Question

Two cars leave towns 750 kilometers at the same time and travel toward each other. One car's rate is 12 kilometers per hour less than the other's. If they meet hours , what is the rate of the slower car ? Do not do any rounding .

Answer 1

If the distance between the two twons equals 750 km, and the cars met at some point on the road then the distance of the first car added to the distance of the second car equals 750 km, like this:

d1 + d2 = 750

Where d1 is the driven distance by the first car and d2 is the driven distance by the second car.

Distance, is the product of speed and time, then we can rewrite the above equation like this:

s1×t + s2×t = 750

S1 is the speed of the first car and s2 is the speed of the second time, t is the time. We already know that the time it took the cars to meet each other equals 5 hours, then, by replacing 5 for t into the above expression, we get:

s1×5 + s2×5 = 750

5(s1 + s2) = 750

We are also told that the speed of the first car is 12 less than the speed of the second car, this means that the speed of the first car equals the speed of the second one minus twelve. Then we can formulate the following expression for the speeds of the cars:

s1 = s2 - 12

By replacing s2 - 12 for s1 into 5(s1 + s2) = 750 we get:

5(s1 + s2) = 750

5(s2 - 12 + s2) = 750

5(s2 + s2 - 12) = 750

5(2s2 - 12) = 750

By dividing both sides by 5, we get:

(2s2 - 12) = 750/5

2s2 - 12 = 150

By adding 12 on both sides:

2s2 = 150 + 12

2s2 = 162

Finally, by dividing by 2 on both sides:

s2 = 162/2

s2 = 81

Now that we know that the speed of the second car, equals 81, we can find the speed of the first car by means of the formula s1 = s2 - 12, then we get:

s1 = s2 - 12

s1 = 81 - 12

s1 = 69

Then, the speed of the slower car equals 69 kilometers per hour