Secondary 2 Mathematics Algebra Functions Quiz

Secondary 2 Mathematics Quiz - Algebra Functions: Answer Key

Total Marks: 50

Section A: Multiple-Choice Questions (4 marks)

1. (A) $7x + 5$ [1]

• Explanation: "Seven times a number $x$" is $7x$. "The sum of ... and five" means add 5. So the expression is $7x + 5$.
• Common mistake: Choosing (C) $7(x+5)$ which means "seven times the sum of $x$ and five", a different operation order.

2. (B) 3 [1]

• Method: Substitute $x = 2$ into $f(x) = 2x^2 - 3x + 1$.
• $f(2) = 2(2)^2 - 3(2) + 1 = 2(4) - 6 + 1 = 8 - 6 + 1 = 3$.
• Common mistake: Forgetting to square the 2 before multiplying by 2 (doing $2 \times 2^2$ as $4^2 = 16$).

3. (A) $4a - 14b$ [1]

• Method: $3(2a - 4b) - 2(a + b) = 6a - 12b - 2a - 2b = 4a - 14b$.
• Key step: Remember to distribute the negative sign: $-2(a + b) = -2a - 2b$.
• Common mistake: Writing $-2(a + b) = -2a + 2b$ (sign error).

4. (B) $x = 4$ [1]

• Method: $5x - 3 = 2x + 9$ $5x - 2x = 9 + 3$ $3x = 12$ $x = 4$
• Key concept: Bring variable terms to one side and constant terms to the other.

Section B: Short-Answer Questions (20 marks)

5. $3(x - 4) = 2x + 5$ [2]

• Step 1: Expand brackets: $3x - 12 = 2x + 5$.
• Step 2: Bring $x$ terms to LHS, constants to RHS: $3x - 2x = 5 + 12$.
• Step 3: Simplify: $x = 17$.
• Marking: M1 for correct expansion or rearrangement, A1 for correct answer.
• Common mistake: Forgetting to multiply the -4 inside the bracket by 3.

6. $ab + c^2$ where $a=2$, $b=-3$, $c=4$ [2]

• Step 1: Substitute: $(2)(-3) + 4^2$.
• Step 2: Evaluate: $-6 + 16 = 10$.
• Marking: M1 for correct substitution, A1 for correct evaluation.
• Teaching note: $c^2 = 4^2 = 16$, NOT $4 \times 2 = 8$. Also note $a \times b = 2 \times (-3) = -6$, not $6$.

7. $\frac{2x}{3} + \frac{x}{4}$ [2]

• Step 1: Find LCM of denominators (LCM of 3 and 4 is 12).
• Step 2: Convert each fraction: $\frac{2x}{3} = \frac{8x}{12}$, $\frac{x}{4} = \frac{3x}{12}$.
• Step 3: Add: $\frac{8x}{12} + \frac{3x}{12} = \frac{11x}{12}$.
• Marking: M1 for correct common denominator, A1 for correct simplified answer.

8. Factorise $6x^2 + 9x$ completely. [2]

• Step 1: Find the highest common factor (HCF) of $6x^2$ and $9x$. The HCF is $3x$.
• Step 2: Factor out $3x$: $6x^2 + 9x = 3x(2x + 3)$.
• Marking: M1 for identifying a common factor, A1 for fully factorised form.
• Common mistake: Only factoring out $3$ or only factoring out $x$, giving $3(2x^2 + 3x)$ or $x(6x + 9)$. These are not completely factorised.

9. Expand and simplify $(2x + 3)(x - 5)$. [2]

• Step 1: Use FOIL or distributive property: $(2x + 3)(x - 5) = 2x(x - 5) + 3(x - 5)$
• Step 2: Expand: $2x^2 - 10x + 3x - 15$.
• Step 3: Simplify: $2x^2 - 7x - 15$.
• Marking: M1 for correct expansion (allow one sign error), A1 for fully simplified expression.
• Common mistake: $-10x + 3x = -7x$, not $-13x$ (combining signs incorrectly).

10. $P = 2(l + w)$, make $l$ the subject. [2]

• Step 1: Divide both sides by 2: $\frac{P}{2} = l + w$.
• Step 2: Subtract $w$ from both sides: $\frac{P}{2} - w = l$.
• Final: $l = \frac{P}{2} - w$.
• Marking: M1 for dividing by 2, A1 for final answer with $l$ isolated.
• Common mistake: Trying to subtract $w$ before dividing by 2.

11. Cost $= 12x + 8y$, find cost when $x=3$, $y=5$. [2]

• Step 1: Substitute: $12(3) + 8(5) = 36 + 40$.
• Step 2: Evaluate: $36 + 40 = 76$.
• Final: Total cost is $76.
• Marking: M1 for correct substitution, A1 for correct answer with unit.

12. $\frac{2y}{5} - 3 = 7$ [2]

• Step 1: Add 3 to both sides: $\frac{2y}{5} = 10$.
• Step 2: Multiply both sides by 5: $2y = 50$.
• Step 3: Divide by 2: $y = 25$.
• Marking: M1 for correct first step (either adding 3 or multiplying by 5), A1 for correct answer.
• Common mistake: Forgetting to multiply the 10 by 5 (thinking $\frac{2y}{5} = 10 \Rightarrow 2y = 10$).

13. Find gradient of line through (0, 2) and (3, 0). [2]

• Method: Gradient $= \frac{\text{change in } y}{\text{change in } x} = \frac{0 - 2}{3 - 0} = \frac{-2}{3} = -\frac{2}{3}$.
• Marking: M1 for correct formula/substitution, A1 for correct value.
• Teaching note: A line sloping downwards (left to right) has a negative gradient. The gradient is the rate at which $y$ changes per unit change in $x$.

14. $p \propto q$, so $p = kq$. When $p=20$, $q=5$, so $20 = k(5) \Rightarrow k = 4$. [2]

• $p = 4q$. When $q = 12$, $p = 4(12) = 48$.
• Marking: M1 for finding constant $k=4$, A1 for correct answer 48.
• Common mistake: Writing $p = kq$ but calculating $k = q/p = 5/20 = 0.25$ (inverting the relationship).

Section C: Structured-Response Questions (26 marks)

15. (a) $x^2 + 5x - 14 = 0$ [2]

• Step 1: Find factor pairs of -14 that sum to 5: $7 \times (-2) = -14$, and $7 + (-2) = 5$.
• Step 2: Factorise: $(x + 7)(x - 2) = 0$.
• Step 3: Solve: $x + 7 = 0$ or $x - 2 = 0 \Rightarrow x = -7$ or $x = 2$.
• Marking: M1 for correct factorisation, A1 for both solutions.

(b) $(3y - 1)^2 + 5(3y - 1) - 14 = 0$ [2]

• Method: Let $X = 3y - 1$. Then the equation becomes $X^2 + 5X - 14 = 0$.
• From part (a): $X = -7$ or $X = 2$.
• Solve: $3y - 1 = -7 \Rightarrow 3y = -6 \Rightarrow y = -2$. $3y - 1 = 2 \Rightarrow 3y = 3 \Rightarrow y = 1$.
• Final: $y = -2$ or $y = 1$.
• Marking: M1 for substitution method or expanding and factorising, A1 for correct solutions.
• Teaching note: This is a "quadratic in disguise" — recognising the structure $(3y - 1)$ as a single variable simplifies the problem.

16. $x + 2y = 7$ (1), $3x - y = 7$ (2) [3]

• Method 1 (Elimination): Multiply (2) by 2: $6x - 2y = 14$. Add to (1): $7x = 21 \Rightarrow x = 3$.
• Substitute $x = 3$ into (1): $3 + 2y = 7 \Rightarrow 2y = 4 \Rightarrow y = 2$.
• Method 2 (Substitution): From (1): $x = 7 - 2y$. Substitute into (2): $3(7 - 2y) - y = 7 \Rightarrow 21 - 6y - y = 7 \Rightarrow 21 - 7y = 7 \Rightarrow -7y = -14 \Rightarrow y = 2$. Then $x = 7 - 2(2) = 3$.
• Marking: M1 for correct elimination or substitution setup, M1 for solving for first variable, A1 for both correct values.
• Common mistake: Sign errors when adding/subtracting equations.

17. (a) Width $= x - 5$ (since length is 5 m longer than width). [1]

• Marking: A1 for correct expression.

(b) Area $= \text{length} \times \text{width} = x(x - 5) = 84$. [2]

• Expand: $x^2 - 5x = 84$.
• Rearrange: $x^2 - 5x - 84 = 0$. (Shown)
• Marking: M1 for forming correct equation $x(x-5) = 84$, A1 for showing the required form.

(c) $x^2 - 5x - 84 = 0$ [2]

• Step 1: Factor pairs of -84 summing to -5: $(-12) \times 7 = -84$, and $-12 + 7 = -5$.
• Step 2: Factorise: $(x - 12)(x + 7) = 0$.
• Step 3: Solve: $x - 12 = 0 \Rightarrow x = 12$ or $x + 7 = 0 \Rightarrow x = -7$.
• Step 4: Since length must be positive, $x = 12$.
• Final: Length of field is 12 m.
• Marking: M1 for correct factorisation, A1 for correct positive solution with context check.
• Teaching note: Always check solutions against the problem context. A negative length is impossible, so we reject $x = -7$.

18. (a) Gradient $= \frac{3 - (-1)}{2 - 0} = \frac{4}{2} = 2$. [2]

• Marking: M1 for correct formula/substitution, A1 for correct value 2.
• Teaching note: The gradient is the "rise over run". From (0, -1) to (2, 3), the rise is $3 - (-1) = 4$ and the run is $2 - 0 = 2$.

(b) Equation: $y = mx + c$, where $m = 2$ (gradient) and $c = -1$ (y-intercept, where line crosses y-axis at $x=0$). [2]

• So $y = 2x - 1$.
• Marking: M1 for correct gradient used in equation, A1 for correct y-intercept and final equation.
• Check: When $x = 0$, $y = 2(0) - 1 = -1$ ✓. When $x = 2$, $y = 2(2) - 1 = 3$ ✓.

19. (a) $t \propto \frac{1}{y^2}$, so $t = \frac{k}{y^2}$. [2]

• When $t = 3$, $y = 2$: $3 = \frac{k}{2^2} = \frac{k}{4} \Rightarrow k = 12$.
• Equation: $t = \frac{12}{y^2}$.
• Marking: M1 for correct form $t = k/y^2$, A1 for correct constant $k=12$.

(b) When $y = 4$: $t = \frac{12}{4^2} = \frac{12}{16} = \frac{3}{4}$ or 0.75. [1]

• Marking: A1 for correct value.
• Teaching note: Inverse proportionality means as $y$ increases, $t$ decreases (and vice versa). Doubling $y$ from 2 to 4 makes $t$ one-quarter of its original value (since $y$ is squared in the relationship).

20. (a) Total amount $= 2p + 5n$. [1]

• Marking: A1 for correct expression.

(b) Total items: $p + n = 10$ (1) [4] Total cost: $2p + 5n = 29$ (2)

• Method: From (1): $p = 10 - n$. Substitute into (2): $2(10 - n) + 5n = 29 \Rightarrow 20 - 2n + 5n = 29 \Rightarrow 20 + 3n = 29 \Rightarrow 3n = 9 \Rightarrow n = 3$.
• Then $p = 10 - 3 = 7$.
• Final: John bought 7 pens and 3 notebooks.
• Check: 7 pens cost 7 \times \2 = $14$, 3 notebooks cost$3 \times $5 = $15. Total: \14 + $15 = $29 ✓. Total items: $7 + 3 = 10$ ✓.
• Marking: M1 for forming two correct equations, M1 for correct elimination/substitution, M1 for solving for first variable, A1 for both correct values with units.

Total Marks: 50

Common marking principles applied throughout:

• Method marks (M) are awarded for correct mathematical procedures even if the final answer has an arithmetic error.
• Accuracy marks (A) require the correct final answer.
• For multi-part questions, answers from earlier parts can be used in later parts even if incorrect (error carried forward).
• Units must be stated where required for full marks.