Qormaata galuumsa Herregaa, bara 2016 A.L.I (2024)

The topic is the classification of natural numbers by how many divisors they have.

Count divisors:
A prime such as \(7\) has exactly two divisors, 1 and 7.
A composite such as \(4\) has the divisors 1, 2 and 4, which is three.
A composite such as \(6\) has 1, 2, 3 and 6, which is four.

So a composite number is one that has more divisors than a prime, that is three or more different divisors. That is option D.

The classic wrong turn is option C. Saying two or more factors still lets every prime in, because a prime has exactly two. Option A is the definition of a prime, and option B fails for \(121\), which is composite yet is divisible by none of the numbers 2 to 9.

Textbook: Mathematics Grade 9, Unit 1 The Number System, section 1.1 (page 18).

Rationalizing the denominator means rewriting a fraction so that no radical is left underneath.

You multiply the numerator and the denominator by the same well chosen expression, so the value of the fraction never changes:
\( \frac{1}{\sqrt{5}} = \frac{1}{\sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}} = \frac{\sqrt{5}}{5} \)
With a two term denominator you use the conjugate:
\( \frac{1}{3-\sqrt{2}} = \frac{3+\sqrt{2}}{(3-\sqrt{2})(3+\sqrt{2})} = \frac{3+\sqrt{2}}{7} \)

In both cases the denominator ends up rational, so option C states the purpose correctly.

The classic wrong turn is option D. The multiplier is not always irrational, and multiplying the whole number by it would change the value. You multiply by a fraction equal to 1.

Textbook: Mathematics Grade 9, Unit 1 The Number System, section 1.3 (page 39).

The division algorithm says that for a dividend \(a\) and a divisor \(b>0\) there are unique numbers \(q\) and \(r\) with \(a = qb + r\) and \(0 \le r < b\). The remainder must be non negative and smaller than the divisor.

Divide 142 by 13:
\(13 \times 10 = 130\)
\(142 - 130 = 12\)
\(0 \le 12 < 13\), so the remainder is allowed.

Therefore \(142 = 10(13) + 12\), which is option A.

The classic wrong turn is option B. It is true as arithmetic, since \(11 \times 13 - 1 = 142\), but \(-1\) is not a legal remainder, so it is not the division algorithm form.

Textbook: Mathematics Grade 9, Unit 1 The Number System, section 1.1 (page 18).

This is a system of two linear equations in two variables. Put both equations in the form \(y = mx + c\) and compare them.

First equation:
\(2x - y = 5\)
\(y = 2x - 5\)

Second equation:
\(3y + 15 = 6x\)
\(3y = 6x - 15\)
\(y = 2x - 5\)

The two equations describe the same line, so the system is dependent and has infinitely many solutions. Write them with a parameter: let \(x = t\), then \(y = 2t - 5\).

The solution set is \(\{(t,\ 2t-5) : t \in \mathbb{R}\}\), which is option C.

The classic wrong turn is option A. Students see two equations, subtract, get \(0 = 0\) and read that as no solution. \(0 = 0\) means every point of the line works, not that the set is empty.

Textbook: Mathematics Grade 9, Unit 2 Solving Equations, section 2.2 (page 91).

Two plane figures are similar when they have the same shape but not necessarily the same size. Formally, corresponding angles are congruent and corresponding sides are in the same ratio.

Saying one figure is an enlargement (or a reduction) of the other is the same statement in everyday words: every length is multiplied by one scale factor \(k\), and the angles stay as they are. So option B is the definition.

Why the others fail:
Option A: angles are equal, not in a ratio.
Option C: the ratios of corresponding sides must be equal, not different.
Option D: same size with different shape is not similarity. Figures with the same shape and the same size are congruent, which is the special case \(k = 1\).

Textbook: Mathematics Grade 9, Unit 6 Congruency and Similarity, section 6.2 (page 214).

In a right angled triangle the sine of an acute angle is the opposite side over the hypotenuse.

Here \(A = 30^\circ\) and the side opposite \(A\) is 11 cm. Let the hypotenuse be \(c\).

\( \sin 30^\circ = \frac{11}{c} \)
\( \frac{1}{2} = \frac{11}{c} \)
\( c = 2 \times 11 = 22 \)

The hypotenuse is 22 cm, which is option A. A quick check: in a 30, 60, 90 triangle the side facing \(30^\circ\) is always half the hypotenuse.

The classic wrong turn is using \(\cos 30^\circ\) or \(\tan 30^\circ\), which gives the answers with \(\sqrt{3}\) in them. Those belong to the other two sides, not to the pair given here.

Textbook: Mathematics Grade 9, Unit 4 Introduction to Trigonometry, section 4.2 (page 167).