How Big How Heavy - Class 5 Maths - Extra Questions

Find the number of cubes in the base layer of the following figure:
1793108_036e2a9ac28d43c1a64b1dc4dfdd0971.png

The number of cubes in the base layer of the given figure = 6
1653694_1793108_ans_dffa9cd2cd27438099ba179bb4815359.png


Count the number of cubes in the given shape
1792825_9ffc0d27e30e40b7aef2bf80aae9b00b.png

Total cubes in given figure are :11


Count the number of cubes in the given shape
1792821_9eda59d9a454480e9eb74d0926e4ec0f.png

Total cubes in given figure are :9


Count the number of cubes in the given shape.


1792828_703dd90a390641c689448d36e2912323.png

Total cubes in given figure are :110


Count the number of cubes in the given shape
1792832_2d96052aeb9c4ccf8eb65d457481f593.png

Total cubes in given figure are :113


Count the number of cubes in the given shape
1792843_6bfcf203cc774d33bde123c880b0b0b0.png

Total cubes in given figure are :14


Count the number of cubes in the given shape
1792835_cf8253e8e5a74fa19048067a2950f6af.png

Total cubes in given figure are :66


A cardboard box is $$1.2\ m$$ long, $$72\ cm$$ wide and $$54\ cm$$ high. How many bars of soap can be put into it if each bar measures $$6 cm \times 4.5 cm \times 4 cm$$?

Volume of box $$=$$ Volume of a cuboid$$ = lbh = 1.2\times 0.72\times 0.54 = m^{3}$$                 $$ (100cm = 1 m)$$
Volume of soap bar $$ = \frac{6}{100}\times \frac{4.5}{100}\times \frac{4}{100} = 24\times 4.5\times 10^{-6}m^{3}$$
$$ \Rightarrow $$ Number of soap bar that can be filled in the box;
$$ = \frac{1.2\times 72\times 54\times 10^{-4}}{24\times 4.5\times 10^{-6}} = \frac{216}{5}\times 100 = 4320 $$ Bars



Marbles of diameter $$1.4\ cm$$ each, are dropped into a cylindrical beaker of diameter $$7\ cm$$ containing some water. Find the number of marbles that should be dropped into the beaker so that the water level rises by $$5.6cm$$.

Let $$m$$ be the number of marbles needed, Let $${r}_{1}$$ and $${r}_{2}$$ be the radii of the marbles and cylindrical beaker respectively.
Marbles

Diameter, $$2{r}_{1}=1.4cm$$

Radius $${r}_{1}=0.7cm$$

Cylindrical Beaker

Diameter, $$2{r}_{2}=7cm$$

Radius, $${r}_{2}=\cfrac{7}{2}cm$$

Let $$h$$ be the height of the water level raised. Then, $$h=5.6cm$$
After the marbles are dropped into the beaker,

Volume of water raised $$=$$ Volume of $$n$$ marbles

$$\Rightarrow \pi { r }_{ 2 }^{ 2 }h=n\times \cfrac { 4 }{ 3 } \pi { r }_{ 1 }^{ 3 }\quad $$

Thus, $$n=\cfrac { 3{ r }_{ 2 }^{ 2 }h }{ 4{ r }_{ 1 }^{ 3 } } $$

$$n=\cfrac { 3\times \cfrac { 7 }{ 2 } \times \cfrac { 7 }{ 2 } \times 5.6 }{ 4\times \cfrac { 7 }{ 10 } \times \cfrac { 7 }{ 10 } \times \cfrac { 7 }{ 10 }  } =150$$

The number of marbles needed is $$150$$.

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A box measures $$80\text{ cm}$$ in length, $$60\text{ cm}$$ in breadth and $$40\text{ cm}$$ in height. How many cubical boxes of $$20\text{ cm}$$ length can be arranged in it ?

Given:-
Length of box, $$l=80\text{ cm}$$
Breadth of box, $$b=60\text{ cm}$$
Height of box, $$h=40\text{ cm}$$
Length of each side of cubical boxes, $$a=20\text{ cm}$$

Now,
$$\text{No. of cubical boxes} = \cfrac{\text{Volume of cuboidal boxes}}{\text{Volume of cubical boxes}}$$

                                      $$ = \cfrac{{l \times b \times h}}{{{a^3}}}$$
                                      $$ = \cfrac{{80 \times 60 \times 40}}{{20 \times 20 \times 20}}$$
                                      $$ = 6 \times 4$$ 
                                      $$=24\text{ boxes}$$

Hence, the number of cubical boxes which can be arranged in the cuboidal boxes is equal to $$24$$.


Metal spheres, each of radius 2 cm, are packed into a rectangular box of internal dimensions 16cm x 8cm x 8cm. When 16 spheres are packed the box is filled with preservative liquid. Find the volume of this liquid (in $$cm^3$$). Give your answer to the nearest integer. [Use $$\pi=3.14$$]

Volume of rectangular box$$=lbh= 16(64)=1024 cm^3 $$
Volume of sphere$$=\frac{4}{3} \pi r^3=33.5238 cm^3$$
16 sphere$$=16(33.5238)=536.3808$$
Volume of liquid$$=1024-536.3808=488  cm^3$$


Count the number of cubes in the given shape

1792786_044d3a7452b64f23ac7a504fc9d17e11.png


Total cubes in given figure are :10


How many bricks will be required to construct a wall $$8\ m$$ long, $$6\ m$$ high and $$22.5\ cm$$ thick if each brick measures $$(25\ cm\times 11.25\ cm\times 6\ cm)$$?

The dimension of the wall are
Length $$=8m=800cm$$
Breadth $$=6m=600cm$$
Height $$=22.5cm$$

We know that

Volume of the wall$$=1\times b\times h$$

By substituting the values 

Volume of the wall $$=800\times 600\times 22.5$$

By multiplication

Volume of the wall $$=10800000\ cm^3$$

The dimension of the brick are

Length $$=25\ cm$$

Breadth $$=11.25\ cm$$

Height $$=5cm$$

We know that

Volume of one brick $$=1\times b\times h$$

By substituting the values

Volume of one brick $$=25\times 11.25\times 6$$

By multiplication

Volume of one brick $$=1687.5\ cm^3$$

So the number of brick required $$=\dfrac{Volume \ \ of \ \ wall }{ Volume \ \ of \ \ brick}$$

By substituting the values

Number of brick required $$=\dfrac{10800000}{1687.5}$$

By division

Number of brick requited $$=6400$$

Therefore, the number of bricks required to construct a wall is $$6400$$.



Count the number of cubes and find the volume in cu. cm.
(Let the side of each cube $$=a$$ )
1342414_fa6a28a06f2445ebbd0244137bdd773b.png

We observe that the number of cubes present $$=10$$

Given:  Side of each cube $$=a$$

Total volume $$=10\times { a }^{ 3 }$$

                       $$=10a^3$$


How many lead cubes of side $$2\text{ cm}$$ could be made from a lead cube of side $$8\text{ cm}$$?

Let $$x$$ be the number of lead cubes of side $$2\text{ cm}$$ 
Since all the smaller lead cubes are formed from the bigger lead cube of side $$8\text{ cm}$$
$$\implies$$ Number of smaller lead cubes $$\times $$ volume of each lead cube $$=$$ volume of bigger lead cube
$$\implies x\times (2)^3=(8)3$$   $$[\because$$ Volume of cube of edge length $$a$$ is $$a^3]$$
$$\implies x=\dfrac{8\times 8 \times 8}{2\times 2\times 2}$$
$$\implies x=64$$
$$\therefore ~~64$$ smaller lead cubes can be made form the bigger lead cube


A godown measures $$40\,\,m\,\, \times \,\,25\,\,m\,\, \times \,\,15\,\,m.$$ Find the maximum number of wooden crates each measuring $$1.5\,\,m\,\, \times \,\,1.25\,\,m\,\, \times \,\,0.5\,\,m$$ that can be stored in the godown.

It is given that
Length of the godown $$=40m$$
Breadth of the godown $$=25m$$
Height of the godown $$=15m$$

We know that

Volume of the godown $$=1\times b\times h$$

By substituting the values

Volume of godown $$=40\times \times 25\times 15$$

On further calculation

Volume of godown $$=15000\ m^3$$

It is given that

Length of wooden crate $$=1.5m$$
Breadth of wooden crate $$=1.25m$$
Height wooden crate $$=0.5m$$

We know that

Volume of each wooden crate $$=1\times b\times h$$

By substituting the values

Volume of each wooden crate $$=1.5\times 1.25\times 0.5$$

On further calculation

Volume of each wooden crate $$=0.9375\ m^3$$

So we get

Number of wooden crates that can be stored in godown $$= \dfrac{Volume \ \ of \ \ godown}{ Volume \ \ of \ \ each  \ \ wooden \ \ crate}$$  

By substituting the values

Number of wooden crates that can be stored in godown $$=\dfrac{15000}{0.9375}$$

So we get

Number of wooden crates that can be stored in godown $$=16000$$

Therefore, the number of wooden crates that can be stored in the godown are $$16000$$.  


Solid spheres of diameter 6 cm are dropped into a cylindrical beaker containing some water and are fully submerged. If the diameter of the beaker is 18cm  and the water rises by 40cm, find the number of solid spheres dropped in the water.

By volume conservation we can say that the amount of water displaced by solid sphere is equal to the volume of spheres being dropped inside the cylindrical beaker.
$$\therefore x\times {\dfrac{4\pi}{3}}\times r^3 = \pi\times R^2\times h$$              where $$x$$ is the number of spheres
$$\Rightarrow x\times {\dfrac{4\pi}{3}}\times 3^3=\pi\times 9^2\times 40$$          where $$r = 3$$ and $$R = 9$$
On solving, 
$$x  = 90$$ 


Count the number of cubes in the given shape

1792790_8dee8a396aa64966bd6499944fb8239e.png

Total cubes in given figure are :10


There is water to a height of $$14 \ cm$$ in a cylindrical glass jar of radius $$8 \ cm$$. Inside the water there is a sphere of diameter $$12 \ cm$$ completely immersed. By what height will the water go down when the sphere is removed?


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Class 5 Maths Extra Questions