
Three tables are purchased for Rs.\[2500\] each. First is sold at a profit of \[8\% \], the second is sold at a loss of \[3\% \]. If their average selling price is Rs.\[2575\], find the profit percent on the third.
Answer
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Hint: The given problem is based on profit and loss concept; in this problem we can find profit percentage using the formula \[\left( {\dfrac{{profit}}{{cp}}} \right) \times 100\]and to find the selling price of first two tables we can make use of the formula \[sp = \left( {\dfrac{{100 + profit\% }}{{100}}} \right) \times cp\] and \[sp = \left( {\dfrac{{100 - loss\% }}{{100}}} \right) \times cp\] as per the requirement of the problem.
Complete step by step solution:
Given The cost of \[1\] table = Rs.\[2500\].
Then, the Cost price of the three tables = Rs. \[2500 \times 3 = 7500\]
Given The average Selling price of the \[3\] tables = Rs. \[2575\].
Then, the total Selling price = Rs. \[2575 \times 3 = 7725\]
Now we have to calculate selling price of the first table using the formula \[sp = \left( {\dfrac{{100 + profit\% }}{{100}}} \right) \times cp\]
\[ \Rightarrow \] given The Selling price of the first table at \[8\% \] profit by using above formula
we get =\[\left( {\dfrac{{100 + 8}}{{100}}} \right) \times 2500\]
\[\therefore \] The Selling price of the first table=2700 Rs
Now we have to calculate selling price of the first table using the formula \[sp = \left( {\dfrac{{100 - loss\% }}{{100}}} \right) \times cp\]
\[ \Rightarrow \] given The Selling price of the second table at \[3\% \] loss by using above formula
we get =\[\left( {\dfrac{{100 - 3}}{{100}}} \right) \times 2500\]
\[\therefore \] The Selling price of the second table=2425 Rs
Then, the sum of the Selling price of two tables = \[(2700 + 2425) = 5125\]
So, the Selling price of the third table = The total of the Selling price − the sum of the Selling price of two tables
\[ = (7725 - 5125) = 2600\]Rs
Profit on the third tables \[ = (2600 - 2500) = 100\]Rs
$ = \dfrac{{100}}{{2500}} \times 100\% $
$
= \dfrac{{100}}{{2500}} \times 100\% \\
= 4\% \\
$
$ = 4\% $
Therefore, profit percent on third table is \[4\% \]
So, the correct answer is “ \[4\% \]”.
Note: To solve the above problem we have used the direct formula along with unitary method (if the value of one unit is given then multiply the value of a single unit to the number of units to get necessary value) where ever necessary because as in the problem cost of each table is given.
Complete step by step solution:
Given The cost of \[1\] table = Rs.\[2500\].
Then, the Cost price of the three tables = Rs. \[2500 \times 3 = 7500\]
Given The average Selling price of the \[3\] tables = Rs. \[2575\].
Then, the total Selling price = Rs. \[2575 \times 3 = 7725\]
Now we have to calculate selling price of the first table using the formula \[sp = \left( {\dfrac{{100 + profit\% }}{{100}}} \right) \times cp\]
\[ \Rightarrow \] given The Selling price of the first table at \[8\% \] profit by using above formula
we get =\[\left( {\dfrac{{100 + 8}}{{100}}} \right) \times 2500\]
\[\therefore \] The Selling price of the first table=2700 Rs
Now we have to calculate selling price of the first table using the formula \[sp = \left( {\dfrac{{100 - loss\% }}{{100}}} \right) \times cp\]
\[ \Rightarrow \] given The Selling price of the second table at \[3\% \] loss by using above formula
we get =\[\left( {\dfrac{{100 - 3}}{{100}}} \right) \times 2500\]
\[\therefore \] The Selling price of the second table=2425 Rs
Then, the sum of the Selling price of two tables = \[(2700 + 2425) = 5125\]
So, the Selling price of the third table = The total of the Selling price − the sum of the Selling price of two tables
\[ = (7725 - 5125) = 2600\]Rs
Profit on the third tables \[ = (2600 - 2500) = 100\]Rs
$ = \dfrac{{100}}{{2500}} \times 100\% $
$
= \dfrac{{100}}{{2500}} \times 100\% \\
= 4\% \\
$
$ = 4\% $
Therefore, profit percent on third table is \[4\% \]
So, the correct answer is “ \[4\% \]”.
Note: To solve the above problem we have used the direct formula along with unitary method (if the value of one unit is given then multiply the value of a single unit to the number of units to get necessary value) where ever necessary because as in the problem cost of each table is given.
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