Mathematics

The problem asks for the smallest number such that when increased by 1, it is divisible by 12, 18, 24, 32, and 40. This means the number plus 1 is the least common multiple (LCM) of these numbers. The LCM of 12, 18, 24, 32, and 40 is 1440, so the number is 1440 - 1 = 1439.

The bells toll at intervals of 5, 10, 15, 20, and 25 seconds respectively. To find when they will toll together again, we calculate the least common multiple (LCM) of these intervals. The LCM of 5, 10, 15, 20, and 25 is 300 seconds, which equals 5 minutes. Therefore, the bells will toll together again after 5 minutes, making option A correct.

To find the HCF of fractions, find the HCF of the numerators and the LCM of the denominators. Numerators: 3, 5, 7; HCF is 1. Denominators: 16, 12, 8; LCM is 48. Therefore, HCF = 1/48.

The volume of a sphere is proportional to the cube of its diameter. Given the diameters are in ratio 1:2, the volume ratio is 1³:2³ = 1:8.

The volume V of a sphere is given by V = (4/3)πr³ and the surface area A is given by A = 4πr². Setting volume equal to surface area, we have (4/3)πr³ = 4πr². Dividing both sides by 4πr² (assuming r ≠ 0) gives (1/3)r = 1, which simplifies to r = 3. Therefore, the radius of the sphere is 3 units, corresponding to option C.

Given two numbers with sum 55, HCF 5, and LCM 120, let the numbers be 5a and 5b where a and b are co-prime. Then, 5a + 5b = 55 implies a + b = 11. Also, LCM = 5 * a * b = 120, so a * b = 24. The sum of reciprocals is 1/(5a) + 1/(5b) = (a + b) / (5ab) = 11 / (5 * 24) = 11/120.

The surface area of a cube is 6a² = 24 cm², so a² = 4 cm² and a = 2 cm. The volume is a³ = 2³ = 8 cubic centimeters.

The surface area of a sphere is proportional to the square of its radius (A ∝ r²), and the volume is proportional to the cube of its radius (V ∝ r³). Given the area ratio 1:4, the radius ratio is √1:√4 = 1:2. Therefore, the volume ratio is 1³:2³ = 1:8.

Given the HCF of 3240, 3600, and the third number is 36, and their LCM is 2^4 × 3^5 × 5^2 × 7^2, we use the relationship HCF × LCM = product of the three numbers. Prime factorization: - 3240 = 2^3 × 3^4 × 5 - 3600 = 2^4 × 3^2 × 5^2 - HCF = 36 = 2^2 × 3^2 - LCM = 2^4 × 3^5 × 5^2 × 7^2 The third number must have the prime factors to ensure the HCF is 2^2 × 3^2 and the LCM is as given. This means the third number must include 2^2 × 3^5 × 7^2. Calculating the third number: 2^2 × 3^5 × 7^2 = 4 × 243 × 49 = 47628. Option A corresponds to this factorization (interpreting '22' as 2^2, '35' as 3^5, and '72' as 7^2). Therefore, the third number is 47628, confirming option A is correct.

The volume of a cube with side length a is a³. The largest sphere that fits inside has diameter a, so radius r = a/2. The sphere's volume is (4/3)πr³ = (4/3)π(a/2)³ = (πa³)/6. The ratio of cube volume to sphere volume is a³ / (πa³/6) = 6/π.