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July 31st, 2019, 06:56 AM  #1 
Senior Member Joined: Aug 2014 From: India Posts: 476 Thanks: 1  Explain this solution for this profit/loss problem?
By selling an article at 80% of its MP, a merchant makes a loss of 12% what will be the profit /loss percent made by the merchant if he sells the article at 95% of its M.P? Sol: 80% is equated to 88 95% is equated to 104.5 104.5  100 = 4.5 Why 80% is equated to 88 & 95% is equated to 104.5?? Last edited by Ganesh Ujwal; July 31st, 2019 at 06:58 AM. 
July 31st, 2019, 07:19 AM  #2 
Global Moderator Joined: Dec 2006 Posts: 21,028 Thanks: 2259 
80% of M.P. corresponds to 88% (i.e. 100%  12%) of C.P. 88 × 95/80 = 104.5 exactly, so the answer is 4.5% (profit). 
July 31st, 2019, 07:36 AM  #3 
Senior Member Joined: Aug 2014 From: India Posts: 476 Thanks: 1 
How did you write this equation: 88 × 95/80 = 104.5?

July 31st, 2019, 08:15 AM  #4 
Global Moderator Joined: Dec 2006 Posts: 21,028 Thanks: 2259 
When the selling price is changed from 80% of the marked price to 95% of the same marked price, it is multiplied by 95/80. The original selling price is 88% of the cost price (because the merchant is making a 12% loss). To get the corresponding percentage of the cost price after the selling price is multiplied by 95/80, one just multiplies that 88% by 95/80, which gives 104.5%. Anything over 100% of the cost price is profit, so there is a profit of 4.5% for the merchant after the price change. 
July 31st, 2019, 07:39 PM  #5  
Senior Member Joined: Aug 2014 From: India Posts: 476 Thanks: 1  Quote:
Explain this fraction; 95/80 ;if something is changed into new thing, then fraction becomes something/new thing. in profit and loss problems??  
July 31st, 2019, 10:54 PM  #6 
Global Moderator Joined: Dec 2006 Posts: 21,028 Thanks: 2259 
If v is the original selling price, v = 0.8 M.P. If w is the new selling price, w = 0.95 M.P. Hence both w = (0.95/0.8)v and w = v + 0.15 M.P. As the value of M.P. isn't given, the first equation is more useful. For a different profit/loss problem, the second equation might be more useful. 

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