P6 Math: Flour and Sugar

Shalih bought some bags of flour and sugar with the mass and prices shown below. The total mass of the flour is twice the total mass of the sugar. He paid a total amount of $127.20. Find the total mass of the flour and sugar bought by Shalih.

(M1) In this type of question there are two units involved, mass and price.

Total mass of flour = 2u

Total mass of sugar = u

Number of flour bags bought = f

Number of sugar bags bought = s

So,

2u = 0.7f and u= 0.2s (mass equations)

127.20 = 1.8f + 2s (price equation)

Solve the mass equation:

2u = 0.7f and 2u = 2 x 0.2s = 0.4s 

--> s : f = 7 : 4

Substitute to the price equation:

127.20 = 1.8f + 2 x (7/4f) = 5.3f --> f = 127.20 / 5.3 =  24

There were 24 flour bags.

s = (7/4)f = 42 

There were 42 sugar bags.

Total mass = (24 x 0.7) + (42 x 0.2) = 25.2 kg

(M2) To ease this kind of problem for Primary school level, the mass of floor and sugar should be always multiple of the units (e.g bags, sacks, ...) shown for each item.

For example:

Mass of fluor has to be : 0.7, 1.4, 2.1 ...
Mass of sugar has to be : 0.2, 0.4, 0.6 ...

then find the combination of mass of fluor and mass of sugar that satisfy the condition that mass of the fluor is twice mass of the sugar. It can be done by iterating multiple of 0.7 and 0.2, and check when mass of fluor is twice mass of sugar AND the total price is $127.20. 

0.7, 1.4, 2.1, 2.8, 3.5, 4.2 ...

0.2, 0.4. 0.6, 0.8, 1.0, 1.2, 1.4 ... 

It shows the first combination is 2.8 kg of fluor and 1.4 kg of sugar.

2.8 F + 1.4 S = 4x1.80 + 7x2 = 21.2 <--- 1st set (4 fluor bags and 7 sugar bags).

1 set is $21.2

In $127.20 there are 6 sets.

It means 6x4 = 24 bags of fluor and 6x7 = 42 bags of sugar.