A field is 80 m long and 50 m broad. In one corner of the field a pit which is 10 m long, 7.5 m broad and 8 m deep has been dug out. The earth taken out of it is evenly spread over the remaining part of the field. If the rise in the level of the field is x(in cm), then x-0.3 =

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Solution

Area of the field = (80 × 50) $m^{2} = 4000 m^{2}$

Area of the pit = (10 × 7.5) $m^{2} = 75 m^{2}$

Area over which the earth is spread out = (4000 - 75) $m^{2} = 3925 m^{2}$

Volume of earth dug out = (10 × 7.5 × 8) $m^{3} = 600 m^{3}$ .

Rise in level = $\frac{V o l u m e}{A r e a} = \frac{600}{3925} m = \frac{(600 \times 100)}{3925} c m = 15.3 c m$

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A field is 80 m long and 50 m broad. In one corner of the field a pit which is 10 m long, 7.5 m broad and 8 m deep has been dug out. The earth taken out of it is evenly spread over the remaining part of the field. If the rise in the level of the field is x(in cm), then x-0.3 = ___

Area of the field = (80 × 50) m2=4000m2 Area of the pit = (10 × 7.5) m2=75m2 Area over which the earth is spread out = (4000 - 75) m2=3925m2 Volume of earth dug out = (10 × 7.5 × 8) m3=600m3. Rise in level = VolumeArea=6003925m=(600×100)3925cm=15.3cm

A field is 80 m long and 50 m broad. In one corner of the field a pit which is 10 m long, 7.5 m broad and 8 m deep has been dug out. The earth taken out of it is evenly spread over the remaining part of the field. If the rise in the level of the field is x(in cm), then x-0.3 =

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