A can do a work in 5 , text{days} less than t | vedclass

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(C) Let the time taken by $B$ to complete the work be $x \, 	ext{days}$.Then, the time taken by $A$ to complete the work is $(x-5) \, 	ext{days}$.Gi

Vedclass · Accepted Answer

$25$

Vedclass · Answer

(C) Let the time taken by $B$ to complete the work be $x \, 	ext{days}$.Then, the time taken by $A$ to complete the work is $(x-5) \, 	ext{days}$.Given that both together take $11 \frac{1}{9} = \frac{100}{9} \, 	ext{days}$.The work done by $A$ and $B$ in one day is $\frac{1}{x-5} + \frac{1}{x} = \frac{9}{100}$.Solving the equation: $\frac{x + x - 5}{x(x-5)} = \frac{9}{100}$.$\frac{2x - 5}{x^2 - 5x} = \frac{9}{100}$.$100(2x - 5) = 9(x^2 - 5x)$.$200x - 500 = 9x^2 - 45x$.$9x^2 - 245x + 500 = 0$.Factoring the quadratic equation: $9x^2 - 225x - 20x + 500 = 0$.$9x(x - 25) - 20(x - 25) = 0$.$(9x - 20)(x - 25) = 0$.So, $x = \frac{20}{9}$ or $x = 25$.Since $x$ must be greater than $5$ (as $A$ takes $x-5$ days), we have $x = 25$.Therefore, the time taken by $B$ alone is $25 \, 	ext{days}$.

$A$ can do a work in $5 \, \text{days}$ less than the time taken by $B$ to do it. If both of them together take $11 \frac{1}{9} \, \text{days}$, then the time taken by $B$ alone to do the same work (in $\text{days}$) is

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