Question 288736
Let V1 = the volume of the 10% solution.
Let V2= the volume of the 20% solution.

Working in Litres (1 Litre = 1000mL), and decimals instead of percents (i.e. 12%=0.12)

Eqn 1:  0.10*V1 + 0.20*V2 = 0.12*1.0
Eqn 2: V1 + V2 = 1.0 L


Isolate V2 in equation 2:


V2 = 1 - V1


Substitute this value for V2 into Eqn 1:

0.10*V1 + 0.20(1-V1) = 0.12
-0.10*V1 + 0.20 = 0.12
-0.10*V1 = -0.08
V1 = 0.8


Eqn 2 tells us that V1 + V2 must equal 1.00 Litres

Therefore V2=1-0.8

V2=0.2 L


So you would need 800 mL (0.8 L) of the 10% solution and 200mL (0.2L) of the 20% solution.


Hope this helps. If you have any questions, just let me know!