SAT Math II

To make this problem more concrete, make up a number for the circumference of the racetrack. It doesn't really matter what number you use; I'll use 75 feet. Since $speed=\frac{distance}{time}$ , the speed of car A is $\frac{75feet}{15seconds}=5\frac{feet}{second}$ . (I picked 75 mostly because it is divided evenly by 15 and 25.) Every second, car A gains 2 feet on car B. To pass ca B, car A must gain 75 feet on car B. This will require $\frac{75}{2}=\boxed{37.5}$ seconds.

You may be thinking "Whoa, tricky solution!" Here is the mostly straightforward but somewhat tedious algebraic solution. Once again, I'll use 75 feet for the circumference of the track. Suppose that you count time from when car A first passes car B. Then, car A travels a distance (75/15)t=5t feet after t seconds. (Remember that distance = speed x time.) For example, after 15 seconds, car A has traveled a distance 5 15=75feet, and after 30 seconds, car A has traveled a distance 5 30=150 feet. Similarly, car B travles a distance (75/25)t=3t feet after t seconds. When the two cars pass again, car has traveled 75 feet more than car B: 5t=3t+75. Solving for t gives: 2t=75, or t=75/2= $\boxed{37.5}$ seconds.

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General solution for any length of racetrack ${C}$ :

NOTE: Do NOT do this during the SAT! (unless you'd like $t_{you}=0$ .)

Let's define the following:

$s_{car_A}=\frac{{C}}{15sec}$

$s_{car_B}=\frac{{C}}{25sec}$ .

And so,

$\frac{s_{car_A}}{s_{car_B}}=\frac{\frac{{C}}{15sec}}{\frac{{C}}{25sec}}$

$\Rightarrow \frac{s_{car_A}}{s_{car_B}}=\frac{5}{3}$ .

Thus, car A's speed to car B's speed is 5 to 3. Or, car A gains 2 units of speed distance (or, the units of distance per time, for every unit of time, i.e. distance. $^i$ ) on car B for every unit of time $t$ .

And so, once again, we're looking for the time when the distance of car A equals the distance of car B. In other words, car A has to gain ${C}$ units of distance on car B. Hence,

$5s_{car_B}t=3s_{car_A}t+{C}$ .

Looks familiar? Good. You take it on from here.

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$i$

$speed=\frac{distance}{time}=\frac{{C}}{time}$ .

Apply this to each individual car's time. Good luck.

For each lap Car A takes 10 seconds on Car B. Car B spends 25 seconds for every lap i travels. Therefor it will take Car A 2 1/2 laps before it will once again pass Car B. It takes 37,5 seconds for Car A to drive 2 1/2 laps therefor the answer is 37,5

Track distance = d ==> r(A) = d/15 and r(B) = d/25

From point of Car A passing Car B, it will travel distance d + x, while Car B travels x. Both cars travel for some time, t.

d + x = (d/15) * t and x = (d/25) * t

Substituting: d + (d/25) * t = (d/15)*t

Simplifying: d = ( d/15 - d/25) * t

==> d = (10 * d/375) * t

Multiplying by reciprocal: d * (37.5/d) = t

Canceling leaves 37.5 = t