The areas of three fields are 165 m^2,195 m | vedclass

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(D) Let the length of the flower bed be $L$ meters. Since the width is $3 \ m$,the area of the flower bed in each field is $3L \ m^2$.Since $3L$ must

Vedclass · Accepted Answer

None of these

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(D) Let the length of the flower bed be $L$ meters. Since the width is $3 \ m$,the area of the flower bed in each field is $3L \ m^2$.Since $3L$ must be a common divisor of the areas $165, 195,$ and $85$,we first find the $H.C.F.$ of $165, 195,$ and $85$.Prime factorization:$165 = 3 	imes 5 	imes 11$$195 = 3 	imes 5 	imes 13$$85 = 5 	imes 17$The $H.C.F.$ of $165, 195,$ and $85$ is $5$.Thus,the maximum area of the flower bed that can be carved out of each field is $5 \ m^2$.Given the width is $3 \ m$,we have $3 	imes L = 5$.Therefore,$L = \frac{5}{3} \ m \approx 1.67 \ m$.Since $\frac{5}{3}$ is not among the options,the correct answer is "None of these".

The areas of three fields are $165 \ m^2$,$195 \ m^2$,and $85 \ m^2$ respectively. In each of the fields,a flower bed of equal length has to be made. If the flower bed in each of the fields is $3 \ m$ wide,then what is the maximum length of the flower bed in each of the fields? (in $m$)

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