2009年AMC10竞赛A第20题
$$2009-AMC10A-20$$  Andrea and Lauren are $$20$$ kilometers apart. They bike toward one another with Andrea traveling three times as fast as Lauren, and the distance between them decreasing at a rate of~ $$1$$ kilometer per minute. After $$5$$ minutes, Andrea stops biking because of a flat tire and waits for Lauren. After how many minutes from the time they started to bike does Lauren reach Andrea?  安德里亚和劳伦相距 20 公里。 他们骑自行车相向而行，$$Andrea$$ 的速度是 Lauren 的三倍，而且他们之间的距离以每分钟 1 公里的速度减少。 5 分钟后，安德里亚因为轮胎漏气而停止骑自行车并等待劳伦。 在他们开始骑自行车几分钟后，劳伦与安德里亚相遇？
课内体系->知识点->等式与不等式->等式->方程组的解集 美国AMC10/12->Knowledge Point->Algebra->Application->Equation Word Problems
Let their speeds in kilometers per hour be $$v_A$$ and $$v_L$$. We know that $$vA=3vL$$ and that $$v_A+v_L=60$$. (The second equation follows from the fact that $$1\rm km/min=60\rm km/h)$$. This solves to $$v_A=45$$ and $$v_L=15$$.  As the distance decreases at a rate of $$1$$ kilometer per minute, after $$5$$ minutes the distance between them will be $$20-5=15$$ kilometers.  From this point on, only~ Lauren will be riding her bike. As there are $$15$$ kilometers remaining and $$vL=15$$, she will need exactly an hour to get to Andrea. Therefore the total time in minutes is $$5+60= 65$$.  Because the speed of Andrea is $$3$$ times as fast as Lauren and the distance between them is decreasing at a rate of $$1$$ kilometer per minute, Andrea\textquotesingle s speed is $$\dfrac{3}{4}\rm km/min$$,and Lauen\textquotesingle s $$\dfrac{1}{4}\rm km/min$$. Therefore, after $$5$$ minutes, Andrea will have biked $$\dfrac{3}{4}\cdot 5= \dfrac{15}{4}\rm km$$. In all, Lauren will have to bike $$20- \dfrac{15}{4}= \dfrac{80}{4}- \dfrac{15}{4}= \dfrac{65}{4}\rm km$$. Because her speed is$$\dfrac{1}{4}\rm km/\min$$, the time elapsed will be$$\dfrac{65}{\dfrac{4}{\dfrac{1}{4}}}=\left( \rm D\right)65$$. 