A trader sells goods to a customer at a profit of k% over the cost price, besides it he cheats his customer by giving 880 g only instead of 1 kg. Thus his overall profit percentage is 25%. Find the value of k?

Cost price of 1000 gm is 1000
In that sense cp should have been 880 for 880 gm
But he makes a profit of 25 percent on cost,880*125/100=1100
Initial cost 1000.Therefore profit =100/1000*10=‌‌10 percent

let, CP of 1kg = 100
SP = 100+k
so, CP of one goods = 100/1000 = 1/10
the trader cheats his customer by giving 880 g only instead of 1 kg
so, SP of one goods = (100+k)/880
so, gain = [(100+k)/880 -1/10] = (100+k-88)/880 = (k+12)/880
APQ, [[(k+12)/880]/(1/10)] * 100 = 25
or, (k+12)/88 = 1/4
or, k+12 = 22
or, k = 10
so, value of k is 10%.

Short and simple...
Cp=100 , profit = k% => SP = 100+ K
NOW
ACTUAL CP IS FOR 880g not for 1000g
CP = 88
P%(given)=25%=(sp-cp)/(cp)
25/100=(100+k-88)/88
1/4=(12+k)/88
22=12+k
[K=10%]

===
Applying MF
100×(100+k)/100×100/88-100= 25
=> (100+k)×(25/22)= 125
=> (100+k)×25= 125×22
=> 2500+25k= 125×22
=> 25k= 125×22-2500= 250
=> k= 10
So the value of k= 10[Answer.]
===

Let C.P= 100
S.p=
880gm- 100+k
so,(1000-880)=120gm s.p is (300+3k)/22
Now,Total profit= 100+K+(300+3k)/22= (2500+25K)/22
A/Q,
(2500+25K)/22-100=25
so,K= 10%

Let C.P= 100
S.p=
880gm- 100+k
so,(1000-880)=120gm s.p is (300+3k)/22
Now,Total profit= 100+K+(300+3k)/22= (2500+3K)/22
A/Q,
(2500+3K)/22-100=25
so,K= 8.33%

it would be 25% instead of 255

can anyone explain this solution?

Pleas explain this question according to formula