When load P is removed
$\Sigma M_A = 0$
$Qx = 20(5 + 1)$
$Qx = 120$ → Equation (1)
When load P is applied
$\Sigma M_B = 0$
$Q(x + 5) = 20(1) + 20(10)$
$Qx + 5Q = 220$
From Equation (1), Qx = 120, thus,
$120 + 5Q = 220$
$5Q = 100$
$Q = 20 \, \text{ tons}$ answer
Substitute Q = 20 tons to Equation (1)
$20x = 120$
$x = 6 \, \text{ ft}$ answer