Bunuel wrote:

Radina and Russell are participating the Greystone Annual Run/Walk. Radina leaves a half hour after Russell and runs 1 mile per hour slower than twice Russell's rate. If Radina arrives at the finish line 4 hours after she started and finishes at the same time as Russell, approximately how many miles long is the race?

A. 4/3

B. 7/2

C. 9/2

D. 32/7

E. 36/7

Set up (R*T) = D for each runner. Set distances equal to each other, which will yield one actual rate. Use that rate to find distance.

1) TIME for each runner

Radina: 4 hours

Russell: 30 minutes MORE:

\(4\frac{1}{2}=\frac{9}{2}\) hours

2) RATE for each in variables

Let Russell's rate = R

Radina's rate = (2R - 1)

3) Find Russell's actual rate from D = D (they run the same distance)

\((2R - 1)(4) = (R)(\frac{9}{2})\)

\((2R - 1)(8) = 9R\)

\(16R - 8 = 9R\)

\(7R = 8\)

\(R = \frac{8}{7}\) mph

4) Actual distance?

Use Russell's rate * time:

\(D = (\frac{8}{7}mph * \frac{9}{2}hrs) = \frac{36}{7}\) miles

Answer E

If D = \(\frac{36}{7}\), then Radina's rate is D/T: \(\frac{\frac{36}{7}}{4} = \frac{36}{28} = \frac{9}{7}\)

Radina's rate is also (2R - 1)

Radina's rate: \([(\frac{8}{7}*2) - 1] = (\frac{16}{7} - \frac{7}{7}) = \frac{9}{7}\)

That's a match. The answer is correct.
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