CMAT Mensuration Questions [Download PDF]

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CMAT Mensuration Questions [Download PDF]

Download Mensuration Questions for CMAT PDF – CMAT Mensuration questions PDF by Cracku. Practice CMAT solved Mensuration Questions paper tests, and these are the practice question to have a firm grasp on the Mensuration topic in the XAT exam. Top 20 very Important Mensuration Questions for XAT based on asked questions in previous exam papers. Click on the link below to download the Mensuration Questions for CMAT PDF with detailed solutions.

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Instructions

Read the following information and answer the questions based on it.
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

1) Answer (A)

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

2) Answer (D)

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

3) Answer (B)

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

4) Answer (E)

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 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

5) Answer (B)

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 6: 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

6) Answer (A)

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 7: 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

7) Answer (E)

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 8: 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

8) Answer (A)

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 9: 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

9) Answer (D)

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 10: 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

10) Answer (C)

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

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