# 2013 Wep Mathematics Challenge

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Quizzes Created: 1 | Total Attempts: 97
Questions: 10 | Attempts: 97

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Welcome to the 2013 WEP Mathematics Challenge. Just answer as many of the 10 questions as you can, as fast as you can, within two weeks of the start of the Challenge at 12 noon on Sunday, 13 January 2013. If you want to skip a question just enter 0 as an answer (since the software will not accept your submission unless all questions have been answered). You will have the opportunity to review all your answers at the end before submitting them. See rules and further information below. First prize is an i-Pad 3 (64GB Read moreWiFi + 3G) and there are five i-Pods (4th generation 32GB) for five runners up.

• 1.

### You have 12 balls which look identical, but one of them is either slightly heavier or slightly lighter than the others. You have a set of scales, but no weights, and you are able to weigh any selection of the 12 balls against any selection of the remaining balls. You can also mark individual balls to keep track of them without affecting their weight. What is the minimum number of weighings you need to perform to ensure that you can identify the odd ball and determine whether it is heavier or lighter than the others.

Explanation
In order to identify the odd ball and determine its weight, we can use the strategy of dividing the 12 balls into three groups of 4. Weigh any two of these groups against each other. If the scales balance, then the odd ball is in the third group. In the second weighing, take three balls from the third group and weigh two of them against each other. If they balance, then the odd ball is the remaining one. If they don't balance, then the odd ball is the lighter one. If the scales don't balance in the first weighing, then the odd ball is in the heavier group. In the second weighing, take three balls from the heavier group and weigh two of them against each other. If they balance, then the odd ball is the remaining one. If they don't balance, then the odd ball is the heavier one. Therefore, a minimum of 3 weighings is needed to identify the odd ball and determine its weight.

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• 2.

### You are presented with four boxes. One of them contains a prize, the other three contain nothing. Dr Know invites you to guess which box contains the prize and if you guess correctly you win the prize. You make your selection, but before opening the box Dr Know (who knows which box contains the prize) offers you a further option. He opens two of the three other boxes and shows you that they are both empty. He then offers you the option of changing your mind and opting for the fourth box instead of your initial choice. What is the probability that you will win the prize if you accept his offer and now opt to open the fourth box?

Explanation
When you initially selected a box, there was a 1/4 (25%) chance that it contained the prize. After Dr. Know opens two empty boxes, the remaining unopened box has a 3/4 (75%) chance of containing the prize. This is because the initial 1/4 probability gets redistributed among the three remaining boxes, making each box now have a 1/3 chance of containing the prize. Therefore, if you switch your choice to the fourth box, you have a 3/4 (75%) chance of winning the prize.

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• 3.

### A regular 5-point star is inscribed in another regular 5-point star, as shown. What, in terms of the golden ratio Φ = ½ (1+√5), is the ratio of the area of the smaller star to the area of the larger star?

Explanation
The ratio of the area of the smaller star to the area of the larger star is 1/Φ^4, which is equivalent to 1/(Φ^4) or Φ^(-4). This can be explained by considering the relationship between the side lengths of the stars. The golden ratio is often found in geometric shapes, and in this case, it relates the side lengths of the larger and smaller star. The ratio of the side lengths is Φ, so the ratio of the areas is Φ^2. Since the area is a two-dimensional measure, we square the golden ratio to get Φ^2. To find the ratio of the smaller star to the larger star, we square Φ^2, which gives us Φ^4. Therefore, the correct answer is 1/Φ^4.

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• 4.

### A chocolate orange consists of a sphere of smooth uniform chocolate.  It has mass M and radius r, sliced into equal wedges by planes through its vertical axis. It stands on a horizontal table held together by a narrow weightless ribbon round its equator. Given that the centre of mass of a segment of angle θ lies at a distance of [3πr sin(θ/2)]/(8θ) from the vertical axis, what is the minimum tension in the ribbon needed to hold the chocolate orange together in a gravitational field of strength G.

Explanation
The minimum tension in the ribbon needed to hold the chocolate orange together is (3/32)MG. This can be determined by considering the forces acting on the chocolate orange. The weight of the chocolate orange, which is equal to its mass M multiplied by the gravitational field strength G, acts vertically downwards. The tension in the ribbon acts horizontally and prevents the chocolate orange from falling apart. By analyzing the forces and moments acting on the chocolate orange, it can be determined that the minimum tension in the ribbon needed is (3/32)MG.

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• 5.

### N beetles (points) are located at the corners of a regular polygon with n sides each of length L. Each beetle crawls directly towards the next beetle, all at the same speed and all crawling anti-clockwise with respect to the center of the polygon. They each travel a distance s before meeting. Another beetle starting at a vertex of the polygon and traveling at the same speed as all the others walks in a straight line (starting along one of the sides) and also covers a distance s. If the lone beetle walking in a straight line starts at point A and ends at point T, and if the center of the polygon is point O, what is the angle AOT in radians?

Explanation
The angle AOT in radians is π/2 or pi/2 or (pi)/2 or ½π or ½ π.

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• 6.

• 7.

### A cube has edges of unit length. A laser cuts the largest possible hole with square cross-section through the cube, without splitting the cube into pieces. What is the size of the largest possible cube that can pass through the hole (ie what is the length of this cube’s sides)?

Explanation
The size of the largest possible cube that can pass through the hole is (3/4)√2. This can be determined by considering that the diagonal of the square cross-section of the hole is equal to the space diagonal of the cube. Using the Pythagorean theorem, the diagonal of the square is √2 times the length of its sides. Therefore, the length of the sides of the largest possible cube is (3/4)√2.

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• 8.

### A round hole is drilled through the center of a sphere. Sitting on a flat surface with the hole vertical, the resulting shape has height h. What is its volume in terms of h.

Explanation
The volume of a sphere with radius r is given by V = (4/3)πr^3. In this case, the sphere has a hole drilled through the center, resulting in a shape with height h. Since the hole is drilled through the center, the radius of the resulting shape is also h. Therefore, the volume of the resulting shape can be calculated as V = (4/3)πh^3. Simplifying this expression gives V = (1/6)πh^3, which matches the given answer choices.

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• 9.

### Consider a square S0 with sides of length L. Now consider a figure S1 created by adding squares with areas of 1/9 of S0 externally and centrally to each side of S0For n>0, the shape Sn is created by adding squares with areas of 1/9n of S0 centrally and externally to each side of Sn-1. (for n ε Z). If An is the area of Sn, what is Lim An as n approaches infinity.

Explanation
The given answer options all represent the same value, which is 2 times the area of the original square S0, represented as 2L^2. This can be determined by observing the pattern in the construction of the figures Sn. As n approaches infinity, the added squares become increasingly smaller in comparison to the original square S0. Therefore, their total area becomes negligible compared to the area of S0, resulting in the limit of 2L^2 as the final answer.

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• 10.

### A  relay runner drops her baton on the straight part of an athletics track. It lands in a random position on the track. The running lanes are all of equal width apart and the baton’s length is equal to exactly half this width. Assuming you can ignore the widths of the baton and of the lane dividers, what is the probability that the baton crosses one of the lane dividers

Explanation
The probability that the baton crosses one of the lane dividers can be calculated by considering the length of the track and the length of the baton. Since the baton's length is equal to exactly half the width of the lanes, if the baton lands anywhere within the distance of one lane width from the start or end of the track, it will cross one of the lane dividers. The total length of the track is equal to the width of one lane multiplied by the number of lanes. Therefore, the probability is equal to the length of one lane divided by the total length of the track, which is 1/π.

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• Mar 17, 2023
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• Nov 23, 2012
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