Question. $x$ pencils cost $y$ dollars and 4 cents. Find the cost, in cents, of one pencil.

Solution.

Step 1 — Total cost in cents. $y$ dollars $= 100y$ cents. Adding the extra 4 cents:

$$\text{Total cost} = (100y + 4) \text{ cents}$$

Step 2 — Cost per pencil.

$$\text{Cost per pencil} = \frac{100y + 4}{x} \text{ cents}$$

Answer: $\dfrac{100y + 4}{x}$ cents.

Lesson: convert all quantities to the same unit (cents here) before dividing. Forgetting the $\times 100$ on $y$ would lose every mark.

Question. Find an expression in terms of $r$ for the volume of the largest cube that fits inside a sphere of radius $r$.

Solution.

Step 1 — Spot the geometry. The largest inscribed cube has its vertices on the sphere, so the space diagonal of the cube equals the diameter of the sphere.

Step 2 — Relate cube side to space diagonal. For a cube of side $s$:

$$\text{space diagonal} = s\sqrt{3}$$

Setting this equal to $2r$:

$$s\sqrt{3} = 2r \Rightarrow s = \frac{2r}{\sqrt{3}}$$

Step 3 — Cube volume.

$$V = s^3 = \left(\frac{2r}{\sqrt{3}}\right)^3 = \frac{8r^3}{3\sqrt{3}}$$

Rationalise:

$$= \frac{8r^3}{3\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} = \frac{8\sqrt{3}\, r^3}{9}$$

Answer: $V = \dfrac{8\sqrt{3}\, r^3}{9}$.

Lesson: for "largest inscribed" problems, the constraint is usually a diameter equation. Then volume/area falls out.

Question. Express $x^2 + 5x + 7$ in the form $(x + a)^2 + b$.

Solution.

Step 1 — Half the coefficient of $x$: $a = \dfrac{5}{2}$.

Step 2 — Compensate. $\left(x + \dfrac{5}{2}\right)^2 = x^2 + 5x + \dfrac{25}{4}$, so add and subtract $\dfrac{25}{4}$:

$$x^2 + 5x + 7 = \underbrace{x^2 + 5x + \frac{25}{4}}_{(x + 5/2)^2} - \frac{25}{4} + 7$$

Step 3 — Combine constants: $-\dfrac{25}{4} + 7 = -\dfrac{25}{4} + \dfrac{28}{4} = \dfrac{3}{4}$.

$$\therefore x^2 + 5x + 7 = \left(x + \frac{5}{2}\right)^2 + \frac{3}{4}$$

Answer: $a = \dfrac{5}{2}$, $b = \dfrac{3}{4}$.

Lesson: the constant $b$ equals the original constant minus the square of half the $x$-coefficient.

Question. Make $h$ the subject of $s = \sqrt{2h^2 - r}$.

Solution.

Step 1 — Square both sides to remove the surd:

$$s^2 = 2h^2 - r$$

Step 2 — Isolate the $h^2$ term:

$$2h^2 = s^2 + r$$ $$h^2 = \frac{s^2 + r}{2}$$

Step 3 — Square-root both sides (take $\pm$):

$$h = \pm \sqrt{\frac{s^2 + r}{2}}$$

Answer: $h = \pm\sqrt{\dfrac{s^2 + r}{2}}$.

Lesson: always square-root with $\pm$ unless the question specifies (e.g. "$h > 0$"). Forgetting the $\pm$ is a common 1-mark trap.

Question. Show that $(2n - 1)^2 + 3$ is a multiple of 4 for all integer $n$.

Solution.

Step 1 — Expand the square:

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

Step 2 — Factor out 4:

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

Step 3 — Conclude. Since $n$ is an integer, $n^2 - n + 1$ is also an integer, so $4(n^2 - n + 1)$ is a multiple of 4.

$$\therefore (2n - 1)^2 + 3 \text{ is always a multiple of 4.} \quad \checkmark$$

Lesson: expand → group → factor → cite "integer × integer = integer". Three lines and you're done.

Question. Find the integers satisfying $-5 \leq 7x + 2 < 3$.

Solution.

Step 1 — Subtract 2 from all three parts:

$$-7 \leq 7x < 1$$

Step 2 — Divide all three parts by 7 (positive — no flip):

$$-1 \leq x < \frac{1}{7}$$

Step 3 — List the integers in this range. $\dfrac{1}{7} \approx 0.143$, so integers $x$ with $-1 \leq x < 0.143$ are:

$$x = -1 \text{ or } x = 0$$

Answer: $x \in \{-1, 0\}$.

Lesson: apply each operation to all three parts; list integer solutions only after the algebra is fully simplified.

Question. $T = \sqrt{\dfrac{w}{d^3}}$. If $w$ increases by 20% and $d$ decreases by 10%, find the percentage change in $T$.

Solution.

Step 1 — New values as multipliers.

• $w_{\text{new}} = 1.20\,w$
• $d_{\text{new}} = 0.90\,d$

Step 2 — Compute $T_{\text{new}}$:

$$T_{\text{new}} = \sqrt{\dfrac{1.20\,w}{(0.90\,d)^3}} = \sqrt{\dfrac{1.20\,w}{0.729\,d^3}} = \sqrt{\dfrac{1.20}{0.729}} \cdot \sqrt{\dfrac{w}{d^3}}$$

So:

$$\frac{T_{\text{new}}}{T_{\text{old}}} = \sqrt{\dfrac{1.20}{0.729}} = \sqrt{1.6461\ldots} \approx 1.2830$$

Step 3 — Percentage change:

$$(1.2830 - 1) \times 100\% \approx 28.3\%$$

Answer: $T$ increases by approximately $28.3\%$.

Lesson: powers amplify percentage changes. The $d^3$ in the denominator is what drives the big jump.

Question. By inserting brackets, make $2 + 3 \times 4 - 1$ equal $19$.

Solution.

Step 1 — Baseline. Without extra brackets:

$$2 + 3 \times 4 - 1 = 2 + 12 - 1 = 13$$

Step 2 — Target. Need $19$. Difference is $+6$.

Step 3 — Try grouping the $+$ before the $\times$:

$$(2 + 3) \times 4 - 1 = 5 \times 4 - 1 = 20 - 1 = 19 \quad \checkmark$$

Answer: $(2 + 3) \times 4 - 1 = 19$.

Lesson: bracketing changes precedence — adding before multiplying converts $2 + 12$ into $5 \times 4$, a big jump. Try the addition next to the largest multiplier first.