AMC 10 Weekly Practice Round 2

Complete problem set with solutions and individual problem pages

Problem 22 Medium

Person A starts walking from A to B, and 5.5 minutes later, Person B starts walking from B to A. Person B walks 30 meters per minute faster than Person A.They meet at point C along the way. The time taken by Person A to travel from A to C is 4 minutes longer than from C to B. The time taken by Person B to travel from C to A is 3 minutes longer than from B to C. What is the distance between A and B in meters?

  • A.

    1080

  • B.

    1440

  • C.

    1250

  • D.

    1880

  • E.

    2140

Answer:B

Let A walk x meters per minute, then B’s speed is x + 30 meters per minute.

Let t minutes be the time B takes to walk from B to C.

From the problem, we have:

\begin{cases}\dfrac{5.5x + tx}{x} = \dfrac{t(x + 30)}{x} + 4 \\[6pt]\dfrac{5.5x + tx}{x + 30} = t + 3\end{cases}

Solving the system gives:

\begin{cases}x = 90 \\t = 4.5\end{cases}

Verification shows x = 90, t = 4.5 satisfy the original equations.

Therefore, x = 90, t = 4.5.

The time A spends from A to C is 5.5 + 4.5 = 10 minutes, and the distance AB is: 5.5 \times 90 + 4.5 \times 90 + 4.5 \times 120 = 1440 \ \text{meters}.

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