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Saturday 15 June 2013

Singapore Math Olympiad (Senior) 2013

  1. A shop sells two kind of products A and B. One day a salesman sold both A and B at the same price, $2100 to a customer. Suppose A makes a profit of 20% and B makes a loss of 20%. Then the deal
    (A) make a profit of $70; (B) make a loss of $70;(C) make a profit of $175; (D) make a loss of $70;
  2. Let f and g be functions such that for all real numbers x and y g(f(x+y)) = f(x) + (x+y)g(y).. Find the value of g(0) + g(1) + ... + g(2013)
  3. Each chocolate costs 1 dollar, each licorice stick costs 50 cents and each lolly costs 40 cents. How many different combinations of these three items cost a total of 10 dollars.
  4. Let A= {1, 2, 3, 4, 5, 6}. Find the number of distinct functions f: A → A such that f(f(f(n))) = n for all n ∈ A.
  5. Find the number of triangles whose sides are formed by the sides of the diagonals of a regular heptagon (7 sided polygon). (Note: the vertices of the triangle need not be vertices of the heptagon).
  6. Six seats are arranged in a circular table. Each seat is to be painted in red, blue or green such that any two adjacent seats have different colors. How many ways are there to paint the seats?
  7. ΔABC is an equilateral triangle with side length 30. Fold the triangle so that A touches a point X on BC. If BX = 6, find the value of k, where √k is the length of the crease obtained from folding.

  8. As shown in the figure below, circles C1 and C2 of radius 360 are tangent to each other, and both tangent to the straight line l. If the circle C3 is tangent to C1 , C2 and l, and circle C4 is tangent to C1 , C3 and l, find the radius of C4
      
  9. Set {x} = x - ⌊x⌋, where ⌊x⌋ denotes the largest integer less than or equal to x. Find the number of real solutions to the equation {x} + {x2 } = 1 , |x| ≤ 10.
  10. Find the value of   ⌊\( ( \frac {3 + \sqrt 17 } {2} )^6 \)⌋.
  11. A regular dodecagon (12 sided polygon) is inscribed in a circle of radius 10. Find it's area.
  12. to be continued...

4 comments:

  1. 2. for all real numbers x and y g(f(x+y)) = f(x) + (x+y)g(y)
    putting y=-x, g(f(0))=f(x) for all x.
    for any y, choose x such that x+y=1, g(f(1))=f(x)+g(y)
    f(0)=f(1)=g(f(0))
    g(f(1))=g(f(0)), giving g(y)=0
    hence the sum=0

    ReplyDelete
  2. 3.10c+5r+4y=100
    this means y=0,5
    when y=0, 2c+r=20, giving 11 sets of values
    when y=5, 2c+r=16, giving 9 sets of values
    total 20.

    ReplyDelete
  3. 4. for an element n ∈ A, either f(n)=n or f(n)=b, f(b)=c, f(c)=n, b,c,n being all distinct (like a triangle whose sides have a 0 vector sum).
    so, total no. of f's= 1+[6p3/3]+[6p3/3*2]=1+40+80=121

    ReplyDelete
  4. I think all the options in the 1st question are wrong. The correct answer should be that the deal makes a loss of $175.
    CPa + 0.2CPa = SPa = $2100
    CPb - .2CPb = SPb = $2100
    => CPa = $2100/1.2 = $1750
    and CPb = $2100/0.8 = $2625
    so, Total CPdeal = $4375
    and Total SPdeal = $4200
    So, Total LOSS = $175 (as SPdeal<CPdeal)

    ReplyDelete