Answer
Verified
388.2k+ views
Hint:In this problem they give the percentage of item MP a shopkeeper fixes above the CP. We have to find the percentage of discount allowed to gain \[8\% \]. We know that, if the cost price of an object is \[a\]. If the price increase \[b\% \], the price of the object will become \[a \times \dfrac{{100 + b}}{{100}}\].The selling price of an object is the sum of the cost price and the gain amount. Here first we consider the cost price of the item in some variable. Using that we calculate the market price and selling price with the gain of \[8\% \]. Subtracting the both to find the discount. Then we can convert it into percentage. This is our required result.
Complete step-by-step answer:
It is given that; a shopkeeper fixes the MP of an item \[35\% \] above the CP. We have to find the percentage of discount that the shopkeeper allowed to gain \[8\% \].
Let us consider, the cost price of the item is \[Rs.{\rm{ }}x\].
The market price of the item is \[35\% \] above the cost price.
That is marked price of the given item is found using the formula \[\dfrac{{100 + b}}{{100}} \times {\rm{cost price}}\] where b is the percentage of the marked price.
By using the formula the marked price of the item will be\[\dfrac{{100 + 35}}{{100}} \times x = \dfrac{{135x}}{{100}}\].
Now, it is given that the shopkeeper wants to gain \[8\% \].
Hence the selling price of the item is given using the formula \[\dfrac{{100 + b}}{{100}} \times {\rm{cost price}}\] where b is the percentage of the gain.
By using the above formula we get,
The selling price of the item will be\[\dfrac{{100 + 8}}{{100}} \times x = \,\,\dfrac{{108x}}{{100}}\].
Now we have to find the amount of discount using the formula,
\[{\rm{marked price - selling price = discount}}\]
Therefore the discount amount of the item is\[\dfrac{{135x}}{{100}} - \dfrac{{108x}}{{100}} = \dfrac{{27x}}{{100}}\].
Now we have to find the percentage of discount for that we use the following formula,
\[{\rm{discount\% = }}\dfrac{{{\rm{discount}}}}{{{\rm{marked price}}}} \times 100\]
Therefore the percentage of discount is \[\dfrac{{\dfrac{{27x}}{{100}}}}{{\dfrac{{135x}}{{100}}}} \times 100\% \]
On solving the percentage we get,
\[\dfrac{{\dfrac{{27x}}{{100}}}}{{\dfrac{{135x}}{{100}}}} \times 100\% = \dfrac{{27x}}{{100}} \times \dfrac{{100}}{{135x}} \times 100 = \dfrac{{100}}{5} = 20\% \]
The percentage of discount offered to gain \[8\% \] profit is \[20\% .\]
So, the correct answer is “Option C”.
Note:The problem can be solved by another process.
Let us consider, the cost price of the item is \[Rs.{\rm{ }}100\].The marked price of the item is \[35\% \] above the cost price.
So, the marked price of the item will be \[Rs.{\rm{ }}135\].
Now, the shopkeeper wants to gain \[8\% \].
So, the selling price of the item will be \[Rs.{\rm{ }}100 + 8 = Rs.{\rm{ }}108\]
So, he has to give the discount of \[Rs.\,135 - 108 = Rs.{\rm{ }}27\]
Then, the percentage of discount is \[\dfrac{{27}}{{135}} \times 100\% \]
Simplifying we get, the percentage of discount is \[20\% .\]
Complete step-by-step answer:
It is given that; a shopkeeper fixes the MP of an item \[35\% \] above the CP. We have to find the percentage of discount that the shopkeeper allowed to gain \[8\% \].
Let us consider, the cost price of the item is \[Rs.{\rm{ }}x\].
The market price of the item is \[35\% \] above the cost price.
That is marked price of the given item is found using the formula \[\dfrac{{100 + b}}{{100}} \times {\rm{cost price}}\] where b is the percentage of the marked price.
By using the formula the marked price of the item will be\[\dfrac{{100 + 35}}{{100}} \times x = \dfrac{{135x}}{{100}}\].
Now, it is given that the shopkeeper wants to gain \[8\% \].
Hence the selling price of the item is given using the formula \[\dfrac{{100 + b}}{{100}} \times {\rm{cost price}}\] where b is the percentage of the gain.
By using the above formula we get,
The selling price of the item will be\[\dfrac{{100 + 8}}{{100}} \times x = \,\,\dfrac{{108x}}{{100}}\].
Now we have to find the amount of discount using the formula,
\[{\rm{marked price - selling price = discount}}\]
Therefore the discount amount of the item is\[\dfrac{{135x}}{{100}} - \dfrac{{108x}}{{100}} = \dfrac{{27x}}{{100}}\].
Now we have to find the percentage of discount for that we use the following formula,
\[{\rm{discount\% = }}\dfrac{{{\rm{discount}}}}{{{\rm{marked price}}}} \times 100\]
Therefore the percentage of discount is \[\dfrac{{\dfrac{{27x}}{{100}}}}{{\dfrac{{135x}}{{100}}}} \times 100\% \]
On solving the percentage we get,
\[\dfrac{{\dfrac{{27x}}{{100}}}}{{\dfrac{{135x}}{{100}}}} \times 100\% = \dfrac{{27x}}{{100}} \times \dfrac{{100}}{{135x}} \times 100 = \dfrac{{100}}{5} = 20\% \]
The percentage of discount offered to gain \[8\% \] profit is \[20\% .\]
So, the correct answer is “Option C”.
Note:The problem can be solved by another process.
Let us consider, the cost price of the item is \[Rs.{\rm{ }}100\].The marked price of the item is \[35\% \] above the cost price.
So, the marked price of the item will be \[Rs.{\rm{ }}135\].
Now, the shopkeeper wants to gain \[8\% \].
So, the selling price of the item will be \[Rs.{\rm{ }}100 + 8 = Rs.{\rm{ }}108\]
So, he has to give the discount of \[Rs.\,135 - 108 = Rs.{\rm{ }}27\]
Then, the percentage of discount is \[\dfrac{{27}}{{135}} \times 100\% \]
Simplifying we get, the percentage of discount is \[20\% .\]
Recently Updated Pages
Three beakers labelled as A B and C each containing 25 mL of water were taken A small amount of NaOH anhydrous CuSO4 and NaCl were added to the beakers A B and C respectively It was observed that there was an increase in the temperature of the solutions contained in beakers A and B whereas in case of beaker C the temperature of the solution falls Which one of the following statements isarecorrect i In beakers A and B exothermic process has occurred ii In beakers A and B endothermic process has occurred iii In beaker C exothermic process has occurred iv In beaker C endothermic process has occurred
The branch of science which deals with nature and natural class 10 physics CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
Define absolute refractive index of a medium
Find out what do the algal bloom and redtides sign class 10 biology CBSE
Prove that the function fleft x right xn is continuous class 12 maths CBSE
Trending doubts
How do you solve x2 11x + 28 0 using the quadratic class 10 maths CBSE
Which type of bond is stronger ionic or covalent class 12 chemistry CBSE
What organs are located on the left side of your body class 11 biology CBSE
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Difference Between Plant Cell and Animal Cell
How fast is 60 miles per hour in kilometres per ho class 10 maths CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
When people say No pun intended what does that mea class 8 english CBSE