
Naseem runs a readymade garments shop. He marks the garments at such a price that even after allowing a discount of 12.5% he makes a profit of 10% What will be the marked price of the suit which costs him 1470?
Answer
542.1k+ views
Hint: Here in this question we need to find the marked price of a suit which costs Naseem \[1470\ ]and a discount percentage as well as profit percentage is given. So here firstly calculate the selling point and then later on find the marked point by using the relation between selling price and marked price of product
Formula used:
The formula used to calculate the selling price is
\[SP = \left( {\dfrac{{100 + Gain}}{{100}} \times CP} \right)\]
While to calculate the marked price the formula used is
\[MP = \left( {\dfrac{{100 \times SP}}{{100 - Discount\% }}} \right)\]
Complete step by step answer:
Firstly as we now that the cost price \[(CP)\] of the suit is \[1470Rs\] and profit gained is \[10\% \]
So
\[
{\text{CP}} = 1470Rs \\
{\text{Profit = 10\% }} \\
\]
Now we will find the selling price \[(SP)\] of the suit
\[
\Rightarrow SP = 1470 \times \dfrac{{110}}{{100}} \\
\Rightarrow SP = 1617Rs \\
\]
Now as we know that discount given is \[12.5\% \] so we will find the marked price \[(MP)\]
\[
\Rightarrow MP = 1617 \times \dfrac{{100}}{{87.5}} \\
\Rightarrow MP = 1848Rs \\
\]
This means that marked price of suit is \[1848Rs\]
Note: When discount is offered then \[MP > SP\] while if the discount is not offered then \[MP < SP\]where \[MP\]is marked price and \[SP\] is selling price.
Always keep in mind that the selling price of the product is always less than the marked price. In the above given type of formula the case is based on profit and we are aware that profit or loss formula to be substituted and to solve such scenarios extract the solution from specific formula. We can find this by the relation between selling price and marked price
\[
SP = CP + {\text{Profit}} \\
SP = MP - {\text{Discount}} \\
\]
Formula used:
The formula used to calculate the selling price is
\[SP = \left( {\dfrac{{100 + Gain}}{{100}} \times CP} \right)\]
While to calculate the marked price the formula used is
\[MP = \left( {\dfrac{{100 \times SP}}{{100 - Discount\% }}} \right)\]
Complete step by step answer:
Firstly as we now that the cost price \[(CP)\] of the suit is \[1470Rs\] and profit gained is \[10\% \]
So
\[
{\text{CP}} = 1470Rs \\
{\text{Profit = 10\% }} \\
\]
Now we will find the selling price \[(SP)\] of the suit
\[
\Rightarrow SP = 1470 \times \dfrac{{110}}{{100}} \\
\Rightarrow SP = 1617Rs \\
\]
Now as we know that discount given is \[12.5\% \] so we will find the marked price \[(MP)\]
\[
\Rightarrow MP = 1617 \times \dfrac{{100}}{{87.5}} \\
\Rightarrow MP = 1848Rs \\
\]
This means that marked price of suit is \[1848Rs\]
Note: When discount is offered then \[MP > SP\] while if the discount is not offered then \[MP < SP\]where \[MP\]is marked price and \[SP\] is selling price.
Always keep in mind that the selling price of the product is always less than the marked price. In the above given type of formula the case is based on profit and we are aware that profit or loss formula to be substituted and to solve such scenarios extract the solution from specific formula. We can find this by the relation between selling price and marked price
\[
SP = CP + {\text{Profit}} \\
SP = MP - {\text{Discount}} \\
\]
Recently Updated Pages
Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 Business Studies: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

There are 720 permutations of the digits 1 2 3 4 5 class 11 maths CBSE

Discuss the various forms of bacteria class 11 biology CBSE

Draw a diagram of a plant cell and label at least eight class 11 biology CBSE

State the laws of reflection of light

Explain zero factorial class 11 maths CBSE

