ALGEBRAIC OPERATIONS

QUESTION 1 [3 marks] — Criterion A Medium

Given $a=4$, $b=-3$, $c=2$, evaluate each expression:

a. $2a+3b$

[1]

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$$2(4)+3(-3)=8-9=-1$$

b. $a^2-bc$

[1]

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$$4^2-(-3)(2)=16+6=22$$

c. $\dfrac{a+b}{c}$

[1]

Show Solution

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

QUESTION 2 [4 marks] — Criterion A Medium

The formula for the area of a trapezium is $A=\frac{1}{2}(a+b)h$, where $a,b$ are the parallel side lengths and $h$ is the height.

a. Find the area when $a=8$ cm, $b=12$ cm, and $h=5$ cm.

[2]

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$$A=\frac{1}{2}(8+12)(5)=\frac{1}{2}(20)(5)=50 \text{ cm}^2$$

b. A second trapezium has $a=6$ cm, $h=4$ cm, and an area of 32 cm$^2$. Find $b$.

[2]

Show Solution

$32=\frac{1}{2}(6+b)(4) \Rightarrow 32=2(6+b) \Rightarrow 16=6+b \Rightarrow b=10$ cm.

QUESTION 3 [6 marks] — Criterion B Medium

Investigate how the value of $2x^2$ changes as $x$ increases, compared to how $(2x)^2$ changes.

a. Evaluate $2x^2$ and $(2x)^2$ for $x=1,2,3$.

[3]

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$x=1$: $2x^2=2$, $(2x)^2=4$. $x=2$: $2x^2=8$, $(2x)^2=16$. $x=3$: $2x^2=18$, $(2x)^2=36$.

b. Notice that $(2x)^2$ is always exactly double $2x^2$ in each case. Explain algebraically why this happens.

[3]

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$(2x)^2 = 2^2 \times x^2 = 4x^2$, while $2x^2$ is just $2\times x^2$. Since $4x^2 = 2\times(2x^2)$, $(2x)^2$ will always be exactly double $2x^2$, for any value of $x$.

QUESTION 4 [4 marks] — Criterion C Medium

A student substitutes $x=3$ into $5-2x$ and writes: $5-2\times3 = 3\times3=9$.

a. Identify the error in the student's working, and give the correct answer.

[2]

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The student incorrectly combined $5-2$ into $3$ before multiplying — but order of operations requires multiplication before subtraction. Correct working: $5-2\times3=5-6=-1$.

b. Explain, using the order of operations rules, exactly why the multiplication must be done before the subtraction here.

[2]

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In the expression $5-2\times3$, there are no brackets, so standard order of operations applies: multiplication and division are always performed before addition and subtraction, regardless of the order they're written left to right.

QUESTION 5 [5 marks] — Criterion D Medium

The speed of a moving object is given by $v = u + at$, where $u$ is initial speed (m/s), $a$ is acceleration (m/s$^2$), and $t$ is time (s).

a. A car starts at $u=5$ m/s and accelerates at $a=2.5$ m/s$^2$. Find its speed after $t=8$ seconds.

[2]

Show Solution

$$v=5+2.5(8)=5+20=25 \text{ m/s}$$

b. The car's speed limit is 90 km/h. Convert this to m/s (1 km/h $\approx$ 0.278 m/s), and determine after how many seconds (from the same start) the car would reach this limit.

[3]

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$90 \times 0.278 \approx 25.0$ m/s. Since $v=25$ m/s was already reached at $t=8$ s (from part a), the car reaches the speed limit at approximately $t=8$ seconds.

QUESTION 6 [6 marks] — Criterion A Hard

Given $a=-3$, $b=4$, $c=-2$, evaluate each expression:

a. $2a^2-3bc$

[2]

Show Solution

$$2(-3)^2-3(4)(-2) = 18+24=42$$

b. $\dfrac{a-c}{b+c}$

[2]

Show Solution

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

c. $(a+b)^2-c^2$

[2]

Show Solution

$$(-3+4)^2-(-2)^2=1-4=-3$$

QUESTION 7 [6 marks] — Criterion A Hard

The formula for the distance between two points is $d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$.

a. Find the distance between the points $(1,4)$ and $(3,7)$.

[2]

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$$d=\sqrt{(3-1)^2+(7-4)^2}=\sqrt{4+9}=\sqrt{13}\approx3.61$$

b. A third point, $(x, 10)$, is exactly $\sqrt{13}$ units from $(3,7)$ as well, with $x>3$. Set up and solve an equation to find $x$.

[4]

Show Solution

$\sqrt{(x-3)^2+(10-7)^2}=\sqrt{13} \Rightarrow (x-3)^2+9=13 \Rightarrow (x-3)^2=4 \Rightarrow x-3=\pm2$. Since $x>3$: $x=5$.

QUESTION 8 [6 marks] — Criterion B Hard

Investigate how the value of $a^3+2a^2b-b^2$ changes when $a$ and $b$ swap their VALUES (not their positions in the formula).

a. Evaluate $a^3+2a^2b-b^2$ at $a=2,b=-1$, then again at $a=-1,b=2$ (values swapped).

[3]

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At $a=2,b=-1$: $8+2(4)(-1)-1=8-8-1=-1$. At $a=-1,b=2$: $-1+2(1)(2)-4=-1+4-4=-1$.

b. Interesting — both give $-1$! Test ONE more swap-pair, e.g. $a=3,b=1$ then $a=1,b=3$, to see if this 'swap gives the same result' pattern continues, or if the first case was a coincidence.

[3]

Show Solution

$a=3,b=1$: $27+2(9)(1)-1=27+18-1=44$. $a=1,b=3$: $1+2(1)(3)-9=1+6-9=-2$. These are DIFFERENT ($44\ne-2$) — so the earlier match was just a coincidence for those SPECIFIC numbers, not a general property of this expression.

QUESTION 9 [5 marks] — Criterion C Hard

A student substitutes $x=-2$ into $x^2$ and gets $-4$ (calculating $-2\times2$ instead of $(-2)\times(-2)$).

a. Explain the student's error precisely, and give the correct value of $x^2$ when $x=-2$.

[3]

Show Solution

$x^2$ means $x\times x$ — substituting $x=-2$ gives $(-2)\times(-2)$, NOT $-2\times2$. The student seems to have only used ONE negative sign instead of squaring the entire negative value. Correct: $(-2)^2=(-2)\times(-2)=4$.

b. Contrast this with $-x^2$ (note: no brackets around the base) when $x=2$. Explain why $-x^2$ and $(-x)^2$ can give DIFFERENT results, using this example.

[2]

Show Solution

$-x^2$ at $x=2$ means $-(x^2)=-(4)=-4$ (square FIRST, then negate). $(-x)^2$ at $x=2$ means $(-2)^2=4$ (negate first, THEN square). These differ because of where the negative sign sits relative to the squaring operation — without brackets, exponents apply before the leading negative sign, following order of operations.

QUESTION 10 [6 marks] — Criterion D Hard

A projectile's height (in metres) after $t$ seconds is modelled by $h=20t-5t^2$.

a. Find the height after 1 second and after 3 seconds.

[2]

Show Solution

$h(1)=20-5=15$m. $h(3)=60-45=15$m.

b. Interesting — both give 15m! Find the height after 2 seconds, and use ALL THREE results (at $t=1,2,3$) to describe the general SHAPE of the projectile's flight path, and estimate approximately when it reaches its highest point.

[4]

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$h(2)=40-20=20$m — higher than both $t=1$ and $t=3$. This confirms the height rises then falls (a parabolic arc), with the peak somewhere between $t=1$ and $t=3$ — and since $h(1)=h(3)=15$m exactly (equal heights on either side), the peak must occur exactly halfway between them, at $t=2$ seconds, matching where we found the maximum height of 20m.

Algebraic substitution