Advertisements
Advertisements
Question
There is a square field whose side is 44m. A square flower bed is prepared in its centre leaving a gravel path all round the flower bed. The total cost of laying the flower bed and graving the path at Rs 2. 75 and Rs. 1.5 per square metre, respectively, is Rs 4,904. Find the width of the gravel path.
Solution
Let the side of flower bed be a and that of gravel path be b.
Then a + 2b = 44 (as 44 is overall size of field and it contains side of flower bed and double side of gravel path) .... (i)
Area of Flower bed = a2
Area of Gravel path = Area of Square - Area of flower bed = 44 x 44 - a2
⇒ Area of Gravel path = 1936 - a2
Cost of laying flower bed + Gravel path = Area x cost of laying per sq.m
⇒ 4904 = (a2 x 2.75) + (1936 - a2) x 1.5
⇒ 4904 = 2.75a2 -1.5 a2 + 2904
⇒ 1.25 a2 = 2000
⇒ a2 = 1600, Hence a = 40.
Hence gravel path width = `(44 - 40)/2` m = 2 m
APPEARS IN
RELATED QUESTIONS
Solve the following quadratic equations by factorization:
25x(x + 1) = -4
Solve the given quadratic equation for x : 9x2 – 9(a + b)x + (2a2 + 5ab + 2b2) = 0 ?
Determine whether the values given against the quadratic equation are the roots of the equation.
x2 + 4x – 5 = 0 , x = 1, –1
Solve the following quadratic equation by factorisation.
2m (m − 24) = 50
Solve the following quadratic equation by factorisation.
m2 - 11 = 0
If y = 1 is a common root of the equations \[a y^2 + ay + 3 = 0 \text { and } y^2 + y + b = 0\], then ab equals
Solve equation using factorisation method:
`x = (3x + 1)/(4x)`
By increasing the speed of a car by 10 km/hr, the time of journey for a distance of 72 km. is reduced by 36 minutes. Find the original speed of the car.
Use the substitution y = 3x + 1 to solve for x : 5(3x + 1 )2 + 6(3x + 1) – 8 = 0
Find a natural number whose square diminished by 84 is equal to thrice of 8 more than the given number.