Algebra — Diagnostic Tests

Unit Tests

UT-1: Solving Linear and Quadratic Equations

Question: (a) Solve $4(2x - 3) = 5(x + 7)$. (b) Solve $x^2 - 5x + 6 = 0$ by factorisation. (c) Solve $2x^2 + 7x - 4 = 0$ using the quadratic formula. (d) Solve the inequality $3x - 5 \gt 2x + 3$ and represent the solution on a number line.

Solution:

(a) $8x - 12 = 5x + 35$. $3x = 47$. $x = \frac{47}{3} = 15\frac{2}{3}$.

(b) $x^2 - 5x + 6 = (x - 2)(x - 3) = 0$. $x = 2$ or $x = 3$.

(c) $x = \frac{-7 \pm \sqrt{49 - 4(2)(-4)}}{2(2)} = \frac{-7 \pm \sqrt{49 + 32}}{4} = \frac{-7 \pm \sqrt{81}}{4} = \frac{-7 \pm 9}{4}$.

$x = \frac{2}{4} = 0.5$ or $x = \frac{-16}{4} = -4$.

(d) $3x - 5 \gt 2x + 3$. $x \gt 8$.

On a number line: open circle at 8, shading to the right.

UT-2: Simultaneous Equations

Question: (a) Solve the simultaneous equations: $3x + 2y = 16$ and $x - y = 1$. (b) Solve: $y = 2x + 1$ and $x^2 + y^2 = 25$. (c) A shop sells small and large boxes. Small boxes cost $\pounds 3$ and large boxes cost $\pounds 5$. If 20 boxes are sold for a total of $\pounds 76$How many of each type were sold?

Solution:

(a) From equation 2: $x = y + 1$. Substitute into equation 1: $3(y + 1) + 2y = 16$. $3y + 3 + 2y = 16$. $5y = 13$. $y = 2.6$. $x = 3.6$.

(b) Substitute $y = 2x + 1$ into $x^2 + y^2 = 25$: $x^2 + (2x + 1)^2 = 25$. $x^2 + 4x^2 + 4x + 1 = 25$. $5x^2 + 4x - 24 = 0$.

$x = \frac{-4 \pm \sqrt{16 + 480}}{10} = \frac{-4 \pm \sqrt{496}}{10} = \frac{-4 \pm 22.27}{10}$.

$x = 1.827$ or $x = -2.627$. $y = 4.654$ or $y = -4.254$.

(c) Let $s$ = small boxes, $l$ = large boxes. $s + l = 20$ and $3s + 5l = 76$. From equation 1: $s = 20 - l$. Substitute: $3(20 - l) + 5l = 76$. $60 - 3l + 5l = 76$. $2l = 16$. $l = 8$. $s = 12$.

12 small boxes and 8 large boxes.

UT-3: Rearranging Formulas

Question: (a) Rearrange $v = u + at$ to make $a$ the subject. (b) Rearrange $A = \pi r^2 + 2\pi rh$ to make $h$ the subject. (c) Rearrange $s = ut + \frac{1}{2}at^2$ to make $t$ the subject (this is a quadratic in $t$). (d) The formula for the surface area of a sphere is $A = 4\pi r^2$. Rearrange to make $r$ the subject and calculate $r$ when $A = 200$ cm$^2$Giving your answer to 3 significant figures.

Solution:

(a) $v = u + at$. $v - u = at$. $a = \frac{v - u}{t}$.

(b) $A = \pi r^2 + 2\pi rh$. $A - \pi r^2 = 2\pi rh$. $h = \frac{A - \pi r^2}{2\pi r}$.

(c) $\frac{1}{2}at^2 + ut - s = 0$. This is $at^2 + 2ut - 2s = 0$ (multiplying by 2). $t = \frac{-2u \pm \sqrt{4u^2 + 8as}}{2a} = \frac{-2u \pm 2\sqrt{u^2 + 2as}}{2a} = \frac{-u \pm \sqrt{u^2 + 2as}}{a}$.

Taking the positive root (time is positive): $t = \frac{-u + \sqrt{u^2 + 2as}}{a}$.

(d) $r^2 = \frac{A}{4\pi}$. $r = \sqrt{\frac{A}{4\pi}}$. $r = \sqrt{\frac{200}{4\pi}} = \sqrt{\frac{50}{\pi}} = \sqrt{15.915} = 3.99$ cm (3 s.f.).

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Integration Tests

IT-1: Quadratic Functions and Graphs (with Geometry)

Question: A ball is thrown and its height $h$ metres after $t$ seconds is given by $h = -5t^2 + 20t + 1$. (a) Find the maximum height of the ball and the time at which it occurs. (b) Calculate when the ball hits the ground. (c) Sketch the graph of $h$ against $t$Labelling the maximum point, the $t$-intercept, and the $h$-intercept. (d) Find the height of the ball after 3 seconds and use this to determine whether the ball is rising or falling at that moment.

Solution:

(a) Maximum occurs at the vertex. $t = \frac{-b}{2a} = \frac{-20}{2(-5)} = 2$ seconds. $h(2) = -5(4) + 20(2) + 1 = -20 + 40 + 1 = 21$ metres.

Maximum height is 21 metres at $t = 2$ s.

(b) Ball hits ground when $h = 0$: $-5t^2 + 20t + 1 = 0$. $t = \frac{-20 \pm \sqrt{400 + 20}}{-10} = \frac{-20 \pm \sqrt{420}}{-10} = \frac{-20 \pm 20.494}{-10}$.

$t = \frac{-20 + 20.494}{-10} = -0.0494$ (reject, negative time) or $t = \frac{-20 - 20.494}{-10} = 4.049$.

The ball hits the ground after approximately 4.05 seconds.

(c) The graph is a downward-opening parabola with:

• $h$-intercept: $(0, 1)$
• Vertex: $(2, 21)$
• $t$-intercept: $(4.05, 0)$

(d) $h(3) = -5(9) + 20(3) + 1 = -45 + 60 + 1 = 16$ metres.

At $t = 3$The ball is at 16 metres. Since the maximum is at $t = 2$ (21 m), the ball is falling at $t = 3$. This is confirmed by the negative coefficient of $t^2$ and the fact that $3 \gt 2$.

IT-2: Algebraic Proof and Sequences (with Number)

Question: (a) Prove that the sum of any three consecutive odd numbers is always a multiple of 3. (b) The $n$Th term of a sequence is given by $u_n = n^2 + 3n$. Find the first 4 terms and show that the difference between consecutive terms forms a linear sequence. (c) Prove that $\frac{n(n+1)}{2} + (n+1) = \frac{(n+1)(n+2)}{2}$. (d) A student claims “the product of two even numbers is always a multiple of 8.” Is this true? Prove or disprove with a counterexample.

Solution:

(a) Three consecutive odd numbers: 2k + 1$$2k + 3$$2k + 5 (where $k$ is an integer).

Sum $= (2k + 1) + (2k + 3) + (2k + 5) = 6k + 9 = 3(2k + 3)$.

Since $2k + 3$ is an integer, the sum is a multiple of 3.

(b) $u_1 = 1 + 3 = 4$. $u_2 = 4 + 6 = 10$. $u_3 = 9 + 9 = 18$. $u_4 = 16 + 12 = 28$.

Differences: 10 - 4 = 6$$18 - 10 = 8$$28 - 18 = 10. The differences are 6, 8, 10 — a linear sequence with common difference 2. This confirms the sequence is quadratic (second differences are constant).

(c) $\frac{n(n+1)}{2} + (n+1) = \frac{n(n+1) + 2(n+1)}{2} = \frac{(n+1)(n + 2)}{2} = \frac{(n+1)(n+2)}{2}$. Proven.

(d) False. Counterexample: $2 \times 2 = 4$Which is not a multiple of 8. Another: $2 \times 6 = 12$Not a multiple of 8. The product of two even numbers $2a \times 2b = 4ab$ is always a multiple of 4, but not necessarily 8. It is only a multiple of 8 if at least one of the numbers is itself a multiple of 4 (i.e., $a$ or $b$ is even).

IT-3: Real-World Algebraic Modelling (with Ratio and Proportion)

Question: A taxi company charges a fixed hiring fee plus a cost per kilometre. A journey of 10 km costs $\pounds 18$ and a journey of 25 km costs $\pounds 33$. (a) Find the fixed fee and cost per kilometre. (b) Write a formula for the cost $C$ of a journey of $d$ km. (c) A customer has $\pounds 50$. What is the longest journey they can afford? (d) A competing company charges $\pounds 1.50$ per km with no fixed fee. For what range of distances is the first company cheaper?

Solution:

(a) Let fixed fee $= f$Cost per km $= p$. $f + 10p = 18$ and $f + 25p = 33$. Subtracting: 15p = 15$$p = \pounds 1 per km. $f = 18 - 10 = \pounds 8$.

(b) $C = 8 + d$ (where $d$ is in km and $C$ in pounds).

(c) $50 = 8 + d$. $d = 42$ km.

(d) First company: $C_1 = 8 + d$. Second company: $C_2 = 1.5d$. $C_1 \lt C_2$ when $8 + d \lt 1.5d$. $8 \lt 0.5d$. $d \gt 16$.

The first company is cheaper for journeys longer than 16 km.

Summary

The key principles covered in this topic are linked in the sub-pages above. Focus on understanding the definitions, applying the formulas or frameworks, and evaluating strengths and limitations of each approach.

Worked Examples

Worked examples demonstrating the application of key concepts are covered in the detailed sub-pages linked above.

Common Pitfalls

• Confusing terminology or concepts that appear similar but have distinct meanings.
• Overlooking key assumptions or boundary conditions that limit applicability.