SSC CGL 19 Oct 2014 Shift 2 Paper

Fixed distance = x km and certain speed = y kmph (let).
Case I,
$\frac{\mathit{x}}{\mathit{y}+10}$ = $\frac{\mathit{x}}{\mathit{y}}-1$
⇒ $\frac{\mathit{x}}{\mathit{y}+10}$ + 1 = $\frac{\mathit{x}}{\mathit{y}}$ ... (i)
Case II,
$\frac{\mathit{x}}{\mathit{y}+20}$ = $\frac{\mathit{x}}{\mathit{y}}$ - 1 - $\frac{3}{4}$
= $\frac{\mathit{x}}{\mathit{y}}-\frac{4+3}{4}$
⇒ $\frac{\mathit{x}}{\mathit{y}+20}$ + $\frac{7}{4}$ = $\frac{\mathit{x}}{\mathit{y}}$ ... (ii)
From equations (i) and (ii),
$\frac{\mathit{x}}{\mathit{y}+10}+1$ = $\frac{\mathit{x}}{\mathit{y}+20}+\frac{7}{4}$
⇒ $\frac{\mathit{x}}{\mathit{y}+10}-\frac{\mathit{x}}{\mathit{y}+20}$ = $\frac{7}{4}$ - 1
⇒ x $\left(\frac{\mathit{y}+20-\mathit{y}-10}{\left(\mathit{y}+10\right)\left(\mathit{y}+20\right)}\right)$
= $\frac{7-4}{4}$ = $\frac{3}{4}$
⇒ $\frac{\mathit{x}\times 10}{\left(\mathit{y}+10\right)\left(\mathit{y}+20\right)}$ = $\frac{3}{4}$
⇒ 3 (y + 10) (y + 20) = 40 x
⇒ $\frac{3\left(\mathit{y}+10\right)\left(\mathit{y}+20\right)}{40}$ = x ... (iii)
From equation (i),
$\frac{3\left(\mathit{y}+10\right)\left(\mathit{y}+20\right)}{40\left(\mathit{y}+10\right)}$ + 1 = $\frac{3\left(\mathit{y}+10\right)\left(\mathit{y}+20\right)}{40\mathit{y}}$
⇒ 3 (y + 20) + 40 = $\frac{3\left(\mathit{y}+10\right)\left(\mathit{y}+2\right)}{\mathit{y}}$
⇒ $3{\mathit{y}}^2$ + 60y + 40 y = 3(${\mathit{y}}^2$ + 30y + 200)
⇒ $3{\mathit{y}}^2$ + 100y = $3{\mathit{y}}^2$ + 90y + 600
⇒ 10y = 600 ⇒ y = 60
Again from equation (i),
$\frac{\mathit{x}}{\mathit{y}+10}$ + 1 = $\frac{\mathit{x}}{\mathit{y}}$
⇒ $\frac{\mathit{x}}{60+10}$ + 1 = $\frac{\mathit{x}}{60}$
⇒ $\frac{\mathit{x}}{70}$ + 1 = $\frac{\mathit{x}}{60}$
⇒ $\frac{\mathit{x}+70}{70}$ = $\frac{\mathit{x}}{60}$
⇒ 6x + 420 = 7x
⇒ 7x – 6x = 420
⇒ x = 420 km.