A motorist drove 180 mi before running out of gas and walking 6 mi to a gas station. The motorist's driving rate was twelve times the walking rate. The time spent walking was 3 h less than the time spent driving. How long did it take for the motorist to drive the 180 mi?
since time = distance/speed, if the man's walking speed was w, then we have
180/12w - 3 = 6/w
w = 3
so, he drove at 36 mi/hr, doing the 180 miles in 5 hours.
Walking rate = r mi/h.
Time = (T-3) hrs.
r*(T-3) = 6, r = 6/(T-3).
Driving rate = 12r mi/h.
Time = T hrs.
12r*T = 180, Replace r with 6/(T-3): (12^6/(T-3))*T = 180,
72T/(T-3) = 180, 72T = 180*(T-3),
72T = 180T-540, T = 5 hrs.
r = 180mi/5h = 36 mi/h.
Let's assume the driving rate of the motorist is D mph and the walking rate is W mph.
Given that the motorist drove 180 miles before running out of gas and walking 6 miles, we can set up the equation:
180 = D * T1, where T1 is the time spent driving in hours.
We also know that the motorist's driving rate was twelve times the walking rate:
D = 12W
The time spent walking was 3 hours less than the time spent driving:
T1 - 3 = T2, where T2 is the time spent walking in hours.
Since we have two equations with two variables, we can solve the system of equations.
Let's substitute the value of D from the second equation into the first equation:
180 = 12W * T1
Simplifying the equation, we get:
15 = W * T1
Now, substitute T2 = T1 - 3 into the equation:
15 = W * (T2 + 3)
Simplifying further:
15 = W * T2 + 3W
Since we know that W = 15 / T1, substitute this value into the equation:
15 = (15 / T1) * T2 + 3 * (15 / T1)
Multiplying through by T1 to get rid of the fraction:
15T1 = 15T2 + 45
Rearranging the terms:
15T1 - 15T2 = 45
Divide both sides by 15:
T1 - T2 = 3
Now, we can solve the system of equations formed by combining the equations T1 - T2 = 3 and 180 = 15T1:
180 = 15T1
T1 = 180 / 15 = 12
Substitute this value of T1 into the equation T1 - T2 = 3:
12 - T2 = 3
T2 = 12 - 3 = 9
Therefore, it took the motorist 12 hours to drive the 180 miles.
To solve this problem, we can start by setting up a system of equations.
Let's denote the driving rate as "r" and the walking rate as "w".
From the problem statement, we know that the driving rate is twelve times the walking rate:
r = 12w ...(1)
We also know that the motorist drove 180 miles before running out of gas. The time spent driving can be calculated by dividing the distance by the driving rate:
Time spent driving = Distance / Driving rate
= 180 / r ...(2)
The motorist then walked 6 miles to the gas station. The time spent walking can be calculated by dividing the distance by the walking rate:
Time spent walking = Distance / Walking rate
= 6 / w ...(3)
The problem also states that the time spent walking was 3 hours less than the time spent driving:
Time spent walking = Time spent driving - 3 ...(4)
Now, we have a system of equations consisting of equations (1), (2), (3), and (4). We can solve this system to find the driving rate and the time spent driving, which will give us the answer to the question.
First, let's substitute equation (1) into equations (2) and (3):
Time spent driving = 180 / (12w) = 15/w ...(5)
Time spent walking = 6 / w ...(6)
Next, we substitute equations (5) and (6) into equation (4):
6 / w = (15/w) - 3
Multiplying both sides of the equation by "w", we get:
6 = 15 - 3w
Rearranging the equation, we get:
3w = 15 - 6
3w = 9
w = 9/3
w = 3
Now that we know the walking rate is 3 mph, we can substitute it back into equation (1) to find the driving rate:
r = 12w
r = 12(3)
r = 36
Therefore, the driving rate is 36 mph.
Finally, we can use equation (2) to find the time spent driving:
Time spent driving = 180 / r
= 180 / 36
= 5
Therefore, it took the motorist 5 hours to drive the 180 miles.