Mathematics Questions and Answers – Surface Area of a Cuboid and a Cube

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This set of Mathematics Multiple Choice Questions & Answers (MCQs) focuses on “Surface Area of a Cuboid and a Cube”.

1. A box of dimensions 25cm * 30cm * 45cm is to be painted. A container which contains colour has a capacity to paint 1075cm2. How much containers are required to paint the box completely?
a) 6
b) 5
c) 4
d) 3
View Answer

Answer: a
Explanation: For the box, l = 25, b = 30 and h = 45 (given)
Total surface area of the box = 2(lb + bh + lh)
= 2{(25*30) + (30*45) + (25*45)}
= 2{3225}
= 6450 cm2.
1075 cm2 of area can be painted by one container.
Therefore, to paint 6450 cm2 of area, number of containers requires are = \(\frac{6450}{1075}\) = 6.
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2. A water tank of dimensions 2m * 3m * 1.5m is to be painted leaving its base. Find the total cost of painting the tank if rate of painting = ₹20/m2.
a) ₹540
b) ₹500
c) ₹420
d) ₹400
View Answer

Answer: c
Explanation: For the tank, l = 2m, b = 3m and h = 1.5m (given).
The total area of the tank = 2{(2*3) + (3*1.5) + (2*1.5)}
= 2{13.5}
= 27 m2
But it is given that all sides except the base are to be painted.
Hence, total area to be painted = 27 – (2*3)
= 21 m2
Now, the cost of painting 1 m2 = ₹20
Hence, the cost of painting 21 m2 = 21 * 20 = ₹420.

3. A man has built a cubical water tank in his house. All sides of it are equal to 1.8m. All sides except the base are to be covered by tiles having equal sides of 30cm. How much money will he spend on tiles if cost of tiles is ₹400/dozen?
a) ₹7000
b) ₹7200
c) ₹6000
d) ₹6500
View Answer

Answer: b
Explanation: Total surface area of the cubical tank = 6l2
= 6(1.8)2
= 19.44 m2
= 194400 cm2
It is given that tiles are to be covered leaving the base.
Hence, the total area that is to be covered by tiles = 19.44 – (1.8*1.8)
= 16.2 m2
= 162000 cm2
Area of one tile = 30*30 = 900 cm2
Hence, total number of tiles required = \(\frac{Total \,area \,that \,is \,to \,be \,covered \,by \,tiles}{Area \,of \,one \,tile}\)
= \(\frac{162000}{900}\)
= 180
180 tiles = \(\frac{180}{12}\) = 15 dozens of tiles
Total cost of tiles = Rate of tiles/dozen * Total number of dozens tiles required
= 400*18
= ₹7200.
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Sanfoundry Global Education & Learning Series – Mathematics – Class 9.

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Manish Bhojasia - Founder & CTO at Sanfoundry
Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn | Youtube | Instagram | Facebook | Twitter