    Mathematical Puzzles  Puzzle 1 Which square, which circle, and which triangle has the closest area to the doughnut shape on the left? The drawings are to scale, so you might be able to judge it, as well as working the actual areas out. [Puzzle Code = ZKRC] Puzzle Copyright © Kevin Stone    Direct Link: www.brainbashers.com?ZKRC Hint: The area of a circle is π x Radius2. Answer: Circle: 3.5. Square: 3.1. Triangle: 4.7. Doughnut The area of a circle is π x Radius2. The larger circle has diameter = 4, therefore the radius is 2, and the area is π x 22 = 4π. The smaller circle has diameter = 2, therefore the radius is 1, and the area is π x 12 = π. Therefore the shaded area is 4π - π = 3π ≈ 9.42. Circle The area of a circle is π x Radius2. The circle with diameter 3.1 has a radius of 1.55 and an area of π x 1.552 = 7.55. The circle with diameter 3.5 has a radius of 1.75 and an area of π x 1.752 = 9.62 (** closest match). The circle with diameter 3.9 has a radius of 1.95 and an area of π x 1.952 = 11.95. Square The area of a square is Side x Side. The square with side 3.1 has an area of 3.1 x 3.1 = 9.61 (** closest match). The square with side 3.5 has an area of 3.5 x 3.5 = 12.25. The square with side 3.9 has an area of 3.9 x 3.9 = 15.21. Triangle The area of a triangle is ½ x Base x Height. Using Pythagoras' theorem it can be shown that the area of an equilateral triangle is Sqrt(3) x Base2 ÷ 4. The triangle with side 3.9 has an area of Sqrt(3) x 3.9 x 3.9 ÷ 4 = 6.59. The triangle with side 4.3 has an area of Sqrt(3) x 4.3 x 4.3 ÷ 4 = 8.01. The triangle with side 4.7 has an area of Sqrt(3) x 4.7 x 4.7 ÷ 4 = 9.57 (** closest match). Puzzle 2 Starting in the bottom left corner and moving either up or right, one square at a time, adding up the numbers along the way, what is the largest sum which can be made once you have reached the top right corner? [Puzzle Code = ZNGR] Puzzle Copyright © Kevin Stone    Direct Link: www.brainbashers.com?ZNGR Hint: The answer is over 35. Answer: 37.  Puzzle 3 At the local sweet shop, three particularly nice sweets are on special offer. A Sparkle, a Wibbler and a Nobbler together cost 40p. A Nobbler is over three times the price of a Sparkle. Six Sparkles are worth more than a Wibbler. A Nobbler, plus two Sparkles costs less than a Wibbler. Can you determine the price of each type of sweet? [Puzzle Code = ZQVP] Puzzle Copyright © Kevin Stone    Direct Link: www.brainbashers.com?ZQVP Hint: Which sweet is the least expensive? Answer: Sparkle =  4p Wibbler = 23p Nobbler = 13p If we label the clues. A Nobbler is over three times the price of a Sparkle.      Six Sparkles are worth more than a Wibbler.                A Nobbler, plus two Sparkles costs less than a Wibbler.    A Sparkle, a Wibbler and a Nobbler together cost 40p.      By  a Nobbler, plus two Sparkles costs less than a Wibbler, a Wibbler must be the most expensive sweet. By  a Nobbler is over three times the price of a Sparkle, a Sparkle is cheaper than a Nobbler and hence the cheapest sweet. So the order of sweets, from the least to most expensive, is Sparkle, Nobbler, Wibbler. If a Sparkle was 1p, by  a Wibbler could only be up to 5p, by  a Nobbler would cost at least 34p, which is more than a Wibbler and isn't allowed as the Wibbler is the most expensive sweet. If a Sparkle was 2p, by  a Wibbler could only be up to 11p, by  a Nobbler would cost at least 27p, which is more than a Wibbler and isn't allowed as the Wibbler is the most expensive sweet. If a Sparkle was 3p, by  a Wibbler could only be up to 17p, by  a Nobbler would cost at least 20p, which is more than a Wibbler and isn't allowed as the Wibbler is the most expensive sweet. So a Sparkle must be at least 4p. If a Sparkle was 4p, by  a Nobbler must be at least 13p, by  a Wibbler would cost 23p - this combination matches all of the clues and is a possible solution. If a Sparkle was 4p and a Nobbler 14p, by  a Wibbler would cost 22p. This would not satisfy . And if we increase the price of a Nobbler,  is never satisfied. If a Sparkle was 5p, by  a Nobbler must be at least 16p, by  making a Wibbler at most 19p. This would not satisfy . If we increase the price of a Sparkle or Nobbler further,  is will never be satisfied. Therefore the only solution we came across must be the correct one. Puzzle 4 A numismatist decides to divide their coin collection between their children. The eldest gets 1/2 of the collection, the next gets 1/4, the next gets 1/5, and the youngest gets the remaining 49 coins. How many coins are in the collection? [Puzzle Code = ZXNW] Puzzle Copyright © Kevin Stone    Direct Link: www.brainbashers.com?ZXNW Hint: What do the fractions add to? Answer: There are 980 coins in the collection. To check this we add:    1÷2 + 1÷4 + 1÷5 + 49 this works out as    490 + 245 + 196 + 49 = 980, as required. To work the answer out from the question, we observe that: 1÷2 + 1÷4 + 1÷5 = 19÷20 and we know that there were 49 coins left, so 1÷20 of the collection is 49 coins. Which means that the entire collection is 20 x 49 = 980 coins.      