Question
Question: By selling an article of Rs. 240, Nilesh gained 20%. Find his profit % if he sells it for Rs. 272....
By selling an article of Rs. 240, Nilesh gained 20%. Find his profit % if he sells it for Rs. 272.
Solution
Hint: By using the first condition, as it is given a profit you can find the cost of the article which be bought for. Now by using the cost and selling price in the second condition find the profit percentage in that case. If CP is cost price, SP is selling price and p is profit. The formula of profit is given by:
profit=CPSP−CP×100
Complete step-by-step answer:
In the first, we will see definitions of selling price, cost price, profit.
Cost price: - The price given to an article which is paid by us to take it is called cost price.
Selling Price: - The price we get by giving away the article to who pays the price is called selling price of the article.
Profit: - If we get more selling then buying it, then the difference is called profit.
Loss: - If the selling price is less than lost price then difference is called Loss.
Case-1: The selling price of an article is given as 240 Rs. Given in the question the profit P is 20%.
By general knowledge, we can say formula of P as
profit=CPSP−CP×100
By substituting value of SP, profit into formula we get:
20=(CP240−CP)×100
By simplifying the above equation, we get:
20CP = 24000-100CP
By adding 100CP on both sides, we get:
120CP = 24000
By dividing with 120 on both sides, we get:
CP = 200
So, the CP of the article is 200 Rs.
Case-2: The CP will remain constant. Now it is given that article is sold for 272 Rs.
So, the selling price = 272
By general knowledge, we can say formula of P as
profit=CPSP−CP×100
By substituting values of SP, CP; we get
P=200272−200×100
By simplifying the term, we get: P=36 .
So, by selling the article at the cost of Rs. 272, the man gets 36% profit.
Note: The formula can be derived as SP is the sum of CP and profit (profit is percentage of CP)
SP = CP + (100P. (CP)). This expression is nothing but our formula. Be careful while substituting the values of SP, P. The idea of taking CP of article constant in both the cases plays a crucial role in the solution.