1. Problem 12: We need to build a wall 100 feet long and 7 feet high using 1-foot-high blocks that are either 1 or 2 feet long. The vertical joints must be staggered, and the wall must be even at the ends. 2. Since the wall is 7 feet high, it has 7 rows of blocks. 3. Each row is 100 feet long. The blocks can be 1 or 2 feet long, and the vertical joints must be staggered, meaning the joints in one row must not line up with the joints in the row above or below. 4. To stagger joints, rows alternate between starting with a 1-foot block and a 2-foot block. 5. For even ends, the total length of blocks in each row must be exactly 100 feet. 6. Consider the pattern for two consecutive rows: - Row 1: starts with a 2-foot block, then alternating 1-foot and 2-foot blocks. - Row 2: starts with a 1-foot block, then alternating 2-foot and 1-foot blocks. 7. Calculate the number of blocks per row: - For rows starting with 2-foot blocks: The pattern is 2,1,2,1,... - For rows starting with 1-foot blocks: The pattern is 1,2,1,2,... 8. Each pair of blocks in a row sums to 3 feet. 9. For a 100-foot row, the number of pairs is $\frac{100}{3} = 33\frac{1}{3}$, which is not an integer, so we must adjust. 10. Instead, use a repeating pattern of blocks that sum to 4 feet over 2 blocks: (2+2) or (1+3) is invalid since only 1 or 2 feet blocks allowed. 11. The only way to have staggered joints and even ends is to alternate rows of all 2-foot blocks and rows of all 1-foot blocks. 12. Calculate the number of blocks: - Rows with 2-foot blocks: $\frac{100}{2} = 50$ blocks per row. - Rows with 1-foot blocks: 100 blocks per row. 13. Since the wall is 7 feet high, and to stagger joints, rows alternate: - 4 rows of 2-foot blocks: $4 \times 50 = 200$ blocks - 3 rows of 1-foot blocks: $3 \times 100 = 300$ blocks 14. Total blocks = 200 + 300 = 500 blocks, which is too large and not among the options. 15. Reconsider the staggering: The problem states vertical joints must be staggered as shown, so the pattern is likely alternating rows of blocks starting with 1-foot and 2-foot blocks. 16. For rows starting with 2-foot blocks: - Number of 2-foot blocks = $x$ - Number of 1-foot blocks = $x-1$ Total length: $2x + (x-1) = 3x -1 = 100$ feet Solve for $x$: $$3x - 1 = 100 \Rightarrow 3x = 101 \Rightarrow x = \frac{101}{3}$$ Not an integer, so try rows starting with 1-foot blocks: - Number of 1-foot blocks = $y$ - Number of 2-foot blocks = $y$ Total length: $y + 2y = 3y = 100$ feet Solve for $y$: $$3y = 100 \Rightarrow y = \frac{100}{3}$$ Not an integer either. 17. Since 100 is not divisible by 3, the pattern must include some rows with an extra 1-foot block at the end. 18. The minimal number of blocks is achieved by minimizing the number of 1-foot blocks because 2-foot blocks cover more length per block. 19. The minimal number of blocks per row is when the row is made of all 2-foot blocks: 50 blocks. 20. Since the wall is 7 feet high, total blocks = $7 \times 50 = 350$ blocks. 21. This matches option (C) 350. --- 22. Problem 13: Triangle CAT with $\angle ACT = \angle ATC$ and $\angle CAT = 36^\circ$. TR bisects $\angle ATC$. Find $\angle CRT$. 23. Since $\angle ACT = \angle ATC$, triangle CAT is isosceles with $CT = CA$. 24. Sum of angles in triangle CAT: $$\angle CAT + \angle ACT + \angle ATC = 180^\circ$$ $$36^\circ + 2\times \angle ACT = 180^\circ$$ $$2\times \angle ACT = 144^\circ \Rightarrow \angle ACT = 72^\circ$$ 25. TR bisects $\angle ATC$, so each part is $\frac{72^\circ}{2} = 36^\circ$. 26. In triangle CRT, $\angle CRT$ is the angle opposite side TR. 27. Using angle sum in triangle CRT: $$\angle CRT + \angle CTR + \angle RTC = 180^\circ$$ 28. $\angle CTR = 36^\circ$ (since TR bisects $\angle ATC$), and $\angle RTC = 36^\circ$ (isosceles property). 29. So, $$\angle CRT = 180^\circ - 36^\circ - 36^\circ = 108^\circ$$ 30. Answer is (E) 108°. --- 31. Problem 14: Find the units digit of $19^{19} + 99^{99}$. 32. Units digit of $19^{19}$ depends on units digit of 9 raised to 19. 33. Units digit pattern of 9 powers: 9,1,9,1,... 34. Since 19 is odd, units digit of $9^{19}$ is 9. 35. Units digit of $99^{99}$ depends on units digit of 9 raised to 99. 36. 99 is odd, so units digit of $9^{99}$ is 9. 37. Sum units digit: $9 + 9 = 18$, units digit is 8. 38. Answer is (D) 8. --- 39. Problem 15: Triangles ABC, ADE, and EFG are equilateral. D and G are midpoints of AC and AE. AB = 4. Find perimeter of figure ABCDEFG. 40. Since ABC is equilateral with side 4, perimeter is $3 \times 4 = 12$. 41. D is midpoint of AC, so AD = DC = 2. 42. ADE is equilateral, so each side is 2. 43. G is midpoint of AE, so AG = GE = 2. 44. EFG is equilateral with side equal to GE = 2. 45. The figure ABCDEFG includes sides of ABC plus sides of ADE and EFG, but some sides overlap. 46. The perimeter is sum of outer edges: AB + BC + CD + DE + EF + FG + GA. 47. Calculate lengths: - AB = 4 - BC = 4 - CD = 2 - DE = 2 (side of ADE) - EF = 2 (side of EFG) - FG = 2 (side of EFG) - GA = 2 48. Sum: $$4 + 4 + 2 + 2 + 2 + 2 + 2 = 18$$ 49. Answer is (D) 18. --- 50. Problem 16: Mateen walks 1000m by walking length 25 times or perimeter 10 times. Find area of rectangular backyard. 51. Let length = $L$, width = $W$. 52. Walking length 25 times: $25L = 1000 \Rightarrow L = 40$. 53. Walking perimeter 10 times: $10(2L + 2W) = 1000 \Rightarrow 2L + 2W = 100$. 54. Substitute $L=40$: $$2(40) + 2W = 100 \Rightarrow 80 + 2W = 100 \Rightarrow 2W = 20 \Rightarrow W = 10$$ 55. Area = $L \times W = 40 \times 10 = 400$. 56. Answer is (C) 400. --- Final answers: 12: 350 13: 108° 14: 8 15: 18 16: 400