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A field is 30 m long and 18 m broad. A pit 6 m long, 5 m wide and 3 m deep, is dug out from the middle of the field and the earth removed in evenly spread over the remaining area of the field. - Mathematics

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Question

A field is 30 m long and 18 m broad. A pit 6 m long, 5 m wide and 3 m deep, is dug out from the middle of the field and the earth removed in evenly spread over the remaining area of the field. Find the rise in the level of the remaining part of the field in centimeters correct to one decimal place.

Sum
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Solution

Given: The field is 30 m by 18 m and a pit 6 m by 5 m by 3 m is dug and its earth evenly spread over the remaining area of the field.

Step-wise calculation:

1. Area of whole field

= 30 × 18

= 540 m2

2. Footprint area of pit

= 6 × 5

= 30 m2

3. Remaining area

= 540 – 30

= 510 m2

4. Volume of earth removed

= 6 × 5 × 3

= 90 m3

5. Rise in level (in meters)

= `"Volume"/"Remaining area"` 

= `90/510` 

= `3/17`

= 0.176470588 m

6. Convert to centimeters:

0.176470588 × 100

= 17.6470588 cm

7. Round to one decimal place:

17.6 cm

The rise in the level of the remaining part of the field is 17.6 cm correct to one decimal place.

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Chapter 16: Mensuration - Exercise 16D [Page 344]

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Nootan Mathematics [English] Class 9 ICSE
Chapter 16 Mensuration
Exercise 16D | Q 14. | Page 344
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