Question 788263
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Let f=how much volume of 90% acid
Let y=how much water to add to get 60% acid.


{{{90*f/(y+f)=60}}}
{{{90f=60(y+f)}}}
{{{90f=60y+60f}}}
{{{9f-6f=6y}}}
{{{3f=6y}}}
{{{highlight(y=f/2)}}} liters of water


The second dilution now starts with (f+f/2) liters of 60% acid.
You want now to dilute to 40% acid.
Let x=volume of water to add for diluting to 40%.


{{{60*(f+f/2)/(x+f+f/2)=40}}}
Solve for x.