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(B) Let the faster pipe fill the reservoir in $x$ hours.Then,the slower pipe will fill it in $(x + 5)$ hours.Given that when both pipes work together,

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$10$ $hours$

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(B) Let the faster pipe fill the reservoir in $x$ hours.Then,the slower pipe will fill it in $(x + 5)$ hours.Given that when both pipes work together,the reservoir is filled in $6$ hours.Therefore,the rate of the faster pipe is $1/x$ and the rate of the slower pipe is $1/(x + 5)$.The combined rate is $1/x + 1/(x + 5) = 1/6$.Solving the equation: $(x + 5 + x) / (x(x + 5)) = 1/6$.$6(2x + 5) = x^2 + 5x$.$12x + 30 = x^2 + 5x$.$x^2 - 7x - 30 = 0$.$(x - 10)(x + 3) = 0$.Since time cannot be negative,$x = 10$.Thus,the faster pipe takes $10$ hours to fill the reservoir.

If two pipes function simultaneously,the reservoir will be filled in $6$ $hours$. One pipe fills the reservoir $5$ $hours$ faster than the other. How many hours does the faster pipe take to fill the reservoir?

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