Wednesday, November 4, 2015

The 50 cent exam question



The 50 cent exam question
PHOTO: The 50 cent exam question
Photo: Facebook
By Mark Molloy, 03 November 2015 at 9:34AM GMT
https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEh_GfjH9Q1sQUF5TW1W_6eWLp5lUjAqYgochKVITRA7foprHpSs9H6mrEQFZWTY5xiitP_vAnZIftCT6vjaWsK6sIsAsPt78q8cCaRiciNblCnF1sqhAIOzTYUXrOrPtWrxet9QVwtf5L4/s1600/Screen_Shot_2015-1_3490454b.jpg
http://i.telegraph.co.uk/multimedia/archive/03490/Screen_Shot_2015-1_3490454b.jpg
http://www.telegraph.co.uk/education/11971692/Can-you-solve-the-50-cent-maths-exam-question-that-is-dividing-the-internet.html?utm_campaign=Echobox&utm_medium=Social&utm_source=Facebook


Can you solve the 50 cent maths exam question that is dividing the internet?
High school students have been complaining that the 50 cent piece exam question is too difficult.

A maths exam question for high school students about a 50 cent piece has created a debate online after complaints it was too difficult.

Australian students sat the VCE Further Maths exam which posed the controversial question on November 30 and now adults are having a go at answering it online.

 "A 50 cent coin has 12 sides of equal length. Two 50 cent coins are balanced next to each other on a table so they meet along one edge," the question reads.

Students were then asked to work out the degree of the angle between the coins and given multiple choice options ranging from 12 to 72 degrees. Scroll down for the answer.



From https://www.youtube.com/embed/gY7BKdNxjeQ


Daphne Fourniotis, a student at Shelford Girls’ Grammar in Melbourne, Victoria, said she found the question "quite different to the ones we had been doing".

"I did find it challenging … under the stress of the exam," she explained, "I found it a little ambiguous and confusing. I think I was overthinking it.

"You had to read between the lines and analyse the question, people would have jumped at that question too soon."
By Mark Molloy (Telegraph Media Group), 03 November 2015 at 9:34AM GMT


Alternative answer by lionel17 on 04 November 2015 at 01:16 AM

A polygon with n number of sides
PHOTO: A polygon with n number of sides
https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEi5PlVn6MNo7DqqR7rLPkskroosnQkbi3gj18fKBj89GLMZgKcLQNw0nhiuC_67WCkKqXZ0tUCxFTDVqLuRNJHkd_xON7IU6OTpZYTuMEdP0DYqbEb7svG8oEbclKSSDCJijyeiidJtH3c/s1600/pentri.gif
http://www.regentsprep.org/regents/math/geometry/gg3/pentri.gif
www.regentsprep.org/regents/math/geometry/gg3/lpoly1.htm


Sum of Interior Angles of a Polygon
The number of triangles formed will be 2 less than the number of sides. This pattern is constant for all polygons.

Representing the number of sides of a polygon as n, the number of triangles formed is (n - 2). Since each triangle contains 180°, the sum of the interior angles of a polygon is 180(n - 2).

Step 1 : Use the theory "Sum of Interior Angles of a Polygon"
Sum of Interior Angles of the 12 sides 50¢ coin = 180(n - 2) (where n = number of sides) = 180 (12-2) = 1800°

Step 2 : Find each Interior Angle of the Polygon. Since all sides are equal = 1800/12 = 150°.

Step 3 : Since the 2 coins Interior Angles + the gap = 360°. Therefore 360° - (2*150°) = 60°.

Answered by lionel17 on 04 November 2015 at 01:16 AM



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