Question

The cost of bananas is increased by Re.1 per dozen, one can get 2 dozen less for Rs. 840. Find the original cost of one dozen of banana.

Answer 1

Let the original cost of one dozen of banana is Rs. x. Then, the number of the dozen of bananas in Rs. 840 is$\frac{840}{x}$

Now, the cost of the bananas is increased by Re. 1. Then, the number of the dozen of bananas in Rs. 840 is $\frac{840}{x+1}$
Thus, by the given condition, we get

$$\begin{gathered} \frac{840}{x}-\frac{840}{x+1}=2 \\ \Rightarrow 840\left[\frac{1}{x}-\frac{1}{x+1}\right]=2 \\ \Rightarrow \frac{1}{x}-\frac{1}{x+1}=\frac{2}{840} \\ \Rightarrow \frac{x+1-x}{x(x+1)}=\frac{1}{420} \\ \Rightarrow \frac{1}{x(x+1)}=\frac{1}{420} \\ \Rightarrow x(x+1)=420 \\ \Rightarrow x^2+x-420=0 \end{gathered}$$
On splitting the middle term x as 21x – 20x, we get
$\Rightarrow$ x² + 21x – 20x – 420 = 0
$\Rightarrow$ x(x + 21) – 20(x + 21) = 0
$\Rightarrow$ (x + 21)(x – 20) = 0
$\Rightarrow$ x + 21 = 0 or x – 20 = 0
$\Rightarrow$ x = –21 or x = 20
Since the cost cannot be negative, so we get x = 20
Thus, the original cost of the bananas is Rs. 20 per dozen.