For the above-mentioned: You need to first understand what the question asks you. The percentage of alcohol in the total mixture. For example, 100ml of drinking water has 2 elements which are A and B and the percentage of A is 30%, it means out of 100 ml, 30% or 30 ml is element A.

Now coming back to the question. It tells you that the solutions contain 10%, 14%, and 20%. Now out of 100, 10g is alcohol, 14 g and 20g in A, B, and C.
But here the mentioned quantity is 80g. So (8010)/100 would result in 8g. Similarly, you can calculate for 20g of C for Sol B: This will be (x14)/100

Calculation QA) (0.14x+8+4)/100+x = 13/100
The total percentage of the alcohol in the mixed solution should be 13%

Regarding question 2:
You can just plug in the quantity B and it makes :

80 gm of 10 % alcohol soln

80 gm of 14% soln

20 gm of 20% soln – let’s dilute it to make 80 gm of each which is : alcohol = 20 % of 20 = 4 => 4/80 * 100 = 80 gm of 5 % alcohol.
Now we have :

80 gm of 10 % alcohol soln

80 gm of 14% soln

80 gm of 5% soln
You can either solve it:
Since the amounts are same, avg = (10+14+5)/3 = 9.66 % of 240 gm solution
OR
just simply apply the logic: one 14 % of 80gm cannot compensate for the remaining 160gm to make 13 % solution mix

So, more amount of x that is solution B is required. So the option A is correct.

Thank you so much. Actually I thought this was a mixtures sum where greg used the Tug of war method to find out
he would take the two quantities
like
suppose its 80 100
then he wud find the distance and select the value closer to it and later flip the ratios