(c) 10
Let B take x hours to complete a work. Then, A shall take (x + 5) hours to complete the same work.
B’s 1 hours’ work = \(\frac{1}{X}\), A’s 1 hours’ work = \(\frac{1}{X+5}\)
Given, \(\frac{1}{X}\)+ \(\frac{1}{(X+5)}\) = \(\frac{1}{6}\)
\(\Rightarrow\) \(\frac{X+5+X}{X(X+5)}\) = \(\frac{1}{6}\) \(\Rightarrow\) \(\frac{2X+5}{X^2+5X}\) = \(\frac{1}{6}\)
\(\Rightarrow\) 12x + 30 = x2 + 5x
\(\Rightarrow\) x2 – 7x – 30 = 0
\(\Rightarrow\) x2 – 10x + 3x – 30 = 0
\(\Rightarrow\) x(x – 10) + 3(x – 10) = 0
\(\Rightarrow\) (x – 10) = 0 or (x + 3) = 0
x = 10 or – 3
Neglecting negative value x = 10.