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bhanwar lal kuldeep

· started a discussion

· 1 Months ago

improve solution method (Question: 109911.0 )

please improve solution method not use basic methods

Question:
An article is sold for \(\unicode{x20B9} \)500 and hence lost something. Had the article sold for \(\unicode{x20B9} \)700, the merchant would have gained three times the former loss. Find the cost price of the article.
Options:
A) \(\unicode{x20B9} \)525
B) \(\unicode{x20B9} \)550
C) \(\unicode{x20B9} \)600
D) \(\unicode{x20B9} \)650
Solution:
Ans: (b) Let loss in the first case = \(\unicode{x20B9} \)x

Then profit in second case = \(\unicode{x20B9} \)3x

\(\therefore\) \(\unicode{x20B9} \)500 = cost price - \(\unicode{x20B9} \)x

and, \(\unicode{x20B9} \) 700 = cost price + \(\unicode{x20B9} \) 3x .........(i)

Solving equations (i) and (ii), we get............... (ii)

4x = \(\unicode{x20B9} \)200 or x = \(\unicode{x20B9} \)50 

\(\therefore\) cost price = selling price (first) + loss

= \(\unicode{x20B9} \)500 + \(\unicode{x20B9} \)50=550

or cost price = selling price (second) - profit

= \(\unicode{x20B9} \)700 - \(\unicode{x20B9} \)3 x \(\unicode{x20B9} \)50 = \(\unicode{x20B9} \)700 - \(\unicode{x20B9} \)150 = \(\unicode{x20B9} \)550

Narender

· commented

· 1 Months ago

CP-500/700-CP=loss/3loss calculate cp
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