10 questions that get harder as you go. No calculator. Write your working on paper, pick A–E, then open the solution. Q1–3 are warm-ups, Q4–6 a step up, Q7–10 ESAT standard.
Q1 · Warm-up
What is the nth term of the sequence 5, 8, 11, 14, …?
A 3n + 2 · B n + 3 · C 3n + 5 · D 5n + 3 · E 3n − 2
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A. The terms go up by 3 each time, so start with 3n: 3, 6, 9, 12. Each term is 2 more than that, so the nth term is 3n + 2. Check: n = 4 gives 14 ✓
💡 C is the trap: adding the first term. The number you add is the “zeroth term”, 5 − 3 = 2.
Q2 · Warm-up
Expand (x + 3)(x − 5).
A x² + 2x − 15 · B x² − 8x − 15 · C x² − 2x + 15 · D x² − 2x − 15 · E x² − 15
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D. x² − 5x + 3x − 15 = x² − 2x − 15.
Q3 · Warm-up
Factorise x² − 9x + 20.
A (x + 4)(x + 5) · B (x − 2)(x − 10) · C (x − 4)(x − 5) · D (x − 4)(x + 5) · E (x + 4)(x − 5)
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C. Find two numbers that multiply to +20 and add to −9: −4 and −5. So (x − 4)(x − 5).
Q4 · Step up
Solve the simultaneous equations 2x + y = 11 and x − y = 1.
A x = 3, y = 5 · B x = 5, y = 1 · C x = 6, y = −1 · D x = 2, y = 7 · E x = 4, y = 3
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E. Add the equations to get rid of y: 3x = 12, so x = 4. Then 4 − y = 1, so y = 3.
💡 Every wrong option fits the first equation. Always check your answer in both.
Q5 · Step up
Solve 5 − 2x < 11.
A x < −3 · B x > −3 · C x < 3 · D x > 3 · E x > −8
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B. Subtract 5: −2x < 6. Divide by −2 and flip the sign: x > −3. (A is the trap: forgetting to flip.)
Q6 · Step up
What is the nth term of the sequence 3, 9, 19, 33, 51, …?
A 4n² − 1 · B n² + 2 · C 2n² + 4n − 3 · D 2n² + 1 · E 4n − 1
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D. First differences: 6, 10, 14, 18. Second differences: 4, 4, 4. Constant second differences mean it’s quadratic, and the n² coefficient is half the second difference: 4 ÷ 2 = 2.
2n² gives 2, 8, 18, 32, 50. The sequence is 1 more each time, so the nth term is 2n² + 1.
💡 A and C are the traps: they both give 3 for n = 1 but fail at n = 2. Always check at least two terms.
Q7 · ESAT standard
The volume of a sphere is V = (4/3)πr³. Make r the subject.
A r = ∛(4V/(3π)) · B r = 3V/(4π) · C r = √(3V/(4π)) · D r = ∛(3V/(4π)) · E r = ∛(3πV/4)
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D. Multiply both sides by 3: 3V = 4πr³. Divide by 4π: r³ = 3V/(4π). Cube root: r = ∛(3V/(4π)).
Q8 · ESAT standard
Simplify 1/(x − 1) − 1/(x + 1).
A 0 · B 2x/(x² − 1) · C 2/(x² − 1) · D −2/(x² − 1) · E 1/(x² − 1)
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C. Common denominator (x − 1)(x + 1) = x² − 1. Top: (x + 1) − (x − 1) = 2. So the answer is 2/(x² − 1).
💡 Check by picking a number. With x = 2: 1/1 − 1/3 = ⅔, and 2/(4 − 1) = ⅔ ✓
Q9 · ESAT standard
What are the coordinates of the minimum point of y = x² − 6x + 13?
A (−3, 4) · B (3, 4) · C (3, 13) · D (6, 13) · E (−3, −4)
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B. Complete the square: x² − 6x + 13 = (x − 3)² − 9 + 13 = (x − 3)² + 4. The squared part is smallest (zero) when x = 3, giving y = 4. Minimum point (3, 4).
Q10 · ESAT standard
x + y = 10 and xy = 21. What is x² + y²?
A 42 · B 79 · C 58 · D 100 · E 142
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C. You don’t need x and y. Use (x + y)² = x² + 2xy + y². So x² + y² = (x + y)² − 2xy = 100 − 42 = 58.
Check: x = 3, y = 7 fits both, and 9 + 49 = 58 ✓
🔥 Challenge: full ESAT level
Five harder questions with up to 8 options, answers in symbols and “which statements are correct?” questions, just like the real test. Aim for 90 seconds each.
C1
Make x the subject of y = (2x + 3)/(x − 1).
A x = (y + 3)/(y − 2) · B x = (y − 3)/(y + 2) · C x = (y + 3)/(y + 2) · D x = (3 − y)/(y − 2) · E x = (y − 3)/(2 − y) · F x = (y + 2)/(y − 3) · G x = (2y + 3)/(y − 1) · H x = (y + 1)/(y − 2)
Show solution
A. Multiply out: y(x − 1) = 2x + 3, so yx − y = 2x + 3. Get the x terms together: yx − 2x = y + 3. Factorise: x(y − 2) = y + 3. So x = (y + 3)/(y − 2).
💡 x appears twice, so the trick is always: collect the x terms, then factorise x out.
C2
Consider the equation x² + 6x + k = 0. Which statements are correct?
I If k = 9, there is exactly one solution.
II If k = 10, there are no real solutions.
III If k = 0, x = 0 is a solution.
A none · B I only · C II only · D III only · E I and II only · F I and III only · G II and III only · H I, II and III
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H. Complete the square: x² + 6x + k = (x + 3)² + (k − 9).
I: k = 9 gives (x + 3)² = 0, so only x = −3 ✓. II: k = 10 gives (x + 3)² = −1, and a square can’t be negative, so no real solutions ✓. III: k = 0 gives x² + 6x = x(x + 6) = 0, so x = 0 works ✓.
C3
A rectangle is 3 cm longer than it is wide. Its area is 40 cm². What is its perimeter?
A 13 cm · B 18 cm · C 22 cm · D 26 cm · E 30 cm · F 40 cm · G 52 cm · H 80 cm
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D. Width w: w(w + 3) = 40, so w² + 3w − 40 = 0, which factorises to (w + 8)(w − 5) = 0. A width can’t be negative, so w = 5 and the length is 8. Perimeter = 2 × (5 + 8) = 26 cm. (A is the trap: forgetting to double.)
C4
Which points are where the line y = x + 1 meets the circle x² + y² = 25?
A (3, 4) only · B (3, 4) and (4, 3) · C (−3, −4) and (4, 3) · D (−4, −3) only · E (4, 5) and (−3, −2) · F (3, 4) and (−3, −4) · G (3, 4) and (−4, −3) · H (0, 1) and (5, 6)
Show solution
G. Substitute y = x + 1: x² + (x + 1)² = 25, so 2x² + 2x + 1 = 25, so x² + x − 12 = 0, which factorises to (x + 4)(x − 3) = 0.
x = 3 gives y = 4. x = −4 gives y = −3. So (3, 4) and (−4, −3).
💡 Quick check: every option point must satisfy y = x + 1. That rules out B, C and F immediately.
C5
A ball’s height in metres after t seconds is h = 20t − 5t². For how long is the ball more than 15 m above the ground?
A 1 s · B 2 s · C 3 s · D 4 s · E 0.5 s · F 1.5 s · G 2.5 s · H 3.5 s
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B. Solve 20t − 5t² > 15. Divide by −5 (and flip the sign): t² − 4t + 3 < 0. Factorise: (t − 1)(t − 3) < 0, which is true between t = 1 and t = 3. That’s 2 seconds.
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