A man sold 2 pens, one at a loss of L% and one at a profit of P%. What

Consider pen A and pen B.

Essentially, we want to find profit/loss% = [CP(A) + CP(B) - SP (A) - SP(B) ] *100/ [CP(A) + CP(B)] .............(1)

1) if SP (A) =SP (B)

we know, CP(A) -L% CP(A) = SP(A)
so, CP(A) = SP(A)/(1-L%)

similarly, CP (B) = SP(B)/(1+P%)

So, eq(1) becomes: profit/loss% = [SP(A)/(1-L%) +SP(B)/(1+P%) -SP(A) - SP(B)]*100 /[SP(A)/(1-L%) +SP(B)/(1+P%)] .........(2)

Solving will bring down to: [2+ P% -L% -2(1+P%)(1-L%)] *100 / [2+P%-L%] {remember: SP(A) = SP(B)} ...........(3)
We can see that there will be no unique value, so statement I is not sufficient.

2) if P=L
we can make use of eq(2), [SP(A)/(1-L%) +SP(B)/(1+P%) -SP(A) - SP(B)]*100 /[SP(A)/(1-L%) +SP(B)/(1+P%)]

put L=P, you will realise that you can't solve the equation to give one unique value.

Now, put both statements together, make use of eq(3)
profit/loss% = [2+ P% -L% -2(1+P%)(1-L%)]*100 / [2+P%-L%]

here, SP(A) = SP(B); now simply put P=L;
you will get P^2 or L^2 %

again there is no unique value.

So, answer E.

As I don't know how to make use of editing tools, so the equations may look scary. Believe me, they are not.
Just take pen & paper and you can see similar expressions. If you still have doubt, please let me know.