Answer

Verified

345.9k+ views

**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.

Recently Updated Pages

How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE

Why Are Noble Gases NonReactive class 11 chemistry CBSE

Let X and Y be the sets of all positive divisors of class 11 maths CBSE

Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE

Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE

Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE

Trending doubts

Which are the Top 10 Largest Countries of the World?

The 3 + 3 times 3 3 + 3 What is the right answer and class 8 maths CBSE

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

How many crores make 10 million class 7 maths CBSE

Difference Between Plant Cell and Animal Cell

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Change the following sentences into negative and interrogative class 10 english CBSE

Write a letter to the principal requesting him to grant class 10 english CBSE