2025 AMC 10B Problem 7

Attempt Problem 7 of the 2025 AMC 10B below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2025 AMC 10B solutions, or check the answer key.

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7.

Frances stands 1515 meters directly south of a locked gate in a fence that runs east-west. Immediately behind the fence is a box of chocolates, located xx meters east of the locked gate. An unlocked gate lies 99 meters east of the box, and another unlocked gate lies 88 meters west of the locked gate. Frances can reach the box by walking toward an unlocked gate, passing through it, and walking toward the box. It happens that the total distance Frances would travel would be the same via either unlocked gate. What is the value of x?x?

3273\tfrac{2}{7}

3373\tfrac{3}{7}

3473\tfrac{4}{7}

3573\tfrac{5}{7}

3673\tfrac{6}{7}

Answer: C
Concepts:coordinate geometryPythagorean Theoremradical
Difficulty rating: 1500
Solution:

Put the fence on the xx-axis, the locked gate at the origin, and Frances at (0,15).(0, -15). Then the box is at (x,0),(x, 0), the east gate at (x+9,0),(x + 9, 0), and the west gate at (8,0).(-8, 0). The east route is (x+9)2+152+9;\sqrt{(x + 9)^2 + 15^2} + 9; the west route is 82+152\sqrt{8^2 + 15^2} +(x+8)=17+x+8.+ (x + 8) = 17 + x + 8. Set them equal: (x+9)2+225=x+16.\sqrt{(x + 9)^2 + 225} = x + 16. Square and simplify to get 18x+306=32x+256,18x + 306 = 32x + 256, so x=5014=257=347.x = \tfrac{50}{14} = \tfrac{25}{7} = 3\tfrac{4}{7}. Thus, C is the correct answer.

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