A rectangle tank measuring $5 m \times 4.5 m \times 2.1 m$ is dog in the centre of the field measuring $13.5 m \times 2.5 m$ . The earth dug out is spread over the remaining portion the of the field. How much of the field raised?

3.2 m

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5.2 m

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4.2 m

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6.2 m

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Solution

The correct option is C $4.2 m$

Earth dug out = Volume of tank
$= 5 \times 4.5 \times 2.1$
= 47.25 $m^{2}$
Remaining area of field $= (13.5 \times 2.5) - (5 \times 4.5)$
= 11.25 $m^{2}$
Let the field is raised by 'h' m
then $h \times 11.25 = 47.25$
h $= \frac{47.25}{11.25} = 4.2$ m

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A rectangle tank measuring 5m×4.5m×2.1m is dog in the centre of the field measuring 13.5m×2.5m. The earth dug out is spread over the remaining portion the of the field. How much of the field raised?

The correct option is C 4.2mEarth dug out = Volume of tank=5×4.5×2.1= 47.25 m2Remaining area of field =(13.5×2.5)−(5×4.5) = 11.25 m2Let the field is raised by 'h' m then h×11.25=47.25h =47.2511.25=4.2 m

A rectangle tank measuring $5 m \times 4.5 m \times 2.1 m$ is dog in the centre of the field measuring $13.5 m \times 2.5 m$ . The earth dug out is spread over the remaining portion the of the field. How much of the field raised?

The correct option is C $4.2 m$

Earth dug out = Volume of tank
$= 5 \times 4.5 \times 2.1$
= 47.25 $m^{2}$
Remaining area of field $= (13.5 \times 2.5) - (5 \times 4.5)$
= 11.25 $m^{2}$
Let the field is raised by 'h' m
then $h \times 11.25 = 47.25$
h $= \frac{47.25}{11.25} = 4.2$ m