1. **Problem statement:** A tea dealer mixes two brands of tea, X and Y, to obtain 42 kg of a mixture worth 68 per kg. Brand X costs 74 per kg and brand Y costs 59 per kg. 2. **(a) Calculate the ratio in which brands X and Y are mixed:** Let the mass of brand X be $x$ kg and brand Y be $y$ kg. We know: $$x + y = 42$$ The value equation for the mixture is: $$74x + 59y = 68 \times 42$$ Calculate the total value of the mixture: $$68 \times 42 = 2856$$ Substitute $y = 42 - x$ into the value equation: $$74x + 59(42 - x) = 2856$$ $$74x + 2478 - 59x = 2856$$ $$15x + 2478 = 2856$$ $$15x = 2856 - 2478 = 378$$ $$x = \frac{378}{15} = 25.2$$ Then, $$y = 42 - 25.2 = 16.8$$ The ratio of X to Y is: $$25.2 : 16.8 = \frac{25.2}{16.8} : 1 = 1.5 : 1 = 3 : 2$$ 3. **(b) Find the mass of X in the mixture:** From above, mass of X is $25.2$ kg. 4. **(c) Expand and simplify $(2 - x)^3$:** Use the binomial expansion formula: $$(a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3$$ Here, $a=2$, $b=x$: $$ (2 - x)^3 = 2^3 - 3 \times 2^2 \times x + 3 \times 2 \times x^2 - x^3 = 8 - 12x + 6x^2 - x^3 $$ 5. **Use the first few terms of the expansion to approximate $(2.2)^3$:** Rewrite $2.2$ as $2 - (-0.2)$, so $x = -0.2$. Using the first three terms of the expansion: $$8 - 12x + 6x^2 = 8 - 12(-0.2) + 6(-0.2)^2 = 8 + 2.4 + 6 \times 0.04 = 8 + 2.4 + 0.24 = 10.64$$ The exact value of $2.2^3$ is $10.648$, so the approximation is close. **Final answers:** (a) Ratio of X to Y is $3 : 2$. (b) Mass of X is $25.2$ kg. (c) Expansion: $8 - 12x + 6x^2 - x^3$. Approximate value of $2.2^3$ using first three terms: $10.64$.