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Grade 12 Mathematics entrance exam, 2009 E.C. (2017)

Real questions from the Grade 12 university entrance exam in Mathematics, as students sat it in 2009 E.C. Every question comes with the correct answer and a worked explanation.

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  • 2009 E.C. (2017)

    Exam year

  • 61 questions with answers

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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 1

If A=(aij)3×3A = (a_{ij})_{3\times 3} is a square matrix with A−1=[132113045]A^{-1} = \begin{bmatrix} 1 & 3 & 2 \\ 1 & 1 & 3 \\ 0 & 4 & 5 \end{bmatrix}, then what is the cofactor of a23a_{23}?.

  1. A.−3/14-3/14
  2. B.−2/7-2/7
  3. C.2/72/7
  4. D.−3/7-3/7
Show the answer and explanation

Answer: B −2/7-2/7

Why

The cofactor of a23a_{23} is C23=det⁡(A)(A−1)32C_{23} = \det(A)(A^{-1})_{32}. Since det⁡(A−1)=−14\det(A^{-1}) = -14, we have det⁡(A)=−1/14\det(A) = -1/14, so C23=(−1/14)(4)=−2/7C_{23} = (-1/14)(4) = -2/7.

Question 2

A team of 10 researchers consist 4 biologists and 6 chemists. If 3 persons are chosen randomly from the team, what is the probability that at least one is a biologist?

  1. A.2/32/3
  2. B.2/52/5
  3. C.5/65/6
  4. D.7/107/10
Show the answer and explanation

Answer: C 5/65/6

Why

Using the complement, the probability that none of the 3 chosen researchers is a biologist (all are chemists) is (63)/(103)=20/120=1/6\binom{6}{3}/\binom{10}{3} = 20/120 = 1/6, so the probability of at least one biologist is 1−1/6=5/61 - 1/6 = 5/6.

Question 3

The probability that an electronic device produced by a company does not function properly is equal to 0.1. If 2 devices are bought, then what is the probability that at least one device function properly?

  1. A.0.81
  2. B.0.09
  3. C.0.18
  4. D.0.99
Show the answer and explanation

Answer: D 0.99

Why

The probability that at least one device functions properly is 1−P(both fail)=1−(0.1)2=1−0.01=0.991 - P(\text{both fail}) = 1 - (0.1)^2 = 1 - 0.01 = 0.99.

Question 5

In how many ways can a committee of 3 members be formed from 7 candidates?

  1. A.7
  2. B.21
  3. C.28
  4. D.35
Show the answer and explanation

Answer: D 35

Why

The number of ways to choose a committee of 3 from 7 candidates is (73)=7×6×53×2×1=35\binom{7}{3} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35.

Question 6

The following is a frequency distribution table of a grouped data with variable X. What is the mean (xˉ\bar{x}) and the variance (s2s^2) of the data respectively? [Figure: A two-row bordered frequency table with five columns. Top row: first cell labelled 'X', then the class intervals 3-7, 8-12, 13-17, 18-22. Bottom row: first cell labelled 'Frequency', then the values 4, 6, 8, 2 under those intervals respectively.]

  1. A.xˉ=12, s2=21\bar{x} = 12,\ s^2 = 21
  2. B.xˉ=12, s2=25\bar{x} = 12,\ s^2 = 25
  3. C.xˉ=13, s2=9\bar{x} = 13,\ s^2 = 9
  4. D.xˉ=13, s2=16\bar{x} = 13,\ s^2 = 16
Show the answer and explanation

Answer: A xˉ=12, s2=21\bar{x} = 12,\ s^2 = 21

Why

With midpoints 5,10,15,205, 10, 15, 20 and frequencies 4,6,8,24, 6, 8, 2, the mean is xˉ=240/20=12\bar{x} = 240/20 = 12, and the variance is s2=∑f(x−xˉ)2/N=420/20=21s^2 = \sum f(x - \bar{x})^2 / N = 420/20 = 21.

Question 7

The expenditure of 100 families is given below. The mode of the data is 23.5, what are the values of f1 and f2 [Figure: A two-row bordered table with six columns. Top row: first cell labelled 'Expenditure', then the classes 0-9, 10-19, 20-29, 30-39, 40-49. Bottom row: first cell labelled 'No. of Families', then the values 14, 23, f1f_1, 21, f2f_2 under those classes respectively.]

  1. A.f1=27, f2=15f_1 = 27,\ f_2 = 15
  2. B.f1=15, f2=27f_1 = 15,\ f_2 = 27
  3. C.f1=25, f2=17f_1 = 25,\ f_2 = 17
  4. D.f1=17, f2=25f_1 = 17,\ f_2 = 25
Show the answer and explanation

Answer: A f1=27, f2=15f_1 = 27,\ f_2 = 15

Why

The mode is 23.523.5, which lies in 19.5−29.519.5 - 29.5. Using the mode formula 23.5=19.5+f1−23(f1−23)+(f1−21)×1023.5 = 19.5 + \frac{f_1 - 23}{(f_1 - 23) + (f_1 - 21)} \times 10 yields f1=27f_1 = 27, and since the total is 100, f2=100−(14+23+27+21)=15f_2 = 100 - (14 + 23 + 27 + 21) = 15.

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Mathematics papers from other years

2018 E.C. (2026)2017 E.C. (2025)2016 E.C. (2024)2015 E.C. (2023)2014 E.C. (2022)2013 E.C. (2021)2011 E.C. (2019)2008 E.C. (2016)2007 E.C. (2015)2006 E.C. (2014)2005 E.C. (2013)

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Are these the real 2009 E.C. Mathematics entrance exam questions?

Yes. They come from the national university entrance exam in Mathematics that students sat in 2009 E.C. (2017). Our set holds 61 of its questions, each with its answer and an explanation that points back to the textbook.

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