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?
A. 8.33%
B. 8.25%
C. 10%
D. 12.5%
Answer: Option C
Solution(By Examveda Team)
% Profit = $$\frac{{25}}{{100}}$$ = $$\frac{{120 + {\text{k}}}}{{880}}$$ → k = 100 Thus, Net % profit = $$\frac{{100 \times 100}}{{1000}}$$ = 10%This Question Belongs to Arithmetic Ability >> Profit And Loss
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Related Questions on Profit and Loss
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B. 45 : 51
C. 47 : 56
D. 47 : 51
A. Rs. 2600
B. Rs. 2700
C. Rs. 2800
D. Rs. 3000
A. A neither losses nor gains
B. A makes a profit of 11%
C. A makes a profit of 20%
D. B loses 20%
A. Rs. 3,750
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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