An algebra problem by Kshetrapal Dashottar

$\text{Let } S = f\left( \dfrac{1}{100} \right)+f\left( \dfrac{2}{100} \right)+\cdots+f\left( \dfrac{99}{100} \right) \\ \quad S = f\left( \dfrac{99}{100} \right)+f\left( \dfrac{98}{100} \right)+\cdots+f\left( \dfrac{1}{100} \right) \\ \text{Add the Two equations , we get } \\ \quad 2S = \left( f\left( \dfrac{1}{100} \right)+f\left( \dfrac{99}{100} \right) \right) +\left( f\left( \dfrac{2}{100} \right)+f\left( \dfrac{98}{100} \right) \right) +\cdots+\left( f\left( \dfrac{99}{100} \right)+f\left( \dfrac{1}{100} \right) \right) \\ \quad 2S = 99\cdot10 \implies \boxed{ S = 495 }$

so f(x)+f(1-x)=10
to find f(1/100)+f(2/100)+f(3/100)+.....+f(99/100)
so by making pair of first and last term , second and second last term and so on we find that first term is f(x) and second is f(1-x) So, f(1/100)+f(99/100),f(2/100)+f(98/100),f(3/100)+f(97/100),.........,f(49/100)+f(51/100) all equal to 10 so sum of these 49 terms each value 10
will be 49 times10=490
now the term left is f(50/100) now this is f(x) also and f(1-x) also
(because 50/100 =1-50/100) so f(50/100)+f(50/100)=10
2*f(50/100)=10 , f(50/100)=5
so the ans is 490+5=495