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Grade 12 Mathematics entrance exam, 2018 E.C. (2026), version 1

Real questions from the Grade 12 university entrance exam in Mathematics, as students sat it in 2018 E.C. Every question comes with the correct answer and a worked explanation. The 2018 E.C. exam went out in several versions. This page holds version 1; the others are listed further down.

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  • Both streams

    Who sits it

  • 2018 E.C. (2026), version 1

    Exam year

  • 29 questions with answers

    In this set

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Answer each question yourself before opening the answer. The explanations point to the exact place in the textbook, so you know which page to reread when you miss one.

Question 3

After rationalizing the expression 3/(3√3 - 2√2), the denominator is equal to

  1. A.19
  2. B.57
  3. C.3(3√3 - 2√2)/19
  4. D.3(3√3 + 2√2)/19
Show the answer and explanation

Answer: A 19

Why

To rationalize, multiply the top and the bottom by the conjugate of the denominator, 33+223\sqrt{3}+2\sqrt{2}. The bottom then becomes a difference of two squares: (33)2−(22)2=9×3−4×2=27−8=19(3\sqrt{3})^2-(2\sqrt{2})^2 = 9\times 3 - 4\times 2 = 27-8 = 19. So the denominator the question asks for is 1919. Option C is tempting because it is the value of the whole fraction after rationalizing, but the question asks only for what sits underneath.

Question 5

The solution set of the system {3x - y = 5, 3y + 15 = 9x} is

  1. A.{(0,-5),(1,-2),(2,-1),...}
  2. B.R
  3. C.∅
  4. D.{(t,3t-5): t∈R}
Show the answer and explanation

Answer: D {(t,3t-5): t∈R}

Why

The first equation gives y=3x−5y = 3x-5. Rearrange the second: 3y=9x−153y = 9x-15, so y=3x−5y = 3x-5 as well. The two equations describe the same line, so every point on that line solves the system. Writing xx as any real number tt, the solutions are the pairs (t, 3t−5)(t,\,3t-5). Option B is tempting because there really are infinitely many solutions, but those solutions are points on one line, not every point of the plane.

Question 6

The solution set of |2x+1| = x-7 is

  1. A.{-8,-2}
  2. B.{-8,2}
  3. C.{8,2}
  4. D.{}
Show the answer and explanation

Answer: D {}

Why

An absolute value is never negative, so ∣2x+1∣=x−7|2x+1| = x-7 can only work when x−7≥0x-7 \ge 0, that is when x≥7x \ge 7. Take the two cases anyway: 2x+1=x−72x+1 = x-7 gives x=−8x = -8, and 2x+1=−(x−7)2x+1 = -(x-7) gives 3x=63x = 6, so x=2x = 2. Neither −8-8 nor 22 reaches 77, so both fail the check and the solution set is empty. Option B collects those two candidates, but a candidate only becomes a solution after it survives substitution.

Question 8

Suppose N is the set of natural numbers and A = {1,2,{3}}, then which of the following is true?

  1. A.{3} ⊆ A
  2. B.N ∪ A = N
  3. C.A\{3} = A
  4. D.A ⊆ N
Show the answer and explanation

Answer: C A\{3} = A

Why

In A={1, 2, {3}}A = \{1,\,2,\,\{3\}\} the third element is the set {3}\{3\}, not the number 33. Since 33 itself is not an element of AA, taking it away removes nothing, so A∖{3}=AA \setminus \{3\} = A. Option A says {3}⊆A\{3\} \subseteq A, which would need the number 33 to sit inside AA. Here {3}\{3\} is a member of AA, and being a member is a different thing from being a subset.

Question 10

Let A={0,1,2,3,4} and R be a relation from A to A given by R={(x,y): y=x+2, for x,y∈A}. Which one is true about R?

  1. A.(2,3) is an element of R
  2. B.(3,5) is an element of R
  3. C.The domain of R is {1,2,3,4}
  4. D.The range of R is {2,3,4}
Show the answer and explanation

Answer: D The range of R is {2,3,4}

Why

The rule is y=x+2y = x+2, with both xx and yy taken from A={0,1,2,3,4}A=\{0,1,2,3,4\}. Only x=0,1,2x=0,1,2 keep yy inside AA, so R={(0,2),(1,3),(2,4)}R = \{(0,2),(1,3),(2,4)\}. The second coordinates are 2,3,42, 3, 4, which makes the range {2,3,4}\{2,3,4\}. Option B looks right if you forget that yy must also come from AA: 55 is not in AA, so (3,5)(3,5) never enters RR.

Question 12

The graph of f(x)=x²+6x+7 is obtained from the graph of g(x)=x² by shifting

  1. A.6 units to the right and 9 units upward
  2. B.3 units to the right and 2 units downward
  3. C.3 units to the left and 2 units downward
  4. D.3 units to the left and 2 units upward
Show the answer and explanation

Answer: C 3 units to the left and 2 units downward

Why

Complete the square: x2+6x+7=(x2+6x+9)−9+7=(x+3)2−2x^2+6x+7 = (x^2+6x+9)-9+7 = (x+3)^2-2. When x2x^2 becomes (x+3)2(x+3)^2, the graph has moved 3 units to the left, because the +3+3 sits inside the bracket with xx. The −2-2 outside the bracket then pulls the whole curve 2 units down. Option B has the right sizes but the wrong horizontal direction: (x+3)2(x+3)^2 shifts left, while (x−3)2(x-3)^2 would shift right.

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The other Mathematics papers of 2018 E.C.

2018 E.C. (2026)2018 E.C. (2026), social stream2018 E.C. (2026), version 22018 E.C. (2026), version 32018 E.C. (2026), version 42018 E.C. (2026), version 52018 E.C. (2026), version 6

Mathematics papers from other years

2017 E.C. (2025)2017 E.C. (2025), natural stream2017 E.C. (2025), version 12017 E.C. (2025), version 22017 E.C. (2025), version 42016 E.C. (2024)2016 E.C. (2024), natural stream2016 E.C. (2024), social stream2016 E.C. model exam2015 E.C. (2023)2015 E.C. (2023), natural stream2015 E.C. (2023), social stream2014 E.C. (2022)2014 E.C. (2022), social stream2013 E.C. (2021)2013 E.C. (2021), version 12011 E.C. (2019)2009 E.C. (2017)2008 E.C. (2016)2007 E.C. (2015)2006 E.C. (2014)2005 E.C. (2013)

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Are these the real 2018 E.C. Mathematics entrance exam questions?

Yes. They come from the national university entrance exam in Mathematics that students sat in 2018 E.C. (2026). Our set holds 29 of its questions, each with its answer and an explanation that points back to the textbook.

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