The dimensions of a field are 15 m and 12 m. A 8 m long, 2.5 m wide an

1.	The dimensions of a field are 15 m and 12 m. A 8 m long, 2.5 m wide and 2 m deep pit is dug in one corner of the field and the earth removed from this pit is evenly spread over the remaining area of the field. Calculate by how much the level of field is raised.
Answer» QUESTION:- The dimensions of a field are 15 m and 12 m. A 8 m long, 2.5 m WIDE and 2 m deep pit is dug in one corner of the field and the EARTH removed from this pit is evenly SPREAD over the remaining area of the field. Calculate by how much the level of field is raised. ANSWER :- LET :- length of the field → L breadth of the field → B So, L = 15 m and B = 12 m Area of the field → L × B → 15 m × 12 m = 180 m² LET :- length of the pit → l breadth of the pit → b height of the pit → h So, l = 8 m , b = 2.5 m and h = 2 m Area of the pit's base → l × b → 8 m × 2.5 m → 20 m² Area over which the earth is spread → ( 180 - 20 ) m² → 160 m² Volume of earth dug → l × b × h → ( 8 × 2.5 × 2 ) m³ → 40 m³ Rise in the level of field → $→ \frac{volume \: of \: earth \: dug}{area \: of \: which \: earth \: is \: spread}$ $→ \frac{ {40m}^{2} }{ {160m}^{2} }$ $→ \frac{1}{4} m$ → 25 cm

The dimensions of a field are 15 m and 12 m. A 8 m long, 2.5 m wide and 2 m deep pit is dug in one corner of the field and the earth removed from this pit is evenly spread over the remaining area of the field. Calculate by how much the level of field is raised.

Answer»

QUESTION:-

The dimensions of a field are 15 m and 12 m. A 8 m long, 2.5 m WIDE and 2 m deep pit is dug in one corner of the field and the EARTH removed from this pit is evenly SPREAD over the remaining area of the field. Calculate by how much the level of field is raised.

ANSWER :-

LET :-

length of the field → L
breadth of the field → B

So, L = 15 m and B = 12 m

Area of the field

→ L × B

→ 15 m × 12 m = 180 m²

LET :-

length of the pit → l
breadth of the pit → b
height of the pit → h

So, l = 8 m , b = 2.5 m and h = 2 m

Area of the pit's base

→ l × b

→ 8 m × 2.5 m

→ 20 m²

Area over which the earth is spread

→ ( 180 - 20 ) m²

→ 160 m²

Volume of earth dug

→ l × b × h

→ ( 8 × 2.5 × 2 ) m³

→ 40 m³

Rise in the level of field →

$→ \frac{volume \: of \: earth \: dug}{area \: of \: which \: earth \: is \: spread}$

$→ \frac{ {40m}^{2} }{ {160m}^{2} }$

$→ \frac{1}{4} m$