Factorisation in algebra

Question

c. you are laying the yellow and green brick road. the road is 1m wide and each brick is 500mm x 250mm. there are 3 yellow bricks for every green brick. the road will be built with 12000 yellow bricks. how long will the road be?

Answer 1

There are several good ways to solve this problem, so I am glad that you mentioned:

factorisation - (mathematics) the resolution of an entity into factors such that when multiplied together they give the original entity

 

First, read and re-read the problem until you understand it and can put it into your own words.  How's this?  Using a total or (12000 yellow + ? green) bricks, how long will the road be? (brick size given; 3y:1g bricks)

 

Solution plan:  It looks like the total number of bricks must be calculated.  Then, we find the area of one brick.  The area of the road will be the area of 12000 bricks.  The length of the road will be the area of the road divided by 1 m wide.  We also need the information that 1 m = (100 cm/m)*(10 mm/cm) = 1000 mm.

 

O.K., let's do the math!

  There are 3 yellow bricks for every green brick and there are 12000 yellow bricks.  That's 12000 yellow bricks for 4000 green bricks.  (note: this is a proportion is you want to set up an equation).  The total number of bricks is 16000.

 

  The area taken up by 16000 bricks is:

    A = (16000 bricks)*((500 mm)*(250 mm)/brick)

     [note: now we convert units without changing values by using the multiplicative identity; that is, 1]

    A = (16000 bricks)*(500 mm)*1*(250 mm)*1

    A = (16000 bricks)*(500 mm)*((1 m)/(1000 mm))*(250 mm)*((1 m)/(1000 mm))

     [note: bricks cancels out; mm cancels out two times; leaving only square meters]

    A = (16000)*(500)*(1/1000)*(250)*(1/1000) m²

     [note: I write it that way to paste into Excel]

    A = 2000 m²

 

   The length of the road is:

    L = (2000 m²)/(1 m)

    L = 2000 m

 

[also note:  2000 m = 2 km]