RRB Group D 8-Dec-2013 Shift 1 Solved Paper

Let the price of the chair be $\mathit{₹}$ x.
Price of tables $=\mathit{₹}\mathit{y}$
According to the question:
$5\mathit{x}+2\mathit{y}=1350$
$2\mathit{x}=\mathit{y}$
By substituting the value of $\mathit{y}$ in equation (1):
$5\mathit{x}+2\mathit{y}=1350$
$\Rightarrow 5\mathit{x}+2\times 2\mathit{x}=1350$
$\Rightarrow 5\mathit{x}+4\mathit{x}=1350$
$\Rightarrow 9\mathit{x}=1350$
$\therefore \mathit{x}=150$
cost of one chair $=\mathit{₹}150$
Cost of one table = 2 (150) = ₹ 300
Cost of one chair and one table =150 + 300 = ₹ 450