(Q)The total cost price of two watches is Rs 840. One is sold at a profit of 16% and the other at a loss of 12%. There is no loss or gain in the whole transition. Then their cost price is respectively.
Solution: BY CONCEPT
Let C.P of 1st watch=x
C.P of 2nd watch=840-x
S.P of 1st watch:
x×(100+16)/100=S.P1
116x/100=S.P1
S.P of 2nd watch:
(840-x)×(100-12)/100=S.P2
(840-x)×88/100=S.P2
Now According to question: There is no profit or loss
SO,
S.P1+S.P2=840
116x/100+(840-x)×88/100=840
116x+840×88-88x=840×100
28x=84000-73920
28x=10080
∴ x=360 Ans.
C.P of 2nd watch=840-x
=Rs(840-360)
=Rs 480 Ans.
2nd Method:
C.P of 1st watch at a profit 0f 16%=(3/7)×840
=Rs 360 Ans.
C.P of 2nd watch at a Loss of 12%=(4/7)×840
=Rs 480 Ans
OR
C.P of 2nd watch at a Loss of 12%=Rs(840-360)
=Rs 480 Ans.
(Q) A man bought a horse and carriage for Rs 40000. He sold the horse at a gain of 10% and the carriage at a loss of 5%. He gained 1% on his whole transaction. The cost price of the horse was.
Solution: BY CONCEPT
Let C.P of horse=x
C.P of a carriage=40000-x
S.P of a horse:
x×(100+10)/100=S.P
11x/10=S.P
S.P of a carriage:
(40000-x)×95/100=S.P
Now According to question: he gained 1% on his whole transaction
So,
S.P of a horse+S.P of a carriage=40000×101/100
11x/10+(40000-x)×95/100=40000×101/100
11ox+40000×95-95x=40000×101
15x=40000×101-40000×95
=40000(101-95)
15x=240000
∴ x=Rs 16000 Ans
2nd Method:
C.P of a horse at a profit 0f 10%=(2/5)×40000
=Rs 16000 Ans.
If two article are sold at equal prices, the first one is sold at P1% Profit and whereas the second one at a Profit of P2% And the sum of the cost price of two article is x then
For Example:
(Q) A trader bought two watches for Rs 2300. He sold one at a profit of 10% and the other at a profit of 20%. If the selling price of each watch is the same, then their cost price are respectively.
Solution: By Trick:
C.P of watch at a profit of 10%=((100+20)/(200+10+20))×2300
=(120/230)×2300
=Rs 1200 Ans.
C.P of watch at a profit of 20%=Rs(2300-1200)
=Rs 1100 Ans.
BY CONCEPT:
Let C.P of 1st watch=x
C.P of 2nd watch=2300-x
S.P of 1st watch:
x×(100+10)/100=S.P1
11x/10=S.P1
S.P of 2nd watch:
(2300-x)×(100+20)/100=S.P2
(2300-x)×120/100=S.P2
(2300-x)×12/10=S.P2
Now According to question: selling price of each watch is the same
SO,
S.P1=S.P2
11x/10=(2300-x)×12/10
11x=2300×12-12x
23x=2300×12
x=Rs 1200 Ans
C.P of 2nd watch=2300-x
=Rs(2300-1200)
=Rs 1100 Ans.
If two article are sold at equal prices, the first one is sold at a Loss of L1% and whereas the second one at a Loss of L2%, And the sum of the cost price of two article is x then
For Example:
(Q) A Man purchases two T.V sets for Rs 21,500. He sold one at a Loss of 12% and the other at a loss of 16%. If the selling price of each T.V is the same, then their cost price are respectively.
Solution: By Trick:
C.P of T.V at a Loss of 12%=((100-16)/(200-12-16))×21500
=(84/172)×21500
=Rs 10500 Ans.
C.P of T.V at a Loss of 16%=Rs(21500-10500)
=Rs 11000 Ans.
BY CONCEPT:
Let C.P of 1st T.V =x
C.P of 2nd T.V=21500-x
S.P of 1st T.V:
x×(100-12)/100=S.P1
88x/100=S.P1
S.P of 2nd T.V:
(21500-x)×(100-16)/100=S.P2
(21500-x)×84/100=S.P2
(21500-x)×84/100=S.P2
Now According to question: selling price of each T.V is the same
SO,
S.P1=S.P2
88x/100=(21500-x)×84/100
88x=21500×84-84x
172x=21500×84
x=Rs 1o500 Ans
C.P of 2nd T.V=21500-x
=Rs(21500-10500)
=Rs 11000 Ans.
If two article are sold at equal prices, the first one is sold at a Profit of P% and whereas the second one at a Loss of L%, And the sum of the cost price of two article is x then
For Example:
(Q) A Man purchases two horses for Rs 16,800. He sold one at a Profit of 25% and the other at a loss of 15%. If the selling price of each horse is the same, then their cost price are respectively.
Solution: By Trick:
C.P of horse at a profit of 25%=((100-15)/(200+25-15))×16800
=(85/210)×16800
=(85/210)×16800
=Rs 6800 Ans.
C.P of horse at a Loss of 15%=Rs(16800-6800)
= Rs 10000 Ans.
BY CONCEPT:
Let C.P of 1st horse=x
C.P of 2nd horse=16800-x
S.P of 1st horse:
x×(100+25)/100=S.P1
125x/100=S.P1
S.P of 2nd horse:
(16800-x)×(100-15)/100=S.P2
(16800-x)×85/100=S.P2
(16800-x)×85/100=S.P2
Now According to question: selling price of each horse is the same
SO,
S.P1=S.P2
125x/100=(16800-x)×85/100
125x=16800×85-85x
210x=16800×85
x=Rs 6800 Ans
C.P of 2nd horse=16800-x
=Rs(16800-6800)
=Rs 10000 Ans.
If two article are sold at equal prices, the first one is sold at a Loss of L% and whereas the second one at a Profit of P%, And the sum of the cost price of two article is x then
For Example:
(Q) A trader bought two horses for Rs 19,500. He sold one at a Loss of 20% and the other at a Profit of 15%. If the selling price of each horse is the same, then their cost price are respectively.
Solution: By Trick:
C.P of horse at a Loss of 20%=((100+15)/(200-20+15))×19500
=(115/195)×19500
=Rs 11500 Ans.
C.P of horse at a Profit of 15%=Rs(19500-11500)
=Rs 8000 Ans.
The common and frequently asked problem of Profit and Loss question with solution and Shortcut Tricks for Bank PO, SSC CGL, SSC CHSL, Railway Exam Type-4.1.