Question

The dimensions of a rectangular field are 50 m and 40 m. A flower bed is prepared inside this field leaving a gravel path of uniform width all around the flower bed. The total cost of laying the flower bed and gravelling the path at Rs. 30 and Rs. 20 per square metre, respectively, is Rs. 52,000. Find the width of the gravel path.

Answer 1

Let the width of the gravel path be w m.

Length of the rectangular field = 50 m

Breadth of the rectangular field = 40 m

Let the length and breadth of the flower bed be x m and y m respectively.

Therefore, we have:

x + 2w = 50  ...(1)

y + 2w = 40  ...(2)

Also, area of rectangular field = 50 m 40 m = 2000 m²

Area of the flower bed = xy m²

Area of gravel path = Area of rectangular field – Area of flower bed = (2000 – xy) m²

Cost of laying flower bed + Gravel path = Area × cost of laying per sq. m

52000 = 30 xy + 20 (2000 – xy)

52000 = 10xy + 40000

xy = 1200

Using (1) and (2), we have:

(50 – 2w)(40 – 2w) = 1200

2000 – 180w + 4w² = 1200

4w² – 180w + 800 = 0

w² – 45w + 200 = 0

w² – 5w – 40w + 200 = 0

w(w – 5) – 40(w – 5) = 0

(w – 5)(w – 40) = 0

w = 5, 40

If w = 40, then x = 50 – 2w = –30, which is not possible.

Thus, the width of the gravel path is 5 m.

Answer 2

Dimensions of the outer rectangle (field) = 50 m × 40 m

Total area = 50 × 40 = 2000 m²

Length = 50 − 2x

Width = 40 − 2x

A_bed​ = (50 − 2x) (40 − 2x)

A_path​ = Total Area − Area of flower bed

= 2000 − (50 − 2x) (40 − 2x)

Total Cost

• Cost of flower bed = ₹30/m²

• Cost of gravel path = ₹20/m²

• Total cost = ₹52,000

Total cost = 30 × Area of bed + 20 × Area of path

52000 = 30(50 − 2x) (40 − 2x) + 20[2000 − (50 − 2x) (40 − 2x)]

Let A = (50 − 2x) (40 − 2x)

52000 = 30A + 20(2000 − A)

= 30A + 40000 − 20A = 10A + 40000

⇒ 10A = 52000 − 40000 = 12000

⇒ A = 1200

(50 − 2x) (40 − 2x) = 1200

(50 − 2x) (40 − 2x) = 1200

2000 − 100x − 80x + 4x² = 1200

4x² − 180x + 2000 = 1200

4x² − 180x + 800 = 0

4x² − 180x + 800 = 0

x² − 45x + 200 = 0

`x = 45 +- sqrt((-45)^2 - 4(1)(200))/(2(1)) = (45 +- sqrt(2025-800))/2`

`= (45 +- sqrt1225)/2 = (45 +- 35)/2`

so

`x= (45+35)/2 = 40`

`x = (45-35)/2 = 5`

Width of the gravel path is 5 metres.