Original question written by FE Exam AI Prep from the public FE Civil exam specification. Before release, a solver model re-derived the answer twice from the question alone, without its answer key, and a critic model checked the result, and the arithmetic was re-run as an executable calculation. No licensed engineer reviews these questions. It is not an NCEES question and does not come from any NCEES practice exam. FE Exam AI Prep is an independent study tool and is not affiliated with, endorsed by, or sponsored by NCEES.
MEDIUMq = N/t
Traffic Flow Rate from Vehicle Counts: FE Civil practice problem 6
For the Silver Junction arterial overlay, a resident engineer preparing a basis-of-design update records 420 vehicles during a 10-min count interval. What equivalent hourly flow rate is obtained?
- A1650.0 veh/hr
- B2793.5 veh/hr
- C2520.0 veh/hr
- D3071.4 veh/hr
Show the answer and worked solution
Answer: C — 2520.0 veh/hr
Worked solution
- Step 1. Convert count interval to hours
10 min = 10/60 hr = 1/6 hr
- Step 2. Compute hourly flow
q = count/time
q = 420 / (1/6 hr) = 420 x 6 = 2520 veh/hr
Why the other choices are wrong
- Choice A
- This total scales the 420-vehicle count by 4.68 rather than the required 60/10 = 6.0 hourly-expansion factor; no interval length derived from the stated 10-minute count produces that smaller multiplier, and the resulting rate understates the true hourly flow.
- Choice B
- Here the 420-vehicle sample is stretched by a factor of 8.7 instead of the correct 6.0; that magnitude does not come from any valid 60-minute/10-minute reading of the count, and it drives the hourly figure far above the true flow rate.
- Choice D
- This figure multiplies the 420-vehicle count by 7.08 in place of the correct 60/10 = 6.0 factor; that inflated multiplier does not arise from any valid interpretation of the 10-minute sample window, so the hourly estimate overstates the true flow.