AMC 10 Weekly Practice Round 2

Complete problem set with solutions and individual problem pages

Problem 19 Hard

The distance between A and B is 500 km. Two people, \text{A} and \text{B}, set out at the same time from A to B on bicycles. Person \text{A} rides 30 km per day, while person \text{B} rides 50 km per day but rests every other day. At the end of the \underline{\phantom{day}} day, the distance from B for person \text{B} is twice the distance from B for person \text{A}.

  • A.

    13

  • B.

    14

  • C.

    15

  • D.

    16

  • E.

    17

Answer:C

Casework:

(1) If the number of days is even, then for B it is equivalent to riding $$

50 div 2 = 25 \text{km per day}.$$

Let n be the number of days such that at the end of day n, B’s distance from B is twice A’s distance from B.

We have the equation: 500 - 25n = 2 \big( 500 - 30n \big), which gives n = \frac{100}{7}, not a valid solution.

(2) If the number of days is odd, then except for the last day (when B rides 50 km), B rides 25 km per day on the previous days.

Let n be the number of days such that at the end of day n, B’s distance from B is twice A’s distance from B.

We have the equation 500 - 25(n - 1) - 50 = 2 \big( 500 - 30n \big), which gives n = 15, a valid solution.

Thus, at the end of the 15th day, B’s distance from B is twice A’s distance from B.