\( \begin{array}{ll}\text { q) } 2 n^{2} \cdot n^{4} p^{6} \cdot 5 n^{5} p^{2} & \text { r) }\left(2 a^{3} b^{4}\right)^{0} \times\left(2 a^{5} b^{3}\right)^{2} \\ \text { s) }\left(2 p^{5} q^{4} r^{7}\right)^{3} & \text { t) } 2 k^{2} m \times 8 k^{9} m^{2} \\ \text { u) } 10 a^{4} \cdot a^{6} \cdot a^{3} & \text { v) } 4^{0}+3 \times(-2)^{0}+6(8-3)^{0} \\ \text { w) } 10 x^{2} y^{4} \cdot 2 x y^{6} \cdot x y & \text { x) } \frac{18 x^{6} \times y^{3}}{9 x^{4} y}\end{array} \)

When tackling expressions like \( 2n^{2} \cdot n^{4} p^{6} \cdot 5n^{5} p^{2} \), remember to first combine the coefficients (numbers in front) and then apply the rules of exponents. The coefficients \( 2 \) and \( 5 \) multiply to \( 10 \), while the \( n \) terms add their exponents together (i.e., \( n^{2 + 4 + 5} = n^{11} \)), giving you \( 10n^{11}p^{8} \). For the expression \( (2a^{3}b^{4})^{0} \times (2a^{5}b^{3})^{2} \), it’s key to recall that any number raised to the power of zero is \( 1 \). Therefore, the first term contributes nothing to the product. In the second term, you apply the power of a product rule, giving \( (2^2)(a^{5 \cdot 2})(b^{3 \cdot 2}) = 4a^{10}b^{6} \).