A radiator contains 10 quarts of 30% of which is antifreeze. How much fluid should be drained and replaced with pure antifreeze so that the new mixture 40% antitreeze?

Current amount of antifreeze in quarts is `30/100xx10 =3`

Target is `40% -> 4` quarts
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Let the amount drained of and replaced with pure antifreeze be `x`

The amount left after draining off is `10-x`

The amount of antifreeze in this is `30/100(10-x)`

The replacement is pure antifreeze which is the amount `x`

so `30/100(10-x)+x=4` quarts

`3-3/10x+x=4`

`3+x(1-3/10)=4`

`x=(4-3)/(1-3/10)`

`x=1xx10/7 = 1 3/7 "quarts"`

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`color(blue)("Check")`

`10 -1 3/7 = 8 4/7`

`=(30/100xx8 4/7) + 1 3/7`

`=(3/10xx60/7)+10/7`

`=(3xx6)/7+10/7`

`=28/7 =4 larr" 4 litres of pure antifreeze mixed into 10 quarts"`

`color(red)("Correct")`

`color(blue)("Preamble")`

This is a very effective method

You are blending two liquid with different concentration of antifreeze. One liquid has 30% concentration and the other 100% concentration.

If you have all of the original the antifreeze is 30%. If you have all the other the antifreeze content is 100%.

Connect the two points on a graph and you have a model for those conditions and all the other blend concentrations in between. From this we can and may determine the blend that gives us 40%. Sometimes it takes a little thought about how so set up the x-axis which represents the quantities of the blend.
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`color(blue)("Answering the question")`

The slope of part of the graph is the same as the slope of all of it.

`("change in y")/("change in x")->("change blend concentration")/("change in added pure antifreeze")`

`(100-30)/(10) -=(40-30)/x`

As we need `x` turn everything upside down:

`10/(100-30)-=x/(40-30)`

`10/70=x/10`

`x=100/70 ->(100-:10)/(70-:10) = 10/7 = 1 3/7` in quarts