Answer:
Trip A from House to Shed is 0.062 km
Trip B from Shed to School is 4.47 km
Explanation:
Set up a table (attached) that summarizes and organizes the data. Let A be the distance from the hosue (I assume) to the shed, and B the distance from the shed to the school. We know that distance = (speed)*(time).
The total time for a trip is the sum of the two segments, A and B, divided by the speed for each segment.
TRIPS
1 2
House/Shed A 6 18 km/hr
Shed/School B 15 27 km/hr
18'30" 10'10" min
18.5 10.16 min
0.308 0.169 hr [Convert the minutes into hours]
[The distance divided by the speed will give the time required for each segment, A and B].
Trip 1: 1) A/6 + B/15 = 0.308
Trip 2: 2) A/18 + B/27 = 0.169
Trip 1: A/6 + B/15 = 0.308
A + (6/15)B = 1.85 [Multiply each side by 6]
A = 1.85 - (6/15)B [Rearrange]
A = 1.85 - (0.4)B [This will be the defination of A we'll use in the second equation]
Trip2: A/18 + B/27 = 0.169
(1.85 - (0.4)B)/18 + B/27 = 0.169 [Use A= 1.85 - (0.4)B from above]
27(1.85 - 0.4B)/18 + B = 27*(0.169) [Multiply each side by 27]
1.5*(1.85-0.4B) + B = 4.58
2.78 - 0.6B + B = 4.58
0.4B = 1.85
B = 4.47 km
Now use B = 4.47 km in either original equation to find A.
A/18 + B/27 = 0.169
A + (18/27)B = 18*(0.169) {Multiply both sides by 18]
A + (0.67)B = 3.042
A + (0.67)(4.47) = 3.042
A + (2.98) = 3.042
A = 0.062 km
Trip A from House to Shed is 0.062 km
Trip B from Shed to School is 4.47 km