Question

plzzz solve it fastt​

Answer 1

$\: \: \boxed{\sf{\blue{Let's \: understand \: the \: concept \: first \: }}}$

Here the concept of Volume of Cuboid has been used. We know that the packets of soap are in shape of cuboid. Now if, the total number of packets each if same dimensions, are packed in a a box, then the volum of box will be equal to the number of packets of soap into the volume of each soap. Using this concept, let us find our answer.

$\: ➮ \: \large{\underline{\boxed{\rm{\green{Volume \: of \: Cuboid \: = \: Length(L) \: \times \: Breadth(B) \: \times \: Height(H)}}}}}$

$\: ➮ \: \large{\underline{\boxed{\rm{\green{Volume \: of \: box \: = \: Volume \: of \: each \: packet \: \times \: Number \: of \: packets}}}}}$

_____________________________________

★ Question :-

960 identical packets of soaps each of 10 cm long and 5 cm wide can be kept inside a rectangular box of dimension 80 cm*60 cm*40 cm, what is the height of each packet of soap ?

_____________________________________

★ Solution :-

Given,

» Number of packets of soap = n = 960

» Length of each packet of soap = l = 10 cm

» Breadth of each packet of soap = b = 5 cm

» Length of each packet of soap = L = 80 cm

» Breadth of each packet of soap = B = 60 cm

» Height of each packet of soap = H = 40 cm

• Let the height of the packet of soap be 'h' cm. Then, using the concept, we get,

⟹ l × b × h × n = L × B × H

⟹ 10 × 5 × h × 960 = 80 × 60 × 40

⟹ 50 × 960 × h = 80 × 60 × 40

$\: \: \: \: \huge{\longrightarrow \:\:\bold{h \: = \: \dfrac{80 \: \times \: 60 \: \times \: 40}{50 \: \times \: 960}}}$

⟹ h = 4 cm

$\: \: ➮ \: \: \underline{\boxed{\sf{\red{Hence, \: the \: height \: of \: each \: packet \: of \: soap \: is \: \underline{4 \: cm}}}}}$

_______________________________

$\: \: \boxed{\rm{\purple{Have \: a \: glimpse \: here \: for \: better \: understanding \: }}}$

• Volume of cuboid = (side)³

• Volume of cylinder = πr²h

where, r is radius and h is height.

• Volume of cone = ⅓(πr²h)

• Total Surface Area of Cuboid = 2(lb + bh + lh)

• Total Surface Area of Cube = 4a²

• Total Surface Area of cylinder = 2πr(r + h)

• Total Surface Area of Cone = πr(r + l)

Answer 2

$\: \: \boxed{\sf{\blue{Let's \: understand \: the \: concept \: first \: }}}$

Here the concept of Volume of Cuboid has been used. We know that the packets of soap are in shape of cuboid. Now if, the total number of packets each if same dimensions, are packed in a a box, then the volum of box will be equal to the number of packets of soap into the volume of each soap. Using this concept, let us find our answer.

$\: ➮ \: \large{\underline{\boxed{\rm{\green{Volume \: of \: Cuboid \: = \: Length(L) \: \times \: Breadth(B) \: \times \: Height(H)}}}}}$

$\: ➮ \: \large{\underline{\boxed{\rm{\green{Volume \: of \: box \: = \: Volume \: of \: each \: packet \: \times \: Number \: of \: packets}}}}}$

_____________________________________

★ Question :-

960 identical packets of soaps each of 10 cm long and 5 cm wide can be kept inside a rectangular box of dimension 80 cm*60 cm*40 cm, what is the height of each packet of soap ?

_____________________________________

★ Solution :-

Given,

» Number of packets of soap = n = 960

» Length of each packet of soap = l = 10 cm

» Breadth of each packet of soap = b = 5 cm

» Length of each packet of soap = L = 80 cm

» Breadth of each packet of soap = B = 60 cm

» Height of each packet of soap = H = 40 cm

• Let the height of the packet of soap be 'h' cm. Then, using the concept, we get,

⟹ l × b × h × n = L × B × H

⟹ 10 × 5 × h × 960 = 80 × 60 × 40

⟹ 50 × 960 × h = 80 × 60 × 40

$\: \: \: \: \huge{\longrightarrow \:\:\bold{h \: = \: \dfrac{80 \: \times \: 60 \: \times \: 40}{50 \: \times \: 960}}}$

$\: \: ➮ \: \: \underline{\boxed{\sf{\red{Hence, \: the \: height \: of \: each \: packet \: of \: soap \: is \: \underline{4 \: cm}}}}}$

_______________________________

$\: \: \boxed{\rm{\purple{Have \: a \: glimpse \: here \: for \: better \: understanding \: }}}$

• Volume of cuboid = (side)³

• Volume of cylinder = πr²h

where, r is radius and h is height.

• Volume of cone = ⅓(πr²h)

• Total Surface Area of Cuboid = 2(lb + bh + lh)

• Total Surface Area of Cube = 4a²

• Total Surface Area of cylinder = 2πr(r + h)

• Total Surface Area of Cone = πr(r + l)