Problem on JEE and not for JEE

Since the marking scheme is $+4 \ , \ -2$ , a person can never get odd marks in JEE. The maximum marks one can get is $360$ and minimum is $-180$ . Thus the sum is sum of an arithmetic progression with first term $-180$ and last term $360$ and common difference $2$ .Let the total terms be $n$ . So ,

$360=-180+(n-1)2 \\ \Rightarrow 540=(n-1)2 \\ \Rightarrow n-1=270 \\ \Rightarrow n=271$

Thus the sum is given by:

$\dfrac{271}{2}[2(-180)+(271-1)2] \\ \Rightarrow S=\dfrac{271}{2}[-360+540] \\ \Rightarrow S=\dfrac{271}{2}[180] \\ \Rightarrow S=271 \times 90 \\ \Rightarrow S=24390$

But why is this not the answer? The problem is with the mischievous $358$ . We would think it as $360-2$ but we have only $90$ questions and not $91$ questions. Hence , the required sum is $=24390-358=\boxed{24032}$ .