
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
416.1k+ 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.
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
The correct geometry and hybridization for XeF4 are class 11 chemistry CBSE

Water softening by Clarks process uses ACalcium bicarbonate class 11 chemistry CBSE

With reference to graphite and diamond which of the class 11 chemistry CBSE

A certain household has consumed 250 units of energy class 11 physics CBSE

The lightest metal known is A beryllium B lithium C class 11 chemistry CBSE

What is the formula mass of the iodine molecule class 11 chemistry CBSE

Trending doubts
On which river Salal project is situated A River Sutlej class 8 social science CBSE

When Sambhaji Maharaj died a 11 February 1689 b 11 class 8 social science CBSE

What is the Balkan issue in brief class 8 social science CBSE

When did the NonCooperation Movement begin A August class 8 social science CBSE

In Indian rupees 1 trillion is equal to how many c class 8 maths CBSE

Sketch larynx and explain its function in your own class 8 physics CBSE
