AMC 8 Daily Practice Round 8

Complete problem set with solutions and individual problem pages

Problem 20 Hard

Point B is to the east of point A. Addy and Brandon start simultaneously from A and B respectively, both moving east at a constant speed. A dog starts from point A at the same time as Addy, running back and forth between them (i.e., running from Addy toward Brandon, turning around after catching up with Brandon to run toward Addy, meeting Addy and then turning around again toward Brandon, and so on). When the dog returns to Addy for the first time, Addy has traveled exactly 140 meters. When the dog returns to Addy for the second time, Addy has traveled a total of 350 meters. If the dog’s speed is 3 times the speed of Brandon, then when the dog catches up with Brandon for the third time, what is the distance between A and B?

  • A.

    1000 meters

  • B.

    1080 meters

  • C.

    500 meters

  • D.

    800 meters

  • E.

    10000 meters

Answer:B

The first time the dog caught up, Addy had traveled 140 meters.

The second time the dog caught up, Addy had traveled 350 meters.

140 : (350 - 140) = 2 : 3

This shows that when the dog returned to Addy the first time, the distance between Addy and Brandon had become \frac{3}{2} times the original.

 

Consider the initial AB distance as 2 parts.

Time analysis:

\begin{cases}\text{Dog chasing Brandon: } 2 \div (3 - 1) = 1 \ \text{(part)} \\ \text{Dog meeting Addy: } \text{dog gains 3 parts on Brandon} \end{cases}

3 \div (3 + 1) = \frac{3}{4} \ \text{(part)}

Distance Addy travels: 3 - \frac{3}{4} \times 3 = \frac{3}{4} \ \text{(part)}

So the speed ratio is: V_{\text{Addy}} : V_{\text{dog}} = \frac{3}{4} : \left( 3 \times 2 - \frac{3}{4} \right) = 1 : 7

At the first return, the Addy–Brandon distance is: 140 \div \frac{3}{4} \times 2 \times \frac{3}{2} = 560 \ \text{meters}

At the second return, the Addy–Brandon distance is:

560 \times \frac{3}{2} = 840 \ \text{meters}

The speed ratio is: V_{\text{Addy}} : V_{\text{Brandon}} : V_{\text{dog}} = 3 : 7 : 21

Assume Addy’s speed is 3 m/s, then Brandon’s speed is 7~\text{m/s}, and the dog’s speed is 21~\text{m/s}.

The time for the dog to catch Brandon the third time is:

840 \div (21 - 7) = 60 \ \text{seconds}

In this time, Brandon travels 60 \times (7 - 3) = 240 meters more than Addy.

So the Addy–Brandon distance is: 840 + 240 = 1080 \ \text{meters}

Therefore, the answer is \text{B}.

Related Problems
Problem 18 AMC 8 Daily Practice Round 8
Problem 19 AMC 8 Daily Practice Round 8
Problem 21 AMC 8 Daily Practice Round 8
Problem 22 AMC 8 Daily Practice Round 8
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