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(D) Let the time taken by mason $A$ alone to complete the work be $x$ days. Then, mason $B$ alone takes $x+24$ days.Work done by $A$ in $1$ day $= \fr

Vedclass · Accepted Answer

$36$

Vedclass · Answer

(D) Let the time taken by mason $A$ alone to complete the work be $x$ days. Then, mason $B$ alone takes $x+24$ days.Work done by $A$ in $1$ day $= \frac{1}{x}$.Work done by $B$ in $1$ day $= \frac{1}{x+24}$.Work done by $A+B$ in $1$ day $= \frac{1}{22.5} = \frac{1}{45/2} = \frac{2}{45}$.So, $\frac{1}{x} + \frac{1}{x+24} = \frac{2}{45}$.$\frac{x+24+x}{x(x+24)} = \frac{2}{45} \implies \frac{2x+24}{x^2+24x} = \frac{2}{45}$.$45(x+12) = x^2+24x \implies 45x + 540 = x^2 + 24x$.$x^2 - 21x - 540 = 0$.$(x-36)(x+15) = 0$.Since time cannot be negative, $x = 36$ days.

$A$ building construction work can be completed by two masons $A$ and $B$ together in $22.5$ days. Mason $A$ alone can complete the work in $24$ days less than mason $B$ alone. Then mason $A$ alone will complete the work in: (in $days$)

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