A takes 10, text{days} less than the time tak | vedclass

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(A) Let the time required by $B$ to finish the whole work be $x\, 	ext{days}$.Then, $A$ will take $(x-10)\, 	ext{days}$.According to the problem, th

Vedclass · Accepted Answer

$30$

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(A) Let the time required by $B$ to finish the whole work be $x\, 	ext{days}$.Then, $A$ will take $(x-10)\, 	ext{days}$.According to the problem, the combined work rate of $A$ and $B$ is $\frac{1}{12}$ of the work per day.So, $\frac{1}{x} + \frac{1}{x-10} = \frac{1}{12}$.$\frac{(x-10) + x}{x(x-10)} = \frac{1}{12} \Rightarrow \frac{2x-10}{x^2-10x} = \frac{1}{12}$.Cross-multiplying gives $12(2x-10) = x^2-10x$, which simplifies to $24x - 120 = x^2 - 10x$.Rearranging into a quadratic equation: $x^2 - 34x + 120 = 0$.Factoring the equation: $(x-30)(x-4) = 0$.This gives $x = 30$ or $x = 4$.Since $x$ must be greater than $10$ (as $A$ takes $x-10$ days), $x = 4$ is rejected.Therefore, the time taken by $B$ alone is $30\, 	ext{days}$.

$A$ takes $10\, \text{days}$ less than the time taken by $B$ to finish a piece of work. If both $A$ and $B$ can do it in $12\, \text{days}$, then the time taken by $B$ alone to finish the work is (in $\text{days}$)

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