A man sells article A at a profit of 12.5% ​​and article B at a loss of 8$$\frac13$$​% and gains ₹ 250 in the whole transaction. If he sells A at 8$$\frac13$$​% loss and B at 12.5% ​​profit, he neither gains nor loses. What is the cost price of B?

Correct option is C

Solution:
Cost Price of A = x
Cost Price of B = y
First Transaction:
Profit on A = $$12.5\% \text{ of } x = \frac{1}{8} x$$​​
Loss on B = $$8\frac{1}{3}\% \text{ of } y = \frac{1}{12} y$$​​
​$$\frac{1}{8} x - \frac{1}{12} y = 250$$​​
Second Transaction:
Loss on A = $$8\frac{1}{3}\% \text{ of } x = \frac{1}{12} x$$​​
Profit on B = $$12.5\% \text{ of } y = \frac{1}{8} y$$​​
Net result is 0
​$$\frac{1}{8} y - \frac{1}{12} x = 0$$​​
Equation 1: $$\quad \frac{1}{8} x - \frac{1}{12} y = 250$$​

Equation 2: $$\quad \frac{1}{8} y - \frac{1}{12} x = 0$$​​
Multiply Equation (2) by 12:
​$$\frac{12}{8} y - x = 0\\\frac{3}{2} y = x\\x = \frac{3}{2} y$$​​
Substitute x in Equation (1)
​$$\frac{1}{8} \times \frac{3}{2} y - \frac{1}{12} y = 250\\\frac{3}{16} y - \frac{1}{12} y = 250$$​​
LCM of 16 and 12 = 48
​$$\frac{9}{48} y - \frac{4}{48} y = 250\\\frac{5}{48} y = 250\\y = 250 \times \frac{48}{5} = 2400\\$$​​
The cost price of B is ₹ 2400.