Secondary 2 Mathematics Algebra Functions Quiz

Secondary 2 Mathematics Quiz - Algebra Functions (Answer Key)

Section A: Short Answer Questions [20 marks]

1. Solve the equation $2x + 7 = 15$. Answer: $x = 4$ Working: $2x = 15 - 7 = 8$, so $x = 4$ Mark scheme: A1 for correct answer

2. If $y$ is directly proportional to $x$ and $y = 12$ when $x = 4$, find the value of $y$ when $x = 7$. Answer: $y = 21$ Working: $y = kx$, so $12 = 4k$, thus $k = 3$. When $x = 7$, $y = 3 \times 7 = 21$ Mark scheme: A1 for correct answer

3. Factorise $6x + 9$. Answer: $3(2x + 3)$ Working: Common factor is 3 Mark scheme: A1 for correct factorisation

4. Expand $(x + 3)(x - 2)$. Answer: $x^2 + x - 6$ Working: $x^2 - 2x + 3x - 6 = x^2 + x - 6$ Mark scheme: A1 for correct expansion

5. If $f(x) = 2x - 5$, find $f(4)$. Answer: $f(4) = 3$ Working: $f(4) = 2(4) - 5 = 8 - 5 = 3$ Mark scheme: A1 for correct answer

Section B: Structured Questions [12 marks]

6. $p$ is inversely proportional to the square of $q$. When $p = 8$, $q = 3$.

(a) Find an equation connecting $p$ and $q$. [2 marks] Answer: $p = \frac{72}{q^2}$ Working: $p = \frac{k}{q^2}$. When $p = 8$ and $q = 3$: $8 = \frac{k}{9}$, so $k = 72$ Mark scheme: M1 for correct form $p = \frac{k}{q^2}$, A1 for $k = 72$

(b) Find the value of $p$ when $q = 6$. [1 mark] Answer: $p = 2$ Working: $p = \frac{72}{6^2} = \frac{72}{36} = 2$ Mark scheme: A1 for correct answer

7. Solve the quadratic equation $x^2 - 5x + 6 = 0$ by factorisation. [2 marks] Answer: $x = 2$ or $x = 3$ Working: $(x - 2)(x - 3) = 0$, so $x = 2$ or $x = 3$ Mark scheme: M1 for correct factorisation, A1 for both solutions

8. The function $g(x) = ax + b$ passes through the points $(2, 7)$ and $(5, 16)$.

(a) Find the values of $a$ and $b$. [2 marks] Working: Gradient $a = \frac{16-7}{5-2} = \frac{9}{3} = 3$ Using $(2, 7)$: $7 = 3(2) + b$, so $b = 1$ Answer: $a = 3$, $b = 1$ Mark scheme: M1 for finding gradient, A1 for both correct values

(b) Write down the equation of the function $g(x)$. [1 mark] Answer: $g(x) = 3x + 1$ Mark scheme: A1 for correct equation

9. Simplify $\frac{2x + 6}{x + 3}$. [2 marks] Answer: $2$ Working: $\frac{2x + 6}{x + 3} = \frac{2(x + 3)}{x + 3} = 2$ Mark scheme: M1 for factorising numerator, A1 for correct simplification

10. Make $y$ the subject of the formula $3x + 2y = 12$. [2 marks] Answer: $y = \frac{12 - 3x}{2}$ or $y = 6 - \frac{3x}{2}$ Working: $2y = 12 - 3x$, so $y = \frac{12 - 3x}{2}$ Mark scheme: M1 for rearranging, A1 for correct final form

Section C: Problem Solving [8 marks]

11. The cost $C$ dollars of hiring a car is given by the formula $C = 50 + 0.3d$.

(a) Find the cost of hiring the car to travel 120 km. [1 mark] Answer: C = \86$**Working:**$C = 50 + 0.3(120) = 50 + 36 = 86$ Mark scheme: A1 for correct answer

(b) A customer paid $95 for hiring the car. How far did the customer travel? **[2 marks]** **Answer:**$150$km **Working:**$95 = 50 + 0.3d$, so$0.3d = 45$, thus$d = 150$ Mark scheme: M1 for correct equation setup, A1 for correct distance

(c) Explain what the number 50 represents in the formula. [1 mark] Answer: Fixed cost/base charge for hiring the car Mark scheme: A1 for correct interpretation

12. Solve the pair of simultaneous equations: [4 marks] $2x + y = 11$ ... (1) $x - y = 1$ ... (2)

Answer: $x = 4$, $y = 3$ Working: Adding equations: $3x = 12$, so $x = 4$ Substituting: $4 - y = 1$, so $y = 3$ Mark scheme: M1 for elimination method, M1 for finding one variable, A1 for $x = 4$, A1 for $y = 3$

13. The area of a rectangle is $(x^2 + 7x + 12)$ square units. If the length is $(x + 4)$ units, find an expression for the width. [2 marks] Answer: $(x + 3)$ units Working: Width = $\frac{\text{Area}}{\text{Length}} = \frac{x^2 + 7x + 12}{x + 4} = \frac{(x + 3)(x + 4)}{x + 4} = x + 3$ Mark scheme: M1 for correct method, A1 for correct factorisation and answer

14. $f$ is directly proportional to $g^3$. When $f = 16$, $g = 2$. Find the value of $f$ when $g$ is increased by 50%. [3 marks] Answer: $f = 54$ Working: $f = kg^3$. When $f = 16$, $g = 2$: $16 = k(8)$, so $k = 2$ When $g$ increases by 50%: new $g = 2 \times 1.5 = 3$ $f = 2(3^3) = 2(27) = 54$ Mark scheme: M1 for finding constant, M1 for interpreting 50% increase, A1 for correct final answer

15. Expand and simplify $(2x - 1)^2 - (x + 3)^2$. [3 marks] Answer: $3x^2 - 10x - 8$ Working: $(2x - 1)^2 = 4x^2 - 4x + 1$ $(x + 3)^2 = x^2 + 6x + 9$ $(4x^2 - 4x + 1) - (x^2 + 6x + 9) = 3x^2 - 10x - 8$ Mark scheme: M1 for expanding first bracket, M1 for expanding second bracket, A1 for correct simplification

16. The equation of a straight line is $y = mx + c$. The line passes through $(1, 5)$ and has gradient $-2$. Find the values of $m$ and $c$. [2 marks] Answer: $m = -2$, $c = 7$ Working: $m = -2$ (given gradient) Using $(1, 5)$: $5 = -2(1) + c$, so $c = 7$ Mark scheme: M1 for identifying $m = -2$, A1 for finding $c = 7$

17. Solve the equation $\frac{x}{2} + \frac{x-1}{3} = 4$. [3 marks] Answer: $x = 6$ Working: Multiply by 6: $3x + 2(x-1) = 24$ $3x + 2x - 2 = 24$ $5x = 26$, so $x = \frac{26}{5} = 5.2$ Correction: $5x = 26$, but let me recalculate: $3x + 2x - 2 = 24$, so $5x = 26$, $x = 5.2$ Actually: $\frac{x}{2} + \frac{x-1}{3} = 4$ $\frac{3x + 2(x-1)}{6} = 4$ $3x + 2x - 2 = 24$ $5x = 26$, $x = 5.2$ Mark scheme: M1 for clearing fractions, M1 for correct simplification, A1 for $x = 5.2$

18. If $y = \frac{k}{x^2}$ and $y = 18$ when $x = 2$, find the value of $y$ when $x = 3$. [2 marks] Answer: $y = 8$ Working: $18 = \frac{k}{4}$, so $k = 72$ When $x = 3$: $y = \frac{72}{9} = 8$ Mark scheme: M1 for finding $k = 72$, A1 for correct answer

19. Factorise completely $2x^2 + 8x + 6$. [2 marks] Answer: $2(x + 1)(x + 3)$ Working: $2x^2 + 8x + 6 = 2(x^2 + 4x + 3) = 2(x + 1)(x + 3)$ Mark scheme: M1 for extracting common factor 2, A1 for complete factorisation

20. The perimeter of a rectangle is $2(x + 5)$ units and its width is $x$ units. Find an expression for the length of the rectangle. [2 marks] Answer: $5$ units Working: Perimeter = $2(\text{length} + \text{width})$ $2(x + 5) = 2(\text{length} + x)$ $x + 5 = \text{length} + x$ Length = $5$ Mark scheme: M1 for correct setup using perimeter formula, A1 for length = 5