2021 AMC 10 A Fall

Complete problem set with solutions and individual problem pages

Problem 11 Medium

Emily sees a ship traveling at a constant speed along a straight section of a river. She walks parallel to the riverbank at a uniform rate faster than the ship. She counts 210 equal steps walking from the back of the ship to the front. Walking in the opposite direction, she counts 42 steps of the same size from the front of the ship to the back. In terms of Emily's equal steps, what is the length of the ship?(2021 AMC Fall 10A, Question #11)

  • A.

    70

  • B.

    84

  • C.

    98

  • D.

    105

  • E.

    126

Answer:A

Solution 1:

Let x be the length of the ship. Then, in the time that Emily walks 210 steps, the ship moves 210-x steps. Also, in the time that Emily walks 42 steps, the ship moves x-42 steps. Since the ship and Emily both travel at some constant rate, \frac{210}{210-x}=\frac{42}{x-42}. Dividing both sides by 42 and cross multiplying, we get 5(x-42)=210-x, so 6 x=420, and x= (A) 70 .

Solution 2:

Let the speed at which Emily walks be 42 steps per hour. Let the speed at which the ship is moving be s. Walking in the direction of the ship, it takes her 210 steps, or \frac{210}{42}=5 hours, to travel. We can create an equation: d=5(42-s), where d is the length of the ship. Walking in the opposite direction of the ship, it takes her 42 steps, or 42 / 42=1 hour. We can create a similar equation: d=1(42+s) . Now we have two variables and two equations. We can equate the expressions for d and solve for s : \begin{aligned} 210-5 s &=42+s \\ s &=28 \end{aligned} Therefore, we have d=42+s= (A) 70 .