### Maths MCQ Class IX Ch-14 | Statistics

Mathematics  MCQ | Class 09 | Chapter 14 STATISTICS Multiple Choice Questions (MCQ) MCQ Based on the Data. MCQ  Based on the Mean of Data. MCQ Based on the Median. MCQ Based on the Mode. Features In the MCQ given below you find the important MCQ which are strictly according to the CBSE syllabus and are very useful for the CBSE Examinations.  Solution Hints are also given to some difficult problems.  Each MCQ contains four options from which one option is correct.  Action Plan First of all students should Learn and write all basic points and Formulas related to the Chapter 14 Statistics. Start solving  the NCERT Problems with examples. Solve the important assignments on the Chapter 14 Class IX. Then start solving the following MCQ. MCQ |Chapter 14 | Statistics | Class IX

# SURFACE AREA & VOLUME

## MCQ Based on the different types of Solid Figures.MCQ  Based on the concept of Cube & Cuboid.MCQ Based on the Right circular Cylinder and Cone.MCQ Based on the Sphere & Hemi-Sphere.

### Features

• In the MCQ given below you find the important MCQ which are strictly according to the CBSE syllabus and are very useful for the CBSE Examinations.
• Solution Hints are also given to some difficult problems.
• Each MCQ contains four options from which one option is correct.

### Action Plan

• First of all students should Learn and write all basic points and Formulas related to the Chapter 13 Surface Area and Volume.
• Start solving  the NCERT Problems with examples.
• Solve the important assignments on the Chapter 13 Class IX.
• Then start solving the following MCQ.

# MCQ BASED ON CUBE AND CUBOID

Cube and cuboid have following properties

Both have 6 faces, 8 vertices and 12 edges.

Faces of cube are squares and faces of cube are rectangles.

Q1) The formula to find the surface area of a cuboid of length (l), breadth (b) and height (h) is:

a) lb + bh + hl

b) 2(lb + bh + hl)

c) 2(lbh)

d) lbh/2

Q2) If the perimeter of one of the faces of a cube is 40 cm, then its volume is:
a) 6000 cm³
b) 1600 cm³
c) 1000 cm³
d) 600 cm³

Q3) The surface area of a cube whose edge equals to 3cm is:

a) 62 sq.cm

b) 30 sq.cm

c) 54 sq.cm

d) 90 sq.cm

Explanation: Given, a = 3 cm

Surface area of cube = 6a2

SA = 6 x 3 x 3 = 54 sq.cm

Q4) The surface area of a cube whose edge equals to 3cm is:
a) 62 sq.cm
b) 30 sq.cm
c) 54 sq.cm
d) 90 sq.cm

Q5) The surface area of cuboid-shaped box having length = 80 cm, breadth = 40cm and height = 20cm is:

a) 11200 sq.cm

b) 13000 sq.cm

c) 13400 sq.cm

d) 12000 sq.cm

Explanation: surface area of the box = 2(lb + bh + hl)

S.A. = 2[(80 × 40) + (40 × 20) + (20 × 80)]

= 2[3200 + 800 + 1600]

= 2 × 5600 = 11200 sq.cm.

Q6) The length of the longest pole that can be put in a room of dimension (10 m × 10 m × 5 m) is
a) 15 m
b) 16 m
c) 10 m
d) 12 m

Q7) The total surface area of a cube is 96 cm2. The volume of the cube is:

a) 8 cm3

b) 512 cm3

c) 64 cm3

d) 27 cm3

Explanation:

We know that the TSA of the cone = 6a2.

6a2 = 96 cm2

a2 = 96/6 = 16

a = 4 cm

The volume of cone = a3 cubic units

V = 43 = 64cm3.

Q8) The lateral surface area of a cube is 256 m³. The volume of the cube is

a) 512 m³

b) 64 m³

c) 216 m³

d) 256 m³

Q9) The length of the longest pole that can be put in a room of dimensions (10 m × 10 m × 5m) is

a) 15m

b) 16m

c) 10m

d) 12m

Explanation:

Given: l = 10m, b = 10m, h = 5m

The length of the longest pole = √[102 + 102 + 52]

= √(100 + 100 + 25) = √225 = 15 m.

Q10) The number of planks of dimensions (4 m × 50 cm × 20 cm) that can be stored in a pit that is 16 m long, 12m wide and 4 m deep is

a) 1900

b) 1920

c) 1800

d) 1840

Explanation:

Volume of Plank = 400 cm × 50cm × 20cm = 400000cm3

Volume of pits = 1600cm × 1200cm × 400cm = 768000000cm3

Number of planks = Volume of planks/Volume of pits

= 768000000/400000

Hence, the number of pits = 1920

Q11) The total surface area of a cube is 96 cm². The volume of the cube is

a) 8 cm³

b) 512 cm³

c) 64 cm³

d) 27 cm³

Q12) The lateral surface area of a cube is 256 m2. The volume of the cube is

a) 512 m3

b) 64 m3

c) 216 m3

d) 256 m3

Explanation:

The lateral surface area of cube = 4a2

4a2 = 256

a2 = 256/4 =64

a = 8 m

Hence, the volume of cube = a3 cube units

V = 8 = 512 m3.

## MCQ BASED ON RIGHT CIRCULAR CYLLINDER

Cylinder is a solid figure which have three faces, two faces at the top and bottom are plan surface and one is curved surface or Lateral surface.

Q13) If the radius of a cylinder is 4cm and height is 10cm, then the total surface area of a cylinder is:

a) 440 sq.cm

b) 352 sq.cm.

c) 400 sq.cm

d) 412 sq.cm

Explanation: Total Surface Area of a Cylinder = 2πr(r + h)

TSA = 2 x 22/7 x 4(4 + 10)

= (2 x 22 x 4 x 14)/7

= (2 x 22x 4 x 2)

= 352 sq.cm

Q14) The radius of a cylinder is doubled and the height remains the same. The ratio between the volumes of the new cylinder and the original cylinder is
a) 1 : 2
b) 3 : 1
c) 4 : 1
d) 1 : 8

Q15) In a cylinder, the radius is doubled and height is halved, the curved surface area will be

a) Halved

b) Doubled

c) Same

d) Four times

Explanation:

We know that the curved surface area of a cylinder is 2πrh

Given that, r = 2R, h = H/2

Hence, the CSA of new cylinder = 2π(2R)(H/2) = 2πRH

Q16) The curved surface area of a right circular cylinder of height 14 cm is 88 cm2. The diameter of the base is:

a) 2 cm

b) 3cm

c) 4cm

d) 6cm

Q17) The curved surface area of a right circular cylinder of height 14 cm is 88 cm2. The diameter of the base is:
a) 2 cm
b) 3cm
c) 4cm
d) 6cm

Explanation: Curved surface area of cylinder = 88 sq.cm

Height = 14 cm

2πrh = 88

r = 88/2πh

r =1 cm

Diameter = 2r = 2cm

Q18) Volume of hollow cylinder is

a) π(R² – r²)h

b) πR²h

c) πr²h

d) πr²(h1 – h1)

Q19) The Curved surface area of a right circular cylinder is 4.4 sq.cm. The radius of the base is 0.7 cm. The height of the cylinder will be:

a) 2 cm

b) 3 cm

c) 1 cm

d) 1.5 cm

Explanation: Curved surface area of cylinder = 2πrh

2πrh = 4.4

h = 4.4/(2π x 0.7)

h = 1 cm

Q20) The radii of two cylinders are in the ratio of 2 : 3 and their heights are in the ratio of 5 : 3. The ratio of their volumes is

a) 10 : 17

b) 20 : 27

c) 17 : 27

d) 20 : 37

Q 21) The radii of two cylinders are in the ratio of 2 : 3 and their heights are in the ratio of 5 : 3. The ratio of their volumes is:

a) 10 : 17

b) 20 : 27

c) 17 : 27

d) 20 : 37

Explanation:

Given that, the radii of two cylinders are in the ratio of 2 : 3

Hence, r1= 2r, r2 = 3r

Also, given that, the height of two cylinders are in the ratio 5 : 3.

Hence, h1 = 5h, h2 = 3h

The ratio of the volume of two cylinders = V1/V2

= πr12h1/πr22h2

= [(2r)2(5h)]/[(3r)2(3h)]

Ratio of their volumes = (20r2h)/(27r2h) = 20/27 = 20 : 27.

Q 22) If the radius of a cylinder is 4cm and height is 10cm, then total surface area of cylinder is:

a) 440 sq.cm

b) 352 sq.cm.

c) 400 sq.cm

d) 412 sq.cm

## MCQ BASED ON RIGHT CIRCULAR CONE

Cone is a solid figure which has two faces one is plane surface at the bottom and another is a curved surface.

Q 23) The diameter of the base of a cone is 10.5 cm, and its slant height is 10 cm. The curved surface area is:

a) 150 sq.cm

b) 165 sq.cm

c) 177 sq.cm

d) 180 sq.cm

Explanation: Diameter = 10.5, Radius = 10.5/2

Slant height, l = 10cm

Curved surface area of cone = πrl = π(5.25)(10)

CSA = 165 sq.cm

Q 24) A cone is 8.4 cm high and the radius of its base is 2.1 cm. It is melted and recast into a shape. The radius of the sphere is

a) 4.2 cm

b) 2.1 cm

c) 2.4 cm

d) 1.6 cm

Q 25) If slant height of the cone is 21cm and the diameter of the base is 24 cm. The total surface area of a cone is:

a) 1200.77 sq.cm

b) 1177 sq.cm

c) 1222.77 sq.cm

d) 1244.57 sq.cm

Explanation: Total surface area = πr(l + r)

r = 24/2 = 12 cm

l = 21 cm

TSA = π(12)(21 + 12) = 1244.57 sq.cm

Q 26) A cone is 8.4 cm high and the radius of its base is 2.1 cm. It is melted and recast into a sphere. The radius of the sphere is

a) 4.2 cm

b) 2.1 cm

c) 2. 4 cm

d) 1.6 cm

Explanation:

Given that the height of cone = 8.4 cm

Radius of cone = 2.1 cm

Also, given that the volume of cone = volume of a sphere

(⅓)πr2h = (4/3)πr3

(⅓)π(2.1)2(8.4) = (4/3)πr3

37.044 = 4r3

r3= 37.044/4

r3= 9.261

r = 2.1

Therefore, the radius of the sphere is 2.1 cm.

Q27) Diameter of the base of a cone is 10.5 cm and its slant height is 10 cm. The curved surface area is:

a) 150 sq.cm

b) 165 sq.cm

c) 177 sq.cm

d) 180 sq.cm

Q28) The total surface area of a cone whose radius is r/2 and slant height 2l is

a) 2πr(l + r)

b) πr(l + (r/4))

c) πr(l + r)

d) 2πrl

Explanation:

The total surface area of cone = πr(l + r) square units.

If r = r/2 and l = 2l, then the TSA of cone becomes,

TSA of cone = π(r/2)[(2l + (2/r)]

=π[(rl) + (r2/4)]

TSA of new cone =πr[l + (r/4)]

## MCQ BASED ON SPHERE AND HEMI-SPHERE

Sphere is solid figure which has only one curved surface.
Hemi sphere is a solid figure which has two faces, one is plane surface at the top and another is curved surface.

Q29) The volume of a hemisphere whose radius is r is:

a) 4/3 πr3

b) 4πr3

c) 2πr3

d) 2/3 π r3

Q30) The surface area of a sphere of radius 14 cm is:

a) 1386 sq.cm

b) 1400 sq.cm

c) 2464 sq.cm

d) 2000 sq.cm

Explanation: Radius of sphere, r = 14 cm

Surface area = 4πr2

= 4 x 22/7 x (14)2 = 2464 sq.cm.

Q 31)The radius of a hemispherical balloon increases from 6 cm to 12 cm as air is being pumped into it. The ratios of the surface areas of the balloon in the two cases is
a) 1 : 4
b) 1 : 3
c) 2 : 3
d) 2 : 1

Q32) The radius of a sphere is 2r, then its volume will be

a) (4/3) πr3

b) 4πr3

c) (8/3) πr3

d) (32/3) πr3

Explanation:

Given : r = 2r

The volume of sphere = (4/3)πr= (4/3)π(2r)3

V = (4/3)π(8r3) = (32/3)πr3.

Q 33) The radius of a hemispherical balloon increases from 6 cm to 12 cm as air is being pumped into it. The ratio of the surface areas of the balloon in the two cases is

a) 1 : 4

b) 1 : 3

c) 2 : 3

d) 2 : 1

Explanation:

We know that the total surface area of the hemisphere = 3πr2 square units.

If r= 6cm, then TSA = 3π(6)2 = 108π

If r = 12 cm, then TSA = 3π(12)2= 432π

Then the ratio = (108π)/(432π)

Ratio = ¼, which is equal to 1:4.