GATE Civil Engineering (CE) 2018 Shift 2 Solved Paper

A + B + C = C + D + E = E + F + G
= G + H + K =13
If we add all, we will get $=4\times 13=52$
$=45=\frac{9\times 10}{2}$
But sum of all natural number 1 to 9
A + B + C + C + D + E + E + F + G + G + H + K
$=52$
A + B + C + D + E + F + G + H + K = 45
Subtraction eq. (ii) from (i)
Hence, $\mathit{C}+\mathit{E}+\mathit{G}=7$
Also, $\mathit{C}+\mathit{D}+\mathit{E}=13$
Subtraction eq. (iii) from (iv)
$\mathit{D}-\mathit{G}=6$
$\mathit{E}=4$
Alternative Method
By checking other equations