AMC10 2008 A

AMC10 2008 A · Q15

AMC10 2008 A · Q15. It mainly tests Systems of equations, Rates (speed).

Yesterday Han drove 1 hour longer than Ian at an average speed 5 miles per hour faster than Ian. Jan drove 2 hours longer than Ian at an average speed 10 miles per hour faster than Ian. Han drove 70 miles more than Ian. How many more miles did Jan drive than Ian?

昨天汉比伊恩多开1小时，平均速度比伊恩快5英里每小时。简比伊恩多开2小时，平均速度比伊恩快10英里每小时。汉比伊恩多开70英里。简比伊恩多开多少英里？

(A) 120 120

(B) 130 130

(D) 150 150

(E) 160 160

Answer

Correct choice: (D)

正确答案：（D）

Solution

Answer (D): Suppose that Jan drove for $t$ hours at an average speed of $r$ miles per hour. Then he covered a distance of $rt$ miles. The number of miles Han covered by driving 5 miles per hour faster for 1 additional hour is $(r+5)(t+1)=rt+5t+r+5$. Since Han drove 70 miles more than Jan, $70=(r+5)(t+1)-rt=5t+r+5$, so $5t+r=65$. The number of miles Jan drove more than Ian is consequently $(r+10)(t+2)-rt=10t+2r+20=2(5t+r)+20=2\cdot 65+20=150$.

答案（D）：假设 Jan 以平均速度 $r$ 英里/小时行驶了 $t$ 小时，那么他行驶的距离为 $rt$ 英里。Han 比他每小时快 5 英里并且多开 1 小时，他行驶的英里数为 $(r+5)(t+1)=rt+5t+r+5$。因为 Han 比 Jan 多行驶了 70 英里， $70=(r+5)(t+1)-rt=5t+r+5$，所以 $5t+r=65$。因此，Jan 比 Ian 多行驶的英里数为 $(r+10)(t+2)-rt=10t+2r+20=2(5t+r)+20=2\cdot 65+20=150$。

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