Question

An ice-cream brick measures 20cm by 10cm by 7cm. how much such bricks can be stored in deep fridge whose inner dimensions are 100 cm by 50 cm cm by 42 ?​

Answer 1

$\large\underline{\sf{Solution-}}$

Dimensions of ice-cream brick

• Length of ice-cream brick = 20 cm

• Breadth of ice-cream brick = 10 cm

• Height of ice-cream brick = 7 cm

We know, ice-cream brick is in the shape of cuboid and volume of cuboid of length l, breadth b and height h is given by

$\boxed{ \rm{ \:Volume_{(Cuboid)} \: = \: l \times b \times h \: }}$

So, using this result, we get

$\rm \: Volume_{(ice-cream\:brick)} = 20 \times 10 \times 7$

$\rm\implies \:\boxed{ \rm{ \:Volume_{(ice-cream\:brick)} = 1400 \: {cm}^{3} \: \: }}$

Inner dimensions of deep - fridge

• Length of deep - fridge brick = 100 cm

• Breadth of deep - fridge cream brick = 50 cm

• Height of deep - fridge = 42 cm

We know, deep - fridge is in the shape of cuboid and volume of cuboid of length l, breadth b and height h is given by

$\boxed{ \rm{ \:Volume_{(Cuboid)} \: = \: l \times b \times h \: }}$

So, using this result, we get

$\rm \: Volume_{(deep \: - \:fridge)} = 100 \times 50 \times 42$

$\rm\implies \:\boxed{ \rm{ \:Volume_{(deep \: - \:fridge)} \: = \: 210000 \: {cm}^{3} \: }}$

Now, Let assume that n ice-cream bricks be stored in deep fridge.

So,

$\rm \: n \times Volume_{(ice-cream\:brick)} = Volume_{(deep \: - \:fridge)}$

$\rm \: n \times 1400 = 210000$

$\rm\implies \:\boxed{ \rm{ \:n \: = \: 150 \: \: }}$

So, it means 150 such bricks can be stored in deep fridge whose inner dimensions are 100 cm by 50 cm cm by 42.

$\rule{190pt}{2pt}$

Additional information :-

$\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{ \red{More \: Formulae}}}} \\ \\ \bigstar \: \bf{CSA_{(cylinder)} = 2\pi \: rh}\\ \\ \bigstar \: \bf{Volume_{(cylinder)} = \pi {r}^{2} h}\\ \\ \bigstar \: \bf{TSA_{(cylinder)} = 2\pi \: r(r + h)}\\ \\ \bigstar \: \bf{CSA_{(cone)} = \pi \: r \: l}\\ \\ \bigstar \: \bf{TSA_{(cone)} = \pi \: r \: (l + r)}\\ \\ \bigstar \: \bf{Volume_{(sphere)} = \dfrac{4}{3}\pi {r}^{3} }\\ \\ \bigstar \: \bf{Volume_{(cube)} = {(side)}^{3} }\\ \\ \bigstar \: \bf{CSA_{(cube)} = 4 {(side)}^{2} }\\ \\ \bigstar \: \bf{TSA_{(cube)} = 6 {(side)}^{2} }\\ \\ \bigstar \: \bf{Volume_{(cuboid)} = lbh}\\ \\ \bigstar \: \bf{CSA_{(cuboid)} = 2(l + b)h}\\ \\ \bigstar \: \bf{TSA_{(cuboid)} = 2(lb +bh+hl )}\\ \: \end{array} }}\end{gathered}\end{gathered}\end{gathered}$

Answer 2

$\huge{ \pink{\dag}} \large \underline{ \bf{ Solution:-}}$

Dimensions of ice cream brick

$\sf{ = \: 20 \: cm×10 \: cm×7cm}$

$\sf{∴ Volume = 20×10×7cm ^{3} =1400cm ^{3} }$

Dimensions of inner of fridge

$\sf{= 100 \: cm×50 \: cm×42 \: cm}$

$\sf{= 210000 \: cm³}$

∴ Number of bricks to be kept in the fridge

$\bf{ \red{ = }\frac{210000}{1400} = \green{150}}$

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