To go to school in the morning. I first walk to the garden shed at 6 km/h to collect my bicycle. I then cycle to school at 15 km/h. The total journey normally takes 18.5 minutes, One day, I am lote and I run to the shed at 18 km/h and cycle at 27 km/h. The journey takes me 10 min 10 s. How far is it from my house to the garden shed?

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