AMC10 2010 A

AMC10 2010 A · Q13

AMC10 2010 A · Q13. It mainly tests Systems of equations, Rates (speed).

Angelina drove at an average rate of 80 kph and then stopped 20 minutes for gas. After the stop, she drove at an average rate of 100 kph. Altogether she drove 250 km in a total trip time of 3 hours including the stop. Which equation could be used to solve for the time $t$ in hours that she drove before her stop?

Angelina 以平均速度 80 千米/时行驶，然后停下 20 分钟加油。停下后，她以平均速度 100 千米/时行驶。总共她行驶了 250 千米，总行程时间为 3 小时（包括停顿）。哪一个方程可以用来解她在停下前行驶的时间 $t$（小时）？

(A) $80t + 100(\frac{8}{3} - t) = 250$ $80t + 100(\frac{8}{3} - t) = 250$

(B) $80t = 250$ $80t = 250$

(D) $90t = 250$ $90t = 250$

(E) $80(\frac{8}{3} - t) + 100t = 250$ $80(\frac{8}{3} - t) + 100t = 250$

Answer

Correct choice: (A)

正确答案：（A）

Solution

Answer (A): Angelina drove $80t$ km before she stopped. After her stop, she drove $\left(3-\frac{1}{3}-t\right)$ hours at an average rate of 100 kph, so she covered $100\left(\frac{8}{3}-t\right)$ km in that time. Therefore $80t+100\left(\frac{8}{3}-t\right)=250$. Note that $t=\frac{5}{6}$.

答案（A）：安吉丽娜在停下来之前行驶了 $80t$ 千米。停下后，她以平均时速 100 千米/小时行驶了 $\left(3-\frac{1}{3}-t\right)$ 小时，因此在这段时间内她行驶了 $100\left(\frac{8}{3}-t\right)$ 千米。所以有 $80t+100\left(\frac{8}{3}-t\right)=250$。注意 $t=\frac{5}{6}$。

Topics