1. The areas of the three adjacent faces of a rectangular box which meet in a point are known. The product of these areas is equal to :

A. the volume of the box B. twice the volume of the box C. the square of the volume of the box D. the cube root of the volume of the box

Answer Answer: Option C Explanation: Let length = l, breadth = b and height = h, Then,
Product of areas of 3 adjacent faces = (lb * bh * lh) = (lbh)^{2} = (Volume)^{2}.

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

A. Rs. 31. 50 B. Rs 31.68 C. Rs 63 D. Rs. 63. 36

Answer Answer: Option D Explanation: Let the dimensions be 3x, 2x and x respectively, Then,
3x * 2x * x = 10368
= x^{3} = (^{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)] dm^{2}
= [2 * 144(6 + 2 + 3)] dm^{2} = 3168 dm^{2}.
Cost of polishing = Rs. (^{2 * 3168}⁄_{100}) = Rs. 63. 36

3. An open box is made of wood 3 cm thick. It external dimensions are 1.46 m, 1.16m and 8.3 dm. The cost of painting the inner surface of the box at 50 paise per 100 sq. cm is :

A. Rs. 138.50 B. Rs 277 C. Rs. 415.50 D. Rs 554

Answer Answer: Option B Explanation: Internal length = (146 - 6) cm= 140 cm.
Internal breadth (116 - 6) cm = 110 cm.
Internal depth = (83 - 3) cm= 80 cm.
Area of inner surface = [2(l + b) * h] + lb.
= [2(140 + 110) * 80 + 140 * 110] cm^{2} = 55400 cm^{2}.
Cost of painting = Rs. (^{1}⁄_{2} * ^{1}⁄_{100} * 55400) = Rs. 277