CTET Paper II (Mathematics and Science) Sep 2016 Solved Paper

Concept:The product of cube roots equals the cube root of the product. Factor each number into perfect cubes to simplify.Explanation:We have $\sqrt[3]{500} \times \sqrt[3]{3456} = 40 \times A$. Combine the cube roots: $\sqrt[3]{500 \times 3456}=40A$. Factor 500 = $5^3 \times 4$ and 3456 = $4^3 \times 6^3$. Actually, 500 = $5^3 \times 4$ (since $5^3=125$), and 3456 = $6^3 \times 4^3$? Check: $6^3=216$, $216 \times 16 = 3456$, but $16=4^2$, not $4^3$. Alternatively, $3456 = 4^3 \times 6^3$? $4^3=64$, $6^3=216$, $64 \times 216 = 13824$, too big. Let's factor correctly: $500 = 5^3 \times 4$; $3456 = 2^?$ Better: $3456 = 4 \times 864 = 4 \times 2^?$ Simpler: use prime factorization: $500 = 2^2 \times 5^3$; $3456 = 2^7 \times 3^3$ (since $3456/2=1728$, $1728=12^3= (2^2 \times 3)^3 = 2^6 \times 3^3$, so $3456=2^7 \times 3^3$). Then product: $500 \times 3456 = 2^{2+7} \times 5^3 \times 3^3 = 2^9 \times 5^3 \times 3^3 = (2^3)^3 \times 5^3 \times 3^3 = (2^3 \times 5 \times 3)^3 = (8 \times 15)^3 = 120^3$. So cube root is 120. Then $120 = 40 \times A$ gives $A = 3$. But to keep simple, use the grouping from the existing solution: they wrote $5^3 \times 4^3 \times 6^3$ which product is $120^3$? Actually $5 \times 4 \times 6 = 120$, so that works. So stepwise: $\sqrt[3]{500 \times 3456} = \sqrt[3]{ (5 \times 5 \times 5) \times (4 \times 4 \times 4) \times (6 \times 6 \times 6) } = \sqrt[3]{5^3 \times 4^3 \times 6^3} = 5 \times 4 \times 6 = 120$. Then $120 = 40 \times A$ implies $A = 120/40 = 3$.Answer:$A = 3$ (Option C).