B would have taken 10 , text{hours} more than | vedclass

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Question

(B) Let the time taken by $A$ to complete the task be $x \, 	ext{hours}$.Then,the time taken by $B$ to complete the task is $(x + 10) \, 	ext{hours}

Vedclass · Accepted Answer

$15$

Vedclass · Answer

(B) Let the time taken by $A$ to complete the task be $x \, 	ext{hours}$.Then,the time taken by $B$ to complete the task is $(x + 10) \, 	ext{hours}$.Working together,their combined rate of work is $\frac{1}{x} + \frac{1}{x + 10} = \frac{1}{12}$.Solving for $x$: $\frac{x + 10 + x}{x(x + 10)} = \frac{1}{12} \implies 12(2x + 10) = x^2 + 10x$.$24x + 120 = x^2 + 10x \implies x^2 - 14x - 120 = 0$.Factoring the quadratic equation: $(x - 20)(x + 6) = 0$.Since time cannot be negative,$x = 20 \, 	ext{hours}$.Thus,$B$ takes $x + 10 = 30 \, 	ext{hours}$ to complete the whole task.To complete $50 \%$ of the task,$B$ will take $50 \% 	ext{ of } 30 \, 	ext{hours} = 15 \, 	ext{hours}$.

$B$ would have taken $10 \, \text{hours}$ more than what $A$ would have taken to complete a task if each of them worked alone. Working together they can complete the task in $12 \, \text{hours}$. How many $\text{hours}$ would $B$ take to do $50 \%$ of the task?

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