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