SSC JE Civil Engineering 23-Sep-2019 Shift 1 Solved Paper

$\begin{array}{ccccc}\mathit{T}& \mathit{R}& \mathit{U}& \mathit{C}& \mathit{K}\\ \downarrow & \downarrow & \downarrow & \downarrow & \downarrow \\ 20& 18& 21& 3& 11\end{array}$
20 + 18 + 21 + 3 + 11 = 73
Now 'TRUCK' has 5 letters
∴ 73 × 5 = 365
Similarly, $\begin{array}{ccc}\mathit{F}& \mathit{O}& \mathit{G}\\ \downarrow & \downarrow & \downarrow \\ 6& 15& 7\end{array}$
6 + 15 + 7 = 28
Now 'FOG' has 3 letters
∴ 28 × 3 = 84
Similarly, $\begin{array}{cccc}\mathit{D}& \mathit{A}& \mathit{R}& \mathit{T}\\ \downarrow & \downarrow & \downarrow & \downarrow \\ 4& 1& 18& 20\end{array}$
⇒ 4 + 1 + 18 + 20 = 43
Now 'DART' has 4 letters
∴ 43 × 4 = 172