Mensuration Test

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Mensuration Test - 18

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Weekly Quiz Competition

Question 1
1 / -0

A road roller $$4$$ m wide and with a diameter of $$3.5$$ m makes $$4$$ revolutions in $$1$$ minute. How much time will it take to roll over an area of $$6160$$ m$$^2$$?

A
$$40$$ min

B
$$20$$ min

C
$$35$$ min

D
$$17.5$$ min

Solution

Area in one revolution $$= 2 \pi r \times width = 2\pi \displaystyle \left ( \frac{3.5}{2} \right ) \times 4$$
$$ = \displaystyle \frac{22}{7} \times \frac{35}{10} \times 4 = 44 m^2$$
No. of revolution required to roll area of 6160 m$$^2$$ $$= \displaystyle \frac{6160}{44} = 140$$
Time required $$ =\displaystyle \frac{140}{4} = 35 min$$
Question 2
1 / -0

A cylindrical container with internal radius of its base $$10$$ cm, contains water up to a height of $$7$$ cm. Find the area of the wet surface of the cylinder.

A
$$324.29$$ cm$$^{2}$$

B
$$754.29$$ cm$$^{2}$$

C
$$674.29$$ cm$$^{2}$$

D
$$584.29$$ cm$$^{2}$$

Solution

Given, A cylindrical container with internal radius 10 cm and water filled up to height 7 cm.

Wet surface area of cylinder $$=$$ curved surface area of cylinder up to 7 cm + Area of base of cylinder $$= 2\pi rh+\pi { r }^{ 2 } $$
$$=\pi r(2h+r) $$
$$=\dfrac { 22 }{ 7 } \times 10\times (14+10)\\ =\dfrac { 22 }{ 7 } \times 10\times 24\\ =754.29\quad { cm }^{ 2 }$$
Question 3
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The dimensions of a car petrol tank are $$50$$ cm $$ \times\, 32$$ cm $$ \times\, 24\,$$ cm, which is full of petrol. If car's average consumption is $$15$$ km per litre, find the maximum distance that can be covered by the car.

A
$$576$$ km

B
$$500$$ km

C
$$543$$ km

D
$$567$$ km

Solution

To find the capacity of the petrol tank, we have to find the volume of the tank.
Volume of a car petrol tank $$ =$$ $$lbh$$
With the given measurements, we get
$$V =$$ $$50$$ cm $$\times 32$$ cm$$\times 24$$ cm
$$=$$ $$38400$$ $${ cm }^{ 3 }$$
As $$1$$ litre $$= 1000$$ $${ cm }^{ 3 }$$
So, $$ 38400$$ $${ cm }^{ 3 }$$ $$=$$ $$\dfrac { 38400 }{ 1000 } $$ $$= 38.4$$ Lts of petrol
The quantity of petrol in the tank is $$38.4$$ Lts.
As given, the average consumption of the petrol is $$15$$ kms/ Lts.
The maximum distance which can be covered by the car $$= 38.4$$ $$\times $$ $$15=576$$ kms.
Question 4
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Curved surface of right circular cylinder is $$4.4{m}^{2}$$, radius of base is $$0.7\ m,$$ then the height is $$\left(\text{Take }\pi=\dfrac{22}{7}\right)$$:

A
$$1$$ m

B
$$2$$ m

C
$$3$$ m

D
$$4$$ m

Solution

Curved surface area of a cylinder of radius "$$R$$" and height $$"h$$" is $$ 2\pi Rh$$.
Given, curved surface area $$=44,4$$ $$m^2$$
Therefore, CSA of the given cylinder $$ = 2\times \dfrac {22}{7} \times 0.7\times h
= 4.4 $$ sq.cm
$$\Rightarrow \dfrac {44}{10}=\dfrac {44}{7}\times \dfrac {7}{10}\times h$$
$$\Rightarrow h = 1 $$ m
Hence, option 'A' is correct.
Question 5
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A closed cylindrical tank, made of thin iron sheet, had diameter $$8.4$$ m and height $$5.4$$ m. How much metal sheet to the nearest $$\displaystyle m^{2},$$ is used in making this tank, if $$\displaystyle \frac{1}{15}$$ of the sheet actually used was wasted in making the tank ?

A
$$272$$ $$\displaystyle m^{2}$$

B
$$271$$ $$\displaystyle m^{2}$$

C
$$242$$ $$\displaystyle m^{2}$$

D
$$270$$ $$\displaystyle m^{2}$$
Question 6
1 / -0

A closed cylindrical tank, made of thin iron-sheet, has diameter $$8.4$$ m and height $$5.4$$ m. How much metal sheet to the nearest $$m^2$$ is used in making this tank, if $$\displaystyle \frac{1}{15}$$ of the sheet actually used was wasted in making the tank ?

A
$$272\, m^2$$

B
$$270 \,m^2$$

C
$$235 \,m^2$$

D
$$240 \,m^2$$

Solution

Given diameter $$=8.4$$ m
We know, diameter $$=2 \times $$ radius
$$\therefore $$ radius $$=4.2 $$ m
Also given, height $$=5.4$$ m
$$\therefore $$ total surface area of cylinder $$=2 \pi r(r+h) cm^2$$
$$=2 (3.14)(4.2)(4.2+5.4)=253.44$$

$$\dfrac{1}{15} $$ of the sheet is used was wasted to making the tank $$=\dfrac{1}{15} \times 253.44=16.896$$

$$\therefore $$ total metal sheet used in making tank $$=253.44+16.896=270.336 m^2$$ $$=270 m^2$$
Question 7
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Directions For Questions

A cylinder has a diameter of $$20$$ cm. The area of the curved surface is $$100$$ $$cm^{2}$$ (sq. cm). Find

...view full instructions

the height of the cylinder correct to one decimal place.

A
$$1.5$$ cm

B
$$1.1$$ cm

C
$$1.9$$ cm

D
$$1.6$$ cm

Solution

Given, a cylinder with diameter $$20$$ cm and curved surface area $$100 $$ sq. cm.
Radius $$=\dfrac {d}{2} $$ $$=10 $$ cm
Curved surface area $$=2\pi rh$$
$$\Rightarrow 100= 2\times \dfrac { 22 }{ 7 } \times 10\times h$$
$$ \Rightarrow h= \dfrac { 100\times 7 }{ 2\times 22\times 10 } $$
$$\Rightarrow h= 1.6$$ cm
Height of cylinder $$=1.6$$ cm
Question 8
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A building has $$8$$ right cylindrical pillars whose cross sectional diameter is $$1$$ m and whose height is $$4.2$$ m. Find the expenditure to paint these pillars at the rate of Rs.$$24$$ per $$m^2$$

A
Rs.$$2534.40$$

B
Rs.$$2506.13$$

C
Rs.$$2610.9$$

D
Rs.$$2514.5$$

Solution

Given, diameter $$=1 $$m, radius $$=$$ $$\displaystyle \frac{diameter}{2}=\frac{1}{2}$$
height of cylinder $$=4.2$$ m

Curved surface area of cylinder $$=2 \pi r h=2 \pi \times \dfrac{1}{2} (4.2)=13.2 m^2$$
$$\therefore 8 \times$$ right circular cylinder $$= 8 \times 13.2= 105.6 m^2$$

$$\therefore $$ cost of painting pillar at the rate of Rs. $$24$$ per $$m^2=24 \times 105.6=$$ Rs.$$2534.4$$
Question 9
1 / -0

What is the total surface area of a cube whose side is $$0.5 $$ cm?

A
$$1.5 cm^2$$

B
$$ 0.25 cm^2$$

C
$$0.125 cm^2$$

D
$$ 2.5 cm^2$$

Solution

The surface area of the cube of side '$$a$$' is given by,
$$S=6a^2$$
Here, $$a=0.5 $$ cm
$$\therefore$$ the total surface area of a cube $$=6(0.5)^2=1.5 cm^2$$
Question 10
1 / -0

In a trapezium whose parallel sides measure $$24\ cm$$ and $$15\ cm$$ and the distance between them is $$10\ cm$$. Find the area of trapezium.

A
$$\displaystyle 215\:cm^{2}$$

B
$$\displaystyle 205\:cm^{2}$$

C
$$\displaystyle 195\:cm^{2}$$

D
$$\displaystyle 295\:cm^{2}$$

Solution

Area of trapezium $$\displaystyle =\frac{h}{2}\left ( a+b \right )$$
$$\displaystyle =\frac{10}{2}\left (24+15\right )$$
$$= 5 \times 39 $$
$$= 195 \displaystyle \ cm^{2}$$

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Re-Attempt Weekly Quiz Competition

Mensuration Test - 18

Result

Mensuration Test - 18

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SHARING IS CARING

Question 1

$$40$$ min

$$20$$ min

$$35$$ min

$$17.5$$ min

Solution

Question 2

$$324.29$$ cm$$^{2}$$

$$754.29$$ cm$$^{2}$$

$$674.29$$ cm$$^{2}$$

$$584.29$$ cm$$^{2}$$

Solution

Question 3

$$576$$ km

$$500$$ km

$$543$$ km

$$567$$ km

Solution

Question 4

$$1$$ m

$$2$$ m

$$3$$ m

$$4$$ m

Solution

Question 5

$$272$$ $$\displaystyle m^{2}$$

$$271$$ $$\displaystyle m^{2}$$

$$242$$ $$\displaystyle m^{2}$$

$$270$$ $$\displaystyle m^{2}$$

Question 6

$$272\, m^2$$

$$270 \,m^2$$

$$235 \,m^2$$

$$240 \,m^2$$

Solution

Question 7

$$1.5$$ cm

$$1.1$$ cm

$$1.9$$ cm

$$1.6$$ cm

Solution

Question 8

Rs.$$2534.40$$

Rs.$$2506.13$$

Rs.$$2610.9$$

Rs.$$2514.5$$

Solution

Question 9

$$1.5 cm^2$$

$$ 0.25 cm^2$$

$$0.125 cm^2$$

$$ 2.5 cm^2$$

Solution

Question 10

$$\displaystyle 215\:cm^{2}$$

$$\displaystyle 205\:cm^{2}$$

$$\displaystyle 195\:cm^{2}$$

$$\displaystyle 295\:cm^{2}$$

Solution

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