AMC 10 Weekly Practice Round 2

Complete problem set with solutions and individual problem pages

Problem 25 Easy

A car and a truck depart from City A toward City B at the same time. After reaching City B, the car immediately returns to City A at a different speed, while the truck stops after reaching City B. The graphs shown represent the relationship between the distance y (in km) from City A and the travel time x (in hours) for the truck and the car, respectively. When the car is returning from City B to City A and meets the truck along the way, what is the distance from City A to the meeting point?

 

  • A.

    \frac{280}{3}~\text{km}

  • B.

    70~\text{km}

  • C.

    \frac{7}{3}~\text{km}

  • D.

    90~\text{km}

  • E.

    \frac{70}{3}~\text{km}

Answer:A

From the problem, the speed of the car on its return trip to City A is \frac{120}{3.5 - 2} = 80 \ \text{km/h}.

Let the function representing the truck’s distance from City A with respect to travel time be y = k_{1}x.

Substituting the point (3, 120) gives: 3k_{1} = 120,

so k_{1} = 40.

Thus, the truck’s distance function is y = 40x.

Let the function representing the car’s distance from City A on its return trip be y = k_{2}x + b.

Substituting the points (2, 120) and (3.5, 0) gives:

\left\{ \begin{align} & 2{{k}_{2}}+b=120 \\ & 3.5{{k}_{2}}+b=0 \\ \end{align} \right.

Solving gives:

\begin{cases}k_{2} = -80 \\b = 280\end{cases}

Thus, the car’s distance function on its return trip is

y = -80x + 280.

Solving the system

\begin{cases}y = -80x + 280 \\y = 40x\end{cases} gives:

\begin{cases}x = \frac{7}{3} \\y = \frac{280}{3}\end{cases}

Therefore, when the car is returning from City B to City A and meets the truck, the meeting point is \frac{280}{3}~\text{km} from City A.