* Let  the Sale Price of  both items be x and one item is sold at a profit percent p1 & another at profit per cent p2. Both p1,p2 can assume positive or negative values for profit and loss respectively.

* Then total sale price -> 2x ; Total purchase price -> 100x[ (1/(100+p1)) + (1/(100+p2)) ] = 100x*A=200x/A1

* Total Profit/Loss --> 2x - 100x[ (1/(100+p1)) + (1/(100+p2)) ] = 2x - 100x*A

* % of  Profit / Loss -->( 2/[1/(100+p1)  + 1/(100+p2) ] )  - 100 = A1 - 100

* If one is sold at certain profit per cent and another at same loss per cent, then p1=-p2 and then % of profit/loss is[ (100+p1)(100-p1)/100] -100 = A2-100

* If p1 and p2 are such that in net there is net P/L of zero %, then the conditions are M=(-p1p2)/(p1+p2) = 50 where p1,p2 can have values - 100 to +100. M=0 when p1 or p2 become 0 and M is infinity when p1=-p2. Upper bound of M is infinity and lower bound -50 ( by putting p1 as any figure between 99 and 100 and p2 any figure between 99 to 100). M= 50 when p1=75 and p2=-30 or p1=30 & p2=-75. Other values of p1 and p2 for which M is 50 can be explored. When p1=any fig. between -99 & -100 and p2 is also between -99 and -100 , M -----------> 50 but that is not a point of Break Even.

* It will be observed that % of Net Profit/Loss is independent of sale price.

* If 2 items are sold at sane price and in one item there is a certain % of profit, and in 2nd item there is same % of loss, then the net % is always loss.