"The cooler in a car contains $8$ litres. The coolant fluid contains $\dfrac3{10}$ of glycol and rest is water. To increase the glycol content to $\dfrac35$ you drop some of the coolant fluid and fill on with glycol. How much coolant fluid do you need to drop?"
So I'm having trouble solving this, the cooler contains a total of $0.3\times8 = 2.4$ litres of glycol in the beginning. In the end it has to contain $0.6\times8 = 4.8$ litres. So I thought I could do this:
$2.4 - 0.3x + x = 4.8$
$3.1x = 4.8$
$x = 1.55$ , well kind of.
But the answer should be that you need to drop $3.4$ litres.. some I'm stuck. Any help is appreciated!
If $x$ is the dropped amount, you have $0.3\times(8-x)$ of the original and $x$ added glycol at the end, and you want $0.3(8-x)+x=4.8$. So far you have that, but in the next line you get to $3.1x=4.8$. I have no idea where it's from. 2.4+1-0.3? But the first term doesn't have $x$ in it. It should be $0.7x=2.4$.