RRB Assistant Loco Pilot and Technician Model Paper 8 with solutions

Suppose Speed = x km/h
time taken,
${\mathit{t}}_1=\frac{\mathrm{distance}}{\mathrm{speed}}=\frac{2}{\mathit{x}}\mathit{h}$
if Speed = (x + 1) km/h
then time = ${\mathit{t}}_2=\frac{2}{\mathit{x}+1}\mathit{h}$ ...(i)
but according to question
time,
${\mathit{t}}_2=\left(\frac{2}{\mathit{x}}-\frac{10}{60}\right)\mathit{h}$ ...(ii)
Fromeq. (i) and (ii)
$\frac{2}{\mathit{x}+1}=\frac{2}{\mathit{x}}-\frac{1}{6}$
$\frac{2}{\mathit{x}}-\frac{2}{\mathit{x}+1}=\frac{1}{6}$
$\frac{2\mathit{x}+2-2\mathit{x}}{\mathit{x}\left(\mathit{x}+1\right)}=\frac{1}{6}$
or, $\frac{2}{\mathit{x}\left(\mathit{x}+1\right)}=\frac{1}{6}$
or ${\mathit{x}}^2$ + x = 12
or${\mathit{x}}^2$ + x – 12 = 0
or ${\mathit{x}}^2$ + 4x – 3x – 12 = 0
or x(x + 4) – 3(x + 4) = 0
(x – 3) (x + 4) = 0
∴ x = 3 km/h