2009-2010
Mercury Test Solutions Round 2
1. Simply add the number of second in 3 hours plus the number of second in 21 minutes plus 2 seconds to get the answer: 3(3600) + 21(60) + 2 = 12062
2. To solve for x move the 48 to the other side of the equation and then solve the problem using the quadratic formula: x

 b
 b 2 
4 ac
. After solving, x will be found to be equal to 8 and
2 a asks for all of the positive values of x , the correct answer is 8.

3
2
. However since the question
3. 16(-2+5) = 48
48 =
6 x
15 x=120 (the total distance of Klærtün's trip)
11(120-48) (We subtract 48 from 120 since it asks for the remaining time of travel, and he has already traveled 48 parsecs)
11(72)=792 hours remain
4. This is a work problem. The formula for work is the product of the number of hours for each person to finish the job divided by the sum. After using this formula, you will get: (3)(4) / (3+4) = 12/7.
5. The final answer would be solved by finding the answer to (6 3 )x
{
[(3
(7

6

2
3
)
1)

(2
2
)]
}
. 6 3 comes out to be 216.
[(3
(7


2
3
)
6 -

1)
(2
2
)]
equals
24
41

4
or
41 . So the solution to the final answer would be
6 reduce that to 36

41 , giving you the final answer of 1476 people in the area.
216

41
6
. You could then
12062
8
792
12/7
1476
6. A prism with squares for all sides is a cube. The total surface area of a cube is the sum of the areas of the square faces. There are 6 square faces, and each has an area of 10 2 , so the surface area of the cube is 6*100=600.
600
7. 214/13 =
16
6
13
6=x f(x)= 6
3 
8

2

3 = 216+12=228
228
8. To find the number of squares in a n x n grid, you use the formula n (n+1) (2n+1) / 6 where n is dimensions of the grid. This yields 5 (5+1) (10+1) / 6 = 55.
55
9. This is a circular permutation problem. The formula for circular permutation is (n-1)! where n is the number of people or objects arranged around a table. After using this formula, you get: (4-1)! = 6.
10. Factor the equation and you get (3x+5)(x-6)=0, or use the quadratic formula. Either way will give solutions x=6 and x= 
5
3
. Use the positive x for the side length of the square. The area of the square is equal to side squared, or
36. Now using the Pythagorean theorem (a 2 +b 2 =c 2 ), or the ratios of a 45-45-90 right triangle, you get that the diagonal of the square has a length of
1
2
6 2

3 2
6 2 . The radius of the circle is equal to half the length of the diagonal,
. Now solve for the area of the circle
 r
2  equal to the area of the circle minus the area of the square.

A
3
 

18
 
2 
36
18

.

. The area of the shaded region is
11. Create a chart to model your potential answers, with 2x and y summing to 2010. x y
0
1
2
2010
2008
2006
3
…
1003
2004
…
4
1004 2
1005 0
Since the solutions must be positive, only the solution sets with x is 1 to 1004 work. This means that there are 1004 solutions to this problem.
6
18

36
1004

12.
1 1 1
1
2
3
3
0
4
1
4
4
8
1
4
This is a path-walking problem. The 3 x 3 gird represents the paths Dis must walk upon. The numbers on the grid represent the number of ways that Dis can travel to a particular vertex of the grid. For example, Dis can travel 1 way to the vertex directly right of him. The number of paths of a vertex is found to be equal to the sum of the number of paths of the two vertices that are adjacent to that vertex. However, the number of paths to Point C is 0
8 because Dis must avoid Evil Ed. After using this strategy, the number of paths by which Dis can travel to the post office is 8.
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