Everything covered in this lesson, in one place - useful for revision or printing.
\( \text{Nine topics} \;\rightarrow\; \text{one paper} \)
Your November paper covers the whole year. That sounds like a lot, but it is always built from the same nine strands, in roughly the same order, worth roughly the same marks.
This lesson walks through all nine, one at a time, with the method and the mistake to avoid for each:
\( 24 : 60 \;=\; 2 : 5 \)
To simplify a ratio, divide BOTH parts by their highest common factor.
The HCF of 24 and 60 is 12, so \(24 \div 12 = 2\) and \(60 \div 12 = 5\).
Keep the order the question gives you. \(2:5\) and \(5:2\) are different ratios.
\( 0{,}000\,62 \;=\; 6{,}2 \times 10^{-4} \)
Move the comma until it sits just after the first non-zero digit, then count how far it moved.
Here it moved 4 places to the right, so the exponent is \(-4\).
The rule: a number smaller than 1 always has a NEGATIVE exponent; a number bigger than 10 always has a positive one.
\( 850 \times 1{,}12 \;=\; R952{,}00 \)
A 12% increase on R850 can be done two ways.
Two steps: \(12\% \times 850 = R102\), then \(850 + 102 = R952\).
One step: multiply by \(1{,}12\).
A DECREASE of 12% would be \(\times 0{,}88\).
\( a^m \times a^n = a^{m+n} \qquad a^m \div a^n = a^{m-n} \)
Multiplying powers of the same base ADDS the exponents. Dividing SUBTRACTS them. Raising a power to a power MULTIPLIES them.
All three only work when the base is the same.
\( \left(2x^{2}y^{3}\right)^{3} = 8x^{6}y^{9} \)
Everything inside the bracket gets the outside exponent - including the number.
\(2^{3} = 8\), not 2. Leaving the 2 alone is the single most common exponent error in the November paper.
\( a^{-n} = \dfrac{1}{a^{n}} \qquad a^{0} = 1 \)
A negative exponent means reciprocal, not negative answer.
So \(3^{-1} = \dfrac{1}{3}\), and anything to the power 0 is 1.
If a question says leave your answer with a positive exponent, then \(3^{-1}\) is not yet the final answer.
\( 3x^{2} + 12x = 3x(x+4) \)
Always look for a common factor before anything else. Take out the BIGGEST one.
\(x(3x+12)\) is not wrong, but it is not fully factorised - the bracket still has a common factor of 3.
\( a^{2} - b^{2} = (a-b)(a+b) \)
Two perfect squares with a MINUS between them always factorise this way.
\(x^{2} - 49 = (x-7)(x+7)\), because \(49 = 7^{2}\).
Note there is no such rule for \(a^{2} + b^{2}\).
\( 5x^{2} - 20 = 5(x-2)(x+2) \)
This is the one they like to ask. Take out the 5 first:
\(5x^{2} - 20 = 5(x^{2} - 4)\)
Now the bracket is a difference of two squares, so keep going:
\(= 5(x-2)(x+2)\)
Stopping at \(5(x^{2}-4)\) loses the mark. Fully means keep factorising until you cannot.
\( 3x - 7 = 14 \;\Rightarrow\; x = 7 \)
Whatever you do to one side, do to the other.
Add 7 to both sides: \(3x = 21\). Divide both by 3: \(x = 7\).
Always check by substituting back: \(3(7) - 7 = 14\). Correct.
\( 2(x+4) = 3x - 1 \;\Rightarrow\; x = 9 \)
Expand the bracket FIRST, then collect.
\(2x + 8 = 3x - 1\)
\(8 + 1 = 3x - 2x\)
\(x = 9\)
Move the letters to the side where they stay positive and you will make fewer sign errors.
\( 2^{x} = 32 \;\Rightarrow\; x = 5 \)
Write both sides as a power of the SAME base, then equate the exponents.
\(32 = 2^{5}\), so \(2^{x} = 2^{5}\) and \(x = 5\).
Same for \(3^{x} = 81\): since \(81 = 3^{4}\), \(x = 4\).
\( 4\,;\ 7\,;\ 10\,;\ 13\,;\ \ldots \)
First check what is happening between terms. Here each term is 3 more than the one before - a CONSTANT DIFFERENCE.
So the next two terms are 16 and 19.
\( T_{n} = 3n + 1 \)
When the difference is constant, the rule is \(T_{n} = (\text{difference})n + c\).
Difference is 3, so \(T_{n} = 3n + c\). Use the first term: \(4 = 3(1) + c\), so \(c = 1\).
Always test it on a second term: \(3(2) + 1 = 7\). Correct.
\( 3n + 1 = 61 \;\Rightarrow\; n = 20 \)
Which term equals 61? means solve for \(n\).
\(3n + 1 = 61\), so \(3n = 60\) and \(n = 20\).
The answer is the 20th term, not 20. Read what is being asked for.
Watch for patterns that MULTIPLY instead: \(3;6;12;24\) doubles, so the next term is 48, not 27.
\( y = mx + c \)
\(m\) is the gradient (steepness) and \(c\) is the \(y\)-intercept (where it crosses the \(y\)-axis).
For \(y = 2x - 4\): gradient 2, \(y\)-intercept \(-4\).
A POSITIVE gradient rises left to right; a NEGATIVE gradient falls.
\( x\text{-int}: y = 0 \qquad y\text{-int}: x = 0 \)
This is the pair of substitutions to memorise.
For \(y = 2x - 4\), the \(x\)-intercept: let \(y = 0\), so \(0 = 2x - 4\) and \(x = 2\). The point is \((2\,;\,0)\).
The \(y\)-intercept: let \(x = 0\), giving \((0\,;\,-4)\).
\( (0\,;\,-4) \quad\text{and}\quad (2\,;\,0) \)
Two points are enough to draw a straight line, and the two intercepts are the easiest two to find.
Plot them, join with a RULER, and extend the line past both points.
Label the line with its equation. A freehand line loses marks even when the points are right.
\( \text{mean} = \dfrac{\text{sum}}{\text{how many}} \)
Mean is the average. Median is the middle value once the data is in ORDER. Mode is the most common value.
For \(10;14;18;18;20;22;24;26;28\): the sum is 180 and there are 9 values, so the mean is 20.
\( 10\,;14\,;18\,;18\,;\mathbf{20}\,;22\,;24\,;26\,;28 \)
Order the data first - always. With 9 values the median is the 5th, which is 20.
With an EVEN number of values there is no single middle, so you average the middle two.
The mode here is 18, the only value appearing twice.
\( P(E) = \dfrac{\text{favourable}}{\text{total}} \)
A bag holds 4 red, 6 blue and 5 green marbles.
Total \(= 15\), so \(P(\text{blue}) = \dfrac{6}{15} = \dfrac{2}{5} = 0{,}4\).
The denominator is the TOTAL, not the number of the other colours. Probability is always between 0 and 1.
\( \text{straight line} = 180^\circ \qquad \text{full turn} = 360^\circ \)
Angles on a straight line add to \(180^\circ\). Angles round a point add to \(360^\circ\). Angles in a triangle add to \(180^\circ\).
Vertically opposite angles are equal.
\( \text{corresponding} = \quad \text{alternate} = \quad \text{co-interior} + = 180^\circ \)
When a transversal cuts PARALLEL lines:
\( 2x + 3x + 4x = 180^\circ \;\Rightarrow\; x = 20^\circ \)
If the angles of a triangle are in the ratio \(2:3:4\), call them \(2x\), \(3x\) and \(4x\).
They add to \(180^\circ\), so \(9x = 180^\circ\) and \(x = 20^\circ\).
The LARGEST angle is \(4x = 80^\circ\). Stopping at \(x = 20^\circ\) answers a question that was not asked.
\( V = l \times b \times h \)
Perimeter is the distance round the outside - units cm.
Area is the surface covered - units \(\text{cm}^{2}\).
Volume is the space filled - units \(\text{cm}^{3}\).
A prism \(8 \times 5 \times 3\) has volume \(120\ \text{cm}^{3}\).
Getting the unit wrong loses a mark even when the number is right.
\( c^{2} = a^{2} + b^{2} \)
\(c\) is the HYPOTENUSE - the side opposite the right angle, and always the longest.
Legs 9 and 12: \(c^{2} = 81 + 144 = 225\), so \(c = \sqrt{225} = 15\) cm.
Do not forget the square root at the end. 225 is \(c^{2}\), not \(c\).
\( a^{2} = c^{2} - b^{2} \)
When the missing side is a LEG and not the hypotenuse, you SUBTRACT.
Hypotenuse 17, one leg 8: \(a^{2} = 289 - 64 = 225\), so \(a = 15\).
Decide first which side is the hypotenuse. Adding when you should subtract is the classic Pythagoras error.
\[ \begin{gathered} 6{,}2 \times 10^{-4} \qquad 5(x-2)(x+2) \ T_{n} = 3n + 1 \qquad y = mx + c \ c^{2} = a^{2} + b^{2} \end{gathered} \]
All nine strands, in one place.