Hands-on math!

Washington State Math Olympiad
Hints and Solutions
2011 Grade 8 Algebra
Problem
Solution
1) Two teams enter a contest that has two parts to it. Part A problems are each worth the same amount, and Part B problems are each worth the same amount. The Geeks score 54 points. The Whizzes score 51 points. Their captains find out that the Geeks answered 10 problems in Part A correctly and 12 problems in Part B correctly. The Whizzes answered 11 problems in Part A correctly and 9 in Part B correctly. What would the Whizzes have scored if they had answered 2 more problems in Part B correctly for a total of 11 Part A problems and 11 Part B problems? It is important to note that although each question in part A is worth the same points and the same with point B, that the part A points are not worth the same as point B points. We need to find out what points were worth in each part. This is a 2 equations and 2 unknowns problem. Let x = the number of points for a part A problem and b the number of points for a part B problem. We need to know y to answer this problem.
  1. The Geeks equation: 10x + 12y = 54
  2. The Whizzes equation: 11x + 9y = 51
  3. Turn the first equation into an equation for x:
    x = (54 - 12y) / 10
  4. Substitute this expression for x into the second equation and solve for y:
    11(54 - 12y)/10 + 9y = 51
    11(54 - 12y) + 90y = 510
    594 - 132y + 90y = 510
    -42y = -84
    y = 2
  5. The Whizzes would have scored 51 + 2x2 = 55 if they had answered 2 more part B problems.
    (By the way, substituting back in the first equation,
    the part A answers were worth
    (54 - 24)/10 = 3 points)
2) Alex and Ian are shopping for new clothes. Alex says to Ian, "If you give me $5, we will have an equal amount of money and that would be fair." Ian responds, "That may be true, but if you give me $4, I will have twice as much money as you and that would be much better." How much money do Alex and Ian have together? This is another 2 equations with 2 unknowns problem. Let A = the amount of money Alex has and N the amount of money Ian has.
  1. The first equation is: A + $5 = N - $5
  2. The second equation is : N + $4 = 2(A - 4)
  3. Turn the first equation into an equation for A:
    A = N - 10
  4. Substitute this expression for A into the second equation:
    N + 4 = 2(N - $10 - $4)
    N + 4 = 2N -28
    -N = -32, so Ian has $32
  5. Plug this back into the first equation to get A:
    A = N - $10 = $22
  6. Together, they have $54
3) Mrs. Jyoti was keeping track of how many mystery books and how many adventure books her students were reading from the class collection. If this pattern in book reading continues, how many mystery books will be read in January and how many adventure books?
  1. The number of mystery books doubles every month, so the number of mystery books read in January would be
    32x2 = 64
  2. The number of adventure books is multiplied by 3 every month, so the number of adventure books read in January would be
    54x3 = 162.

Problem
Solution
4) Trina falls off a 300 foot cliff. Her height above the ground is modeled by the equation
T = 300 - 32t, t in seconds.
She's in luck and Superman flies after her. His height above the ground is modeled by the equation
S = 500 - 57t, t in seconds.
How high off the ground are they when he catches Trina? 		We first must find the time at which Trina and Superman meet, so we set the 2 equations equal to each other and solve for t:
300 - 32t = 500 - 57t
25t = 200
t = 8 seconds
Now plug this time into either of the equations to get the height:
T = 300 - 32x8 = 300 - 256 = 44 ft. or, (if you prefer):
S = 500 - 57x8 = 500 - 456 = 44 ft.
5) Jesse creates a pattern of equilateral triangles with rhombuses and triangles, the first three are shown. He sees that each rhombus is two triangles. How many rhombuses will he need if he wants to build the 20th triangle in the pattern?
 Let n = the number of the triangle.
  1. Counting just the small triangles, the sequence is:
    4     9     16,
    which is (n+1)2
  2. So the 20th triangle contains 212 = 441 small triangles
  3. Counting the small triangles that are not part of a rhombus:
    2     3     4
    which is (n+1)
  4. So, the number of small triangles in the 20th triangle is 21.
  5. Subtracting the small triangles and dividing by 2:
    (441 - 21)/2 = 420/2 = 210 rhombuses.