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26.  The number of bricks, each measuring 25 cm * 12.5 cm * 7.5 cm, required to construct a wall 6 m long, 5 m high and 0.5 m thick, while the mortar occupies 5% of the volume of the wall, is :


3040
5740
6080
8120


Answer

 Option

Volume of the bricks = 95% of volume of wall = (95⁄100 * 600 * 500 * 50) cm3
Volume of 1 bricks  = (25 * 12.5 * 7,5) cm3
Number of bricks    = (95⁄100 * 600 * 500 * 50⁄25 * 125 * 7.5)
                    = 6080.

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27.  The perimeter of one face a of cube is 20 cm. Its volume must be :


125 cm3
400 cm3
1000 cm3
8000 cm3


Answer

 Option

Edge of the cube = (20⁄4) cm = 5 cm.
          Volume = (5 * 5 * 5) cm3 = 125 cm3

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28.  The sum of the length, breadth and depth of a cuboid is 19 cm and its diagonal is 55 cm. It surface area is :


125 cm2
236 cm2
361 cm2
486 cm2


Answer

 Option

(l + b + h) = 19 and l2 + b2 + h2 = 55 and so (l2 + b2 + h2) = 125.
Now, (l + b + h)2 = 192 = (l2 + b2 + h2) + 2 (lb + bh + lh) = 361.
                = 2(lb + bh + lh) = (361 - 125) = 236.
Surface area = 236 cm2.

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29.  The volume of a rectangle block of stone is 10368 dm3. Its dimensions are in the ratio of 3 : 2 : 1. If its entire surface is polished at 2 paise per dm2, then the total cost will be :


Rs. 31. 50
Rs 31.68
Rs 63
Rs. 63. 36


Answer

 Option

Let the dimensions be 3x, 2x and x respectively, Then,
3x * 2x * x = 10368 
             = x3 = (10368⁄6) = 1728
             = x = 12.
So, the dimensions of the block are 36 dm, 24 dm, and 12 dm.
Surface area = [2 (36 * 24 + 24 * 12 + 36 * 12)] dm2
             = [2 * 144(6 + 2 + 3)] dm2 = 3168 dm2.
Cost of polishing = Rs. (2 * 3168⁄100) = Rs. 63. 36

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30.  Total surface area of a cube whose side is 0.5 cm is :


1⁄4 cm2
1⁄8 cm2
3⁄4 cm2
3⁄2 cm2


Answer

 Option

Surface area = [ 6 * (1⁄2)2] = 3⁄2 cm2.

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