1. **Problem (a): Explain why Eva's calculation is incorrect for $5 + 3 \times 2^2 = 32$.** 2. According to the order of operations (PEMDAS/BODMAS), we first calculate the exponent, then multiplication, and finally addition. 3. Calculate the exponent: $2^2 = 4$. 4. Multiply: $3 \times 4 = 12$. 5. Add: $5 + 12 = 17$. 6. Eva's answer of 32 is incorrect because she likely added before multiplying or miscalculated the exponent. 7. **Problem (b): Work out $\sqrt{94 - 3^3 \times 5}$.** 8. Calculate the exponent: $3^3 = 27$. 9. Multiply: $27 \times 5 = 135$. 10. Subtract inside the square root: $94 - 135 = -41$. 11. Since the square root of a negative number is not a real number, the expression is undefined in real numbers. 12. **Problem N.10 (a): Draw a ring around all calculations equivalent to $\frac{9}{16} + \frac{3}{4}$.** 13. First, find a common denominator and add: $\frac{9}{16} + \frac{3}{4} = \frac{9}{16} + \frac{12}{16} = \frac{21}{16}$. 14. Check each expression: - $\frac{16}{9} \times \frac{3}{4} = \frac{48}{36} = \frac{4}{3}$ (not $\frac{21}{16}$) - $\frac{9}{16} \times \frac{4}{3} = \frac{36}{48} = \frac{3}{4}$ (not $\frac{21}{16}$) - $\frac{9}{4} \times \frac{1}{3} = \frac{9}{12} = \frac{3}{4}$ (not $\frac{21}{16}$) - $\frac{16}{9} \times \frac{4}{3} = \frac{64}{27}$ (not $\frac{21}{16}$) - $\frac{3}{4} \times 1 = \frac{3}{4}$ (not $\frac{21}{16}$) - $\frac{3}{8} \times 2 = \frac{6}{8} = \frac{3}{4}$ (not $\frac{21}{16}$) 15. None of these products equal $\frac{21}{16}$, so none are equivalent to the sum. 16. **Problem N.10 (b): Calculate $3 \times 1 \frac{5}{6}$ and give the answer as a mixed number in simplest form.** 17. Convert mixed number to improper fraction: $1 \frac{5}{6} = \frac{11}{6}$. 18. Multiply: $3 \times \frac{11}{6} = \frac{33}{6} = \frac{11}{2}$. 19. Convert back to mixed number: $\frac{11}{2} = 5 \frac{1}{2}$. 20. **Problem N.8 (a): Expression for Angelique's sweets if Carlos has $n$ sweets.** 21. Angelique has half as many sweets as Carlos: $\frac{n}{2}$. 22. **Problem N.8 (b): Expression for Safia's sweets.** 23. Safia has 4 more sweets than Angelique: $\frac{n}{2} + 4$. 24. **Problem N.8 (c): Total sweets for all three friends.** 25. Total = Carlos + Angelique + Safia = $n + \frac{n}{2} + \left(\frac{n}{2} + 4\right) = n + \frac{n}{2} + \frac{n}{2} + 4 = n + n + 4 = 2n + 4$. 26. **Problem N.7: Find $p$ and $q$ given $15x + 17 = 3(2x + p) + qx + 2$.** 27. Expand: $3(2x + p) = 6x + 3p$. 28. Sum: $6x + 3p + qx + 2 = (6 + q)x + (3p + 2)$. 29. Equate coefficients: - Coefficient of $x$: $15 = 6 + q \Rightarrow q = 9$. - Constant term: $17 = 3p + 2 \Rightarrow 3p = 15 \Rightarrow p = 5$. **Final answers:** - Eva's calculation is incorrect because the order of operations was not followed; the correct result is 17. - $\sqrt{94 - 3^3 \times 5}$ is undefined in real numbers because the expression inside the root is negative. - None of the given products equal $\frac{9}{16} + \frac{3}{4}$. - $3 \times 1 \frac{5}{6} = 5 \frac{1}{2}$. - Angelique has $\frac{n}{2}$ sweets. - Safia has $\frac{n}{2} + 4$ sweets. - Total sweets: $2n + 4$. - $p = 5$, $q = 9$.