A person solds an article at profit of 15%. If he sold it for Rs 81 less his loss would have been 12℅.Find the cp of the article
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Answered by
7
Because CP of any article is 100%
so SP = 115%
if 12% loss occurs then New SP 88%
so according to question
115% - 88% = 81
then 27% = 81
then 100% = 300
therefore CP = 300 Rs.
so SP = 115%
if 12% loss occurs then New SP 88%
so according to question
115% - 88% = 81
then 27% = 81
then 100% = 300
therefore CP = 300 Rs.
suka75:
Thank you
Answered by
2
i have given same type of answer.
I am giving u same answer u can follow that
he question is simple and can be solved in many ways.
Easy to understand method-
Let the cost be ‘X’
He sold for ‘1.1X’
If he had charged Rs. 45 more, the profit would be 25%. That brings us to the equation- 1.1X+45 = 1.25X
Solving this will get you to X=300
Now, the simple method-
Use this method if you have complete understanding of the concept-
If Rs. 45 fetches you 15% extra what would be the 100% i.e. in terms of equation X+45= 1.15X
Solving this would give you the answer X=300
Lets verify the answer -
So the cost price is Rs. 300.
The man sold it for Rs. 330 i.e., Rs. 30 at 10% profit
and by adding Rs. 45 he could have sold it for Rs. 375 i.e.,Rs. 75 at 25% profit.
Feel free to ask for any clarifications.
I am giving u same answer u can follow that
he question is simple and can be solved in many ways.
Easy to understand method-
Let the cost be ‘X’
He sold for ‘1.1X’
If he had charged Rs. 45 more, the profit would be 25%. That brings us to the equation- 1.1X+45 = 1.25X
Solving this will get you to X=300
Now, the simple method-
Use this method if you have complete understanding of the concept-
If Rs. 45 fetches you 15% extra what would be the 100% i.e. in terms of equation X+45= 1.15X
Solving this would give you the answer X=300
Lets verify the answer -
So the cost price is Rs. 300.
The man sold it for Rs. 330 i.e., Rs. 30 at 10% profit
and by adding Rs. 45 he could have sold it for Rs. 375 i.e.,Rs. 75 at 25% profit.
Feel free to ask for any clarifications.
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