All,
This was a data sufficiency problem, but want to double check my math in case I see it in a problem solver.
If Car X followed Car Y across a certain bridge that is \(\frac{1}{2}\) mile long, how many seconds did it take Car X to travel across the bridge?
(1) Car X drove onto the bridge exactly 3 seconds after Car Y drove onto the bridge and drove off the bridge exactly 2 seconds after Car Y drove off the bridge.
(2) Car Y traveled across the bridge at a constant speed of 30 miles per hour.
C is the correct answer for data sufficiency, but I want to go through the problem in various ways (problem solving) to make sure my math is correct.
Provided all the data; What are the rates of each Car? How many seconds would it take for Car X to catch Car Y? At what distance would Car X catch Car Y?
Rate Car X: ~30.5 mph OR \(\frac{1}{118}\) miles per second
Rate Car Y: 30 mph (given) OR \(\frac{1}{120}\) miles per second
Second for Car X to catch Car Y: 180 seconds OR 3 minutes OR \(\frac{1}{20}\) hour
Distance: 1.5 miles
Explanations:
Rate x Time(t) = Distance
Car Y (given at 30 mph, so find \(t\) to solve for rate of car X)
\(y\) x \(t\) = \(\frac{1}{2}\)
30 x \(t\) = \(\frac{1}{2}\) (need miles per second, not hour)
\(\frac{1}{120}\) x \(t\) = \(\frac{1}{2}\)
\(t\) = 60
Car X
\(x\) x (\(t\) - 1) = \(\frac{1}{2}\) ; 1 second for the time difference (waited 3 seconds after Y, finished 2 seconds after Y: 3 - 2 = 1)
\(x\) x (60 - 1) = \(\frac{1}{2}\)
\(x\) x 59 = \(\frac{1}{2}\)
\(x\) = \(\frac{1}{118}\)
Time (in seconds) Car X catches Car Y:
\(\frac{1}{118}\) (\(t\) - 3) = \(\frac{1}{120}t\) ; 3 = amount of seconds after Y left.
\(\frac{1}{118} t\) - \(\frac{3}{118}\) = \(\frac{1}{120}t\)
\(\frac{60}{7080}t\) - \(\frac{180}{7080}\) = \(\frac{59}{7080}\)
\(\frac{1}{7080} t\) = \(\frac{180}{7080}\)
\(t\) = 180 seconds
180 seconds for Car X to catch Car Y
Distance (in miles) for Car X to catch Car Y
Taking either equation and substituting 180 for \(t\)
X) \(\frac{1}{118}\) x (180-3) = 1.5 miles
Y) \(\frac{1}{120}\) x (180) = 1.5 miles
I have all this set up correctly, right?