> t=1/(1-.309) --> t=1.4472. What do you think?

BoardChemist,QuoteI don't know how to hide an answer, but I doubt this is the answer being looked for."" to the mind guess"" of prediction test in "blue letters", This is post separate from the horse hay one. QuoteThe puzzle states that the hay is tied down. It does not state that the stake is driven into the ground so maybe the horse just walks over to the hay dragging the stake.(Though my first thought was that he chews through, or breaks the rope.)Yessssssss!! The above is the correct answer a little silly I know but many people most often just presume the skate is firmly hamered into the groundRegardsAlan

I don't know how to hide an answer, but I doubt this is the answer being looked for

The puzzle states that the hay is tied down. It does not state that the stake is driven into the ground so maybe the horse just walks over to the hay dragging the stake.(Though my first thought

The horse is standing at the extreme length of the rope, five meters from the stake. The horse is facing the stake, and five meters further away in the same direction is the hay. The horse walks the ten meters across the 'circle' and eats the hay

I suppose it could have been a sawing horse - so no problem, again.

LightArrowI am impressed by your solutions to these difficult problems!I could give you the link by PM, your maths is better than mine.perhaps you could then post some math problems from this siteRegardsAlan

Easy SolutionDavid Wilson suggested the more simple solution which follows...Suppose dog A is pursuing dog B who is pursuing dog C. During the entire pursuit, the dogs remain at the corners of a square, and angle ABC is a constant 90 degrees. That is, B's path towards C is always perpendicular to A's path towards B, and B has zero velocity in the direction along A's path. Since B neither approaches nor recedes from A during the walk, A simply covers the intial separation of 1.

Are you sure there's nothing else you need to tell us?Why couldn't you drill a 6cm diameter hole through the Earth or through a football?

I think this is the problem lightarrow's thinking of:Old Boniface he took his cheer,Then he bored a hole through a solid sphere,Clear through the center, straight and strong,And the hole was just six inches long.Now tell me, when the end was gained,What volume in the sphere remained?Sounds like I haven't told enough,But I have, and the answer isn't tough!This can be worked out mathematically or logically (I solved it via logic).

So we can assume the answer is always the same so an infinitely thin drill through a 6cm diameter sphere would leave 4XPiX3^{3}/4 material behind - i.e. taking nothing out of the sphere.Proving that it works for a range of sphere sizes is a bit harder - I will try it sometime. The question implies that original sphere could be ANY radius greater than 3cm. Wierd.

Hold on, you guys have confused me...If you were to drill a hole through a sphere, right through the centre, and the hole was 6cm long, then the sphere must have a diameter of 6cm , no?

I decided to post the solution, ...The Hardest Logic Puzzle Ever ...

A sample of 100 kg of potatoes has a water content of 99%. It is put under the sun to dry a little, and the water content at the end becomes 98%. How does the potatoes weight now?

[quote author=Alan McDougall link=topic=15562.msg188243#msg188243 date=1217501149Ehm...A sample of 100 kg of potatoes has a water content of 99%. It is put under the sun to dry a little, and the water content at the end becomes 98%. How does the potatoes weight now?