20. Randy scored x marks in mathematics. His marks in biology were 3 more than two-thirds of the marks obtained by him in mathematics. The marks scored by Randy in biology were ?
Answers
Step-by-step explanation:
The container will be emptied in 63 minutes.
Step-by-step-explanation:
We have given that,
For a conical container filled with petrol,
Radius ( r ) = 10 m
Height ( h ) = 15 m
Rate of releasing petrol = 25 m³/min
We have to find the time required for the container to empty.
We know that,
\displaystyle{\boxed{\pink{\sf\:Volume\:of\:cone\:=\:\dfrac{1}{3}\:\pi\:r^2\:h\:}}}
Volumeofcone=
3
1
πr
2
h
\displaystyle{\implies\sf\:Volume\:of\:container\:=\:\dfrac{1}{\cancel{3}}\:\times\:3.14\:\times\:(\:10\:)^2\:\times\:\cancel{15}}⟹Volumeofcontainer=
3
1
×3.14×(10)
2
×
15
\displaystyle{\implies\sf\:Volume\:of\:container\:=\:3.14\:\times\:100\:\times\:5}⟹Volumeofcontainer=3.14×100×5
\displaystyle{\implies\sf\:Volume\:of\:container\:=\:314\:\times\:5}⟹Volumeofcontainer=314×5
\displaystyle{\implies\:\boxed{\blue{\sf\:Volume\:of\:container\:=\:1570\:m^3\:}}}⟹
Volumeofcontainer=1570m
3
Now,
\displaystyle{\boxed{\green{\sf\:Time\:required\:to\:empty\:container\:=\:\dfrac{Volume\:of\:container}{Rate\:of\:releasing\:petrol}}}}
Timerequiredtoemptycontainer=
Rateofreleasingpetrol
Volumeofcontainer
\displaystyle{\implies\sf\:Time\:required\:to\:empty\:container\:=\:\dfrac{\cancel{1570}\:m^3}{\cancel{25}\:m^3\:/\:min}}⟹Timerequiredtoemptycontainer=
25
m
3
/min
1570
m
3
\displaystyle{\implies\sf\:Time\:required\:to\:empty\:container\:=\:\dfrac{\cancel{314}\:\cancel{m^3}}{\cancel{5}\:\dfrac{\cancel{m^3}}{min}}}⟹Timerequiredtoemptycontainer=
5
min
m
3
314
m
3
\displaystyle{\implies\sf\:Time\:required\:to\:empty\:container\:=\:\dfrac{62.8}{\dfrac{1}{min}}}⟹Timerequiredtoemptycontainer=
min
1
62.8
\displaystyle{\implies\sf\:Time\:required\:to\:empty\:container\:=\:62.8\:min}⟹Timerequiredtoemptycontainer=62.8min
\displaystyle{\therefore\:\underline{\boxed{\red{\sf\:Time\:required\:to\:empty\:container\:\approx\:63\:min\:}}}}∴
Timerequiredtoemptycontainer≈63min
∴ The container will be emptied in 63 minutes. pm
Step-by-step explanation:
The container will be emptied in 63 minutes.
Step-by-step-explanation:
We have given that,
For a conical container filled with petrol,
Radius ( r ) = 10 m
Height ( h ) = 15 m
Rate of releasing petrol = 25 m³/min
We have to find the time required for the container to empty.
We know that,
\displaystyle{\boxed{\pink{\sf\:Volume\:of\:cone\:=\:\dfrac{1}{3}\:\pi\:r^2\:h\:}}}
Volumeofcone=
3
1
πr
2
h
\displaystyle{\implies\sf\:Volume\:of\:container\:=\:\dfrac{1}{\cancel{3}}\:\times\:3.14\:\times\:(\:10\:)^2\:\times\:\cancel{15}}⟹Volumeofcontainer=
3
1
×3.14×(10)
2
×
15
\displaystyle{\implies\sf\:Volume\:of\:container\:=\:3.14\:\times\:100\:\times\:5}⟹Volumeofcontainer=3.14×100×5
\displaystyle{\implies\sf\:Volume\:of\:container\:=\:314\:\times\:5}⟹Volumeofcontainer=314×5
\displaystyle{\implies\:\boxed{\blue{\sf\:Volume\:of\:container\:=\:1570\:m^3\:}}}⟹
Volumeofcontainer=1570m
3
Now,
\displaystyle{\boxed{\green{\sf\:Time\:required\:to\:empty\:container\:=\:\dfrac{Volume\:of\:container}{Rate\:of\:releasing\:petrol}}}}
Timerequiredtoemptycontainer=
Rateofreleasingpetrol
Volumeofcontainer
\displaystyle{\implies\sf\:Time\:required\:to\:empty\:container\:=\:\dfrac{\cancel{1570}\:m^3}{\cancel{25}\:m^3\:/\:min}}⟹Timerequiredtoemptycontainer=
25
m
3
/min
1570
m
3
\displaystyle{\implies\sf\:Time\:required\:to\:empty\:container\:=\:\dfrac{\cancel{314}\:\cancel{m^3}}{\cancel{5}\:\dfrac{\cancel{m^3}}{min}}}⟹Timerequiredtoemptycontainer=
5
min
m
3
314
m
3
\displaystyle{\implies\sf\:Time\:required\:to\:empty\:container\:=\:\dfrac{62.8}{\dfrac{1}{min}}}⟹Timerequiredtoemptycontainer=
min
1
62.8
\displaystyle{\implies\sf\:Time\:required\:to\:empty\:container\:=\:62.8\:min}⟹Timerequiredtoemptycontainer=62.8min
\displaystyle{\therefore\:\underline{\boxed{\red{\sf\:Time\:required\:to\:empty\:container\:\approx\:63\:min\:}}}}∴
Timerequiredtoemptycontainer≈63min
∴ The container will be emptied in 63 minutes. pm
Step-by-step explanation:
The container will be emptied in 63 minutes.
Step-by-step-explanation:
We have given that,
For a conical container filled with petrol,
Radius ( r ) = 10 m
Height ( h ) = 15 m
Rate of releasing petrol = 25 m³/min
We have to find the time required for the container to empty.
We know that,
\displaystyle{\boxed{\pink{\sf\:Volume\:of\:cone\:=\:\dfrac{1}{3}\:\pi\:r^2\:h\:}}}
Volumeofcone=
3
1
πr
2
h
\displaystyle{\implies\sf\:Volume\:of\:container\:=\:\dfrac{1}{\cancel{3}}\:\times\:3.14\:\times\:(\:10\:)^2\:\times\:\cancel{15}}⟹Volumeofcontainer=
3
1
×3.14×(10)
2
×
15
\displaystyle{\implies\sf\:Volume\:of\:container\:=\:3.14\:\times\:100\:\times\:5}⟹Volumeofcontainer=3.14×100×5
\displaystyle{\implies\sf\:Volume\:of\:container\:=\:314\:\times\:5}⟹Volumeofcontainer=314×5
\displaystyle{\implies\:\boxed{\blue{\sf\:Volume\:of\:container\:=\:1570\:m^3\:}}}⟹
Volumeofcontainer=1570m
3
Now,
\displaystyle{\boxed{\green{\sf\:Time\:required\:to\:empty\:container\:=\:\dfrac{Volume\:of\:container}{Rate\:of\:releasing\:petrol}}}}
Timerequiredtoemptycontainer=
Rateofreleasingpetrol
Volumeofcontainer
\displaystyle{\implies\sf\:Time\:required\:to\:empty\:container\:=\:\dfrac{\cancel{1570}\:m^3}{\cancel{25}\:m^3\:/\:min}}⟹Timerequiredtoemptycontainer=
25
m
3
/min
1570
m
3
\displaystyle{\implies\sf\:Time\:required\:to\:empty\:container\:=\:\dfrac{\cancel{314}\:\cancel{m^3}}{\cancel{5}\:\dfrac{\cancel{m^3}}{min}}}⟹Timerequiredtoemptycontainer=
5
min
m
3
314
m
3
\displaystyle{\implies\sf\:Time\:required\:to\:empty\:container\:=\:\dfrac{62.8}{\dfrac{1}{min}}}⟹Timerequiredtoemptycontainer=
min
1
62.8
\displaystyle{\implies\sf\:Time\:required\:to\:empty\:container\:=\:62.8\:min}⟹Timerequiredtoemptycontainer=62.8min
\displaystyle{\therefore\:\underline{\boxed{\red{\sf\:Time\:required\:to\:empty\:container\:\approx\:63\:min\:}}}}∴
Timerequiredtoemptycontainer≈63min
∴ The container will be emptied in 63 minutes. pm