Mathematics

Let the original price per notebook be x. Then, the number of notebooks bought is 300/x. When the price increases by Rs.5, the price per notebook is x+5, and the number of notebooks bought is 300/(x+5), which is 10 less than before. So, 300/x - 300/(x+5) = 10. Solving this equation gives x = 15. Hence, the original price per notebook is Rs.15.

The total present age of the family is 26 × 5 = 130 years. Since the youngest member is 10 years old, 10 years ago the total age was 130 - (5 × 10) = 130 - 50 = 80 years. Therefore, the average age at that time was 80 ÷ 5 = 16 years.

Let the sides of the rectangle be x and y. The perimeter is 2(x + y) = 28, so x + y = 14. The area of the rectangle is xy. The square formed on the diagonal has side equal to the diagonal length, which is $\sqrt{x^2+y^2}$, so its area is x^2 + y^2. The problem states this area is 108 1/3 % (which is 13/12 times) more than the area of the rectangle, so: x^2 + y^2 = xy + (13/12)xy = (25/12)xy. Using (x + y)^2 = x^2 + y^2 + 2xy, we get 14^2 = (25/12)xy + 2xy = (49/12)xy, so xy = (14^2 * 12)/49 = 48. Then, (x - y)^2 = (x + y)^2 - 4xy = 196 - 192 = 4, so x - y = 2. Hence, the difference between the sides is 2 units.

Let the three consecutive even natural numbers be x, x+2, and x+4. Their squares sum to x^2 + (x+2)^2 + (x+4)^2 = 1460. Simplifying leads to 3x^2 + 12x + 20 = 1460, or 3x^2 + 12x - 1440 = 0. Dividing by 3 gives x^2 + 4x - 480 = 0. Solving this quadratic yields x = 20 (positive root). Thus, the numbers are 20, 22, and 24, which matches option B.

The area of a rhombus is given by the formula: Area = (1/2) × (product of diagonals). Given the side length is 25 cm and one diagonal is 30 cm, we use the Pythagorean theorem to find the other diagonal. Half of the known diagonal is 15 cm. Using the right triangle formed by half the diagonals and the side: (other diagonal/2)^2 + 15^2 = 25^2, so (other diagonal/2)^2 = 625 - 225 = 400, thus other diagonal = 40 cm. The area is then 0.5 × 30 × 40 = 600 cm². Therefore, the correct answer is option C (600 sq.cm). The units in the options should be cm², not sq.m as originally stated.

The volume of the metallic sphere is conserved when it is melted and drawn into a wire. The volume of the sphere is given by $\frac{4}{3}\pi r^3=\frac{4}{3}\pi (12{)}^3=7238.23{\text{ cm}}^3$. The wire is a cylinder with volume $\pi r^{2}h$, where the radius of the wire's cross section is 1.6 cm and $h$ is the length. Equating the volumes: $7238.23=\pi (1.6{)}^{2}h$, solving for $h$ gives $h=\frac{7238.23}{\pi \times 2.56}\approx 900\text{ cm}$ or 9 meters. Since none of the given options (45 cm, 18 cm, 90 cm) are close to 900 cm, the correct choice is "None of these" (option D).

The error in marks is 83 - 63 = 20. The average marks for the class increased by half (0.5) due to this error. Since the increase in total marks divided by the number of pupils equals the increase in average, we have 20 / number_of_pupils = 0.5. Solving for the number of pupils gives number_of_pupils = 20 / 0.5 = 40. Therefore, the number of pupils in the class is 40, which corresponds to option C.

The side length of the square plot is the square root of 289, which is 17 ft. The perimeter is 4 times the side, so 4 × 17 = 68 ft. The cost of building the fence is 68 ft × Rs. 58 = Rs. 3944, which corresponds to option A.

Let the breadth be x meters. Then the length is 3x meters. Area = length × breadth = 3x × x = 3x² = 867. Solving for x, x² = 289, so x = 17 meters. Therefore, the breadth is 17 meters.

The original total sum of the 18 observations is 18 × 124 = 2232. The incorrect observations 46 and 82 sum to 128, while the correct observations 64 and 28 sum to 92. Adjusting the total sum: 2232 - 128 + 92 = 2196. The corrected average is 2196 ÷ 18 = 122. However, since 2196 ÷ 18 = 122, the correct average is 122, which corresponds to option D. Therefore, the correct answer is D, not C.