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2005 AMC 10A Problem 7

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2005 AMC 10A Problem 7AMC 10 Step by Step
Name:                             Date: September 13, 2026
Linear Equations & Word Problems · difficulty 2 of 5 · about 3 min · position P1-10

Josh and Mike live 1313 miles apart. Yesterday Josh started to ride his bicycle toward Mike's house. A little later Mike started to ride his bicycle toward Josh's house. When they met, Josh had ridden for twice the length of time as Mike and at four-fifths of Mike's rate. How many miles had Mike ridden when they met?

  1. A)

    44

  2. B)

    55

  3. C)

    66

  4. D)

    77

  5. E)

    88

Answer

B

Key insight

Distance is rate times time, so Josh rode 2 times 4/5 = 8/5 of Mike's distance; the two distances total 13 miles.

Solution

Suppose Mike rode at rate rr for time tt, covering rtrt miles. Josh rode for time 2t2t at rate 45r\frac{4}{5}r, so his distance is
45r⋅2t=85 rt.\frac{4}{5}r \cdot 2t = \frac{8}{5}\,rt.
Josh's distance is 85\frac{8}{5} of Mike's. When they meet, the two distances make up the whole 1313 miles. With m=rtm = rt for Mike's distance,
m+85m=135m=13⟹m=5.m + \frac{8}{5}m = \frac{13}{5}m = 13 \quad\Longrightarrow\quad m = 5.

The answer is (B) 5\boxed{\textbf{(B)}\ 5}.

Why this works

Two riders approaching each other cover the full separation between them, so the meeting condition is a sum of distances. Each distance is rate times time, and the problem gives both of Josh's factors as multiples of Mike's, so the ratio of distances is just the product 2⋅452 \cdot \frac{4}{5}. The individual rate and time never need to be found.

Alternative approach

Ratio form: distances are in the ratio Mike : Josh =5:8= 5 : 8, so Mike covers 513\frac{5}{13} of 1313 miles, i.e. 55 miles.

The trap

Using Josh's ratios separately (doubling the time but forgetting the slower rate, or vice versa), which gives 13/3 or 65/9.

Common mistakes

  • Using Josh's ratios separately (doubling the time but forgetting the slower rate, or vice versa), which gives 13/3 or 65/9.
  • Reporting Josh's distance (88 miles, choice (E)) instead of Mike's.

Techniques

Set up the equation/formula and compute; no special trick needed