Question 913873
how many liters of water should be added to 20 liters of a 70 percent antifreeze solution to make a 50 percent acid solution?
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"20 liters of a 70 percent antifreeze solution" means we have


20*0.70 = 14 liters of pure antifreeze
20-14 = 6 liters of pure water (alternate calculation: 20*0.3 = 6)


Let x = amount of pure water to add


We have 14 liters of pure antifreeze. This figure does NOT change when we add the pure water.


We have 20 liters of solution. This gets bumped up to 20+x when we add the pure water.


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So we go from {{{14/20}}} to {{{14/(20+x)}}}. The fraction represents the percentage of antifreeze in the solution. 


We want this percentage to be 50%, so set the fraction equal to 0.50 and solve for x.


<pre>
	{{{14/(20+x)=0.50}}}
	
	
	{{{14=0.50(20+x)}}}
	
	
	{{{14=0.50(20)+0.50(x)}}}
	
	
	{{{14=10+0.50x}}}
	
	
	{{{14-10=0.50x}}}
	
	
	{{{4=0.50x}}}
	
	
	{{{4/0.50=x}}}
	
	
	{{{8=x}}}
	
	
	{{{x=8}}}
</pre>


So you need to add <font color="red">8 liters</font> of pure water to make a 50 percent antifreeze solution.


Let me know if you need more help or if you need me to explain a step in more detail.
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Thanks,


Jim