CTET Paper I (Classes I to V) 3 Jan 2022 Solved Paper

Concept:The smallest perfect square divisible by given numbers is obtained by first finding their LCM, then multiplying it by the missing factors to make every prime exponent even.Explanation:Find the LCM of 16, 20 and 24.Prime factorize each:$16 = 2 \times 2 \times 2 \times 2 = 2^4$$20 = 2 \times 2 \times 5 = 2^2 \times 5$$24 = 2 \times 2 \times 2 \times 3 = 2^3 \times 3$LCM takes the highest power of each prime: $2^4$, $3$, $5$. So LCM = $2^4 \times 3 \times 5 = 240$.For a number to be a perfect square, every exponent in its prime factorization must be even.240 = $2^4 \times 3^1 \times 5^1$. The exponents of 3 and 5 are odd (1). Multiply by $3$ and $5$ to make them even:$240 \times 3 \times 5 = 3600$.Now $3600 = 2^4 \times 3^2 \times 5^2 = (2^2 \times 3 \times 5)^2 = 60^2$, a perfect square.Check divisibility: 3600 ÷ 16 = 225, 3600 ÷ 20 = 180, 3600 ÷ 24 = 150. No smaller perfect square works.Answer:Option B, 3600.