Answer
Verified
468.9k+ views
Hint: When a product is purchased in the view of selling it to the consumer in order to do business then, the price in which the product is bought by the seller is known as the cost price of the product and the price in which the seller sells the product to the consumer is known as selling of the product for the seller. If the selling price of the product is greater than the cost price of the product, then the difference in the prices can be termed as the profit or the gain on the product while at the same time if the selling price is less than the cost price of the product, then the difference in the price is known as the loss on the product. Profit percent or loss percent of a product is always calculated on the cost price of the product. In this question, it is already mentioned the salesman has gained a profit on a watch and would have gained more profit if the selling price would have been raised.
Complete step by step solution: Given that the salesman has gained a profit, which means that the selling price is more than the cost price
Let us assume \[S.P = x\] and cost price C.P is fixed in both the case
Hence we can write for the 15% profit.
\[
P\% = \dfrac{{SP - CP}}{{CP}} \times 100 \\
15\% = \dfrac{{x - CP}}{{CP}} \times 100 \\
\dfrac{x}{{CP}} - 1 = \dfrac{{15}}{{100}} \\
\dfrac{x}{{CP}} = 0.15 + 1 \\
CP = \dfrac{x}{{1.15}} - - - - \left( i \right) \\
\]
Now when the selling price is raised by Rs.48, the profit percentage also increases
\[
P'\% = 18\% \\
SP' = x + 48 \\
\]
Hence we can write
\[
P'\% = \dfrac{{SP' - CP}}{{CP}} \times 100 \\
18 = \dfrac{{\left( {x + 48} \right) - CP}}{{CP}} \times 100 \\
\dfrac{{x + 48}}{{CP}} = \dfrac{{18}}{{100}} + 1 \\
\dfrac{{x + 48}}{{CP}} = 1.18 \\
CP = \dfrac{{x + 48}}{{1.18}} - - - - (ii) \\
\]
Since the selling price is being increased on the same cost price, hence we can say cost price is the same in both the cases; hence we can say\[\left( i \right) = \left( {ii} \right)\], by equating both the equations
\[
\left( i \right) = \left( {ii} \right) \\
\dfrac{x}{{1.15}} = \dfrac{{x + 48}}{{1.18}} \\
1.18x = 1.15x + 55.2 \\
0.03x = 55.2 \\
x = Rs.1840 \\
\]
Hence the selling price is Rs.1840
Now put the value of \[x\]which is the selling price in equation (i), we get
\[CP = \dfrac{x}{{1.15}} = \dfrac{{1840}}{{1.15}} = Rs.1600\]
Hence the cost price of the watch is Rs.1600
Note: It is to be noted here that many a time, the marked price has been given in the question instead of selling price. So, be careful while reading the question as the marked price is the price that has been marked on the product by the seller, but the selling price is the price of the product which the seller actually gets for the product after discount.
Complete step by step solution: Given that the salesman has gained a profit, which means that the selling price is more than the cost price
Let us assume \[S.P = x\] and cost price C.P is fixed in both the case
Hence we can write for the 15% profit.
\[
P\% = \dfrac{{SP - CP}}{{CP}} \times 100 \\
15\% = \dfrac{{x - CP}}{{CP}} \times 100 \\
\dfrac{x}{{CP}} - 1 = \dfrac{{15}}{{100}} \\
\dfrac{x}{{CP}} = 0.15 + 1 \\
CP = \dfrac{x}{{1.15}} - - - - \left( i \right) \\
\]
Now when the selling price is raised by Rs.48, the profit percentage also increases
\[
P'\% = 18\% \\
SP' = x + 48 \\
\]
Hence we can write
\[
P'\% = \dfrac{{SP' - CP}}{{CP}} \times 100 \\
18 = \dfrac{{\left( {x + 48} \right) - CP}}{{CP}} \times 100 \\
\dfrac{{x + 48}}{{CP}} = \dfrac{{18}}{{100}} + 1 \\
\dfrac{{x + 48}}{{CP}} = 1.18 \\
CP = \dfrac{{x + 48}}{{1.18}} - - - - (ii) \\
\]
Since the selling price is being increased on the same cost price, hence we can say cost price is the same in both the cases; hence we can say\[\left( i \right) = \left( {ii} \right)\], by equating both the equations
\[
\left( i \right) = \left( {ii} \right) \\
\dfrac{x}{{1.15}} = \dfrac{{x + 48}}{{1.18}} \\
1.18x = 1.15x + 55.2 \\
0.03x = 55.2 \\
x = Rs.1840 \\
\]
Hence the selling price is Rs.1840
Now put the value of \[x\]which is the selling price in equation (i), we get
\[CP = \dfrac{x}{{1.15}} = \dfrac{{1840}}{{1.15}} = Rs.1600\]
Hence the cost price of the watch is Rs.1600
Note: It is to be noted here that many a time, the marked price has been given in the question instead of selling price. So, be careful while reading the question as the marked price is the price that has been marked on the product by the seller, but the selling price is the price of the product which the seller actually gets for the product after discount.
Recently Updated Pages
10 Examples of Evaporation in Daily Life with Explanations
10 Examples of Diffusion in Everyday Life
1 g of dry green algae absorb 47 times 10 3 moles of class 11 chemistry CBSE
If the coordinates of the points A B and C be 443 23 class 10 maths JEE_Main
If the mean of the set of numbers x1x2xn is bar x then class 10 maths JEE_Main
What is the meaning of celestial class 10 social science CBSE
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Which are the Top 10 Largest Countries of the World?
How do you graph the function fx 4x class 9 maths CBSE
Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
Why is there a time difference of about 5 hours between class 10 social science CBSE
Give 10 examples for herbs , shrubs , climbers , creepers