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Let us assume the gravel path’s width to be x m. Since the flower bed is in the centre of the square field with a gravel path surrounding it. So each side of the flower bed is= side of a square field- $2$(width of gravel).

$ \Rightarrow $ Each Side of the flower bed=$44 - 2{\text{x}}$ m

Then the area of flower bed=${\left( {44 - 2{\text{x}}} \right)^2}$ m {as the area of square=${\text{sid}}{{\text{e}}^2}$}

And the area of the square field =$44 \times 44 = 1936$ sq.m.

Then the area of gravel path=area of square field-area of flower bed

$ \Rightarrow $ Area of gravel path=$1936 - {\left( {44 - 2{\text{x}}} \right)^2} = 1936 - 1936 + 176{\text{x - 4}}{{\text{x}}^2}$

On simplifying, we get

$ \Rightarrow $ Area of gravel path=$176 - 4{{\text{x}}^2}$ sq.m.

Now, the cost of laying flower bed=area of flower bed × rate per sq. m. and we can write rate as$2.75 = \dfrac{{275}}{{100}}$ .Then on putting the values, we get-

$ \Rightarrow $ The cost of laying flower bed=${\left( {44 - 2{\text{x}}} \right)^2} \times \dfrac{{275}}{{100}} = \dfrac{{11}}{4}{\left( {44 - 2{\text{x}}} \right)^2} = 11{\left( {22 - {\text{x}}} \right)^2}$

And the cost of laying gravel path=area of gravel path × rate per sq. m. We can write rate as$1.50 = \dfrac{{150}}{{100}}$ .Then on putting values, we get-

$ \Rightarrow $ Cost of laying gravel path=$\left( {176 - 4{{\text{x}}^2}} \right)\dfrac{{150}}{{100}} = 6\left( {44 - {{\text{x}}^2}} \right) = 264 - 6{{\text{x}}^2}$

According to the question we know that the total cost of laying flower bed and gravel path=$4904$ Rs.

So,$ \Rightarrow 11{\left( {22 - {\text{x}}} \right)^2} + 264 - 6{{\text{x}}^2} = 4904$

On solving the eq. we get,

$

\Rightarrow 11\left( {484 + {{\text{x}}^2} - 44{\text{x}}} \right) + 264 - 6{{\text{x}}^2} = 4904 \\

\Rightarrow 5324 + 11{{\text{x}}^2} - 484{\text{x + }}264 - 6{{\text{x}}^2} = 4904 \\

\Rightarrow 5{{\text{x}}^2} - 220{\text{x + 5324 = 4904}} \\

\Rightarrow 5{{\text{x}}^2} - 220{\text{x + 420 = 0}} \\

$

On dividing the eq. by 5, we get-

$ \Rightarrow {{\text{x}}^2} - 44{\text{x + 84 = 0}}$

By factoring, we get-

\[

\Rightarrow {{\text{x}}^2} - 42{\text{x - 2x + 84 = 0}} \\

\Rightarrow {\text{x}}\left( {{\text{x - 42}}} \right) - 2\left( {{\text{x - 42}}} \right) = 0 \\

\Rightarrow \left( {{\text{x - 42}}} \right)\left( {{\text{x - 2}}} \right) = 0 \\

\Rightarrow {\text{x = 42 or x = 2}} \\

\]

But since the side of the square field =$44$ m so the width of gravel path cannot be longer than that.