1). If the total surface area of a cube is 6 sq units, then what is the volume of the cube?
A). 1 cu unit 
B). 2 cu units 
C). 4 cu units 
D). 6 cu units 
Correct Answer: 1 cu unit
Total surface area of a cube = 6 \( \Large a^{2} \)
=> 6 = 6 \( \Large a^{2} \) = \( \Large a^{2} \) = 1
Therefore, a = 1
Now, volume of the cube = \( \Large a^{3} \) = \( \Large 1^{3} \) = 1 cu unit 
2). If the volume of a cube is 729 cu. cm. what is the length of its diagonal?
A). \( \Large 9\sqrt{2} cm \) 
B). \( \Large 9\sqrt{3} cm \) 
C). 18 cm 
D). \( \Large 18\sqrt{3} cm \) 
Correct Answer: \( \Large 9\sqrt{3} cm \)
Volume of cube = \( \Large Side^{3} \)
Therefore, 729 = \( \Large a^{3} \) => a = 9 cm
Therefore, Diagonal of cube = \( \Large Side \times \sqrt{3} \)
\( \Large 9 \times \sqrt{3} \)
\( \Large 9\sqrt{3} \) cm . 
3). Find the surface area of a cuboid 10 m long, 5 In broad and 3 m high.
A). 105 sq m 
B). 104 sq m 
C). 170 sq m 
D). 190 sq m 
Correct Answer: 190 sq m Given that, l = 10 m, b = 5m, h = 3m
lb =\( \Large 10 \times 5 \) = 50, bh = \( \Large 5 \times 3 \) = 15, lh = 10 x 3 = 30
Therefore, Surface area of a cuboid
= \( \Large 2 \left(lb+bh+lh\right) \)
= \( \Large 2 \left(50+15+30\right) \)
= \( \Large 2 \times 95 = 190 sq m. \)

4). The surface area of a cube is 726 sq cm. Find the volume of the cube.
A). 1331 \( \Large cm^{3} \) 
B). 1232 \( \Large cm^{3} \) 
C). 1626 \( \Large cm^{3} \) 
D). 1836 \( \Large cm^{3} \) 
Correct Answer: 1331 \( \Large cm^{3} \) According to the question,
\( \Large 6a^{2} = 726 \) [a = edge of the cube]
=>\( \Large a^{2} = \frac{726}{6} = 121 \)
Therefore, \( \Large a = \sqrt{121} = 11 cm \)
Therefore, Required volume =
\( \Large a^{3} = 11^{3} = 1331 cm^{3} \)

5). The maximum length of a pencil that can be kept in a rectangular box of dimensions 8 cm x 6cm x 2 cm, is
A). \( \Large 2\sqrt{13} cm \) 
B). \( \Large 2\sqrt{14} cm \) 
C). \( \Large 2\sqrt{26} cm \) 
D). \( \Large 10\sqrt{2} cm \) 
Correct Answer: \( \Large 2\sqrt{26} cm \) Length of largest pencil that can be kept in a box
= Diagonal of box = \( \Large \sqrt{l^{2}+b^{2}+h^{2}} \)
where, l = 8 cm, b = 6 cm, h = 2 cm
= \( \Large \sqrt{64 + 36 + 4} \)
= \( \Large \sqrt{104} = 2\sqrt{26 cm} \)

6). Internal length, breadth and height of a rectangular box are 10 cm, 8 cm and 6 cm, respectively. How many boxes are needed which can be packed in a cube whose volume is 6240 cu. cm.?
A). 12 
B). 13 
C). 15 
D). 17 
Correct Answer: 13 Volume of rectangular box
= \( \Large 10 \times 8 \times 6 = 480 cm^{2}\)
= \( \Large 6240 cm^{3} \)
= \( \Large \frac{Volume \ of \ cubes}{Volume \ of \ rectangular \ box} \)
= \( \Large \frac{6240}{480} = 13 \)
Hence, 13 boxes are needed.

7). A cube has each edge 2 cm and a cuboid is 1 cm long, 2 cm wide and 3 cm high. The paint in a certain container is sufficient to paint an area equal to 54 \( \Large cm^{2} \). Which one of the following is correct?
A). Both cube and cuboid can be painted 
B). Only cube can be painted 
C). Only cuboid can be painted 
D). Neither cube nor cuboid can be painted 
Correct Answer: Both cube and cuboid can be painted Surface area of cube which can be painted = \( \Large 6 \left(Side\right)^{2} = 6 \left(2\right)^{2} = 2 cm^{2} \)
Now. surface area of cuboid which can be painted
\( \Large 2 \times \left(lb+bh+lh\right) \)
\( \Large 2 \left(2+6+3\right) = 22 cm^{2} \)
Total surface area of both cube and cuboid
= \( \Large 22 + 24 = 46 cm^{2} < 54 cm^{2} \)
Therefore, both cube and cuboid can be painted.

8). The whole surface area of a rectangular block is 8788 sq cm. If length, breadth and height are in the ratio of 4 : 3 : 2, then find the length.
A). 26 cm 
B). 52 cm 
C). 104 cm 
D). 13 cm 
Correct Answer: 52 cm
Let length. breadth and height be 4x, 3x and 2x respectively.
Whole surface area = \( \Large 2 \left(lb+bh+lh\right) \)
=> \( \Large \left(lb+bh+lh\right) = \frac{8788}{2} = 4394 \)
\( \Large \left(4 \times 3 + 3 \times 2 + 2 \times 4\right)x^{2} = 4394 \)
=> \( \Large 26x^{2} = 4397 \)
=> \( \Large x^{2} = 169 => x = 13 \)
Therefore, Length = \( \Large 4x = 4 \times 13 = 52 cm \) 
9). What are the dimensions (length, breadth and height, respectively) of a cuboid with volume 720 cu cm, surface area 484 sq cm and the area of the base 72 sq cm?
A). 9, 8 and 10 cm 
B). 12, 6 and 10 cm 
C). 18, 4 and 10 cm 
D). 30, 2 and 12 cm 
Correct Answer: 9, 8 and 10 cm Volume of the cuboid = \( \Large 720 cm^{3} \)
Height of the cuboid
= \( \Large \frac{Volume \ of \ the \ cuboid}{Base \ area \ of \ the \ cuboid} \)
= = \( \Large \frac{720}{72} = 10 cm \)
Surface area of the cuboid
= \( \Large 2 \left(lb + bh + hl\right) \)
= \( \Large 2 \left(9 \times 8 + 8 \times 10 + 10 \times 9\right) \)
= \( \Large 2 \left(72 + 80 + 90\right) \)
=\( \Large 2 \times 242 \)
= \( \Large 484 cm^{2} \)
Therefore, It is obvious that length, breadth and height of the cuboid is 9 cm, 8 cm and 10 cm.

10). The volume of a cube is equal to sum of its edges. What is the total surface area in square units?
A). 12 
B). 36 
C). 72 
D). 144 
Correct Answer: 72 Let the edge of a square be x and sum of its edges =12x
Now, by condition, \( \Large x^{3} = 12x \) [Because, x ≠ 0] ...(i)
=> \( \Large x \left(x^{2}  12\right) = 0 \)
=> \( \Large x^{2} = 12 \)
Its total surface area
= \( \Large 6x^{2} = 6 \times \left(12\right) = 72 \ sq \ units \)
