Mensuration Test

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Mensuration Test - 32

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Weekly Quiz Competition

Question 1
1 / -0

What is the volume in cubic centimetres of a $$0.6$$ L bottle of shampoo?

A
$$6000 cm^3$$

B
$$60 cm^3$$

C
$$400 cm^3$$

D
$$600 cm^3$$

Solution

$$1L=1000cm^3$$
$$0.6L=0.6 \times 1000=600cm^3$$
Question 2
1 / -0

The radius of the cylinder whose lateral surface area is $$704\ cm^{2}$$ and height $$8\ cm$$, is ___________.

A
$$6\ cm$$

B
$$4\ cm$$

C
$$8\ cm$$

D
$$14\ cm$$

Solution

Let radius of cylinder be $$r$$.

Lateral surface area$$=2\pi rh$$

$$\Rightarrow 704=2\pi r \times 8$$

$$\Rightarrow r=\dfrac{704}{16\pi}=14$$

$$\therefore $$ Radius of cylinder is $$14$$ $$cm$$.
Question 3
1 / -0

Find the total surface area of a cylinder with diameter of base $$7\ cm$$ and height $$40\ cm$$.

A
$$1016\ cm^{2}$$

B
$$880\ cm^{2}$$

C
$$1540\ cm^{2}$$

D
$$957\ cm^{2}$$

Solution

we k.n.t total surface area of cylinder $$=2\pi r(r+h) $$.

Given $$d=7\ cm$$ and $$h=40\ cm$$ .

Total surface area $$=2\pi r(r+h)$$
$$=2\pi\times 3.5\times(3.5+40)=956.13$$
$$\therefore T.S.A=957$$ $$cm^2$$.
Question 4
1 / -0

The area of a field in the shape of a trapezium measure 1440 $$m^2$$. The perpendicular distance between its parallel sides is $$24 cm. $$ If the ratio of the parallel sides is $$5:3,$$ then the length of the longer parallel sides is

A
75m

B
45m

C
120m

D
60m

Solution

Given that,
Area of trapezium field = 1440 $$m^2$$
Height of field = 24 cm
Ratio of parallel sides 5 : 3
Let the sides be 5x and 3x.
We know that,
Area of trapezium = $$\dfrac{1}{2}$$ (Surn of parallel side) $$\times$$ Height
1440 = $$\dfrac{1}{2}$$ (5x + 3x) $$\times$$ 24 $$\Rightarrow \, \dfrac{1440 \, \times \, 2}{24}$$ = (5x + 3x)
120 = 8x $$\Rightarrow$$ = 15
Hence, longer side = 5 $$\times$$ 15 = 75 cm
Question 5
1 / -0

The total surface area of a closed cylindrical petrol storage tank whose diameter $$4.2$$ m and height $$4.5$$ m.

A
$$87.12$$ sq. m

B
$$78.21$$ sq. m

C
$$69.23$$ sq. m

D
$$65.45$$ sq. m

Solution

Total surface area of cylindrical tank $$=2\pi r(r+h) \\ = 2\times \cfrac{22}{7}\times \cfrac{4.2}{2}\left(\cfrac{4.2}{2}+4.5\right) \\ =13.2\times 6.6 = 87.12 cm^2$$
Question 6
1 / -0

A building has $$25$$ cylindrical shaped poles. Each has a radius of $$28$$ cm and a height of $$4$$ cm. Find the cost of painting curved surface of all poles at the rate of Rs. $$8$$ per $$m^2$$.

A
$$7.23 $$ Rs

B
$$8 $$ Rs

C
$$14.08 $$ Rs

D
$$56.32 $$ Rs

Solution

Given,
Radius of each pillar= 28 cm
Height of each pillar=4 cm
So, as we all know
CSA of a cylinder=$$ 2 \dfrac {22} {7} rh$$

CSA of one pillar
$$= 2 \dfrac {22} {7} (28) (4)$$

$$=704 \ cm^2$$ = $$704 \times 10^{-4} \ m^2$$

$$=704 \times 10^{-4} \times (25)$$

$$=17600 \times 10^{-4} \ m^2$$

Since, cost of painting 1 sq m= Rs. 8
∴Cost of 1.76 sq m =Rs. 8 ( 1.76)
= Rs. 14.08
Question 7
1 / -0

A closed water tank has an internal size of $$10m \times 15m \times 20m$$. It needs to be lined with waterproofing cement on all its internal sides. At the rate of Rs. $$250\ \text {per} \ m^2$$, the total cost of lining will be Rs:

A
$$237, 500$$

B
$$325,000$$

C
$$400,000$$

D
$$475,000$$

Solution

Total surface area of water tank
$$=2(lb+bh+hl)=2(10\times 15+15\times 20+20\times 10) $$

$$= 2\times (150+300+200)=2\times 650 $$

$$ =1300m^2$$

Cost of lining $$=Rs (250\times 1300)=Rs\; 325000$$

Hence option $$'B'$$ is the answer.
Question 8
1 / -0

The shape of the top surface of a table is a trapezium. Find its area if its parallel sides are $$1\ m$$ and $$1.2\ m$$ and perpendicular distance between them is $$0.8\ m$$.

A
$$0.37m^2$$

B
$$0.47m^2$$

C
$$0.88m^2$$

D
$$0.67m^2$$

Solution

Area of trapezium$$=\dfrac{1}{2}\times$$ sum of parallel sides$$\times$$ perpendicular distance
Parallel sides= 1 and 1.2m
Perpendicular distance=0.8m
So,
$$Area=\dfrac{1}{2}\times(1+1.2)\times0.8$$
$$=1.1\times0.8$$
$$=0.88 m^2$$
Question 9
1 / -0

ABCD is a trapezium with AB and CD parallel. If AB$$=6$$, BC$$=5$$, CD$$=3$$, DA$$=4$$, A$$=90^o$$ the area of ABCD is?

A
$$27$$

B
$$12$$

C
$$18$$

D
$$15$$

Solution

as $$\angle A = 90^\circ$$ Hence $$DA$$ is the height of the trapezium as $$AB$$ and $$CD$$ are parallel.
Area of the trapezium = $$\cfrac{6+3}{2} \times 4$$
$$\implies 18 sq.units$$
Question 10
1 / -0

Area of the four walls of a big room measuring $$25\ m \times 18\ m\times 10\ m$$ is

A
$$1000\ m^{2}$$

B
$$940\ m^{2}$$

C
$$860\ m^{2}$$

D
None of these

Solution

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Mensuration Test - 32

Result

Mensuration Test - 32

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Question 1

$$6000 cm^3$$

$$60 cm^3$$

$$400 cm^3$$

$$600 cm^3$$

Solution

Question 2

$$6\ cm$$

$$4\ cm$$

$$8\ cm$$

$$14\ cm$$

Solution

Question 3

$$1016\ cm^{2}$$

$$880\ cm^{2}$$

$$1540\ cm^{2}$$

$$957\ cm^{2}$$

Solution

Question 4

75m

45m

120m

60m

Solution

Question 5

$$87.12$$ sq. m

$$78.21$$ sq. m

$$69.23$$ sq. m

$$65.45$$ sq. m

Solution

Question 6

$$7.23 $$ Rs

$$8 $$ Rs

$$14.08 $$ Rs

$$56.32 $$ Rs

Solution

Question 7

$$237, 500$$

$$325,000$$

$$400,000$$

$$475,000$$

Solution

Question 8

$$0.37m^2$$

$$0.47m^2$$

$$0.88m^2$$

$$0.67m^2$$

Solution

Question 9

$$27$$

$$12$$

$$18$$

$$15$$

Solution

Question 10

$$1000\ m^{2}$$

$$940\ m^{2}$$

$$860\ m^{2}$$

None of these

Solution

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