Question

The central park of the city is 40 metres long and 30 metres wide. The mayor wants to construct two roads of equal width in the park such that the roads intersect each other at right angles and the center of the rectangle coincides with the center of the rectangle formed by the intersection of the two roads. Further, the mayor wants that the area of the two roads to be equal to the remaining area of the park. What should be the width of the roads?

• A. 10 metres
• B. 12.5 metres
• C. 14 metres
• D. 15 metres
• E. 16 metres

Hint:

For crossing roads in a rectangle, use inclusion–exclusion: sum of two strips minus overlap. Always check feasibility against the park's dimensions.

Answer 1

The Correct Option is A

Step 1: Total area of park = \(40 \times 30 = 1200 \ \text{m}^2\). Let width of each road = \(x\). Step 2: Area of two roads (a cross) = Area of horizontal strip + Area of vertical strip \(-\) overlapping square. \[ = (40 \times x) + (30 \times x) - (x \times x) = 70x - x^2. \] Step 3: Given: Area of two roads = remaining park area. Thus \[ 70x - x^2 = 1200 - (70x - x^2). \] \[ ⇒ 2(70x - x^2) = 1200 ⇒ 70x - x^2 = 600. \] \[ ⇒ x^2 - 70x + 600=0. \] Step 4: Solve quadratic: \[ x = \frac{70 \pm \sqrt{70^2 - 4\cdot600}}{2} = \frac{70 \pm \sqrt{4900-2400}}{2} = \frac{70 \pm \sqrt{2500}}{2} = \frac{70 \pm 50}{2}. \] So \(x=60\) or \(x=10\). Width cannot exceed smaller dimension (30), hence \(x=10\). Thus required width is \(\boxed{10 \ \text{metres}}\).