Mathematics

The perimeter of a square is 4 times its side length. The side length of a square can be found from its diagonal using the relation side = diagonal / √2. If one square has a diagonal 3 times the length of the other, then its side length is also 3 times the other (since side is proportional to diagonal). Therefore, the ratio of their perimeters is also 3:1. Among the given options, this corresponds to option D.

The area of a square with side length s is s². The diagonal of this square is s√2, so the square drawn on the diagonal has area (s√2)² = 2s². Therefore, the ratio of the area of the original square to the square on its diagonal is s² : 2s² = 1 : 2.

Given the short side is 30 m and the area is 1200 m², the long side is 1200 ÷ 30 = 40 m. The diagonal is √(30² + 40²) = 50 m. The total fencing length is 30 + 40 + 50 = 120 m. At Rs.10 per meter, the cost is 120 × 10 = Rs.1200. However, the question states fencing one long side, one short side, and the diagonal, so total fencing length is 30 + 40 + 50 = 120 m, costing Rs.1200. The correct cost is Rs.1200, which corresponds to option B, not C. Therefore, the original correct_answer 'B' is correct, but the explanation was missing and is now provided.

The area of a triangle is given by (1/2) × base × height. If the height is decreased by 40%, the new height is 60% of the original. If the base is increased by 40%, the new base is 140% of the original. The new area is (1/2) × 1.4 × base × 0.6 × height = 0.84 × original area, which means the area is 84% of the original. This corresponds to a 16% decrease in area. Therefore, the correct answer is option C: 16% decrease.

To find the greatest four-digit number leaving specified remainders when divided by 10, 15, 21, and 28, we set up congruences and solve using the Chinese Remainder Theorem. The number minus the remainder must be divisible by each divisor. Calculating the least common multiple (LCM) of the divisors and adjusting for remainders leads to 9664 as the greatest such number under 10,000.

To find the H.C.F of fractions, convert each fraction to have a common denominator and then find the H.C.F of the numerators divided by the common denominator. The fractions are 54/9 = 6 = 102/17, 3 9/17 = 60/17, and 36/51 = 12/17. The common denominator is 17. The H.C.F of the numerators 102, 60, and 12 is 6. Therefore, the H.C.F of the fractions is 6/17. This fraction divides each original fraction exactly, making 6/17 the highest common factor of the given fractions. The originally provided answer '6' is incorrect because it ignores the denominators.

The largest possible box size that can evenly pack 408 kg, 468 kg, and 516 kg without mixing is the greatest common divisor (GCD) of these weights. The GCD of 408, 468, and 516 is 24, so the capacity of the largest possible box is 24 kg.

The HCF is the product of the last divisor and all the successive quotients: 41 × 2 × 4 × 2 = 656. Using the Euclidean algorithm backward with these quotients and divisor, the original numbers are 820 and 360, which correspond to option B.

The product of two numbers equals the product of their HCF and LCM. Given HCF = 12, LCM = 396, and one number = 36, the other number = (HCF × LCM) / 36 = (12 × 396) / 36 = 4752 / 36 = 132. Therefore, the other number is 132, which corresponds to option C.

The total area of the paper is 30 cm × 15 cm = 450 cm². The margins reduce the usable width to 15 cm - 2 × 1.25 cm = 12.5 cm, and the usable height to 30 cm - 2 × 2.5 cm = 25 cm. Therefore, the usable area is 12.5 cm × 25 cm = 312.5 cm². The percentage of the paper area available for typing is (312.5 / 450) × 100 ≈ 69.44%, which is approximately 70%. Hence, the correct answer is option B (70%).