A motorist was to travel from town A to town B, a distance of 80 miles. He traveled the first 24 minutes at a certain rate; traffic then increased and for the next 6 minutes he averaged 10 miles per hour less than his original speed; then traffic eased up and he traveled the remaining distance at a rate 50% greater than his original rate. He arrived 22 earlier than he would had he traveled the whole distance at his original rate. Find his original rate.
PLEASE PROVIDE SOME EXPLANATIONS AND A TABLE WOULD BE BEST THANKS!!
let the original rate be x miles/hr
distance covered on first leg = (24/60)x = 2x/5 miles
distance covered on his second leg = (6/60)(x-10) = (x-10)/10
remaining distance = 80 - 2x/5 - (x-10)/10
time for remaining distance = (80 - 2x/5 - (x-10)/10 )/1.5x
time had he gone the whole distance at x mph = 80/x
sum of times of 3 legs + 22/60 = time at one speed
24/60 + 6/60 + (80 - 2x/5 - (x-10)/10 )/(1.5x) + 22/60 = 80/x
2/5 + 1/10 + (80 - 2x/5 - (x-10)/10 )/1.5x + 11/30 = 80/x
13/15 + (80 - 2x/5 - (x-10)/10 )/1.5x ) = 80/x
times 15x
13x + 10(80 - 2x/5 - (x-10)/10 ) = 1200
13x + 800 - 4x - x +10 = 1200
8x = 390
x = 390/8 = 48.75 mph
check:
distance of 1st leg = (24/60)(48.75) = 19.5 miles
distance of 2nd leg = (6/60)(38.75) = 3.875
distance left = 80-19.5-3.875 = 56.625 miles
at speed of 50% more or 73.125 mph, time = 56.625/73.125 hrs or 46.46 min.
total time = 24+10+46.46 or 80.46 min
time going at one speed of 48.75 = 1.6410 hr or 98.46 min
difference = 98.46 - 80.46 = 18
ARGGGG, should have been 22
Can't find my arithmetic error, see if you can find it.
To solve this problem, let's break it down into steps and create a table to organize the information.
Step 1: Define the variables:
Let x be the original speed of the motorist.
Step 2: Calculate the time taken for each segment of the trip:
The first segment - 24 minutes at the original speed (x) = (24/60) hours = 0.4 hours
The second segment - 6 minutes at 10 mph less than the original speed (x-10) = (6/60) hours = 0.1 hours
The third segment - Remaining distance (80 - 24 - 6) at 50% greater than the original speed (1.5x)
Step 3: Create a table to organize the information:
_____________________________
| Segment | Speed (mph) | Time (hours) |
|_______________________|____________|_________________|
| Segment 1 | x | 0.4 |
|_______________________|____________|_________________|
| Segment 2 | x - 10 | 0.1 |
|_______________________|____________|_________________|
| Segment 3 | 1.5x | ? |
|_______________________|____________|_________________|
Step 4: Use the given information to establish equations:
Equation 1: Distance = Speed * Time
For Segment 1: 80 = x * 0.4
For Segment 2: 80 - (x * 0.4) - (x - 10) * 0.1 = 1.5x * ?
(Note: The remaining distance for Segment 2 is 80 minus the distance covered in Segment 1 and Segment 2.)
Step 5: Solve the equations to find the original rate (x):
From Equation 1: x = 200 (after simplifying)
Substitute x = 200 into Equation 2:
80 - (200 * 0.4) - (200 - 10) * 0.1 = 1.5 * 200 * ?
Solve for ?: ?
=? = 0.6
Therefore, the original rate (x) is 200 mph.