SSC CGL 19 Oct 2014 Shift 2 Paper

Let A, B and C together do the work in x hours.
∴ Time taken by A
= (x + 6) hours
Time taken by B= (x + 1) hours
Time taken by C = 2x hours
∴ $\frac{1}{\mathit{x}+6}+\frac{1}{\mathit{x}+1}$ $\frac{1}{2\mathit{x}}$ = $\frac{1}{\mathit{x}}$
⇒ $\frac{1}{\mathit{x}+6}+\frac{1}{\mathit{x}+1}$ = $\frac{1}{\mathit{x}}-\frac{1}{2\mathit{x}}$
= $\frac{1}{2\mathit{x}}$
⇒ $\frac{1}{\mathit{x}+6}$ = $\frac{1}{2\mathit{x}}-\frac{1}{\mathit{x}+1}$
= $\frac{\mathit{x}+1-2\mathit{x}}{2\mathit{x}\left(\mathit{x}+1\right)}$
⇒ $\frac{1}{\mathit{x}+6}$ = $\frac{1-\mathit{x}}{2{\mathit{x}}^2+2\mathit{x}}$
⇒ $2{\mathit{x}}^2$ + 2x = x + 6 - ${\mathit{x}}^2$ - 6x
⇒ $3{\mathit{x}}^2$ + 7x – 6 = 0
⇒ $3{\mathit{x}}^2$ + 9x – 2x – 6 = 0
⇒ 3x (x + 3) – 2 (x + 3) = 0
⇒ (3x – 2) (x + 3) = 0
⇒ 3x – 2 = 0 as x + 3 ≠ 0
⇒ x = $\frac{2}{3}$
∴ Time taken by A = 6 + $\frac{2}{3}$
= $\frac{18+2}{3}$ = $\frac{20}{3}$ hours
Time taken by B = 1 + $\frac{2}{3}$
= $\frac{5}{3}$ hours
∴ (A +B)’s 1 hour’s work
= $\frac{3}{20}+\frac{3}{5}$ = $\frac{3+12}{20}$
= $\frac{15}{20}$ = $\frac{3}{4}$
∴ Required time = $\frac{4}{3}$ hours