Mensuration Test

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Mensuration Test - 19

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Weekly Quiz Competition

Question 1
1 / -0

If the diameter of a cylinder is $$28$$ cm and its height is $$20$$ cm, then total surface area (in $${cm}^{2}$$) is:

A
$$2993$$

B
$$2992$$

C
$$2292$$

D
$$2229$$

Solution

Total surface area of a cylinder of Radius "$$R$$" and height "$$h$$" $$ = 2\pi R(R + h)$$
Radius of the base of the cylinder $$ = \dfrac {28}{2} = 14 $$ cm
Hence, total surface area of the cylinder $$ =2\times \dfrac { 22 }{ 7 } \times 14(14 + 20) $$
$$= 2992 cm^2 $$
Question 2
1 / -0

How many bricks, each measuring $$25 cm \times 11.25 cm \times 6 cm$$ will be needed to build a wall $$8 m \times 6 m \times 22.5 cm$$ ?

A
$$5600$$

B
$$6000$$

C
$$6400$$

D
$$7200$$

Solution

Number of bricks $$=\, \displaystyle \frac{Volume\, of\, wall}{Volume\, of\, 1\, brick}$$

$$=\, \displaystyle \frac{800\, \times\, 600\, \times\, 22.5}{25\, \times\, 11.25\, \times\, 6}$$ ....$$(1m = 100 cm)$$

$$= 6400.$$
Question 3
1 / -0

It costs Rs.$$2200$$ to paint the inner curved surface of a cylindrical vessel $$10\ m$$ deep. If the cost of painting is at the rate of Rs.$$20$$ per $${m}^{2}$$.Find Inner curved surface area of the vessel.

A
$$110\ {m}^{2}$$

B
$$140\ {m}^{2}$$

C
$$210\ {m}^{2}$$

D
$$120\ {m}^{2}$$

Solution

Total cost of point$$=Rs.2200$$
cost of painting $$ 1{ m }^{ 2 }\quad is\quad Rs.20.$$
$$ \therefore$$ total area painted$$=\cfrac { 2200 }{ 20 } =110{ m }^{ 2 }$$
therefore inner curved surface area is $$ 110{ m }^{ 2 }.$$
Question 4
1 / -0

A match box measures $$4\ cm\times 2.4\ cm\times 1.5\ cm$$. What will be the volume of a packet containing $$12$$ such boxes?

A
$$172.8\ {cm}^{3}$$

B
$$120\ {cm}^{3}$$

C
$$280\ {cm}^{3}$$

D
$$360\ {cm}^{3}$$

Solution

Volume of 1 match box$$=(4\times 2.4\times 1.5){ cm }^{ 3 }\\ =14.4{ cm }^{ 3 }$$
volume of 12 such match box$$=14.4\times 12\\ =172.8{ cm }^{ 3 }$$
Question 5
1 / -0

The cross-section of a canal is in the shape of a trapezium. If the canal is $$12$$ m wide at the top and $$8$$ m wide at the bottom and the area of its cross-section is 84 $$m^2$$, determine its depth.

A
$$8.4$$ m

B
$$7.4$$ m

C
$$4.8$$ m

D
$$ 6.4$$ m

Solution

Given :- $$a =12m$$ , $$b=8m$$ , height $$ =h$$ and area of trapezium =$$84m^2$$
So, putting the respective values , we get,
Area of trapezium = $$\dfrac{1}{2} \times (a+b) \times h$$
$$84 = \dfrac{1}{2} \times (8+12) \times h$$
$$84 = \dfrac{20h}{2}$$
$$84 = 10h$$
$$h = \dfrac{84}{10}$$
$$h = 8.4m$$
Question 6
1 / -0

If the curved surface area of a solid right circular cylinder of height $$h$$ and radius $$r$$ is one-third of its total surface area, then

A
$$\displaystyle h\, =\, \frac{1}{3}\, r$$

B
$$\displaystyle h\, =\, \frac{1}{2}\, r$$

C
$$h = r$$

D
$$h = 2r$$

Solution

We know, curved surface area of a cylinder of radius "$$R$$" and height "$$h$$" $$= 2\pi Rh$$
Total surface area of a cylinder of radius "$$r$$" and height "$$h$$" $$
= 2\pi r(r + h)$$
Given, $$ 2\pi Rh = \dfrac { 1 }{ 3 } \times 2\pi r(r+h) $$
$$ \therefore h = \dfrac { 1 }{ 3 } \times (r+h) $$
$$ \therefore \dfrac { 2 }{ 3 } h = \dfrac { 1 }{ 3 } r $$
$$\therefore h =\displaystyle \dfrac { 1 }{ 2 } r$$
Question 7
1 / -0

If a cistern is $$3$$ metres long, $$2$$ metres wide and $$1$$ metre deep, its capacity is

A
$$6000$$ litres

B
$$600$$ litres

C
$$60$$ litres

D
$$6$$ litres

Solution

Capacity of cistern $$=\, l\, \times\, b\, \times\, h$$
$$=\, 3\, m\, \times\, 2\, m\, \times\, 1\, m\, =\, 6\, m^3=\, 6000\, litres$$.

$$(\therefore\, 1\, m^3\, =\, 1000\, L)$$
Question 8
1 / -0

Write the curved surface area of a right circular cylinder whose radius is 3 cm and height is 5 cm.

A
$$30\pi \ {cm}^{2}$$

B
$$20\pi \ {cm}^{2}$$

C
$$35\pi \ {cm}^{2}$$

D
$$40\pi \ {cm}^{2}$$

Solution

Radius of cylinder $$=3 \text{ cm}$$
Height of cylinder $$=5 \text{ cm}$$
Cross sectional area of cylinder $$=2\pi rh=2\times \pi \times 3\times 5=30\pi \text{ cm}^2$$
Question 9
1 / -0

The capacity of a water tank that measures $$9$$ cm $$\times 3.5$$ cm $$\times 7.5$$ cm is

A
$$236.25$$ c$$m^3$$

B
$$189$$ $$cm^3$$

C
$$236.25$$ $$m^3$$

D
$$189$$ $$m^3$$

Solution

Here, length $$= 9$$ cm, breadth $$= 3.5$$ cm and height $$= 7.5$$ cm
$$\therefore $$ capacity of water tank $$= $$ length $$\times$$ breadth $$\times$$ height
$$=9 \times 3.5 \times 7.5 = 236.25$$ $$cm^3$$ .
Question 10
1 / -0

A godown measures 40 m $$\times$$ 25 m $$\times$$ 10 m The maximum number of wooden crates each measuring 1.0 m $$\times$$ 1.25 m $$\times$$ 0.5 m that can be stored in the godown is:

A
14000

B
18000

C
15000

D
16000

Solution

Both godown and wooden crates are of the shape of a cuboid.

Volume of a cuboid of length l, breadth b and height h $$ = l

\times b \times h $$.

Number of wooden crates that can be stored $$ = \dfrac {Volume \quad of \quad the \quad go-down}{Volume \quad of \quad one \quad wooden \quad crate} = \dfrac {40

\times 25 \times 10}{1 \times 1.25 \times 0.5} = 16000 $$

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Mensuration Test - 19

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Mensuration Test - 19

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SHARING IS CARING

Question 1

$$2993$$

$$2992$$

$$2292$$

$$2229$$

Solution

Question 2

$$5600$$

$$6000$$

$$6400$$

$$7200$$

Solution

Question 3

$$110\ {m}^{2}$$

$$140\ {m}^{2}$$

$$210\ {m}^{2}$$

$$120\ {m}^{2}$$

Solution

Question 4

$$172.8\ {cm}^{3}$$

$$120\ {cm}^{3}$$

$$280\ {cm}^{3}$$

$$360\ {cm}^{3}$$

Solution

Question 5

$$8.4$$ m

$$7.4$$ m

$$4.8$$ m

$$ 6.4$$ m

Solution

Question 6

$$\displaystyle h\, =\, \frac{1}{3}\, r$$

$$\displaystyle h\, =\, \frac{1}{2}\, r$$

$$h = r$$

$$h = 2r$$

Solution

Question 7

$$6000$$ litres

$$600$$ litres

$$60$$ litres

$$6$$ litres

Solution

Question 8

$$30\pi \ {cm}^{2}$$

$$20\pi \ {cm}^{2}$$

$$35\pi \ {cm}^{2}$$

$$40\pi \ {cm}^{2}$$

Solution

Question 9

$$236.25$$ c$$m^3$$

$$189$$ $$cm^3$$

$$236.25$$ $$m^3$$

$$189$$ $$m^3$$

Solution

Question 10

14000

18000

15000

16000

Solution

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