Question 1: Find the lateral surface area and total surface area of a cuboid of length $\displaystyle 80 \text{ cm }$, breadth $\displaystyle 40 \text{ cm }$ and height $\displaystyle 20 \text{ cm }$.

Dimension of cuboid: Length $\displaystyle (l) = 80 \text{ cm }$, Breadth $\displaystyle (b) = 40 \text{ cm }$ and Height $\displaystyle (h) = 20 \text{ cm }$

$\displaystyle \text{Lateral surface area } = 2 (l+b) \times h = 2 (80 + 40) \times 20 = 4800 \text{ cm}^2$

$\displaystyle \text{Total surface area } = 2 (lb + bh + hl) = 2(80 \times 40 + 40 \times 20 + 20 \times 80)$

$\displaystyle = 2(3200 + 800 + 1600) = 11200 \text{ cm}^2$

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Question 2: Find the lateral surface area and total surface area of a cube of edge $\displaystyle 10 \text{ cm }$.

Dimension of the cube: Length $\displaystyle (l) = 10 \text{ cm }$

$\displaystyle \text{Lateral surface area } = 4l^2 = 4 \times 10^2 = 400 \text{ cm}^2$

$\displaystyle \text{Total surface area } = 6l^2 = 6 \times 10^2 = 600 \text{ cm}^2$

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Question 3: Find the ratio of the total surface area and lateral surface area of a cube.

Let the length of the side of the cube $\displaystyle = l$

$\displaystyle \text{Lateral surface area } = 4l^2$

$\displaystyle \text{Total surface area } = 6l^2$

$\displaystyle \text{Therefore } \frac{\text{Total surface area}}{\text{Lateral surface area}} = \frac{6l^2}{4l^2} = \frac{3}{2}$

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Question 4: Mary wants to decorate her Christmas tree. She wants to place the tree on a wooden block covered with colored paper with picture of Santa Claus on it. She must know the exact quantity of paper to buy for this purpose. If the box has length, breadth and height as $\displaystyle 80 \text{ cm }$, $\displaystyle 40 \text{ cm }$ and $\displaystyle 20 \text{ cm }$ respectively. How many square sheets of paper of side $\displaystyle 40 \text{ cm }$ would she require?

Dimension of box: Length $\displaystyle (l) = 80 \text{ cm }$, Breadth $\displaystyle (b) = 40 \text{ cm }$ and Height $\displaystyle (h) = 20 \text{ cm }$

$\displaystyle \text{Total surface area } = 2 (lb + bh + hl) = 2(80 \times 40 + 40 \times 20 + 20 \times 80)$

$\displaystyle = 2(3200 + 800 + 1600) = 11200 \text{ cm}^2$

$\displaystyle \text{Therefore No of sheets }= \frac{11200}{40 \times 40} = 7 \text{ sheets }$

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Question 5: The length, breadth and height of a room are $\displaystyle 5 m, 4 \text{ m }$ and $\displaystyle 3 \text{ m }$ respectively. Find the cost of white washing the walls of the room and the ceiling at the rate of $\displaystyle \text{ Rs. } 7.50 \text{ m}^2$.

Dimension of room: Length $\displaystyle (l) = 5 \text{ m }$, Breadth $\displaystyle (b) = 4 \text{ m }$ and Height $\displaystyle (h) = 3 \text{ m }$

Area to be painted $\displaystyle =$ $\displaystyle \text{Lateral surface area } +$ Roof

$\displaystyle = 2 (l+b)h + lb = 2 (5 + 4) \times 3 + 5 \times 4 = 54 + 20 = 74 \text{ m}^2$

Painting rate $\displaystyle = 7.5 \text{ Rs. }/ \text{ m}^2$

Therefore Total cost $\displaystyle = 7.50 \times 74 = 555 \text{ Rs. }$

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Question 6: Three equal cubes are placed adjacently in a row. Find the ratio of total surface area of the new cuboid to that of the sum of the surface areas of the three cubes.

Let the length of the side of the cube $\displaystyle = l$

Total surface area of one cube $\displaystyle = 6l^2$

Total surface area of three cubes $\displaystyle = 3 \times 6l^2 = 18l^2$

Dimension of cuboid: Length $\displaystyle (l) = l$, Breadth $\displaystyle (b) = 3l$ and Height $\displaystyle (h) = l$

$\displaystyle \text{Total surface area } = 2 (lb + bh + hl) = 2(3l^2 + 3l^2 +l^2) = 14l^2$

$\displaystyle \text{Therefore } \frac{\text{Total surface area of cuboid}}{\text{Lateral surface area of 3 cubes}} = \frac{14l^2}{18l^2} = \frac{7}{9}$

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Question 7: A $\displaystyle 4 \text{ cm }$ cube is cut into $\displaystyle 1 \text{ cm }$ cubes. calculate the total surface area of all the small cubes.

Dimension of big cube $\displaystyle = 4 \text{ cm }$

Volume of Big cube $\displaystyle = 4^3 = 64 cm^3$

Dimension of small cube $\displaystyle = 1 \text{ cm }$

Volume of small cube $\displaystyle = 1^3 = 1 cm^3$

Therefore no of small cubes $\displaystyle = \frac{64}{1} = 6$

Therefore Total surface area of $\displaystyle 64$ small cubes $\displaystyle = 64 \times 6 (1) = 384 \text{ cm}^2$

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Question 8: The length of a hall is $\displaystyle 18 \text{ m }$ and the width $\displaystyle 12 \text{ m }$. The sum of the areas of the floor and the flat roof is equal to the sum of the areas of the four walls. Find the height of the hall.

Dimension of hall: Length $\displaystyle (l) = 18 \text{ m }$, Breadth $\displaystyle (b) = 12 \text{ m }$ and Height $\displaystyle (h) = h$

Area of floor $\displaystyle = 18 \times 12 = 216 \text{ m}^2$

Area of Roof $\displaystyle = 18 \times 12 = 216 \text{ m}^2$

Lateral surface area of room $\displaystyle = 2 (l+b)h = 2(18 + 12) h = 60h$

$\displaystyle \text{Hence } 60h = 216 + 216 \Rightarrow h = \frac{36}{5} = 7.2 \text{ m }$

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Question 9: Hameed has built a cubical water tank with lid for his house, with each other edge $\displaystyle 15 \text{ m }$ long. He gets the outer surface of the tank excluding the base covered with square tiles of side $\displaystyle 25 \text{ cm }$. Find how much he would spend for the tiles, if the cost of tiles is $\displaystyle \text{ Rs. } 360$ per dozen.

Dimension of cubical tank: Length $\displaystyle (l) = 1.5 \text{ m }$

Total surface is to be tiles = Lateral surface area + Top

$\displaystyle = 4l^2 + l^2 = 5 \times 1.5^2 = 11.25 \text{ m}^2$

Area of one square tile $\displaystyle = 0.25 \times 0.25 = 0.0625 \text{ m}^2$

Therefore the number of tiles used $\displaystyle = \frac{11.25}{0.0625} = 180$

Hence the total cost of the tiles $\displaystyle = \frac{180}{12} \times 360 = 5400 \text{ Rs. }$

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Question 10: Each edge of a cube is increased by $\displaystyle 50%$. Find the percentage increase in the surface area of the cube.

$\displaystyle \text{Let the initial dimension of the cube } = l$

$\displaystyle \text{Final dimension of the cube } = 1.5l (50\% increase)$

$\displaystyle \text{Total surface area of initial cube } = 6l^2$

$\displaystyle \text{Total surface area of enlarged cube } = 6(1.5l)^2 = 13.5l^2$

$\displaystyle \text{Therefore increase in surface area } = 13.5l^2 - 6l^2 = 7.5l^2$

$\displaystyle \text{Therefore percentage increase } = \frac{7.5l^2}{6l^2} \times 100 = 125$

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Question 11: The dimensions of a rectangular box are int he ratio of $\displaystyle 2 : 3 : 4$ and the difference between the cost of covering it with sheet of paper at the rates of $\displaystyle \text{ Rs. } 8$ and $\displaystyle Rs 9.50 per \text{ m}^2$ is $\displaystyle \text{ Rs. } 1248$. Find the dimensions of the box.

Let the dimension of the box: Length $\displaystyle (l) = 2x$, Breadth $\displaystyle (b) = 3x$ and Height $\displaystyle (h) = 4x$

$\displaystyle \text{Total surface area } = 2 (lb + bh + hl) = 2(6x^2 + 12x^2 + 8x^2) = 52x^2$

$\displaystyle \text{Cost of covering with } \text{ Rs. } 8 / \text{ m}^2 = 52x^2 \times 8 = 416x^2$

$\displaystyle \text{Cost of covering with } \text{ Rs. } 9.5 / \text{ m}^2 = 52x^2 \times 9.5 = 494x^2$

$\displaystyle \text{Therefore } 494 x^2 - 416x^2 = 1248$

$\displaystyle \Rightarrow 78^2 = 1248$

$\displaystyle \Rightarrow x^2 = 16 \Rightarrow x = 4$

Hence the dimensions are: Length $\displaystyle (l) = 8 \text{ m }$, Breadth $\displaystyle (b) = 12 \text{ m }$ and Height $\displaystyle (h) = 16 \text{ m }$

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Question 12: A closed iron tank $\displaystyle 12 \text{ m }$ long, $\displaystyle 9 \text{ m }$ wide and $\displaystyle 4 \text{ m }$ deep is to be made. Determine the cost of iron sheet used at the rate of $\displaystyle \text{ Rs. } 5$ per meter sheet, sheet being $\displaystyle 2 \text{ m }$ wide.

Dimension of the Iron tank: Length $\displaystyle (l) = 12 \text{ m }$, Breadth $\displaystyle (b) = 9 \text{ m }$ and Height $\displaystyle (h) = 4 \text{ m }$

$\displaystyle \text{Total surface area } = 2 (lb + bh + hl) = 2(12 \times 9 + 9 \times 4 + 4 \times 12) = 384 \text{ m}^2$

Therefore length of sheet required $\displaystyle = \frac{384}{2} = 192 \text{ m }$

Hence the cost of the Iron sheet $\displaystyle = 192 \times 5 = 960 \text{ Rs. }$

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Question 13: Ravish wanted to make a temporary shelter for his car by making a box-like structure with tarpaulin that covers all the four sides and the top of the car (with the front face as a flap which can be rolled up). Assuming that the stitching margins are very small, and therefore negligible, how much. tarpaulins would be required to make the shelter of height $\displaystyle 2.5 \text{ m }$ with base dimensions $\displaystyle 4 m \times 3 \text{ m }$?

Dimension of the Car park: Length $\displaystyle (l) = 4 \text{ m }$, Breadth $\displaystyle (b) = 3 \text{ m }$ and Height $\displaystyle (h) = 2.5 \text{ m }$

Area of trampoline $\displaystyle =$ Total $\displaystyle \text{Lateral surface area } +$ Top surface area

$\displaystyle = 2(l+b)h + lb = 2 (4 + 3) \times 2.5 + 4 \times 3 = 47 \text{ m}^2$

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Question 14: An open box is made of wood $\displaystyle 3 \text{ cm }$ thick. Its external length, breadth and height are $\displaystyle 1.48 m, 1.16 \text{ m }$ and $\displaystyle 8.3 d \text{ m }$. Find the cost of painting the inner surface of $\displaystyle 50 per \text{ m}^2$.

Thickness of box $\displaystyle = 3 \text{ cm }$

Dimension of the Box: Length $\displaystyle (l) = 1.48 \text{ m }$, Breadth $\displaystyle (b) = 1.16 \text{ m }$ and Height $\displaystyle (h) = 0.83 \text{ m }$

Height of the painted surface $\displaystyle = 0.83 - 0.03 = 0.80 \text{ m }$

Length of painted surface $\displaystyle = 1.48 - 2 \times 0.03 = 1.42 \text{ m }$

Breadth of the painted surface $\displaystyle = 1.16 - 2 \times 0.03 = 1.10 \text{ m }$

Therefore area to be painted $\displaystyle = 2 (l+b)h + bh$

$\displaystyle = 2 (1.42 + 1.10) \times 0.80 + 1.10 \times 1.42 = 5.594 \text{ m}^2$

Hence cost of painting $\displaystyle = 50 \times 5.594 = 279.70 \text{ Rs. }$

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Question 15: The cost of preparing the walls of a roo \text{ m } $\displaystyle 12 \text{ m }$ long at the rate of $\displaystyle \text{ Rs. } 1.35$ per square meter is $\displaystyle \text{ Rs. } 340.20$ and the cost of matting the floor at $\displaystyle 85$ paise per square meter is $\displaystyle \text{ Rs. } 91.80$. Find the height of the room.

Dimension of the Room: Length $\displaystyle (l) = 12 \text{ m }$, Breadth $\displaystyle (b) = b$ and Height $\displaystyle (h) = h$

$\displaystyle \text{Lateral surface area } = 2 (l+b)h = 2 (12+b)h$

Cost of preparing walls at rate of $\displaystyle \text{ Rs. } 1.35/ \text{ m}^2 = 340 \text{ Rs. }$

$\displaystyle \text{Hence } 340.20 = 2 (12+b)h \times 1.35$

$\displaystyle \Rightarrow 340.20 = 2.70(12+b)h$ … … … … … i)

Cost of matting the floor $\displaystyle = 91.80 \text{ Rs. }$

$\displaystyle \Rightarrow 91.80 = lb \times 0.85$

$\displaystyle \Rightarrow b = \frac{91.80}{12 \times 0.85} = 9$

Therefore substituting this in i) we get

$\displaystyle 340.20 = 2.70(12+9) h$

$\displaystyle \Rightarrow h = \frac{340.20}{2.70 \times 21} = 6$

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Question 16: The dimensions of a room are $\displaystyle 12.5 m \times 9 m \times 7 \text{ m }$. There are $\displaystyle 2$ doors and $\displaystyle 4$ windows in the room; each door measures $\displaystyle 2.5 m \times 1.2 \text{ m }$ and each window $\displaystyle 1.5 m \times 1 \text{ m }$. Find the cost of painting the walls at $\displaystyle \text{ Rs. } 3.50$ per square meter.

Dimension of the Room: Length $\displaystyle (l) = 12.5 \text{ m }$, Breadth $\displaystyle (b) = 9 \text{ m }$ and Height $\displaystyle (h) = 7 \text{ m }$

Number of doors $\displaystyle = 2$

Dimension of door: Breadth $\displaystyle = 1.2 \text{ m }$ and Height $\displaystyle = 2.5 \text{ m }$

Number of windows $\displaystyle = 4$

Dimension of window: Breadth $\displaystyle = 1 \text{ m }$ and Height $\displaystyle = 1.5 \text{ m }$

Lateral surface area of walls $\displaystyle = 2 (l+b)h = 2 (12.5 + 9) \times 7 = 301 \text{ m}^2$

Surface area of doors and windows $\displaystyle = 2 \times 1.2\times 2.5 + 4 \times 1 \times 1.5 = 6 + 6 = 12 \text{ m}^2$

Therefore Area to be painted $\displaystyle = 301 - 12 = 298 \text{ m}^2$

Therefore cost of painting $\displaystyle = 298 \times 3.5 = 1011.50 \text{ Rs. }$

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Question 17: The length and breadth of a hall are in the ratio $\displaystyle 4 : 3$ and its height is $\displaystyle 5.5 \text{ m }$. The cost of decorating its walls (including doors and windows) at $\displaystyle \text{ Rs. } 6.60$ per square meter is $\displaystyle \text{ Rs. } 5082$. Find the length and breadth of the room.

Dimension of the Room: Length $\displaystyle (l) = 4x$, Breadth $\displaystyle (b) = 3x$ and Height $\displaystyle (h) = 5.5 \text{ m }$

$\displaystyle \text{Lateral surface area } = 2 (l+b)h = 2 (4x+3x) \times 5.5 = 77x$

Cost of decorating $\displaystyle = 5082 \text{ Rs. }$

$\displaystyle \text{Therefore } 77x \times 6.6 = 5082$

$\displaystyle \Rightarrow x = \frac{5082}{77 \times 6.6} = 10$

Therefore Length $\displaystyle (l) = 40 \text{ m }$, Breadth $\displaystyle (b) = 30 \text{ m }$

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Question 18: A wooden bookshelf has external dimensions as follows: Height $\displaystyle = 110 \text{ cm }$, Depth $\displaystyle =25 \text{ cm }$, Breadth $\displaystyle = 85 \text{ cm }$. The thickness of the plank is $\displaystyle 5 \text{ cm }$ ever where. The external faces are to be polished and the inner faces are to be painted. If the rate of polishing is $\displaystyle 20 paise per \text{ cm}^2$ and the rate of painting $\displaystyle 10 paise per \text{ cm}^2$. Find the total expenses required for polishing and painting the surface of the bookshelf.

External Dimension of the book shelf: Length $\displaystyle (l) = 85 \text{ cm }$, Breadth $\displaystyle (b) = 25 \text{ cm }$ and Width $\displaystyle (h) = 110 \text{ cm }$

Internal Dimension of the book shelf: Length $\displaystyle (l) = 75 \text{ cm }$, Breadth $\displaystyle (b) = 30 \text{ cm }$ and Width $\displaystyle (h) = 30 \text{ cm }$

External surface are to be polished $\displaystyle = (25 \times 110) \times 2 + (85 \times 25) \times 2 + 110 \times 85 + (5 \times 110) \times 2 + (70 \times 5) \times 4$

$\displaystyle = 5500 + 4250 + 9350 + 1100 + 1400 = 21600 \text{ cm}^2$

Therefore cost of polishing $\displaystyle = 21600 \times 0.20 = 4320 \text{ Rs. }$

Internal surface area $\displaystyle =$ Area of the five faces of 3 cuboids

$\displaystyle = 3 [ 2 (75 + 30) \times 20 ] + 3 [ 30 \times 75] = 19350 \text{ cm}^2$

Cost of painting the internal surface $\displaystyle = 19350 \times 0.10 = 1935 \text{ Rs. }$

Hence the total cost $\displaystyle = 4320 + 1935 = 6255 \text{ Rs. }$

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Question 19: The paint in a certain container is sufficient to paint on area equal to $\displaystyle 9.375 \text{ m}^2$. How many bricks of dimension $\displaystyle 22.5 cm \times 10 cm \times 7.5 \text{ cm }$ can be painted out of this container?

Dimension of the brick: Length $\displaystyle (l) = 22.5 \text{ cm }$, Breadth $\displaystyle (b) = 10 \text{ cm }$ and Width $\displaystyle (h) = 7.5 \text{ cm }$
$\displaystyle \text{Total surface area } = 2 (lb + bh + hl) = 2(22.5 \times 10 + 10 \times 7.5 + 7.5 \times 22.5) = 937.5 cm^2$
$\displaystyle \text{Therefore the number of bricks that can be painted }= \frac{9.375 \times 10000}{937.5} = 100$
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