Question

Question #c3948

Answer 1

`color(green)("Method 1")`

`color(blue)("Fuel at "$3.00 -> 80" gallons")`
`color(blue)("Fuel at "$2.20-> 280-80 = 200" gallons")`

Explanation:

There are at least 2 ways of solving this and I am going to show you the 2 ways I know.

`color(blue)("Method 1")`

As an example, suppose we had 80 gallons if the $3.00 gas. Then it would mean that we have 280 - 80 gallons of the $2.20 gas.

From this you can observe that the quantities of each type of gas are directly linked. So if you consider just one of them the quantity of the other one 'looks after itself'.

I select the more expensive gas at £3.00/ gallon

If we have none of this then we have all of the 280 gallons as the $2.20 gas. If we have only the $3.00 gas then we have none of the $2.20 gas.

Cost if all `$2.20 = 280xx$2.20=$616.00`
Cost of all `$3.00=280xx$3.00=$840.00`

Let `x` be the amount of $3.00 gas in the blend to give £680

Then we have:

The gradient of part is the same as the gradient of the whole

`=>(840-616)/280 = (680-616)/x`

`x=(280(680-616))/(840-616) = 80" gallons"`

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Fuel at "$3.00 -> 80" gallons")`
`color(blue)("Fuel at "$2.20-> 280-80 = 200" gallons")`

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(red)("Check")`

`color(purple)((80xx$3.00)+(200xx2.2)= $680.00)`

Answer 2

`color(green)("Method 2")`

`color(blue)( 80" gallons "$3.00" fuel")`

`color(blue)(200" gallons "$2.20" fuel")`

Explanation:

As stated in method 1 the two quantities are directly linked in that:

`color(blue)("quantity of $2.20 fuel = total quantity - quantity of $3.00 fuel")`

However we need this to include prices

Given: Target price is $680

Let the target quantity of $3.00 fuel be `x`

Then quantity of $2.20 fuel is `280-x`

So now we can write:

`color(blue)($2.20(280-x) +$3.00x=$680)`

dropping the $ sign
.............................................................................................
Mathematically it can be demonstrated why the $ sign disappears but that is outside the scope of this solution
..............................................................................................

`color(blue)(616-2.2x+3x=680)`

`color(blue)(616+0.8x=680)`

`color(blue)(0.8x=64)`

`color(blue)(x=64/0.8=80)`

`color(blue)(=> 80" gallons "$3.00" fuel")`

`color(blue)(=>280-80 = 200" gallons "$2.20" fuel")`