Distance, rate, time problem. Would you help?

Question

On the first part of a 317 mile trip, a salesperson averaged 58 miles per hour. He averaged only 52 miles per hour on the last part of the trip because of an increased volume of traffic. The total time of the trip was 5 hours and 45 minutes. Find the amount of time at each of the two speeds.

Answer 1

time = distance / speed



let x be the first part of the trip, at 58 miles / hr

t for this part = x / 58



so the second part will be (317 - x) at 52 miles / hr

t for this part = (317 - x) / 52



total time = 5.75 hours (45 min = .75 hour)



so the equation is:

x / 58 + (317 - x) / 52 = 5.75

multiply by 58 * 52 to clear the fraction



52x + 58(317 - x) = 5.75(58)(52)

52x + 18386 - 58x = 17342

18386 - 17342 = 6x

1044 = 6x

x = 1044 / 6 = 174



therefore, he drove 174 miles at 58 miles / hr (3 hours) and

317 - 174 = 143 miles at 52 miles / hr (2.75 hours)



total: 317 miles in 5.75 hours (an average of 55.13 mph, by the way)

Answer 2

HINT: Write what you know in a table.



First, convert 5 hours and 45 minutes into hours.

5 hours + [45 minutes * (hour / 60 minutes)] =

5 hours + 0.75 hours =

5.75 hours



................ d ............ r ....... t

first part … d …......... 58 .… t

last part … 317 - d .... 52 .… 5.75 - t

total .....… 317 .......... --- .... 5.75



Remember the distance formula:

d = r * t



Apply this to both parts.



first part:

d = 58 * t

d = 58t



last part:

317 - d = 52(5.75 - t)

317 - d = 52(5.75 - t)

317 - d = 299 - 52t



You have 2 equations.

d = 58t

317 - d = 299 - 52t



Add the two equations together.

d = 58t

317 - d = 299 - 52t

-------------------------------

d + 317 - d = 58t + 299 - 52t

317 = 6t + 299

317 - 299 = 6t + 299 - 299

18 = 6t

18 / 6 = 6t / 6

3 = t



Plug this solved t value into the table.

................ d ............ r ....... t

first part … d …......... 58 .… t = 3

last part … 317 - d .... 52 .… 5.75 - t = 5.75 - 3 = 2.75

total .....… 317 .......... --- .... 5.75



ANSWER: The first part is 3 hours; the second part is 2.75 hours.



------------------------



CHECK:



Use the d = r * t formula to make sure that the distance for the two legs add up to 317 miles.



................ d .................................... r ....... t

first part … d = 58 * 3 ….............. 58 .… 3

last part … 317 - d = 52 * 2.75.... 52 .… 2.75

total .....… 317 .............................. --- ...... 5.75



first part:

d = 58 * 3

d = 174



second part:

317 - d = 52 * 2.75

317 - d =143

d = 174



The two distances add up to 317 miles, so the answer is correct.

Answer 3

let the distance covered in first part = x miles



then the distance covered in last part = (317 - x ) miles



time taken for first part = d /r = x / 58



time taken for last part = (317 - x) / 52



total time taken = x / 58 + (317 - x) / 52 = [26x +9193 - 29x ] /1508



=( 9193 - 3x ) /1508



so ( 9193 - 3x ) /1508 = 5 h 45 m = 5 + 45/60 = 23/ 4 h



9193 - 3x = (23/4)(1508) = 8671



3x = 9193 - 8671 = 522



x = 522 /3 = 174 miles



time taken for first part = 174 / 58 = 3 hours



time taken for last part = 5 h 45 m - 3 = 2 hours 45 minutes

Answer 4

Charts always helped me. I'd make a chart with d, r and t across the top, and down the left I'd put trip out on one line and trip home on the second line. Now you know the formula d=rt, and in this case the distance is the same, or d=d, which means that rt=rt. You are trying to solve for t, or time, so let's try to write an equation that expresses rate in terms of t. The total time is 5 hours and 45 minutes, which is .75 of an hour so lets call it 5.75. We can say he travelled t hours at 58 mph and t hours at 52 mph for a total of 5.75 hours, so

58t + 52t=5.75. 110t=5.75, t=.052(approx). Now put that value back into the rates. The first part of his trip was 58t, so 58*.052=3 hours (approx) and the second part of his trip was 52t, so 52* .052=2.7 (approx). Check by adding the times: 3+2.7=5.7, which is approximately 5.75 hours. Finish with a complete sentence: The salesman completed the first part of his trip in approximately three hours and the second part of his trip in approximately 2.7 hours.