2007 iTest Problems/Problem 8

Problem

Joe is right at the middle of a train tunnel and he realizes that a train is coming. The train travels at a speed of 50 miles per hour, and Joe can run at a speed of 10 miles per hour. Joe hears the train whistle when the train is a half mile from the point where it will enter the tunnel. At that point in time, Joe can run toward the train and just exit the tunnel as the train meets him. Instead, Joe runs away from the train when he hears the whistle. How many seconds does he have to spare (before the train is upon him) when he gets to the tunnel entrance?

$\mathrm{(A)}\,7.2\quad\mathrm{(B)}\,14.4\quad\mathrm{(C)}\,36\quad\mathrm{(D)}\,10\quad\mathrm{(E)}\,12\quad\mathrm{(F)}\,2.4\quad\mathrm{(G)}\,25.2\quad\mathrm{(H)}\,123456789$

Solution

Using the well-known $d=rt$ equation, letting $x$ be half the length of the tunnel, and $y$ being the time Joe needs to run that distance, we see that:

$10y=x$

$\frac{1}{2}=50y$

Solving, $y=\frac{1}{100}$ of an hour, and $x=\frac{1}{10}$ of a mile. Now, we calculate the time it takes for the train to get to the other end of the tunnel, and we subtract $y$ from that. The distance the train has to travel is $\frac{1}{5}+\frac{1}{2}$ miles, which equals $\frac{7}{10}$ miles. Since $d=rt$ , the time it takes is $\frac{7}{500}$ of an hour. $\frac{7}{500}-\frac{1}{100}=\frac{2}{500}$ hours, which is equal to $14.4 \text{ seconds}\Rightarrow \boxed{B}$ .

2007 iTest (Problems, Answer Key)
Preceded by: Problem 7	Followed by: Problem 9
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 • 51 • 52 • 53 • 54 • 55 • 56 • 57 • 58 • 59 • 60 • TB1 • TB2 • TB3 • TB4

Page

Toolbox

Search

2007 iTest Problems/Problem 8

Problem

Solution

See Also