Mass Percentage of an unknown used in acid-base titration

Aqueous K3 contains 5.0g dm-3 mixture NaOH+KOH. Aqueous K4 contains 0.05 mol dm-3 H2 SO4. It was found that 25cm3 K4 is exactly neutralized by 23.80cm3 K3.

a. Calculate the molarity of OH- ions in d K3 mixture.
b. Determine d percentage by mass of NaOH in 1 dm3 of K3.

I need to work out question b. For the life of me I cannot get how to work it out after I have solved for ques a. I have a niggling feeling that the '5g dm-3' plays a role because I did not use it to find the OH- concentration but I can't put my finger on it.
Please help? :(

well, i can try this one

2NaOH + H2SO4 -----------> Na2SO4 + 2H2O
Let grams of NaOH in the mixture = x g

Now that we know that 23.80cc of this 5g/l solution resulted in complete neutralization so, total mass of NaOH + KOH that must have reacted with H2SO4 = 5 x (23.8/1000) = 0.119g

So, 2KOH + H2SO4 -----------> K2SO4 + 2H2O
Let the amount of KOH = 0.119 -x grams

Gram-eq of NaOH = x/40; Gram-eq of KOH = (0.119-x)/56
Gram-eq of H2SO4 = Normality x Vol(liters) = (0.05 x2) x (25/1000) = 2.5 x 10^-3
Now, gram-eq of NaOH + Gram-eq of KOH = Gram-eq of H2SO4
(x/40) + (0.119-x)/56 = 2.5 x 10^-3
([7x + 0.595 - 5x) / 280 = 2.5 x 10^-3
0.595 + 2x = 0.7
2x = 0.105 or x = 0.0525grams
So, 0.0525 grams NaOH are present in 23.80cc solution that we used in the reaction.
So 1000cc will have (0.0525/23.8) x 1000 g NaOH = 2.2g
But as per the strength 1 litre had 5g of NaOH + KOH mixture
So mass percentage of NaOH = (2.2/5) x 100 = 44.11 %