🧠 Maths Practice — Part 2 (20 Questions)

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SOLVING EQUATIONS — Linear & Quadratic

Linear: get \(x\) alone — do the SAME to both sides.
Quadratic: factorise → set each bracket = 0 → solve.
Or use formula: \(x = \dfrac{-b \pm \sqrt{b^2-4ac}}{2a}\)

KEY WORDS: rearrange · factorise · quadratic formula · discriminant · roots · solutions

WORKED EXAMPLE

Solve \(x^2 - 5x + 6 = 0\)

1

Find factors of 6 that add to −5: (−2)(−3) ✓

2

\((x-2)(x-3) = 0\)

3

\(x = 2\) or \(x = 3\)

QUESTION 1 ★ MEDIUM

Solve \(x^2 + x - 12 = 0\)

📖 EXPLANATION

Need two numbers that multiply to −12 and add to +1 → (+4) and (−3):

\((x+4)(x-3) = 0\)

\(x = -4\) or \(x = 3\) ✓

⚠️ TRAP: Signs flip when setting each bracket to zero. If \((x+4)=0\) then \(x = -4\), NOT \(+4\)!

QUESTION 2 ★★ HARD

Solve \(2x^2 - 7x + 3 = 0\)

📖 EXPLANATION

Factorise: \(2x^2 - 7x + 3\). Need: \(ac = 6\), adds to \(-7\) → \(-6\) and \(-1\):

\(2x^2 - 6x - x + 3 = 2x(x-3) - 1(x-3)\)

\(= (2x-1)(x-3) = 0\)

\(x = \frac{1}{2}\) or \(x = 3\) ✓

⚠️ TRAP: When \(a \neq 1\), you CANNOT just find factors of \(c\). Use the \(ac\) method!

QUESTION 3 ★★★ TRICKY

Solve \(3x^2 + 5x - 2 = 0\) using the quadratic formula. Give answers to 2 d.p.

📖 EXPLANATION

\(a=3, b=5, c=-2\). Discriminant: \(b^2-4ac = 25+24 = 49\)

\(x = \dfrac{-5 \pm \sqrt{49}}{6} = \dfrac{-5 \pm 7}{6}\)

\(x = \dfrac{2}{6} = \frac{1}{3} \approx 0.33\) or \(x = \dfrac{-12}{6} = -2.00\) ✓

⚠️ TRAP: \(\sqrt{49} = 7\), not 7.something. Always check if the discriminant is a perfect square first!

simultaneous equations

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SIMULTANEOUS EQUATIONS

Elimination: make one coefficient match → add or subtract.
Substitution: isolate one variable → plug into the other equation.
Always CHECK your answer by substituting back!

KEY WORDS: eliminate · substitute · coefficient · verify · linear pair · intersection

WORKED EXAMPLE

Solve simultaneously: \(2x+y=7\) and \(x-y=2\)

1

Add equations: \(3x = 9 \Rightarrow x=3\)

2

Sub into eq 1: \(6+y=7 \Rightarrow y=1\)

3

Check in eq 2: \(3-1=2\) ✓

QUESTION 4 ★ MEDIUM

Solve simultaneously: \(3x + 2y = 16\) and \(5x - 2y = 8\)

📖 EXPLANATION

Both have \(2y\) — add the equations to eliminate \(y\):

\(8x = 24 \Rightarrow x = 3\)

Sub into eq 1: \(9 + 2y = 16 \Rightarrow 2y = 7 \Rightarrow y = 3.5\)

Check eq 2: \(15 - 7 = 8\) ✓

⚠️ TRAP: The \(2y\) terms have OPPOSITE signs so ADD, not subtract. If you subtract, the \(y\) stays!

QUESTION 5 ★★★ TRICKY

Solve simultaneously: \(y = x^2 - 3\) and \(y = 2x\)

📖 EXPLANATION

Substitute \(y=2x\) into \(y=x^2-3\):

\(2x = x^2 - 3 \Rightarrow x^2 - 2x - 3 = 0\)

\((x-3)(x+1) = 0 \Rightarrow x=3\) or \(x=-1\)

When \(x=3\): \(y=6\). When \(x=-1\): \(y=-2\) ✓

⚠️ TRAP: A line can intersect a curve at TWO points! Don't stop after finding one solution.

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INEQUALITIES

Treat like an equation BUT: if you multiply/divide by a negative number → FLIP the sign!
\(<\) means "less than" · \(\leq\) means "less than OR equal to"
On a number line: ● = included, ○ = NOT included

KEY WORDS: inequality · flip · integer · number line · region · satisfy

QUESTION 6 ★ MEDIUM

Solve \(3x - 7 > 5\) and write the solution on a number line description.

📖 EXPLANATION

\(3x - 7 > 5\)

\(3x > 12\)

\(x > 4\) ✓ — open circle at 4, arrow pointing right

⚠️ TRAP: The original sign is \(>\) (strict), NOT \(\geq\). Adding/subtracting doesn't change the inequality type — only multiplying/dividing by negatives does!

QUESTION 7 ★★ HARD

Solve the inequality \(x^2 - 5x + 6 \leq 0\)

📖 EXPLANATION

Factorise: \((x-2)(x-3) \leq 0\). Roots are \(x=2\) and \(x=3\).

The parabola opens upward (positive \(x^2\)), so it's ≤ 0 BETWEEN the roots:

\(2 \leq x \leq 3\) ✓

⚠️ TRAP: The "outside" answer \(x\leq2\) or \(x\geq3\) is where the parabola is POSITIVE (≥0), not negative. Draw a sketch to see the U-shape!

sequences

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SEQUENCES — nth Term

Arithmetic: \(n\text{th term} = a + (n-1)d\) where \(a\) = first term, \(d\) = common difference.
Quadratic: if 2nd differences are constant → \(an^2 + bn + c\). Find \(a\) first: \(a = \frac{\text{2nd diff}}{2}\)

KEY WORDS: arithmetic · geometric · nth term · common difference · first difference · second difference

WORKED EXAMPLE

Find the nth term of: 5, 8, 11, 14, ...

1

Common difference: \(d = 3\)

2

First term: \(a = 5\)

3

\(n\text{th term} = 3n + 2\) (since \(3(1)+2=5\) ✓)

QUESTION 8 ★ MEDIUM

Find the nth term of the sequence: 7, 11, 15, 19, 23, ...

📖 EXPLANATION

Common difference \(d = 4\), so starts with \(4n\)

When \(n=1\): \(4(1) = 4\), but first term is 7. So add 3.

\(n\text{th term} = 4n + 3\) ✓

Check: \(n=2\): \(4(2)+3=11\) ✓, \(n=3\): \(4(3)+3=15\) ✓

⚠️ TRAP: Option D gives \(4(1)-3=1\) for \(n=1\) — not 7. Always verify with \(n=1\)!

QUESTION 9 ★★★ TRICKY

The nth term of a sequence is \(n^2 + 3n - 1\). Which term equals 49?

📖 EXPLANATION

Set the nth term equal to 49:

\(n^2 + 3n - 1 = 49\)

\(n^2 + 3n - 50 = 0\)

\((n+\ldots)\) — discriminant: \(9 + 200 = 209\)... hmm, let's try values.

\(n=6\): \(36+18-1=53\). \(n=5\): \(25+15-1=39\). \(n=7\): \(49+21-1=69\).

None give exactly 49 by formula here, but let's check \(n=6\): \(36+18-1=53 \neq 49\).

Actually \(n=5\): \(25+15-1 = 39\), \(n=6\): 53. So 49 is NOT in this sequence!

⚠️ The trick: always verify your answer. If no integer \(n\) works, the value is NOT in the sequence — a common exam question!

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PROPORTION — Direct & Inverse

Direct: \(y \propto x \Rightarrow y = kx\). Both go UP together.
Inverse: \(y \propto \frac{1}{x} \Rightarrow y = \frac{k}{x}\). One up, other DOWN.
Method: find \(k\) using the given values, then use it.

KEY WORDS: proportional · constant of proportionality · direct · inverse · substitute · k-value

QUESTION 10 ★★ HARD

\(y\) is directly proportional to \(x^2\). When \(x=3\), \(y=36\). Find \(y\) when \(x=5\).

📖 EXPLANATION

Since \(y \propto x^2\), write \(y = kx^2\).

Find \(k\): \(36 = k(3^2) = 9k \Rightarrow k = 4\)

Now: \(y = 4(5^2) = 4 \times 25 = 100\) ✓

⚠️ TRAP: Some students write \(y = kx\) instead of \(y = kx^2\). Read carefully — "proportional to \(x^2\)" means the SQUARE!

QUESTION 11 ★★★ TRICKY

\(y\) is inversely proportional to \(x\). When \(x=4\), \(y=9\). Find \(x\) when \(y=12\).

📖 EXPLANATION

Inverse proportion: \(y = \dfrac{k}{x}\)

Find \(k\): \(9 = \dfrac{k}{4} \Rightarrow k = 36\)

Now find \(x\): \(12 = \dfrac{36}{x} \Rightarrow x = \dfrac{36}{12} = 3\) ✓

⚠️ TRAP: Don't confuse \(y = kx\) with \(y = \frac{k}{x}\). With inverse proportion, as \(y\) increases, \(x\) DECREASES — so \(x\) should be smaller than 4. Only option A makes sense!

percentages

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PERCENTAGE CHANGE & REVERSE

% increase: multiply by \((1 + \frac{r}{100})\)
% decrease: multiply by \((1 - \frac{r}{100})\)
Reverse %: DIVIDE by the multiplier, NOT subtract the percent!

KEY WORDS: multiplier · percentage change · reverse percentage · original value · compound interest

QUESTION 12 ★★ HARD

A jacket costs £84 after a 30% discount. What was the original price?

📖 EXPLANATION

30% off means the price is 70% of original. Multiplier = 0.7.

\(\text{Original} \times 0.7 = 84\)

\(\text{Original} = \dfrac{84}{0.7} = 120\) ✓

⚠️ BIGGEST TRAP: Students calculate \(84 \times 1.3 = 109.20\) — adding 30% back. This is WRONG! You must DIVIDE by the multiplier, not multiply by the inverse percentage.

QUESTION 13 ★★★ TRICKY

£2000 is invested at 3% compound interest per year. How much after 4 years? (to nearest penny)

📖 EXPLANATION

Compound interest formula: \(A = P(1+r)^n\)

\(A = 2000 \times (1.03)^4\)

\(= 2000 \times 1.12550881 = £2251.02\) ✓

⚠️ TRAP: Option A uses simple interest: \(2000 + 4 \times 60 = 2240\). That's wrong! With COMPOUND interest, you earn interest on the interest too — use the power!

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STRAIGHT LINE GRAPHS

\(y = mx + c\) · \(m\) = gradient (steepness) · \(c\) = y-intercept
Gradient = \(\dfrac{\text{rise}}{\text{run}} = \dfrac{y_2 - y_1}{x_2 - x_1}\)
Parallel lines: SAME gradient. Perpendicular: gradients multiply to \(-1\).

KEY WORDS: gradient · intercept · parallel · perpendicular · negative reciprocal · slope

QUESTION 14 ★ MEDIUM

A line passes through \((2, 5)\) and \((6, 13)\). What is the equation of the line?

📖 EXPLANATION

Gradient: \(m = \dfrac{13-5}{6-2} = \dfrac{8}{4} = 2\)

Use \(y - y_1 = m(x - x_1)\): \(y - 5 = 2(x-2)\)

\(y = 2x - 4 + 5 = 2x + 1\) ✓

⚠️ TRAP: Many students forget to substitute back to find \(c\). Getting the gradient right isn't enough — you must find the y-intercept!

QUESTION 15 ★★ HARD

Line \(L\) has equation \(y = 3x - 2\). What is the equation of a line perpendicular to \(L\) passing through \((6, 1)\)?

📖 EXPLANATION

Perpendicular gradient = negative reciprocal of 3:

\(m_\perp = -\dfrac{1}{3}\)

Through \((6,1)\): \(y - 1 = -\frac{1}{3}(x - 6)\)

\(y = -\frac{1}{3}x + 2 + 1 = -\frac{1}{3}x + 3\) ✓

⚠️ TRAP: "Negative reciprocal" means flip AND negate. The reciprocal of 3 is \(\frac{1}{3}\), so the perpendicular gradient is \(-\frac{1}{3}\), NOT \(-3\) or \(+\frac{1}{3}\).

circle theorems

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CIRCLE THEOREMS

Key theorems:
• Angle at centre = 2 × angle at circumference (same arc)
• Angles in same segment are EQUAL
• Opposite angles in a cyclic quadrilateral add to 180°
• Angle in a semicircle = 90°
• Tangent ⊥ radius at point of contact

KEY WORDS: subtend · cyclic quadrilateral · tangent · chord · arc · circumference · semicircle

QUESTION 16 ★★ HARD

In a circle, the angle at the centre subtended by an arc is 124°. What is the angle at the circumference subtended by the same arc?

📖 EXPLANATION

Circle Theorem: Angle at centre = 2 × angle at circumference.

Angle at circumference \(= 124° \div 2 = 62°\) ✓

⚠️ TRAP: Many students go the wrong way and double it (248°). Remember: the CENTRE angle is BIGGER — it's double the circumference angle!

QUESTION 17 ★★★ TRICKY

ABCD is a cyclic quadrilateral. Angle A = \(3x + 10\)° and angle C = \(2x + 30\)°. Find \(x\).

📖 EXPLANATION

Cyclic quadrilateral: opposite angles ADD to 180°.

\((3x+10) + (2x+30) = 180\)

\(5x + 40 = 180\)

\(5x = 140 \Rightarrow x = 28\) ✓

Check: A = \(3(28)+10 = 94°\), C = \(2(28)+30 = 86°\). Sum = 180° ✓

⚠️ TRAP: Adjacent angles in a cyclic quadrilateral do NOT necessarily sum to 180° — only OPPOSITE ones!

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SURDS — Simplify & Rationalise

Simplify: \(\sqrt{ab} = \sqrt{a} \cdot \sqrt{b}\) — find the largest perfect square factor.
Rationalise: multiply top and bottom by the surd (or conjugate).
\(\sqrt{a} \times \sqrt{a} = a\) — surds disappear when multiplied by themselves!

KEY WORDS: surd · rationalise · conjugate · simplify · irrational · perfect square

QUESTION 18 ★ MEDIUM

Simplify \(\sqrt{72} + \sqrt{50}\)

📖 EXPLANATION

\(\sqrt{72} = \sqrt{36 \times 2} = 6\sqrt{2}\)

\(\sqrt{50} = \sqrt{25 \times 2} = 5\sqrt{2}\)

\(6\sqrt{2} + 5\sqrt{2} = 11\sqrt{2}\) ✓

⚠️ BIGGEST TRAP: \(\sqrt{72} + \sqrt{50} \neq \sqrt{122}\). You CANNOT add surds under the root sign — simplify each one first!

QUESTION 19 ★★ HARD

Rationalise the denominator of \(\dfrac{6}{3-\sqrt{3}}\). Simplify fully.

📖 EXPLANATION

Multiply top and bottom by the CONJUGATE \((3+\sqrt{3})\):

\(\dfrac{6(3+\sqrt{3})}{(3-\sqrt{3})(3+\sqrt{3})}\)

Denominator: \(9 - 3 = 6\) (difference of squares: \(a^2 - b^2\))

\(= \dfrac{6(3+\sqrt{3})}{6} = 3+\sqrt{3}\) ✓

⚠️ TRAP: Option A is the correct intermediate step but forgetting to simplify the 6s. Option C just flips the sign — that doesn't rationalise!

trigonometry

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TRIGONOMETRY — SOH CAH TOA & Beyond

SOH CAH TOA for right-angled triangles.
Sine rule: \(\dfrac{a}{\sin A} = \dfrac{b}{\sin B}\)
Cosine rule: \(a^2 = b^2 + c^2 - 2bc\cos A\)
Use sine/cosine rule when there's NO right angle!

KEY WORDS: opposite · adjacent · hypotenuse · sine rule · cosine rule · ambiguous case · bearing

QUESTION 20 ★★★ TRICKY

In triangle ABC, \(AB = 8\) cm, \(BC = 5\) cm, and angle \(B = 60°\).
Find the length of AC. Give your answer to 1 decimal place.

📖 EXPLANATION

No right angle — use the Cosine Rule: \(AC^2 = AB^2 + BC^2 - 2 \cdot AB \cdot BC \cdot \cos B\)

\(AC^2 = 64 + 25 - 2(8)(5)\cos 60°\)

\(= 89 - 80 \times 0.5 = 89 - 40 = 49\)

\(AC = \sqrt{49} = 7.0\) cm ✓

⚠️ TRAP: Some students use Pythagoras (only valid for right-angled triangles). Since \(\cos 60° = 0.5\) exactly, the numbers work out perfectly here — always know your exact trig values!

Exact values to memorise: \(\cos 60° = 0.5\), \(\sin 30° = 0.5\), \(\sin 45° = \frac{\sqrt{2}}{2}\)

📐 Core Maths Practice — Part 2