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Let us first take the cost price (C.P.) of the motorcycle to be Rs. x.

Now, it is given in the question that he sold the motorcycle at a loss% of 28%.

Now, using the formula for calculating the loss that is given in the hint, we get the following

$\begin{align}

& \Rightarrow loss\%=\dfrac{loss}{C.P}\times 100 \\

& \Rightarrow 28=\dfrac{loss}{x}\times 100 \\

& \Rightarrow loss=\dfrac{28x}{100}=\dfrac{7x}{25} \\

\end{align}$

Now. Using the loss amount, we can calculate the selling price of the motorcycle for Ravish which is as follows

\[\begin{align}

& loss=\text{ }C.P\text{ }-S.P. \\

& \dfrac{7x}{25}=x-S.P. \\

& S.P.=\dfrac{18x}{25} \\

\end{align}\]

Now, this selling price for Ravish would be the cost price for Vineet.

Now, Vineet spent Rs. 1680 for the repair work of the motorcycle, hence, the net cost price for Vineet is Rs. \[\dfrac{18x}{25}\]+1680.

(Because the repair cost will also be taken up by Vineet only)

Now, the selling price for Vineet is given as Rs. 35910 and the profit percent for his transaction is 12.5%.

Hence, we can again use the formula that is given in the hint as follows

\[\begin{align}

& \Rightarrow gain\%=\dfrac{gain}{C.P}\times 100 \\

& \Rightarrow 12.5=\dfrac{S.P.-C.P.}{\dfrac{18x}{25}+1680}\times 100 \\

& \Rightarrow 12.5=\dfrac{35910-\left( \dfrac{18x}{25}+1680 \right)}{\dfrac{18x}{25}+1680}\times 100 \\

& \Rightarrow 12.5=\dfrac{34230\times 25-18x}{18x+1680\times 25}\times 100 \\

& \Rightarrow \dfrac{125}{10}=\dfrac{34230\times 25-18x}{18x+1680\times 25}\times 100 \\

& \Rightarrow \dfrac{1}{8}=\dfrac{34230\times 25-18x}{18x+1680\times 25} \\

& \Rightarrow 68460\times 100-144x=18x+1680\times 25 \\

& \Rightarrow 68460\times 100=162x+1680\times 25 \\

& \Rightarrow x=Rs.\ 42000 \\

\end{align}\]