Help! Math problem of the week

[Tung]

My sister has this problem she can't solve. I can't solve it either! Please help out

You have a 100 ML flash containing 60 ML of 20% acid. You need to make a 40% solution. On the shelf there are three bottles:
A 200 ML bottle full of a 50% solution
A 200 ML bottle half full of a 60% solution
A 200 ML bottle with 10 ML of a 70% solution
Describe how you would make a 40% solution without using another container.

Thanks in advance.

 

[kt]

pour 50ML of 50% into the 10ML of 70% bottle, then add 40ML of 20% and you 100ML of 40% acid in the bottle that originally contained 10ML of 70% acid.

 

[minendo]

<< Describe how you would make a 40% solution without using another container. >>


I think this means you wont be able to make accurate measurments.

 

[dullard]

There are an infinite amount of ways to do this:
You have 12 mL acid and 48 mL water at the moment.

Lets add stuff from flask into the third bottle. The third bottle has 7 mL acid and 3 mL water.

Now you have 19 mL acid and 51 mL water.

Is this the concentration you want?
19/(19+51) = 27.1%.
No.

Lets forget bottle two for now (for fun you can use it).

Add 2*x mL of the stuff from bottle one to your new 200 mL bottle.
Thus you will be adding x mL acid and x mL water.

Now you have (x+19) mL acid and (x+51) mL water.

You want a specific concentration:
(x+19)/(x+19+x+51) = 4/10
or
(x+19)/(2x+70)=4/10

Solve for x:

x+19 = 4/10 * (2x+70)

x+19 = 4/5*x + 28

1/5*x = 9

x = 9*5 = 45.

Check the concentration:
You now have 19+45 = 64 mL acid and 51+45 = 96 mL water.
64/(64+96) = 40%.

Did you overflow the bottle?
64 mL acid + 96 mL water = 160 mL.
No - your problem is solved.

Edit: to accurately measure the 2 x = 90 mL, use your now empty 100 mL flask.

 

[RadMan]

Does the question mention how many mL of 40% solution you want?