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.

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