An agricultural field is in the form of a rectangle of length $20$ m and width $14$ m A pit $6$ m long $3$ m wide and $2.5$ m deep is dug in a corner of the field and the earth taken out of the pit is spread uniformly over the remaining area of the field. The level of the field has been raised by

15.16

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16.17

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17.18

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18.19

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Solution

The correct option is C $17.18$ cm
Volume of pit $= 6 \times 3 \times 2.5 = 45 m^{3}$
Now, volume of earth taken out=volume of earth spread in the remaining field
$\Rightarrow 45 = [(20 \times 14) - (6 \times 3)] \times h$
$\Rightarrow 45 = 262 \times h$
$\Rightarrow h = \frac{45 \times 10^{6}}{262 \times 10^{4}} = \frac{4500}{262} = 17.18 c m$

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An agricultural field is in the form of a rectangle of length 20 m and width 14 m A pit 6 m long 3 m wide and 2.5 m deep is dug in a corner of the field and the earth taken out of the pit is spread uniformly over the remaining area of the field. The level of the field has been raised by

The correct option is C 17.18 cmVolume of pit=6×3×2.5=45m3Now, volume of earth taken out=volume of earth spread in the remaining field⇒45=[(20×14)−(6×3)]×h⇒45=262×h⇒h=45×106262×104=4500262=17.18cm

An agricultural field is in the form of a rectangle of length $20$ m and width $14$ m A pit $6$ m long $3$ m wide and $2.5$ m deep is dug in a corner of the field and the earth taken out of the pit is spread uniformly over the remaining area of the field. The level of the field has been raised by