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A rectangular field is $16$ m long and $10$ m wide. There is a path of equal width all around it, having an area of $120$ sq.m. Find the width of the path.
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Solution: Let the width of the path be $x$ m.
A. T. Q. $$\begin{aligned}& (16 + 2x)(10 + 2x) - 16 \times 10 = 120 \\ & \Rightarrow 4x^2 + 52x - 120 = 0 \text{ or } x^2 + 13x - 30 = 0 \\ & \Rightarrow (x - 2)(x + 15) = 0 \\ & \Rightarrow x = 2 \text{ (Rejecting } x = -15) \\ & \therefore \text{Width of the path is } 2 \text{ m.}\end{aligned}$$
A. T. Q. $$\begin{aligned}& (16 + 2x)(10 + 2x) - 16 \times 10 = 120 \\ & \Rightarrow 4x^2 + 52x - 120 = 0 \text{ or } x^2 + 13x - 30 = 0 \\ & \Rightarrow (x - 2)(x + 15) = 0 \\ & \Rightarrow x = 2 \text{ (Rejecting } x = -15) \\ & \therefore \text{Width of the path is } 2 \text{ m.}\end{aligned}$$