This set of Class 9 Maths Chapter 13 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 1075cm^{2}. How much containers are required to paint the box completely?

a) 6

b) 5

c) 4

d) 3

View Answer

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 cm

^{2}.

1075 cm

^{2}of area can be painted by one container.

Therefore, to paint 6450 cm

^{2}of area, number of containers requires are = \(\frac{6450}{1075}\) = 6.

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/m^{2}.

a) ₹540

b) ₹500

c) ₹420

d) ₹400

View Answer

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 m

^{2}

But it is given that all sides except the base are to be painted.

Hence, total area to be painted = 27 – (2*3)

= 21 m

^{2}

Now, the cost of painting 1 m

^{2}= ₹20

Hence, the cost of painting 21 m

^{2}= 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

Explanation: Total surface area of the cubical tank = 6l

^{2}

= 6(1.8)

^{2}

= 19.44 m

^{2}

= 194400 cm

^{2}

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 m

^{2}

= 162000 cm

^{2}

Area of one tile = 30*30 = 900 cm

^{2}

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.

**Sanfoundry Global Education & Learning Series – Mathematics – Class 9**.

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