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# Mensuration Questions for SSC CHSL and MTS

Here you can download the Mensuration Questions for SSC CHSL and MTS PDF with solutions by Cracku. These are the most important Mensuration questions PDF prepared by various sources also based on previous year’s papers. Utilize this PDF for Mensuration for SSC CHSL and MTS. You can find a list of Mensuration in this PDF which help you to test yourself and practice. So you can click on the below link to download the PDF for reference and do more practice.

Instructions

The length ,breadth and height of a rectangular piece of wood in the 4cm,3cm, 5cm respectively
Opposite side of 5cm x 4 cm pieces are coloured in red colour
Oppsite sides 4cm x 3 cm ,are cloured in blue
Rest 5 cm x 3 cm are coloured in green in both sides
Now the piece is cut in such way that a cuboid of 1cm x 1cm x 1cm will be made

Question 1: How many cuboids shall have all the three colours?

a) 8

b) 10

c) 12

d) 14

e) None of these

Solution:

The number of cuboid which will have all the three colours are the corner pieces.

Thus, 8 cuboids will have all the three colours.

=> Ans – (A)

Question 2: How many cuboids shall not any colour?

a) No any

b) 2

c) 4

d) 6

e) None of these

Solution:

Number of cuboids which do not have any colour = $(5-2) \times (4-2) \times (3-2)$

= $3 \times 2 \times 1=6$

=> Ans – (D)

Question 3: How many cuboids shall have only two colours red and green in their two sides?

a) 8

b) 12

c) 16

d) 20

e) None of these

Solution:

Number of cuboids which have only two colours red and green in their two sides are the middle cuboids at the corner edges. There are 4 such edges which have combination of red and green colour.

Number of required cuboids = $(5-2) \times 4$

= $3 \times 4=12$

=> Ans – (B)

Question 4: How many cuboids shall have only one colour ?

a) 12

b) 16

c) 22

d) 28

e) None of these

Solution:

Number of cuboids which have only 1 colour are the middle cuboids in all the faces. Also, there are 2 types of each faces.

2*(B-2)*(H-2)+2*(H-2)*(L-2).

=2*(4-2)*(3-2)+2*(3-2)*(5-2)+2*(5-2)*(4-2).

= 2*2*1 + 2*1*3 + 2*3*2.= 4 + 6 + 12.

=22.

Question 5: The edge of an ice cube is 14 cm. The volume of the largest cylindrical ice cube that can be formed out of it is

a) 2200 cu. cm

b) 2000 cu. cm

c) 2156 cu. cm

d) 2400 cu. cm

e) None of these

Solution:

Radius of the cylinder = r = $\frac{14}{2}$ = 7

Height of the cylinder = h =14

Volume = Pi x r2 x h

= $\frac{22}{7}$ x 7 x 7 x 14 = 2156

Question 6: The edge of an ice cube is 14 cm. The volume of the largest cylindrical ice cube that can be fit into it?

a) 2200 cu. cm

b) 2000 cu. cm

c) 2156 cu. cm

d) 2400 cu. cm

e) None of these

Solution:

Radius of the cylinder = r = $\frac{14}{2}$ = 7
Height of the cylinder = h =14
Volume = Pi x rx h
= $\frac{22}{7}$ x 7 x 7 x 14 = 2156

Question 7: The sum of the radius and height of a cylinder is 42 cm. Its total surface area is 3696 cm 2. What is the volume of cylinder ?

a) 17428 cubic cm

b) 17248 cubic cm

c) 17244 cubic cm

d) 17444 cubic cm

e) None of these

Solution:

Total surface area of cylinder

=> $2 \pi r h + 2 \pi r^2 = 3696$

=> $2 \pi r (r + h) = 3696$

$\because (r + h) = 42$   [Given]

=> $2 \times \frac{22}{7} \times r \times 42 = 3696$

=> $44 \times 6 \times r = 3696$

=> $r = \frac{3696}{44 \times 6} = 14$ cm

=> $h = 42 – 14 = 28$ cm

$\therefore$ Volume of cylinder = $\pi r^2 h$

= $\frac{22}{7} \times 14 \times 14 \times 28$

= $17248 cm^3$

Question 8: The respective ratio of radii of two right circular cylinders (A and B) is 4 : 5. The respective ratioof volume of cylinders A and B is 12 : 25. What is the respective ratio of the heights of cylinders A and B ?

a) 2 : 3

b) 3 : 5

c) 5 : 8

d) 4 : 5

e) 3 : 4

Solution:

Volume of a cylinder =$\pi r^2 h$
where r and h are the radius and height of the cylinder respectively.
The ratio of volumes and ratio of radii of the two cylinders is given.
Ratio of square of their radii = 16 : 25
Therefore the ratio of their heights $h_1$ : $h_2$ = $12 \times 25$ : $16 \times 25$
where $h_1$ and $h_2$ are the heights of two cylinders.
the ratio of their heights = 12 : 16 = 3 : 4
Option E is the correct answer

Question 9: The respective ratio of radii of two right circular cylinders (A and B) is 4 : 7. The respective ratio of the heights of cylinders A and B is 2 : 1. What is the respective ratio of volumes of cylinders A and B ?

a) 25 : 42

b) 23 : 42

c) 32 : 49

d) 30 : 49

e) 36 : 49

Solution:

Volume of a cylinder = $\pi r^2 h$
where r and h are the radius and height of the cylinder respectively.
The ratio of volumes of the two cylinders will be equal to the ratio of $r^2 h$ of both the cylinders..
For cylinder 1 $r^2 h$ = $4^2 \times 2 = 32$
For cylinder 2 $r^2 h$ = $7^2 \times 1 = 49$
Ratio of their volumes = $\frac{32}{49}$
Option C is the correct answer.

Question 10: The respective ratio of radii of two right circular cylinders (A and B) is 3 : 2. The respective ratio of volumes of cylinders A and B is 9 : 7, then what are the heights of cylinders A and B ?

a) 8 : 5

b) 4 : 7

c) 7 : 6

d) 5 : 4

e) 6 : 5

Solution:

Volume of a cylinder = $\pi r^2 h$
where r and h are radius and height of the cylinder respectively.
Let $r_1$ , $h_1$ , $r_2$ and $h_2$ be the radius and heights of the two cylinders respectively.
$\pi (r_1)^2 h_1$ : $\pi (r_2)^2 h_2$ = 9 : 7 ————- 1
Ratio of radii $r_1 : r_2 = 3 : 2$
Ratio of square of radii = 9 : 4
Replacing the ratio of radii in 1
$9h_1 : 4h_2$ $= 9: 7$
$h_1 : h_2$ $= (9\times 4) : (7\times 9)= 4 : 7$
Option B is the correct answer.

Question 11: If the volume and curved surface area of a cylinder are 616 $m^3$ and 352 $m^2$ respectively what is the total surface area of the cylinder (in $m^2$)

a) 429

b) 419

c) 435

d) 421

e) 417

Solution:

Volume of a cylinder=$\pi \times r^{2} \times h$
where $r$ and $h$ are the radius and height of the cylinder.
$\pi \times r^{2} \times h$ = $616 m^{3}$
Curved Surface Area of Cylinder=$2\times \pi \times r \times h$=$352 m^{2}$
$\pi \times r \times h$=$176$
Replacing $\pi \times r \times h$ in Volume formula we get,
$r \times 176$=$616$
$r=3.5 m$
Total Surface Area = Curved Surface Area + 2$\times$ Area of base
=$352 + 2\times pi \times r^{2}$
=$352 + 2\times pi \times 3.5^{2}$
=$352+77$
=$429 m^{2}.$
Hence Option A is the correct answer.

Question 12: The sum of the radius and height of a cylinder is 18 metre. The total surface area of the cylinder is 792 sq. metre, what is the volume of the cylinder ? (in cubic metre)

a) 1848

b) 1440

c) 1716

d) 1724

e) 1694

Solution:

let the height and radius of cylinder be H mtr and R mtr

R + H = 18

total surface area of cylinder = 2$\barwedge$RH + 2$\barwedge(R)^2$ = 792

R(H + R) = $\frac{792×7}{22×2}$

R = 7 mtr

H = 18-7 = 11 mtr

volume = $\frac{22}{7}(R)^2(H)$

Volume = 1694 cubic mtr

Question 13: The respective ratio of curved surface area and total surface area of a cylinder is 4:5. If the curved surface area of the cylinder is 1232 cm2, what is the height ? (in cm)

a) 28 cm

b) 34 cm

c) 36 cm

d) 30 cm

e) None of these

Solution:

Let R and H be the radius and height of cylinder

Curved Surface area of cylinder = 2$\barwedge$RH = 1232

($\barwedge$RH= 616 sqr cm)

Total surface area of cylinder = 2$\barwedge$RH + 2$\barwedge(R)^2$ = 2$\barwedge$R(H+R)

it is given that CurvedSurfaceArea/TotalSurfaceArea = 4/5

1232/(1232 + 2$\barwedge(R)^2$) = 4/5

R = 7 cm

2$\barwedge$RH = 1232

2$\barwedge$7H = 1232

44H =1232
H = 28 cm

Question 14: A right circular cylindrical tank has the storage capacity of 98808 ml. If the radius of the base of the cylinder is three-fourth of the height, what is the diameter of the base ?

a) 28 cms

b) 56 cms

c) 21 cms

d) 58 cms

e) None of these

Solution:

Volume if cylinder = π$(R)^2$h

R is the radius and h is the height of cylinder

Volume = 98808 ml

It is given that R = $\frac{3}{4}$h

$\frac{4}{3}$ $(R)^3$ π = 98808

R = 28.68

Diameter = 28.68 × 2 ~58

Question 15: A rope makes 125 rounds of a cylinder with base radius 15 cm. How many times can it go round a cylinder with base radius 25 cm?

a) 100

b) 75

c) 80

d) 65

e) None of these

Solution:

Let the required number of rounds be x.

More radius, less rounds (Inverse Proportion)