If two cycles are sold on equal price. He get 10% profit on the first and 10% loss on the second then his overall profit or loss percent = ……………….. 

A) Loss percentage 1%

Correct option is (A) Loss percentage 1%

Let cost price of first cycle is Rs x and cost price of second cycle is Rs y.

\(\therefore\) Selling price of first cycle = cost price of first cycle + profit

= Rs (x + 10% of x)

\(=Rs(x+\frac x{10})=Rs\,\frac{11x}{10}\)

And selling price of second cycle = cost price of second cycle - loss

= Rs (y - 10% of y)

\(=Rs(y-\frac y{10})=Rs\,\frac{9y}{10}\)

Cost price of both cycles = Rs (x + y)

Selling price of both cycles \(=Rs\,(\frac{11x}{10}+\frac{9y}{10})\)

\(\because\) Given that both cycles are sold on equal price.

\(\therefore\) \(\frac{11x}{10}=\frac{9y}{10}\)

\(\Rightarrow\) 11x = 9y

\(\Rightarrow\) \(y=\frac{11x}9\)    ______________(1)

\(\therefore\) Profit in percent \(=\frac{\text{Selling price - cost price}}{\text{cost price}}\times100\%\)

\(=\cfrac{(\frac{11x}{10}+\frac{9y}{10})-(x+y)}{x+y}\times100\%\)

\(=\cfrac{\frac{x}{10}-\frac{y}{10}}{x+y}\times100\%\)

\(=\cfrac{x-\frac{11x}9}{10(x+\frac{11x}9)}\times100\%\)

\(=\cfrac{\frac{9x-11x}9}{\frac{9x+11x}9}\times10\%\)

\(=\frac{-2x}{20x}\times10\%\)

= -1%

Hence, overall loss 1%.