Question

A trader bought a number of articles for Rs.900, five articles were found damaged. He sold each of the remaining articles at Rs.2 more than what he paid for it. He got a profit of Rs. 80 on the whole transaction. Find the number of articles he bought.

Answer 1

Hint :- In this type of question we had to first of all assume the total number of articles. And then we had to assume the cost per article too. These two assumptions will help us a lot in solving such problems. And the other most important thing that we had to keep in mind is the basic format of quadratic equation i.e. \[a{x^2} + bx + c = 0\].

Complete step-by-step answer:
Let us first of all assume that the total number of articles be ‘x’.
Now let us assume that the cost of per article is Rs. ‘a’.
And we know that the trader had bought ‘x’ number of articles in Rs.900
So, we can say that product of total number of articles and cost of per article must be equal to 900 ( i.e. \[x \times a = 900\])
From above we can also say that \[a = \dfrac{{900}}{x}\].
Now as we know that the 5 articles are damaged from the total articles so the remaining good articles must be ( \[x - 5\] )
And the trader sold this \[x - 5\] articles at a profit of Rs. 2 on per article it means the articles were sold at cost of ( \[a + 2\] ) and the total profit on this is Rs. 80.
So now the equation for undamaged articles will be
\[ \Rightarrow (x - 5) \times (a + 2) = 900 + 80\]
\[ \Rightarrow \]\[xa - 5a + 2x - 10 = 980\]
Now putting the value of ‘a’ in above equation
\[ \Rightarrow \]\[x \times \dfrac{{900}}{x} - 5 \times \dfrac{{900}}{x} + 2x - 10 = 980\]
Now solving L.H.S and R.H.S
\[ \Rightarrow \]\[890 - \dfrac{{4500}}{x} + 2x = 980 \Rightarrow 2x - \dfrac{{4500}}{x} = 90\]
\[ \Rightarrow \]\[x - \dfrac{{2250}}{x} = 45\]
So the quadratic equation we get on taking L.C.M and solving above is
\[ \Rightarrow \]\[{x^2} - 45x - 2250 = 0\]
Now solving the above quadratic equation by middle term splitting
\[ \Rightarrow \]\[{x^2} - 75x + 30x - 2250 = 0\]
\[ \Rightarrow \]\[x(x - 75) + 30(x - 75) = 0\]
\[ \Rightarrow \]\[(x - 75)(x + 30) = 0\]
So now comparing the above value with 0 we will get the values of x as
\[x = 75{\text{ }}or{\text{ }}x = - 30\]
And as we had assumed earlier that ‘x’ is the total number of articles so the value or counting of articles can never be a negative term.
So, the value of ‘x’ must be 75.
Hence the total number of articles traders had bought is 75.

Note:- Whenever we come up with such a problem we must know the cost per article and the number of articles if in case it is not given then we have to assume it. And we can solve the quadratic equations by other methods also such as using square roots , completing the square , or by using quadratic formula ( i.e.\[ - b \pm \dfrac{{\sqrt {{b^2} - 4ac} }}{{2a}}\] ).