Question

A soft drink is available in two packs -
(i) a tin can with a rectangular base of length 5 cm and width 4 cm, having a height of 15 cm and
(ii) a plastic cylinder with circular base of diameter 7 cm and height 10 cm.
Which container has greater capacity and by how much?
​

Answer 1

Given :

• A soft drink is available in two packs :-

(i) a tin can with a rectangular base of length 5 cm and width 4 cm, having a height of 15 cm and

(ii) a plastic cylinder with circular base of diameter 7 cm and height 10 cm.

$\rule{200pt}{3pt}$

To Find :

• Which container has greater capacity and by how much ?

$\rule{200pt}{3pt}$

Solution :

❒ Formula Used :

• Cuboid :

$\large{\star \: {\underline{\boxed{\purple{\sf{\pmb{ Volume{\small_{(Cuboid)}} = Length \times Breadth \times Height }}}}}}} \: {\star}$

• Cylinder :

$\large{\star \: {\underline{\boxed{\purple{\sf{\pmb{ Volume{\small_{(Cylinder)}} = \pi r²h }}}}}}} \: {\star}$

$\qquad{\rule{150pt}{1pt}}$

❒ Calculating the Volume of first pack :

Here we have :

• ➙ Length = 5 cm
• ➙ Breadth = 4 cm
• ➙ Height = 15 cm

Calculating :

${\longmapsto{\qquad{\sf{ Volume{\small_{(First \: pack )}} = Length \times Width \times Height }}}} \\ \\ \ {\longmapsto{\qquad{\sf{ Volume{\small_{(First \: pack )}} = 5 \times 4 \times 15 }}}} \\ \\ \ {\longmapsto{\qquad{\sf{ Volume{\small_{(First \: pack )}} = 20 \times 15 }}}} \\ \\ \ {\qquad{\pmb{\sf{ Volume \: of \: the \: first \: pack \: = {\red{\frak{ 300 \: cm³ }}}}}}}$

$\qquad{\rule{150pt}{1pt}}$

❒ Calculating the Volume of second pack :

Here we have :

• ➙ $\pi = \dfrac{22}{7}$

• ➙ Radius = ${\sf{\dfrac{Diameter}{2} = \dfrac{7}{2}}}$

• ➙ Height = 10 cm

Calculating :

${\longmapsto{\qquad{\sf{ Volume{\small_{(Second \: Pack )}} = \pi r²h }}}} \\ \\ \ {\longmapsto{\qquad{\sf{ Volume{\small_{(Second \: Pack )}} = \dfrac{22}{7} \times \dfrac{7}{2} \times \dfrac{7}{2} \times 10 }}}} \\ \\ \ {\longmapsto{\qquad{\sf{ Volume{small_{(Second \: Pack)}} = \dfrac{22}{7} \times \dfrac{7}{2} \times \dfrac{7}{\cancel2} \times \cancel{10} }}}} \\ \\ \ {\longmapsto{\qquad{\sf{ Volume{\small_{(Second \: Pack )}} = \dfrac{11}{7} \times 7 \times 7 \times 5 }}}} \\ \\ \ {\longmapsto{\qquad{\sf{ Volume{\small_{(Second \: Pack )}} = \dfrac{11}{\cancel7} \times \cancel7 \times 7 \times 5 }}}} \\ \\ \ {\longmapsto{\qquad{\sf{ Volume{\small_{(Second \: Pack )}} = 11 \times 35 }}}} \\ \\ \ {\qquad{\pmb{\sf{ Volume \: of \: the \: second \: pack \: = {\orange{\frak{ 385 \: cm³ }}}}}}}$

$\qquad{\rule{150pt}{1pt}}$

❒ Calculating the Difference :

${\dashrightarrow{\qquad{\sf{ Difference = Second \: pack - First \: pack }}}} \\ \\ \ {\dashrightarrow{\qquad{\sf{ Difference = 385 - 300 }}}} \\ \\ \ {\qquad{\pmb{\sf{ Difference \: between \: the \: packs \: = {\green{\frak{ 85 \: cm³ }}}}}}}$

$\qquad{\rule{150pt}{1pt}}$

❒ Therefore :

❝ Second cylinderical pack has greater capacity and by 85 cm³ . ❞

$\\ {\underline{\rule{300pt}{9pt}}}$

${\pmb{\red{\sf{@Advanced}}}{\pmb{\orange{\sf{Nobita}}}}}$

Answer 2

1. Pack (1) :

Volume of Cuboid = L * B * H

⇒ 5 x 4 x 15

⇒ 20 x 15

⇒ 300cm³

2. Pack(2) :

Volume of Cylinder = πr²h

⇒ (22/7) x (3.5)² x 10

⇒ 38.465 x 10

⇒ 384.65cm³

Difference in volume b/w two packs :

⇒ 384.65 - 300

⇒ 84.65 cm³