Question

The total cost price of 2 articles is 10200 and their selling price are equal one of the two articles is sold at a loss of 12% and the other is sold at a loss of 18% then find the cost price of article which is sold at a 18%????e pls solve this question

Answer 1

Answer:

here is your answer

Step-by-step explanation:

loss of 1 =12%

12% = 12/100= 3/25

so cp =25 loss=3 sp=25-3=22

loss 2nd= 18%

18%= 18/100= 9/50

so cp=50 loss=9 sp=50-9=41

but sp is same in both

so sp1= 22*41= 902, cp1=25*41=1025

so sp2=22*41=902, cp2=50*22=1100

cp1+cp2=10200

so (1025+1100)x=10200

x=4.8

so cp1=1025*4.8= 4920

cp2= 1100*4.8=5280

Answer 2

Answer: Rs.5280

Step-by-step explanation:

• GIVEN:- SP of articles are to be equal.
• TO FIND:- SP of article sold at a loss of 18%
• SOLUTION:- Let that the CP of article sold at a loss of 18% is Rs.x.
• So, CP of another article sold at a loss of 12% is Rs.(10200-x)
• SP of 1st article=$x-\frac{18}{100}x$

                                 =Rs.$\frac{82x}{100}$

• SP of 2nd article=$(10200-x)-\frac{12}{100}(10200-x)$

                                    =$\frac{1020000-100x-122400+12x}{100}$

                                    =Rs.$\frac{897600-88x}{100}$

• According to question,$\frac{82x}{100}=\frac{897600-88x}{100}$

                                           ⇒ 82x+88x=897600

                                           ⇒ 170x=897600

                                           ⇒ x=$\frac{897600}{170}$

                                           ∴ x= Rs.5280

• Hence, the required cost of article is Rs.5280.

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