Secondary 2 Mathematics Semestral Assessment 2 (End of Year) Paper 2

TuitionGoWhere Practice Paper — Mathematics Secondary 2

Answer Key — Version 2 of 5

Assessment: SA2 | Total Marks: 60

Section A

Question 1 [2 marks]

$y = kx^2$

Substitute $x = 3$, $y = 45$: $45 = k(3)^2 = 9k$ $k = 5$

Answer: $\boxed{y = 5x^2}$

Marking: [1] for correct proportionality form, [1] for correct substitution and value of $k$.

Question 2 [2 marks]

$P = \dfrac{k}{\sqrt[3]{t}}$

Substitute $t = 8$, $P = 6$: $6 = \dfrac{k}{\sqrt[3]{8}} = \dfrac{k}{2}$ $k = 12$

Answer: $\boxed{P = \dfrac{12}{\sqrt[3]{t}}}$

Marking: [1] for correct inverse proportionality form, [1] for correct $k$.

Question 3 [2 marks]

$m = k\sqrt{n}$

Substitute $n = 16$, $m = 12$: $12 = k\sqrt{16} = 4k$, so $k = 3$.

Equation: $m = 3\sqrt{n}$

(a) When $n = 36$: $m = 3\sqrt{36} = 3 \times 6 = \boxed{18}$

(b) When $m = 3$: $3 = 3\sqrt{n}$, so $\sqrt{n} = 1$, $n = \boxed{1}$

Marking: [1] for each correct part. Award [1] for part (a) if correct method shown even with arithmetic error.

Question 4 [2 marks]

$\frac{6x^2 - 24}{x^2 + x - 6} = \frac{6(x^2 - 4)}{(x + 3)(x - 2)} = \frac{6(x - 2)(x + 2)}{(x + 3)(x - 2)}$

$= \frac{6(x + 2)}{x + 3}$

Answer: $\boxed{\dfrac{6(x + 2)}{x + 3}}$

Marking: [1] for factorising numerator and denominator, [1] for correct simplified form. Do not award full marks if $(x-2)$ is cancelled without showing factorisation.

Question 5 [2 marks]

$\frac{3}{x-2} - \frac{2}{x+5} = \frac{3(x+5) - 2(x-2)}{(x-2)(x+5)}$

$= \frac{3x + 15 - 2x + 4}{(x-2)(x+5)} = \frac{x + 19}{(x-2)(x+5)}$

Answer: $\boxed{\dfrac{x + 19}{(x - 2)(x + 5)}}$

Marking: [1] for correct common denominator and expansion, [1] for correct final simplified numerator.

Question 6 [2 marks]

$y = kx^3$

$81 = k(3)^3 = 27k$

$k = \dfrac{81}{27} = \boxed{3}$

Marking: [2] for correct answer with working. [1] for correct substitution but arithmetic error.

Question 7 [2 marks]

$A = kB^2$

Substitute $B = 5$, $A = 75$: $75 = k(25)$, so $k = 3$.

When $B = 8$: $A = 3(8)^2 = 3 \times 64 = \boxed{192}$

Marking: [1] for finding $k = 3$, [1] for correct final answer.

Question 8 [2 marks]

$4x^2 - 25y^2 = (2x)^2 - (5y)^2$

Answer: $\boxed{(2x - 5y)(2x + 5y)}$

Marking: [2] for correct answer. [1] if student recognises difference of squares but makes sign error.

Question 9 [2 marks]

$R = \dfrac{k}{s^2}$

Substitute $s = 2$, $R = 12$: $12 = \dfrac{k}{4}$, so $k = 48$.

When $s = 6$: $R = \dfrac{48}{36} = \dfrac{4}{3}$

Answer: $\boxed{\dfrac{4}{3}}$ or $\boxed{1.33}$ (to 3 s.f.)

Marking: [1] for finding $k = 48$, [1] for correct final answer.

Question 10 [2 marks]

$\frac{x^2 - 9}{x^2 - 4x + 3} = \frac{(x-3)(x+3)}{(x-1)(x-3)} = \frac{x+3}{x-1}$

Answer: $\boxed{\dfrac{x + 3}{x - 1}}$

Marking: [1] for factorising both numerator and denominator, [1] for correct cancellation and final answer.

Section B

Question 11 [4 marks]

(a) $\boxed{y = k\sqrt{x}}$ [1]

(b) $18 = k\sqrt{9} = 3k$, so $\boxed{k = 6}$ [1]

(c) $y = 6\sqrt{25} = 6 \times 5 = \boxed{30}$ [1]

(d) $6 = 6\sqrt{x}$, so $\sqrt{x} = 1$, $\boxed{x = 1}$ [1]

Marking notes: Award M1 in part (b) for correct substitution even if arithmetic error. Parts (c) and (d) are dependent on correct $k$ — allow follow-through marks.

Question 12 [5 marks]

(a) $(2x - 3)(x + 4) = 2x^2 + 8x - 3x - 12 = \boxed{2x^2 + 5x - 12}$ [2]

Marking: [1] for correct expansion (allow one sign error), [1] for correct simplification.

(b) $3x^2 + 12x = \boxed{3x(x + 4)}$ [1]

(c) $\dfrac{x+1}{3} = \dfrac{2x-5}{4}$

Cross-multiply: $4(x+1) = 3(2x-5)$

$4x + 4 = 6x - 15$

$4 + 15 = 6x - 4x$

$19 = 2x$

$\boxed{x = \dfrac{19}{2} = 9.5}$ [2]

Marking: [1] for correct cross-multiplication and expansion, [1] for correct final answer.

Question 13 [4 marks]

(a) $\boxed{V = \dfrac{k}{w^2}}$ [1]

(b) $5 = \dfrac{k}{16}$, so $\boxed{k = 80}$ [1]

(c) $V = \dfrac{80}{100} = \boxed{0.8}$ [1]

(d) $20 = \dfrac{80}{w^2}$, so $w^2 = 4$, $w = \boxed{2.00}$ (to 3 s.f.) [1]

Marking notes: Part (d) — accept $w = 2$ but award full marks only if expressed to 3 s.f. as requested. Follow-through marks allowed for parts (c) and (d) if $k$ is wrong but method is correct.

Question 14 [5 marks]

(a) $\dfrac{5}{x+3} + \dfrac{3}{x-1} = \dfrac{5(x-1) + 3(x+3)}{(x+3)(x-1)}$

$= \dfrac{5x - 5 + 3x + 9}{(x+3)(x-1)} = \dfrac{8x + 4}{(x+3)(x-1)}$

Answer: $\boxed{\dfrac{8x + 4}{(x + 3)(x - 1)}}$ or $\boxed{\dfrac{4(2x + 1)}{(x + 3)(x - 1)}}$ [3]

Marking: [1] for correct common denominator, [1] for correct expansion of numerator, [1] for correct simplified form.

(b) $\dfrac{2x^2 - 8}{x^2 + 3x + 2} = \dfrac{2(x^2 - 4)}{(x+1)(x+2)} = \dfrac{2(x-2)(x+2)}{(x+1)(x+2)} = \boxed{\dfrac{2(x-2)}{x+1}}$ [2]

Marking: [1] for factorising both, [1] for correct cancellation and final answer.

Question 15 [4 marks]

(a) $y = k\sqrt[3]{x}$; $10 = k\sqrt[3]{27} = k \times 3$, so $k = \dfrac{10}{3}$ ✓ [1]

(b) $y = \dfrac{10}{3} \times \sqrt[3]{64} = \dfrac{10}{3} \times 4 = \boxed{\dfrac{40}{3}}$ or $\boxed{13.3}$ (to 3 s.f.) [1]

(c) $5 = \dfrac{10}{3}\sqrt[3]{x}$, so $\sqrt[3]{x} = \dfrac{15}{10} = \dfrac{3}{2}$

$x = \left(\dfrac{3}{2}\right)^3 = \dfrac{27}{8} = \boxed{3.375}$ [2]

Marking: [1] for correct cube root step, [1] for correct final answer. Follow-through allowed.

Question 16 [5 marks]

(a) $x^2 - 7x + 12 = \boxed{(x - 3)(x - 4)}$ [1]

(b) $2x^2 - 18 = 2(x^2 - 9) = \boxed{2(x - 3)(x + 3)}$ [2]

Marking: [1] for extracting factor of 2, [1] for difference of squares factorisation.

(c) $x^2 - 5x - 14 = 0$

$(x - 7)(x + 2) = 0$

$\boxed{x = 7}$ or $\boxed{x = -2}$ [2]

Marking: [1] for correct factorisation, [1] for both correct solutions.

Question 17 [4 marks]

(a) $T = k\sqrt{L}$

Substitute $L = 0.25$, $T = 1$: $1 = k\sqrt{0.25} = k \times 0.5$, so $k = 2$.

Answer: $\boxed{T = 2\sqrt{L}}$ [2]

Marking: [1] for correct proportionality form, [1] for correct $k$.

(b) $T = 2\sqrt{0.64} = 2 \times 0.8 = \boxed{1.6}$ seconds [1]

(c) $3 = 2\sqrt{L}$, so $\sqrt{L} = 1.5$, $L = 2.25$

Answer: $\boxed{2.25}$ m [1]

Marking notes: Follow-through marks allowed in (b) and (c) if $k$ is wrong.

Question 18 [4 marks]

(a) $\dfrac{3x}{4} - \dfrac{x-2}{6} = \dfrac{9x - 2(x-2)}{12} = \dfrac{9x - 2x + 4}{12} = \boxed{\dfrac{7x + 4}{12}}$ [2]

Marking: [1] for correct common denominator (12) and expansion, [1] for correct simplified numerator.

(b) $\dfrac{2}{x-3} = \dfrac{5}{x+1}$

Cross-multiply: $2(x+1) = 5(x-3)$

$2x + 2 = 5x - 15$

$2 + 15 = 5x - 2x$

$17 = 3x$

$\boxed{x = \dfrac{17}{3}}$ or $\boxed{5.67}$ (to 3 s.f.) [2]

Marking: [1] for correct cross-multiplication and expansion, [1] for correct final answer.

Question 19 [5 marks]

(a) $\boxed{F = \dfrac{k}{d^2}}$ [1]

(b) $36 = \dfrac{k}{25}$, so $\boxed{k = 900}$ [1]

(c) $F = \dfrac{900}{100} = \boxed{9}$ newtons [1]

(d) $9 = \dfrac{900}{d^2}$, so $d^2 = 100$, $\boxed{d = 10}$ cm [1]

(e) If $d$ is doubled, $F = \dfrac{k}{(2d)^2} = \dfrac{k}{4d^2} = \dfrac{1}{4} \times \dfrac{k}{d^2}$

The force becomes one-quarter of the original value. [1]

Marking notes: Part (e) — award [1] for stating "one-quarter" or "reduced by a factor of 4" with valid reasoning. Follow-through marks allowed throughout.

Mark Summary