Question

A retailer cheats his wholesaler and customer both. He purchases $19\% $ more from the wholesaler and sells $15 \% $ less while selling it to its customer. What is the profit percentage of selling the goods at cost price?

Answer 1

Hint:
We can find the profit percentage by multiplying the fraction of goods left by the goods he sold with 100 percent. We can assume the amount of goods he buys without cheating be x. The goods he bought can be found by taking 119 percent of x. Then we can find the goods he sold by subtracting the 15 percentage of x from x. We can find the amount of goods left by subtracting the goods he sold from the goods he bought.

Complete step by step solution:
Let x be the amount of goods the retailer buys without cheating.
It is given that he buys $19\% $ more from the wholesaler.
Then the amount of goods he buys is given by,
  $ \Rightarrow B = x + 19\% \text{x}$
Using \[n\% \text{ of x} = \dfrac{n}{{100}} \times x\] , we get,
  $ \Rightarrow B = x + \dfrac{{19x}}{{100}}$
On simplification we get,
  $ \Rightarrow B = \dfrac{{119x}}{{100}}A$
It is given that he sells $15\% $ less of the original goods.
Then the amount of goods he sells is given by,
 $ \Rightarrow S = x - 15\% x$
Using \[n\% \text{ of x} = \dfrac{n}{{100}} \times x\] , we get,
  $ \Rightarrow S = x - \dfrac{{15x}}{{100}}$
On simplification we get,
 $ \Rightarrow S = \dfrac{{85x}}{{100}}$
Now we can find the goods left after selling. It is given by the difference between the goods he sells and the goods he buys.
  $ \Rightarrow L = B - S$
On substituting the values we get,
 $ \Rightarrow L = \dfrac{{119x}}{{100}} - \dfrac{{85x}}{{100}}$
On simplification we get,
  $ \Rightarrow L = \dfrac{{119 - 85}}{{100}}x$
On doing subtraction we get,
  $ \Rightarrow L = \dfrac{{34}}{{100}}x$
The profit percentage is given by the goods left divided by the goods he sold and the whole multiplied by $100\% $
  $\% = \dfrac{L}{S} \times 100\% $
On substituting the values, we get,
   $\% = \dfrac{{\dfrac{{34}}{{100}}x}}{{\dfrac{{85}}{{100}}x}} \times 100\% $
On cancelling the common terms, we get,
 $\% = \dfrac{{34}}{{85}} \times 100\% $
On simplification, we get,
 $\% = 40\% $

Therefore, the required profit percentage is $40\% $

Note:
Alternate method to solve this problem is given by,
Let us take the amount of goods he buys without cheating as 100 units.
It is given that he buys $19\% $ more from the wholesaler. Then the amount of goods he buys by cheating will be 119 units.
It is given that he sells $15\% $ less of the original goods.
Then the amount of goods he sells is $100 - 15 = 85$ units.
Then the amount of goods after selling is $119 - 85 = 34units$
The profit percentage is given by the goods left divided by the goods he sold and the whole multiplied by $100\% $
 $\% = \dfrac{{34}}{{85}} \times 100\% $
On simplification, we get,
 $\% = 40\% $
Therefore, the required profit percentage is $40\% $