Answer
414.9k+ views
Hint: We are given a field in which a tank is dug out the earth taken out of the field is evenly spread over the field we have to find how much the level of the field raised by the spreading of the earth in this type of problem first we will find the area of field and then we will find the area of tank then we will find out the area of field on which earth needs to be spread then we will find the area of remaining field and the volume of earth taken out from the tank then we will find the height raised.
Formula used: Volume = L×B×H
Here L= Length, B= breadth, H= height
Area = $L \times B$
Here L = Length B = breadth
Complete step-by-step answer:
Step1: We are given field of dimensions length$ = 60$m breadth$ = 20$m
And dimensions tank i.e. dug out is
Length$ = 12$m, breadth$ = 10$m, height$ = 3$m
Step2: Firstly we will find the area of the field using the formula
Area = L×B
Here L$ = 60$m, B$ = 20$m
Substituting the value in formula we get:
Area$ = 60 \times 20$
$ = 1200$m2
Area of the tank dug out is
Area = L × B
Here L$ = 12$m and B$ = 10$m
Area$ = 12 \times 10$
Area of tank$ = 120$$m^2$
Step3: Then we will the area of the field where the earth has to be spread
= (Area of field) - (area tank)
On substituting the value we get
On $ = 1200 - 120$
$ = 1080$$m^2$
Step4: Then we will find the volume of the earth dug out for it
V = L×B×H
Here L$ = 12$m, B$ = 10$m, H$ = 3$m
Substituting the value in the formula
V $ = 12 \times 10 \times 3$
V$ = 360$$m^3$
Step5: Volume of earth that is to be spread = Volume of earth dug out + $\dfrac{1}{8}$(volume of earth dug)$ = 360 + \dfrac{1}{8}(360)$
On solving the equation
$ = 360 + 45$
Adding the number we get
$ = 405$$m^3$
Then we will find the height of the field raised for it and we will use the formula
H= $\dfrac{{Volume}}{{Area}}$
Here V$ = 405$$m^3$
Area$ = 1080$$m^2$
Substituting the value in the formula we get:
Area$ = \dfrac{{405}}{{1080}}$
$ = 0.375$m
The height is $0.375$m
Note: Don’t get confused in applying the formula or where to apply the formula and which one to find the height of we always divide volume by area this should be kept in mind and in case of finding the volume earth dug out it is always equal to the volume of the tank and in all such case.
Formula used: Volume = L×B×H
Here L= Length, B= breadth, H= height
Area = $L \times B$
Here L = Length B = breadth
Complete step-by-step answer:
Step1: We are given field of dimensions length$ = 60$m breadth$ = 20$m
And dimensions tank i.e. dug out is
Length$ = 12$m, breadth$ = 10$m, height$ = 3$m
Step2: Firstly we will find the area of the field using the formula
Area = L×B
Here L$ = 60$m, B$ = 20$m
Substituting the value in formula we get:
Area$ = 60 \times 20$
$ = 1200$m2
Area of the tank dug out is
Area = L × B
Here L$ = 12$m and B$ = 10$m
Area$ = 12 \times 10$
Area of tank$ = 120$$m^2$
Step3: Then we will the area of the field where the earth has to be spread
= (Area of field) - (area tank)
On substituting the value we get
On $ = 1200 - 120$
$ = 1080$$m^2$
Step4: Then we will find the volume of the earth dug out for it
V = L×B×H
Here L$ = 12$m, B$ = 10$m, H$ = 3$m
Substituting the value in the formula
V $ = 12 \times 10 \times 3$
V$ = 360$$m^3$
Step5: Volume of earth that is to be spread = Volume of earth dug out + $\dfrac{1}{8}$(volume of earth dug)$ = 360 + \dfrac{1}{8}(360)$
On solving the equation
$ = 360 + 45$
Adding the number we get
$ = 405$$m^3$
Then we will find the height of the field raised for it and we will use the formula
H= $\dfrac{{Volume}}{{Area}}$
Here V$ = 405$$m^3$
Area$ = 1080$$m^2$
Substituting the value in the formula we get:
Area$ = \dfrac{{405}}{{1080}}$
$ = 0.375$m
The height is $0.375$m
Note: Don’t get confused in applying the formula or where to apply the formula and which one to find the height of we always divide volume by area this should be kept in mind and in case of finding the volume earth dug out it is always equal to the volume of the tank and in all such case.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Why Are Noble Gases NonReactive class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let X and Y be the sets of all positive divisors of class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
At which age domestication of animals started A Neolithic class 11 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Which are the Top 10 Largest Countries of the World?
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give 10 examples for herbs , shrubs , climbers , creepers
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference Between Plant Cell and Animal Cell
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Write a letter to the principal requesting him to grant class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Change the following sentences into negative and interrogative class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)