So the expression becomes \(\dfrac{2+4\sqrt{3}}{\sqrt{22}}\).
Rationalise the denominator by multiplying top and bottom by \(\sqrt{22}\): \(\dfrac{(2+4\sqrt{3})\sqrt{22}}{22} = \dfrac{2\sqrt{22}+4\sqrt{66}}{22}\).
Divide numerator and denominator by the common factor \(2\): \(\dfrac{\sqrt{22}+2\sqrt{66}}{11}\).
Answer: \(\dfrac{\sqrt{22}+2\sqrt{66}}{11}\)
2.3 marks(a)
Rationalise the denominator of \(\frac{1}{\sqrt{3}}\)
(b)
Expand \((2+\sqrt{3})(1+\sqrt{3})\)
Give your answer in the form \(a+b\sqrt{3}\) where \(a\) and \(b\) are integers.
Worked solution
(a)
Multiply numerator and denominator by \(\sqrt{3}\) to remove the surd from the denominator: \(\dfrac{1}{\sqrt{3}} \times \dfrac{\sqrt{3}}{\sqrt{3}} = \dfrac{\sqrt{3}}{3}\).
Rationalise the denominator of \(\frac{1}{\sqrt{7}}\)
(b) (i)
Expand and simplify \((\sqrt{3}+\sqrt{15})^{2}\)
Give your answer in the form \(a+b\sqrt{3}\) where \(a\) and \(b\) are integers.
(ii)
All measurements on the triangle are in centimetres. ABC is a right-angled triangle. \(k\) is a positive integer.
Find the value of \(k\).
Worked solution
(a)
Multiply numerator and denominator by \(\sqrt{7}\): \(\dfrac{1}{\sqrt{7}} \times \dfrac{\sqrt{7}}{\sqrt{7}} = \dfrac{\sqrt{7}}{7}\).
Answer: \(\dfrac{\sqrt{7}}{7}\)
(b)(i)
Note: the question states the answer should be in the form \(a+b\sqrt{3}\), but this is a source typo — the correct form is \(a+b\sqrt{5}\). The mathematically correct expansion is given below.
Use \((x+y)^2 = x^2 + 2xy + y^2\) with \(x=\sqrt{3}\) and \(y=\sqrt{15}\): \((\sqrt{3}+\sqrt{15})^2 = (\sqrt{3})^2 + 2\sqrt{3}\cdot\sqrt{15} + (\sqrt{15})^2\).
Simplify the squares: \((\sqrt{3})^2 = 3\) and \((\sqrt{15})^2 = 15\).
Write \(\sqrt{45}\) in the form \(k\sqrt{5}\), where \(k\) is an integer.
Worked solution
(a)
Recall that \(x^{\frac{1}{2}} = \sqrt{x}\). So \(49^{\frac{1}{2}} = \sqrt{49} = 7\).
Answer: \(7\)
(b)
Find a factor of \(45\) that is a perfect square: \(45 = 9 \times 5\).
Use \(\sqrt{ab} = \sqrt{a}\sqrt{b}\): \(\sqrt{45} = \sqrt{9}\cdot\sqrt{5} = 3\sqrt{5}\).
So \(k = 3\).
Answer: \(3\sqrt{5}\)
6.2 marks
Write \(\frac{\sqrt{18}+10}{\sqrt{2}}\) in the form \(a+b\sqrt{2}\) where \(a\) and \(b\) are integers.
Worked solution
Split the fraction: \(\dfrac{\sqrt{18}+10}{\sqrt{2}} = \dfrac{\sqrt{18}}{\sqrt{2}} + \dfrac{10}{\sqrt{2}}\).
Simplify the first term using \(\dfrac{\sqrt{a}}{\sqrt{b}} = \sqrt{\dfrac{a}{b}}\): \(\dfrac{\sqrt{18}}{\sqrt{2}} = \sqrt{\dfrac{18}{2}} = \sqrt{9} = 3\).
Rationalise the second term: \(\dfrac{10}{\sqrt{2}} \times \dfrac{\sqrt{2}}{\sqrt{2}} = \dfrac{10\sqrt{2}}{2} = 5\sqrt{2}\).
Add the results: \(3 + 5\sqrt{2}\).
So \(a = 3\) and \(b = 5\).
Answer: \(3 + 5\sqrt{2}\)
7.3 marks
Expand and simplify \((2+\sqrt{3})(7-\sqrt{3})\)
Give your answer in the form \(a+b\sqrt{3}\) where \(a\) and \(b\) are integers.
Rationalise the denominator of \(\frac{(4+\sqrt{2})(4-\sqrt{2})}{\sqrt{7}}\)
Give your answer in its simplest form.
Worked solution
Expand the numerator using the difference of two squares \((x+y)(x-y) = x^2 - y^2\): \((4+\sqrt{2})(4-\sqrt{2}) = 4^2 - (\sqrt{2})^2 = 16 - 2 = 14\).
So the expression is \(\dfrac{14}{\sqrt{7}}\).
Rationalise the denominator by multiplying top and bottom by \(\sqrt{7}\): \(\dfrac{14}{\sqrt{7}} \times \dfrac{\sqrt{7}}{\sqrt{7}} = \dfrac{14\sqrt{7}}{7}\).
Simplify: \(\dfrac{14\sqrt{7}}{7} = 2\sqrt{7}\).
Answer: \(2\sqrt{7}\)
9.2 marks
Show that \(\frac{(4-\sqrt{3})(4+\sqrt{3})}{\sqrt{13}}\) simplifies to \(\sqrt{13}\)
Worked solution
Expand the numerator using the difference of two squares \((x-y)(x+y) = x^2 - y^2\): \((4-\sqrt{3})(4+\sqrt{3}) = 4^2 - (\sqrt{3})^2 = 16 - 3 = 13\).
So the expression becomes \(\dfrac{13}{\sqrt{13}}\).
Rationalise by multiplying top and bottom by \(\sqrt{13}\): \(\dfrac{13}{\sqrt{13}} \times \dfrac{\sqrt{13}}{\sqrt{13}} = \dfrac{13\sqrt{13}}{13}\).
Cancel the \(13\): \(= \sqrt{13}\), as required.
Algebraic Proofs
8 questions
Things to remember
Start by expanding the brackets, then factorise.
Remember the following:
2n → even number
2n + 1 → odd number
a(bn + c) → multiple of a
Consecutive numbers are numbers that appear one after the other.
1.6 marks(a)
Expand and simplify \(x(x+1)(x-1)\)
In a list of three consecutive positive integers at least one of the numbers is even and one of the numbers is a multiple of 3.
\(n\) is a positive integer greater than 1.
(b)
Prove that \(n^3 - n\) is a multiple of 6 for all possible values of \(n\).
\(2^n - 1\) is a prime number.
(c)
Explain why \(2^n + 1\) is a multiple of 3.
Worked solution
(a)
First multiply the last two brackets (difference of two squares): \((x+1)(x-1) = x^2 - 1\).
Now multiply by \(x\): \(x(x^2 - 1) = x^3 - x\).
Answer: \(x^3 - x\)
(b)
From part (a), \(n^3 - n = n(n+1)(n-1) = (n-1)\,n\,(n+1)\).
So \(n^3 - n\) is the product of three consecutive integers.
Among any three consecutive integers, at least one is even (so the product is a multiple of 2) and exactly one is a multiple of 3 (so the product is a multiple of 3).
Since the product is a multiple of both 2 and 3, it is a multiple of \(2 \times 3 = 6\).
Therefore \(n^3 - n\) is a multiple of 6 for all positive integers \(n > 1\), as required.
(c)
We are told \(2^n - 1\) is prime. Note that \(n\) must be greater than 1 (since \(2^1 - 1 = 1\) is not prime), so \(2^n - 1 \geq 3\) and in particular \(2^n - 1\) is odd and not equal to 3 only for some \(n\); what matters is that \(2^n - 1\) is not divisible by 3 (otherwise it couldn't be prime, unless it equalled 3 — and if \(2^n - 1 = 3\) then \(2^n + 1 = 5\), but we'll handle the general case).
Consider three consecutive integers \(2^n - 1,\; 2^n,\; 2^n + 1\). Exactly one of any three consecutive integers is a multiple of 3.
\(2^n\) is never a multiple of 3 (its only prime factor is 2).
\(2^n - 1\) is given to be prime, and since \(2^n - 1 > 3\) for \(n \geq 3\) it cannot be a multiple of 3. (For \(n = 2\), \(2^n - 1 = 3\) and \(2^n + 1 = 5\) — but then \(2^n + 1\) is not a multiple of 3, so the premise \(2^n - 1\) prime forces \(n \geq 3\) when we want \(2^n+1\) to be a multiple of 3… in fact for \(n=2\) the claim fails, so the standard exam interpretation takes \(n\) such that \(2^n-1\) is an odd prime greater than 3.)
Therefore the multiple of 3 among the three consecutive numbers must be \(2^n + 1\).
Hence \(2^n + 1\) is a multiple of 3, as required.
2.3 marks
Prove that \((2n+3)^2 - (2n-3)^2\) is a multiple of 8 for all positive integer values of \(n\).
Worked solution
Expand each square: \((2n+3)^2 = 4n^2 + 12n + 9\) and \((2n-3)^2 = 4n^2 - 12n + 9\).
The sum of the two integers is \(n + (n+1) = 2n + 1\).
Both expressions equal \(2n+1\), so the difference between the squares of any two consecutive integers equals the sum of those two integers, as required.
5.6 marks(a)
Factorise \(x^2 + 7x\)
(b)
Factorise \(y^2 - 10y + 16\)
(c) (i)
Factorise \(2t^2 + 5t + 2\)
(ii)
\(t\) is a positive whole number.
The expression \(2t^2 + 5t + 2\) can never have a value that is a prime number. Explain why.
Worked solution
(a)
Take out the common factor \(x\): \(x^2 + 7x = x(x + 7)\).
Answer: \(x(x+7)\)
(b)
Look for two numbers that multiply to \(+16\) and add to \(-10\): these are \(-2\) and \(-8\).
So \(y^2 - 10y + 16 = (y - 2)(y - 8)\).
Answer: \((y-2)(y-8)\)
(c)(i)
Split the middle term: find two numbers that multiply to \(2 \times 2 = 4\) and add to \(5\); these are \(4\) and \(1\).
The product of two consecutive positive integers is added to the larger of the two integers. Prove that the result is always a square number.
Worked solution
Let the two consecutive positive integers be \(n\) and \(n+1\), so the larger is \(n+1\).
Their product is \(n(n+1) = n^2 + n\).
Adding the larger integer: \(n^2 + n + (n + 1) = n^2 + 2n + 1\).
Factorise: \(n^2 + 2n + 1 = (n+1)^2\).
Since \(n+1\) is a positive integer, \((n+1)^2\) is a square number, as required.
Transformations of graphs
4 questions
Things to remember
f(x) means the function of x.
-f(x) is a reflection in the x-axis.
f(-x) is a reflection in the y-axis.
f(x − a) is a translation in the x-axis, a units.
f(x) + b is a translation in the y-axis, b units.
cf(x) is an enlargement in the y-axis, scale factor c.
f(dx) is an enlargement in the x-axis, scale factor 1/d.
1.3 marks
\(y = f(x)\)
The graph of \(y = f(x)\) is shown on the grid.
(a)
On the grid above, sketch the graph of \(y = -f(x)\).
The graph of \(y = f(x)\) is shown on the grid.
The graph G is a translation of the graph of \(y = f(x)\).
(b)
Write down the equation of graph G.
Worked solution
(a)
\(y = -f(x)\) is a reflection of \(y = f(x)\) in the \(x\)-axis, so every point \((x, y)\) on the original curve maps to \((x, -y)\).
The minimum point \((-2, -2)\) maps to the maximum point \((-2, 2)\).
The point \((-4, 2)\) maps to \((-4, -2)\).
The point \((0, 2)\) maps to \((0, -2)\).
Sketch the resulting inverted U-shaped curve through these image points.
Answer: The graph of \(y = -f(x)\) is the original curve reflected in the \(x\)-axis: an inverted U-shape with maximum at \((-2, 2)\), passing through \((-4, -2)\) and \((0, -2)\).
(b)
Compare a key point on \(y = f(x)\) with the corresponding point on graph \(G\). The minimum of \(y = f(x)\) is at \((-2, -2)\), and the minimum of graph \(G\) is at \((4, -2)\).
The \(y\)-coordinate is unchanged and the \(x\)-coordinate has increased by \(4 - (-2) = 6\).
So graph \(G\) is a translation of \(y = f(x)\) by \(6\) units in the positive \(x\)-direction.
A translation of \(a\) units in the \(+x\) direction has equation \(y = f(x - a)\), so here \(a = 6\).
Answer: \(y = f(x - 6)\)
2.2 marks
The graph of \(y = f(x)\) is shown on both grids below.
(a)
On the grid above, sketch the graph of \(y = f(-x)\).
(b)
On this grid, sketch the graph of \(y = -f(x) + 3\).
Worked solution
(a)
\(y = f(-x)\) is a reflection of \(y = f(x)\) in the \(y\)-axis, so every point \((x, y)\) maps to \((-x, y)\).
The minimum point \((-2, -2)\) maps to \((2, -2)\).
The point \((-4, 2)\) maps to \((4, 2)\).
The point \((0, 2)\) is on the \(y\)-axis and maps to itself at \((0, 2)\).
Sketch the U-shaped curve through these image points.
Answer: The graph of \(y = f(-x)\) is the original curve reflected in the \(y\)-axis: a U-shape with minimum at \((2, -2)\), passing through \((0, 2)\) and \((4, 2)\).
(b)
\(y = -f(x) + 3\) applies two transformations to \(y = f(x)\): first reflect in the \(x\)-axis (giving \(-f(x)\)), then translate \(3\) units in the positive \(y\)-direction.
A point \((x, y)\) on the original curve maps to \((x, -y + 3)\).
The minimum \((-2, -2)\) maps to \((-2, -(-2) + 3) = (-2, 5)\); this becomes a maximum.
The point \((-4, 2)\) maps to \((-4, -2 + 3) = (-4, 1)\).
The point \((0, 2)\) maps to \((0, -2 + 3) = (0, 1)\).
Sketch the inverted U-shaped curve through these image points.
Answer: The graph of \(y = -f(x) + 3\) is an inverted U-shape with maximum at \((-2, 5)\), passing through \((-4, 1)\) and \((0, 1)\).
3.4 marks
The graph of \(y = f(x)\) is shown on each of the grids.
(a)
On this grid, sketch the graph of \(y = f(x - 3)\).
(b)
On this grid, sketch the graph of \(y = 2f(x)\).
Worked solution
(a)
\(y = f(x - 3)\) is a translation of \(y = f(x)\) by \(3\) units in the positive \(x\)-direction, so every point \((x, y)\) maps to \((x + 3, y)\).
The minimum point \((0, -1)\) maps to \((3, -1)\).
The point \((-2, 3)\) maps to \((1, 3)\).
The point \((2, 3)\) maps to \((5, 3)\).
Sketch the U-shaped curve through these image points.
Answer: The graph of \(y = f(x - 3)\) is the original curve translated \(3\) units to the right: a U-shape with minimum at \((3, -1)\), passing through \((1, 3)\) and \((5, 3)\).
(b)
\(y = 2f(x)\) is a stretch in the \(y\)-direction with scale factor \(2\), so every point \((x, y)\) maps to \((x, 2y)\). \(x\)-coordinates are unchanged and \(y\)-coordinates are doubled.
The minimum point \((0, -1)\) maps to \((0, -2)\).
The point \((-2, 3)\) maps to \((-2, 6)\).
The point \((2, 3)\) maps to \((2, 6)\).
Sketch the steeper U-shaped curve through these image points.
Answer: The graph of \(y = 2f(x)\) is the original curve stretched vertically by scale factor \(2\): a U-shape with minimum at \((0, -2)\), passing through \((-2, 6)\) and \((2, 6)\).
4.3 marks
The graph of \(y = f(x)\) is shown on the grid.
(a)
On the grid above, sketch the graph of \(y = f(x + 3)\).
The graph of \(y = g(x)\) is shown below.
The graph G is the reflection of \(y = g(x)\) in the x-axis.
(b)
Write down an equation of graph G.
Worked solution
(a)
\(y = f(x + 3)\) is a translation of \(y = f(x)\) by \(3\) units in the negative \(x\)-direction, so every point \((x, y)\) maps to \((x - 3, y)\).
The minimum point \((-2, -2)\) maps to \((-5, -2)\).
The point \((-4, 2)\) maps to \((-7, 2)\).
The point \((0, 2)\) maps to \((-3, 2)\).
Sketch the U-shaped curve through these image points.
Answer: The graph of \(y = f(x + 3)\) is the original curve translated \(3\) units to the left: a U-shape with minimum at \((-5, -2)\), passing through \((-7, 2)\) and \((-3, 2)\).
(b)
A reflection in the \(x\)-axis sends every point \((x, y)\) to \((x, -y)\).
Applied to \(y = g(x)\), this gives \(y = -g(x)\).
This matches graph \(G\), which is shown as the downward-opening reflection of \(y = g(x)\) across the \(x\)-axis.
Answer: \(y = -g(x)\)
Equations of Circles
3 questions
Things to remember
The general equation of a circle is (x − a)² + (y − b)² = r², where (a, b) is the centre and r is the radius.
To calculate the equation of the tangent:
1. Calculate the gradient of the radius of the circle.
2. Calculate the gradient of the tangent of the circle.
3. Substitute the given coordinate and the gradient of the tangent into y = mx + c to calculate the y-intercept.
1.6 marks
The circle \(C\) has radius \(5\) and touches the \(y\)-axis at the point \((0,\ 9)\), as shown in the diagram.
(a)
Write down an equation for the circle \(C\), that is shown in the diagram.
A line through the point \(P(8,\ -7)\) is a tangent to the circle \(C\) at the point \(T\).
(b)
Find the length of \(PT\).
Worked solution
(a)
The circle touches the \(y\)-axis at \((0,\ 9)\), so the radius to that point is horizontal and the centre lies \(5\) units to the right of the tangent point (as shown in the diagram).
Therefore the centre is \((a,\ b) = (5,\ 9)\) and \(r = 5\).
Using the standard form \((x - a)^2 + (y - b)^2 = r^2\):
\((x - 5)^2 + (y - 9)^2 = 25\).
Answer: \((x - 5)^2 + (y - 9)^2 = 25\)
(b)
\(PT\) is a tangent to the circle at \(T\), so the radius \(CT\) is perpendicular to \(PT\). Triangle \(CTP\) is right-angled at \(T\), with hypotenuse \(CP\) and legs \(CT = r = 5\) and \(PT\).
Calculate the distance from the centre \(C(5,\ 9)\) to \(P(8,\ -7)\):
2. Put in brackets with the x and square the brackets.
3. Subtract the half-coefficient squared.
4. Don't forget the constant on the end!
5. Simpify.
For (x − p)² + q = 0, the turning point is (p, q).
1.6 marks(i)
Sketch the graph of \(f(x) = x^2 - 5x + 10\), showing the coordinates of the turning point and the coordinates of any intercepts with the coordinate axes.
(ii)
Hence, or otherwise, determine whether \(f(x+2) - 3 = 0\) has any real roots. Give reasons for your answer.
Worked solution
(i)
Complete the square for \(f(x) = x^2 - 5x + 10\). Half the coefficient of \(x\) is \(-\tfrac{5}{2}\).
So \(f(x) = \left(x - \tfrac{5}{2}\right)^2 + \tfrac{15}{4}\), giving the turning point \(\left(\tfrac{5}{2},\, \tfrac{15}{4}\right)\) (a minimum since the coefficient of \(x^2\) is positive).
y-intercept: \(f(0) = 10\), so the curve crosses the y-axis at \((0, 10)\).
x-intercepts: the minimum value of \(f\) is \(\tfrac{15}{4} > 0\), so the curve lies entirely above the x-axis and there are no x-intercepts. (Equivalently, the discriminant is \((-5)^2 - 4(1)(10) = -15 < 0\).)
Sketch: an upward-opening parabola with minimum at \(\left(\tfrac{5}{2},\, \tfrac{15}{4}\right)\), y-intercept \((0, 10)\), and no x-intercepts.
Answer: Turning point \(\left(\tfrac{5}{2},\, \tfrac{15}{4}\right)\); y-intercept \((0, 10)\); no x-intercepts.
(ii)
The equation \(f(x+2) - 3 = 0\) is equivalent to \(f(x+2) = 3\).
The graph of \(y = f(x+2)\) is the graph of \(y = f(x)\) translated \(2\) units to the left, so it has the same minimum value as \(f\).
From part (i) the minimum value of \(f\) is \(\tfrac{15}{4} = 3.75\), so \(f(x+2) \geq \tfrac{15}{4}\) for all \(x\).
Since \(\tfrac{15}{4} > 3\), the equation \(f(x+2) = 3\) has no solutions.
Answer: No real roots, because the minimum value of \(f(x+2)\) is \(\tfrac{15}{4}\), which is greater than \(3\).
2.4 marks(a)
Write \(2x^2 + 16x + 35\) in the form \(a(x+b)^2 + c\) where \(a\), \(b\), and \(c\) are integers.
(b)
Hence, or otherwise, write down the coordinates of the turning point of the graph of \(y = 2x^2 + 16x + 35\).
Worked solution
(a)
Factor \(2\) from the \(x^2\) and \(x\) terms: \(2x^2 + 16x + 35 = 2(x^2 + 8x) + 35\).
Complete the square inside the brackets. Half the coefficient of \(x\) is \(4\), so \(x^2 + 8x = (x+4)^2 - 16\).
Answer: \(x = 2.87,\ y = -0.87\) or \(x = -0.87,\ y = 2.87\)
Solving Quadratic Inequalities
6 questions
Things to remember
Start by solving the quadratic to find the values of x, then sketch the graph to determine the inequality.
1.3 marks
Solve \(x^2 > 3x + 4\)
Worked solution
Rearrange to get 0 on one side: \(x^2 - 3x - 4 > 0\).
Factorise the corresponding quadratic: \(x^2 - 3x - 4 = (x - 4)(x + 1)\), so the roots are \(x = -1\) and \(x = 4\).
The parabola \(y = x^2 - 3x - 4\) opens upward (coefficient of \(x^2\) is positive) and crosses the \(x\)-axis at \(-1\) and \(4\). It lies above the \(x\)-axis (i.e. \(y > 0\)) outside the roots.
The inequality is already in the form \(x^2 - 2x + 8 < 0\).
Check the discriminant of \(x^2 - 2x + 8 = 0\): \(b^2 - 4ac = (-2)^2 - 4(1)(8) = 4 - 32 = -28 < 0\), so there are no real roots.
Complete the square: \(x^2 - 2x + 8 = (x - 1)^2 + 7\), which is always \(\ge 7 > 0\) for every real \(x\).
The parabola opens upward and lies entirely above the \(x\)-axis, so \(x^2 - 2x + 8 < 0\) has no solutions.
Answer: No solutions (there is no real value of \(x\) for which \(x^2 - 2x + 8 < 0\))
5.3 marks
Solve the inequality \(x^2 - x \geq 12\)
Worked solution
Rearrange: \(x^2 - x - 12 \ge 0\).
Factorise: \(x^2 - x - 12 = (x - 4)(x + 3)\), so the roots are \(x = -3\) and \(x = 4\).
The parabola \(y = x^2 - x - 12\) opens upward and crosses the \(x\)-axis at \(-3\) and \(4\), so \(y \ge 0\) outside the roots (inclusive, because the inequality is non-strict).
The angle at the centre is twice the angle at the circumference.
The angle in a semi-circle is 90°.
Angles subtended by the same arc are equal.
Opposite angles in a cyclic quadrilateral sum to 180°.
Tangents from a point are equal.
A tangent is perpendicular to a radius.
Angles in alternate segments are equal.
1.2 marks
Diagram NOT accurately drawn
\(P\) is a point on the circumference of the circle, centre \(O\).
\(PQ\) is a tangent to the circle.
(i)
Write down the size of angle \(OPQ\).
(ii)
Give a reason for your answer.
Worked solution
\(PQ\) is a tangent to the circle at \(P\) and \(OP\) is a radius.
A tangent to a circle is perpendicular to the radius at the point of tangency, so \(OP \perp PQ\).
Answer: Angle \(OPQ = 90°\). Reason: a tangent is perpendicular to the radius at the point of tangency.
2.4 marks(a)
Diagram NOT accurately drawn
\(A\), \(B\) and \(C\) are points on the circumference of a circle, centre \(O\).
\(AC\) is a diameter of the circle.
(i)
Write down the size of angle \(ABC\).
(ii)
Give a reason for your answer.
(b)
Diagram NOT accurately drawn
\(D\), \(E\) and \(F\) are points on the circumference of a circle, centre \(O\).
Angle \(DOF = 130°\).
(i)
Work out the size of angle \(DEF\).
(ii)
Give a reason for your answer.
Worked solution
(a)
\(AC\) is a diameter of the circle, so triangle \(ABC\) is drawn in a semi-circle with the diameter as one side.
By the theorem 'the angle in a semi-circle is \(90°\)', the angle at \(B\) subtended by the diameter \(AC\) is a right angle.
Answer: Angle \(ABC = 90°\). Reason: the angle in a semi-circle is \(90°\).
(b)
Angle \(DOF = 130°\) is the angle at the centre subtended by arc \(DF\).
Angle \(DEF\) is the angle at the circumference subtended by the same arc \(DF\).
By the theorem 'the angle at the centre is twice the angle at the circumference' (subtended by the same arc), angle \(DEF = \tfrac{1}{2} \times 130° = 65°\).
Answer: Angle \(DEF = 65°\). Reason: the angle at the centre is twice the angle at the circumference (subtended by the same arc).
3.2 marks
Diagram NOT accurately drawn
\(A\) and \(B\) are points on the circumference of a circle, centre \(O\).
\(PA\) and \(PB\) are tangents to the circle.
Angle \(APB\) is \(86°\).
Work out the size of the angle marked \(x\).
Worked solution
\(PA\) and \(PB\) are tangents to the circle, and \(OA\) and \(OB\) are radii to the points of tangency.
A tangent is perpendicular to the radius at the point of tangency, so angle \(OAP = 90°\) and angle \(OBP = 90°\).
The four vertices \(O\), \(A\), \(P\), \(B\) form a quadrilateral \(OAPB\), and the angles in any quadrilateral sum to \(360°\).
In the diagram, \(A\), \(B\), \(C\) and \(D\) are points on the circumference of a circle, centre \(O\).
Angle \(BAD = 70°\).
Angle \(BCD = x°\).
Angle \(BCD = y°\) ⚠
(a) (i)
Work the value of \(x\).
(ii)
Give a reason for your answer.
(b) (i)
Work the value of \(y\).
(ii)
Give a reason for your answer.
Worked solution
(a)
Angle \(BAD = 70°\) is the angle at the circumference subtended by arc \(BD\) (the arc not containing \(A\)).
Angle \(BOD = x°\) is the angle at the centre subtended by the same arc \(BD\).
By the theorem 'the angle at the centre is twice the angle at the circumference', \(x = 2 \times 70 = 140\).
Answer: \(x = 140\). Reason: the angle at the centre is twice the angle at the circumference (subtended by the same arc).
(b)
\(ABCD\) is a cyclic quadrilateral because all four vertices lie on the circumference of the circle.
Opposite angles in a cyclic quadrilateral sum to \(180°\), so angle \(BAD + \text{angle } BCD = 180°\).
Therefore \(y = 180 - 70 = 110\).
Answer: \(y = 110\). Reason: opposite angles in a cyclic quadrilateral sum to \(180°\).
5.5 marks
Diagram NOT accurately drawn
The diagram shows a circle centre \(O\).
\(A\), \(B\) and \(C\) are points on the circumference.
\(DCO\) is a straight line.
\(DA\) is a tangent to the circle.
Angle \(ADO = 36°\)
(a)
Work out the size of angle \(AOD\).
(b) (i)
Work out the size of angle \(ABC\).
(ii)
Give a reason for your answer.
Worked solution
(a)
\(DA\) is a tangent to the circle at \(A\), and \(OA\) is a radius.
A tangent is perpendicular to the radius at the point of tangency, so angle \(OAD = 90°\).
In triangle \(OAD\) the angles sum to \(180°\), so angle \(AOD = 180° - 90° - 36° = 54°\).
Answer: Angle \(AOD = 54°\).
(b)
\(DCO\) is a straight line, so \(C\) lies on segment \(OD\). This means angle \(AOC\) (the angle at the centre between radii \(OA\) and \(OC\)) is the same as angle \(AOD = 54°\).
Angle \(AOC = 54°\) is the angle at the centre subtended by arc \(AC\) (the minor arc not containing \(B\)).
Angle \(ABC\) is the angle at the circumference subtended by the same arc \(AC\).
By the theorem 'the angle at the centre is twice the angle at the circumference', angle \(ABC = \tfrac{1}{2} \times 54° = 27°\).
Answer: Angle \(ABC = 27°\). Reason: the angle at the centre is twice the angle at the circumference (subtended by the same arc).
Vectors
8 questions
Things to remember
Use the letter provided in the question.
Going against the arrow is a negative.
Vectors need to be written in bold or underlined.
They can be manipulated similarly to algebra.
1.7 marks
The diagram shows a regular hexagon ABCDEF with centre O.
In a regular hexagon \(ABCDEF\) with centre \(O\), opposite sides are equal and parallel, and \(EF\) is the side opposite to \(AB\) pointing in the opposite direction; however \(EF\) (traversed \(E \to F\)) is parallel to \(OA\) and equal in magnitude to \(OA\).
Equivalently, by symmetry \(\overrightarrow{OE} = -\overrightarrow{OB} = -6\mathbf{b}\) and \(\overrightarrow{OF} = \overrightarrow{OB} - \overrightarrow{OA} - \overrightarrow{OB} \dots\) — use coordinates: placing \(A\) and \(B\) at consecutive vertices, \(\overrightarrow{EF} = \overrightarrow{OA}\).
Therefore \(\overrightarrow{EF} = 6\mathbf{a}\).
Answer: \(\overrightarrow{EF} = 6\mathbf{a}\)
(b)
First find \(\overrightarrow{OC}\). In a regular hexagon, \(\overrightarrow{OC} = \overrightarrow{AB} = 6\mathbf{b} - 6\mathbf{a}\) (side \(AB\) is parallel and equal to the radius \(OC\)).
\(X\) is the midpoint of \(BC\), so \(\overrightarrow{OX} = \tfrac{1}{2}(\overrightarrow{OB} + \overrightarrow{OC}) = \tfrac{1}{2}(6\mathbf{b} + 6\mathbf{b} - 6\mathbf{a}) = 6\mathbf{b} - 3\mathbf{a}\).
By symmetry of the hexagon, \(\overrightarrow{OE} = -\overrightarrow{OB} = -6\mathbf{b}\).
Route from \(E\) to \(X\) via \(O\): \(\overrightarrow{EX} = \overrightarrow{EO} + \overrightarrow{OX} = 6\mathbf{b} + (6\mathbf{b} - 3\mathbf{a}) = 12\mathbf{b} - 3\mathbf{a}\).
So \(\overrightarrow{EY} = \tfrac{4}{3}\overrightarrow{EX}\): the vectors are parallel and share the common point \(E\), hence \(E\), \(X\) and \(Y\) lie on the same straight line.
2.3 marks
T is the point on PQ for which PT : TQ = 2 : 1.
\(\overrightarrow{OP} = \mathbf{a}\) and \(\overrightarrow{OQ} = \mathbf{b}\).
OPQ is a triangle.
(a)
Write down, in terms of a and b, an expression for \(\overrightarrow{PQ}\).
(b)
Express \(\overrightarrow{OT}\) in terms of a and b. Give your answer in its simplest form.
Worked solution
(a)
Route from \(P\) to \(Q\) via \(O\): \(\overrightarrow{PQ} = \overrightarrow{PO} + \overrightarrow{OQ} = -\mathbf{a} + \mathbf{b}\).
So \(\overrightarrow{OM}\) is a scalar multiple of \(\overrightarrow{OP}\); they are parallel and share the common point \(O\), hence \(O\), \(P\) and \(M\) are collinear (i.e. \(OPM\) is a straight line).
4.5 marks
OPQ is a triangle.
R is the midpoint of OP.
S is the midpoint of PQ.
\(\overrightarrow{OP} = \mathbf{p}\) and \(\overrightarrow{OQ} = \mathbf{q}\)
(i)
Find \(\overrightarrow{OS}\) in terms of p and q.
(ii)
Show that RS is parallel to OQ.
Worked solution
(i)
\(S\) is the midpoint of \(PQ\), so \(\overrightarrow{OS} = \tfrac{1}{2}(\overrightarrow{OP} + \overrightarrow{OQ}) = \tfrac{1}{2}(\mathbf{p} + \mathbf{q})\).
M is the midpoint of PQ and N is the midpoint of OR.
(a)
Find the vector \(\overrightarrow{MN}\) in terms of a and b.
X is the midpoint of MN and Y is the midpoint of QR.
(b)
Prove that XY is parallel to OR.
Worked solution
(a)
\(M\) is the midpoint of \(PQ\), so \(\overrightarrow{OM} = \overrightarrow{OP} + \tfrac{1}{2}\overrightarrow{PQ} = 2\mathbf{b} + \tfrac{1}{2}(2\mathbf{a}) = 2\mathbf{b} + \mathbf{a}\).
\(N\) is the midpoint of \(OR\), so \(\overrightarrow{ON} = \tfrac{1}{2}\overrightarrow{OR} = 3\mathbf{a}\).
Route from \(M\) to \(N\) via \(O\): \(\overrightarrow{MN} = \overrightarrow{MO} + \overrightarrow{ON} = -(2\mathbf{b} + \mathbf{a}) + 3\mathbf{a} = 2\mathbf{a} - 2\mathbf{b}\).
From the diagram, \(P\) lies on \(OA\) with \(\overrightarrow{OP} = 2\mathbf{a}\) and \(\overrightarrow{PA} = \mathbf{a}\), so \(\overrightarrow{OA} = \overrightarrow{OP} + \overrightarrow{PA} = 2\mathbf{a} + \mathbf{a} = 3\mathbf{a}\).
Route from \(A\) to \(B\) via \(O\): \(\overrightarrow{AB} = \overrightarrow{AO} + \overrightarrow{OB} = -3\mathbf{a} + \mathbf{b}\).
From (a)(ii), \(\overrightarrow{PR} = 2\mathbf{b} - 2\mathbf{a}\), and from (a)(iii), \(\overrightarrow{PQ} = \tfrac{1}{2}\mathbf{b} - \tfrac{1}{2}\mathbf{a}\).
So \(\overrightarrow{PR} = 4 \times (\tfrac{1}{2}\mathbf{b} - \tfrac{1}{2}\mathbf{a}) = 4\overrightarrow{PQ}\).
\(\overrightarrow{PR}\) is a scalar multiple of \(\overrightarrow{PQ}\), so they are parallel; as they share the common point \(P\), the points \(P\), \(Q\) and \(R\) lie on the same straight line.
(c)
From (b), \(\overrightarrow{PR} = 4\overrightarrow{PQ}\), so \(PR = 4 \times PQ\).
Given \(PQ = 3\) cm, \(PR = 4 \times 3 = 12\) cm.
Answer: \(PR = 12\) cm
Sine and Cosine Rules
8 questions
Things to remember
For any triangle ABC, a² = b² + c² − 2bc cosA
For any triangle ABC, a/sinA = b/sinB = c/sinC
For any triangle ABC, area = ½ ab sinC
1.5 marks
Diagram NOT accurately drawn
ABCD is a triangle.
D is a point on AC.
Angle BAD = 45°
Angle ADB = 80°
AB = 7.4 cm
DC = 5.8 cm
Work out the length of BC.
Give your answer correct to 3 significant figures.
............... cm
Worked solution
First find \(BD\) in triangle \(ABD\) using the sine rule (we know two angles and side \(AB\)). The third angle is \(\angle ABD = 180^\circ - 45^\circ - 80^\circ = 55^\circ\).
Sine rule: \(\dfrac{BD}{\sin(\angle BAD)} = \dfrac{AB}{\sin(\angle ADB)}\).
\(\dfrac{BD}{\sin 45^\circ} = \dfrac{7.4}{\sin 80^\circ}\), so \(BD = \dfrac{7.4 \sin 45^\circ}{\sin 80^\circ} = \dfrac{7.4 \times 0.70711}{0.98481} = 5.3130\) cm.
In triangle \(BDC\), angle \(BDC\) is the supplement of \(\angle ADB\) (since \(A\), \(D\), \(C\) are collinear): \(\angle BDC = 180^\circ - 80^\circ = 100^\circ\).
Apply the cosine rule with the included angle at \(D\): \(BC^2 = BD^2 + DC^2 - 2 \cdot BD \cdot DC \cdot \cos(\angle BDC)\).
Give your answer correct to 3 significant figures.
............... cm
Worked solution
Split the quadrilateral with diagonal \(BD\). In triangle \(ABD\) we know \(AB = 8.2\), \(AD = 5.6\) and the included angle \(\angle DAB = 76^\circ\), so use the cosine rule.
\(BD^2 = AB^2 + AD^2 - 2 \cdot AB \cdot AD \cdot \cos(\angle DAB)\).
Give your answer correct to 3 significant figures.
............... cm
Worked solution
We know two sides (\(LM = 15.7\), \(MN = 12.8\)) and a non-included angle (\(\angle LNM = 130^\circ\)), so use the sine rule first to find \(\angle MLN\).
\(\sin(\angle MLN) = \dfrac{12.8 \sin 130^\circ}{15.7} = \dfrac{12.8 \times 0.76604}{15.7} = 0.62458\), so \(\angle MLN = 38.648^\circ\) (the obtuse alternative is impossible because the triangle already contains \(130^\circ\)).
The third angle is \(\angle LMN = 180^\circ - 130^\circ - 38.648^\circ = 11.352^\circ\).
Apply the sine rule again for \(LN\): \(\dfrac{LN}{\sin(\angle LMN)} = \dfrac{LM}{\sin(\angle LNM)}\).
The angle between any sloping edge and the plane ABCD is 55°
Calculate the surface area of the pyramid.
Give your answer correct to 2 significant figures.
............... cm²
Worked solution
Place the centre of the square base at \(O\). The diagonal of the square has length \(20\sqrt{2}\) cm, so each base vertex is at distance \(OA = 10\sqrt{2} \approx 14.142\) cm from \(O\).
Each sloping edge \(VA\) makes \(55^\circ\) with the base, so in the right-angled triangle \(VOA\): \(\cos 55^\circ = \dfrac{OA}{VA}\), giving \(VA = \dfrac{10\sqrt{2}}{\cos 55^\circ} = \dfrac{14.142}{0.57358} = 24.656\) cm.
The vertical height of the pyramid is \(h = VO = 10\sqrt{2} \tan 55^\circ = 14.142 \times 1.42815 = 20.196\) cm.
Each triangular face has a base of \(20\) cm. Its slant height \(l\) (from \(V\) to the midpoint \(M\) of a base edge) uses \(OM = 10\): \(l = \sqrt{h^2 + 10^2} = \sqrt{20.196^2 + 100} = \sqrt{407.87 + 100} = \sqrt{507.87} = 22.536\) cm.
Area of one triangular face \(= \dfrac{1}{2} \times 20 \times 22.536 = 225.36\) cm\(^2\). Four faces give \(4 \times 225.36 = 901.45\) cm\(^2\).
Area of the square base \(= 20^2 = 400\) cm\(^2\).
Total surface area \(= 901.45 + 400 = 1301.45\) cm\(^2\).
Answer: Surface area \(\approx 1300\) cm\(^2\) (2 s.f.)
7.7 marks
The diagram shows triangle ABC.
The area of triangle ABC is \(k\sqrt{3}\) cm².
Find the exact value of \(k\).
\(k = \) ...............
Worked solution
Two sides \(AB = x+1\) and \(AC = x-1\) meet at angle \(\angle BAC = 120^\circ\); the opposite side is \(BC = 2x-1\). Apply the cosine rule.
\(BC^2 = AB^2 + AC^2 - 2 \cdot AB \cdot AC \cdot \cos(\angle BAC)\).
Acceleration and deceleration is given by the gradient of the graph (rise/run)
The distance travelled is given by the area under the graph.
1.4 marks
A car has an initial speed of \(u\) m/s.
The car accelerates to a speed of \(2u\) m/s in 12 seconds.
The car then travels at a constant speed of \(2u\) m/s for 10 seconds.
Assuming that the acceleration is constant, show that the total distance, in metres, travelled by the car is \(38u\).
Worked solution
Sketch the speed-time graph. From \(t=0\) to \(t=12\) the speed rises in a straight line from \(u\) to \(2u\); from \(t=12\) to \(t=22\) the speed is constant at \(2u\).
The total distance travelled is the area under the speed-time graph.
Phase 1 (\(0 \le t \le 12\)) is a trapezium with parallel sides \(u\) and \(2u\) and width \(12\): area \(= \tfrac{1}{2}(u + 2u) \times 12 = \tfrac{1}{2} \times 3u \times 12 = 18u\) metres.
Phase 2 (\(12 \le t \le 22\)) is a rectangle of height \(2u\) and width \(10\): area \(= 2u \times 10 = 20u\) metres.
Total distance \(= 18u + 20u = 38u\) metres, as required.
Answer: Total distance \(= 38u\) m (shown).
2.6 marks
Karol runs in a race.
The graph shows her speed, in metres per second, \(t\) seconds after the start of the race.
(a)
Calculate an estimate for the gradient of the graph when \(t = 4\). You must show how you get your answer.
(b)
Describe fully what your answer to part (a) represents.
(c)
Explain why your answer to part (a) is only an estimate.
Worked solution
(a)
The gradient at \(t=4\) is the gradient of the tangent to the curve at that point.
Draw a tangent to the curve at \(t = 4\) (where the speed is about \(6.5\) m/s).
Read two points on the tangent, e.g. \((0,\,2)\) and \((8,\,10)\).
Answer: Gradient \(\approx 1\) m/s\(^2\) (accept values in the range about \(0.9\) to \(1.2\) m/s\(^2\) depending on tangent drawn).
(b)
The gradient of a speed-time graph is the rate of change of speed with respect to time, which is the acceleration.
So the value from (a) tells us how fast Karol's speed is changing at that instant.
Answer: It is Karol's acceleration at \(t = 4\) s: her speed is increasing at approximately \(1\) m/s every second (about \(1\) m/s\(^2\)).
(c)
The tangent has to be drawn by eye, so different people will place it slightly differently, and reading coordinates off the grid introduces small errors.
Answer: Because the tangent is drawn by eye, so its position (and the coordinates read from it) is only approximate.
3.5 marks
Here is a speed-time graph for a car journey.
The journey took 100 seconds.
The car travelled 1.75 km in the 100 seconds.
(a)
Work out the value of \(V\).
(b)
Describe the acceleration of the car for each part of this journey.
Worked solution
(a)
The distance travelled is the area under the speed-time graph. The shape is a trapezium with parallel sides of length \(100\) (the time axis, \(0\) to \(100\)) and \(40\) (the top, from \(t=20\) to \(t=60\)), and height \(V\).
Convert the distance to metres: \(1.75 \text{ km} = 1750\) m.
Area \(= \tfrac{1}{2}(100 + 40) \times V = 70V\).
Set equal to the distance: \(70V = 1750\).
\(V = \dfrac{1750}{70} = 25\).
Answer: \(V = 25\) m/s.
(b)
Acceleration is the gradient of the speed-time graph on each straight section.
From \(t=0\) to \(t=20\): gradient \(= \dfrac{25 - 0}{20 - 0} = 1.25\) m/s\(^2\) (accelerating).
From \(t=20\) to \(t=60\): gradient \(= 0\) (constant speed, no acceleration).
From \(t=60\) to \(t=100\): gradient \(= \dfrac{0 - 25}{100 - 60} = -0.625\) m/s\(^2\) (decelerating).
Answer: For the first \(20\) s the car accelerates at \(1.25\) m/s\(^2\); from \(20\) s to \(60\) s the acceleration is \(0\) (constant speed of \(25\) m/s); from \(60\) s to \(100\) s the car decelerates at \(0.625\) m/s\(^2\).
4.5 marks
The graph shows information about the velocity, \(v\) m/s, of a parachutist \(t\) seconds after leaving a plane.
(a)
Work out an estimate for the acceleration of the parachutist at \(t = 6\). ..................... m/s\(^2\)
(b)
Work out an estimate for the distance fallen by the parachutist in the first 12 seconds after leaving the plane. Use 3 strips of equal width. ..................... m
Worked solution
(a)
The acceleration at \(t = 6\) is the gradient of the tangent to the curve at that point (where \(v \approx 47\) m/s).
Draw a tangent at \(t = 6\) and read two points on it, e.g. \((2,\,35)\) and \((10,\,59)\).
Answer: Acceleration \(\approx 3\) m/s\(^2\) (accept a range of about \(2.5\) to \(3.5\) m/s\(^2\) depending on tangent).
(b)
The distance fallen is the area under the velocity-time graph from \(t=0\) to \(t=12\). Use the trapezium rule with \(3\) strips, each of width \(\Delta t = 4\) s.
Read the velocities at the strip boundaries: \(v(0) = 0\), \(v(4) \approx 40\), \(v(8) \approx 51\), \(v(12) \approx 55\) m/s.
Strip 1 (\(0\) to \(4\)): area \(= \tfrac{1}{2}(0 + 40) \times 4 = 80\) m.
Strip 2 (\(4\) to \(8\)): area \(= \tfrac{1}{2}(40 + 51) \times 4 = \tfrac{1}{2} \times 91 \times 4 = 182\) m.
Strip 3 (\(8\) to \(12\)): area \(= \tfrac{1}{2}(51 + 55) \times 4 = \tfrac{1}{2} \times 106 \times 4 = 212\) m.
Total distance \(\approx 80 + 182 + 212 = 474\) m.
Answer: Distance fallen \(\approx 474\) m (accept approximately \(460\)–\(490\) m depending on readings).
5.4 marks
Here is a speed-time graph for a car.
(a)
Work out an estimate for the distance the car travelled in the first 10 seconds. Use 5 strips of equal width. ..................... m
(b)
Is your answer to (a) an underestimate or an overestimate of the actual distance? Give a reason for your answer.
Worked solution
(a)
Use the trapezium rule with \(5\) strips of width \(\Delta t = 2\) s covering \(t = 0\) to \(t = 10\).
Read the speeds at the strip boundaries: \(v(0) = 0\), \(v(2) \approx 1\), \(v(4) \approx 5\), \(v(6) \approx 10\), \(v(8) \approx 17\), \(v(10) \approx 25\) m/s.
Strip 1 (\(0\) to \(2\)): \(\tfrac{1}{2}(0 + 1) \times 2 = 1\) m.
Strip 2 (\(2\) to \(4\)): \(\tfrac{1}{2}(1 + 5) \times 2 = 6\) m.
Strip 3 (\(4\) to \(6\)): \(\tfrac{1}{2}(5 + 10) \times 2 = 15\) m.
Strip 4 (\(6\) to \(8\)): \(\tfrac{1}{2}(10 + 17) \times 2 = 27\) m.
Strip 5 (\(8\) to \(10\)): \(\tfrac{1}{2}(17 + 25) \times 2 = 42\) m.
Total distance \(\approx 1 + 6 + 15 + 27 + 42 = 91\) m.
Answer: Distance \(\approx 91\) m (accept approximately \(85\)–\(95\) m depending on readings).
(b)
The curve is convex (concave up) — it bends upwards, so the straight chord across each strip lies above the curve.
This means each trapezium has more area than the region under the curve it is approximating.
So the total trapezium area is greater than the true area under the graph.
Answer: Overestimate. The curve is concave up (convex), so the top of each trapezium (the chord) lies above the curve, making each strip area larger than the actual area under the curve.
Histograms
7 questions
Things to remember
Frequency = Frequency Density x Class Width;
The y-axis will always be labelled "frequency density";
The x-axis will have a continuous scale.
1.5 marks
One Monday, Victoria measured the time, in seconds, that individual birds spent on her bird table. She used this information to complete the frequency table.
Time (t seconds)
Frequency
\(0 < t \leq 10\)
8
\(10 < t \leq 20\)
16
\(20 < t \leq 25\)
15
\(25 < t \leq 30\)
12
\(30 < t \leq 50\)
6
(a)
Use the table to complete the histogram.
On Tuesday she conducted a similar survey and drew the following histogram from her results.
(b)
Use the histogram for Tuesday to complete the table.
Time (t seconds)
Frequency
\(0 < t \leq 10\)
10
\(10 < t \leq 20\)
\(20 < t \leq 25\)
\(25 < t \leq 30\)
\(30 < t \leq 50\)
Worked solution
(a)
Compute frequency density (FD) = frequency / class width for each class.
Answer: Draw bars at frequency densities: \(4000 < w \le 6000\) at \(0.008\); \(8000 < w \le 12000\) at \(0.002\).
5.3 marks
The incomplete histogram and table give some information about the distances some teachers travel to school.
(a)
Use the information in the histogram to complete the frequency table.
Distance (d km)
Frequency
\(0 < d \leq 5\)
15
\(5 < d \leq 10\)
20
\(10 < d \leq 20\)
\(20 < d \leq 40\)
\(40 < d \leq 60\)
10
(b)
Use the information in the table to complete the histogram.
Worked solution
(a)
The given \(0 < d \le 5\) bar has frequency \(15\) over width \(5\), so FD \(= 3\); the \(5 < d \le 10\) bar has frequency \(20\) over width \(5\), so FD \(= 4\). Use these to calibrate the y-axis.
\(10 < d \le 20\): read FD \(= 2\), frequency \(= 2 \times 10 = 20\).
\(40 < d \le 60\): frequency \(10\) over width \(20\), FD \(= 0.5\), so this bar is already usable for part (b).
Answer: \(10 < d \le 20\): \(20\); \(20 < d \le 40\) must be read from the histogram. The histogram shows no bar in \(20 < d \le 40\), so the frequency is not determined by (a) alone; for (b) draw it from the table.
(b)
Compute frequency density FD \(=\) frequency / class width for the class that was missing from the histogram.
\(40 < d \le 60\): FD \(= 10 / 20 = 0.5\) (this bar is already drawn).
From part (a), the only unfilled bar is \(20 < d \le 40\). Reading the histogram directly gives its height; a typical reading is FD \(= 1\), giving frequency \(= 1 \times 20 = 20\).
Using this, \(20 < d \le 40\) frequency \(= 20\) and FD \(= 1\).
Answer: Draw the bar \(20 < d \le 40\) at frequency density \(1\) (frequency \(20\)).
6.3 marks
The table gives information about the heights, in centimetres, of some 15 year old students.
Height (h cm)
\(145 < h \leq 155\)
\(155 < h \leq 175\)
\(175 < h \leq 190\)
Frequency
10
80
24
Use the table to draw a histogram.
Worked solution
Compute frequency density FD \(=\) frequency / class width for each class.
Answer: Draw bars at frequency densities: \(145 < h \le 155\) at \(1\); \(155 < h \le 175\) at \(4\); \(175 < h \le 190\) at \(1.6\).
7.2 marks
A teacher asked some year 10 students how long they spent doing homework each night. The histogram was drawn from this information.
Use the histogram to complete the table.
Time (t minutes)
Frequency
\(10 \leq t < 15\)
10
\(15 \leq t < 30\)
\(30 \leq t < 40\)
\(40 \leq t < 50\)
\(50 \leq t < 70\)
Worked solution
The given \(10 \le t < 15\) bar has frequency \(10\) over width \(5\), so FD \(= 10/5 = 2\). Use this to calibrate the y-axis.
Read each bar's frequency density from the histogram, then use Frequency \(=\) FD \(\times\) class width.
\(15 \le t < 30\): FD \(= 3\), frequency \(= 3 \times 15 = 45\).
\(30 \le t < 40\): FD \(= 1.5\), frequency \(= 1.5 \times 10 = 15\).
\(40 \le t < 50\): FD \(= 2.5\), frequency \(= 2.5 \times 10 = 25\).
\(50 \le t < 70\): FD \(= 1\), frequency \(= 1 \times 20 = 20\).
Answer: \(15 \le t < 30\): \(45\); \(30 \le t < 40\): \(15\); \(40 \le t < 50\): \(25\); \(50 \le t < 70\): \(20\).
Capture-Recapture
7 questions
Things to remember
Set up a pair of equivalent fractions → how many out of x were tagged = how many of the second sample are tagged out of how many in the original sample.
This method assumes that the original sample is thoroughly mixed back in.
1.2 marks
A scientist wants to estimate the number of fish in a lake.
He catches 50 fish from the lake and marks them with a dye.
The fish are then returned to the lake.
The next day the scientist catches another 50 fish.
4 of these fish are marked with the dye.
Work out an estimate for the total number of fish in the lake.
Worked solution
Identify the capture-recapture values: \(M = 50\) (fish initially marked), \(n = 50\) (second sample), \(m = 4\) (marked fish in second sample).
Set up equivalent fractions: \(\dfrac{m}{n} = \dfrac{M}{N}\), giving \(\dfrac{4}{50} = \dfrac{50}{N}\).
State one assumption required for the capture-recapture method to give a valid estimate.
Answer: The marked bees mix thoroughly back into the hive, and the population does not change (no bees enter or leave and no tags come off) between the two samples.
5.2 marks
Toga wants to estimate the number of termites in a nest.
On Monday Toga catches 80 termites.
He puts a mark on each termite.
He then puts all 80 termites back in the nest.
On Tuesday Toga catches 60 termites.
12 of these termites have a mark on them.
Work out an estimate for the total number of termites in the nest.
Worked solution
Identify the values: \(M = 80\) (termites marked on Monday), \(n = 60\) (caught on Tuesday), \(m = 12\) (marked on Tuesday).
Set up equivalent fractions: \(\dfrac{m}{n} = \dfrac{M}{N}\), giving \(\dfrac{12}{60} = \dfrac{80}{N}\).
If some tags fell off, then fewer than 8 of the recaptured ducks would actually still appear tagged relative to the true number of previously-tagged ducks still in the park. Equivalently, the recaptured tagged count \(m = 8\) corresponds to fewer than 30 effectively-tagged ducks, so using \(M = 30\) in \(N = \dfrac{Mn}{m}\) makes the numerator too large.
This means the proportion of tagged ducks in the park is actually smaller than \(\tfrac{30}{N}\), so dividing by \(m\) gives a value of \(N\) that is bigger than the true population.
Answer: If some tags fell off, the true number of tagged ducks in the park would be less than 30, so the estimate \(N = \dfrac{30 \times 40}{8} = 150\) would be an overestimate of the actual number of ducks in the park.
Set Theory
3 questions
Things to remember
The intersection is where two sets overlap. A ∩ B. This means A and B.
If you put two sets together, you get the union. A ∪ B. This means A or B.
The complement of A is the region that is not A. A'. This means not A.
1.4 marks
\(\xi = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}\)
\(A = \{\text{multiples of } 2\}\)
\(A \cap B = \{2, 6\}\)
\(A \cup B = \{1, 2, 3, 4, 6, 8, 9, 10\}\)
Draw a Venn diagram for this information.
Worked solution
List the universal set: \(\xi = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}\).
Set A (multiples of 2): \(A = \{2, 4, 6, 8, 10\}\).
The intersection \(A \cap B = \{2, 6\}\), so 2 and 6 must be placed in the overlap of A and B.
Find B: every element of B is in \(A \cup B\), and the elements of B that are not in A lie in \((A \cup B) \setminus A = \{1, 3, 9\}\). Combined with the intersection \(\{2, 6\}\), this gives \(B = \{1, 2, 3, 6, 9\}\).
Elements in A only (\(A \cap B'\)): \(\{2, 4, 6, 8, 10\} \setminus \{2, 6\} = \{4, 8, 10\}\).
Elements in B only (\(A' \cap B\)): \(\{1, 3, 9\}\).
Elements outside both circles (\((A \cup B)'\)): \(\xi \setminus (A \cup B) = \{5, 7\}\).
Answer: Venn diagram with two overlapping circles A and B inside rectangle \(\xi\). In A only: 4, 8, 10. In the intersection \(A \cap B\): 2, 6. In B only: 1, 3, 9. Outside both circles but inside \(\xi\): 5, 7.
2.4 marks
Here is a Venn diagram.
(a)
Write down the numbers that are in set
(i)
\(A \cup B\)
(ii)
\(A \cap B'\)
One of the numbers in the diagram is chosen at random.
(b)
Find the probability that the number is in set \(A'\).
Worked solution
(a)(i)
\(A \cup B\) is every element inside either circle (A only, B only, or the intersection).
A only: \(\{15\}\). Intersection \(A \cap B\): \(\{12, 18\}\). B only: \(\{10, 14, 16\}\).
Combine all these elements.
Answer: \(A \cup B = \{10, 12, 14, 15, 16, 18\}\)
(a)(ii)
\(A \cap B'\) means elements in A but not in B, i.e. the part of A that does not overlap with B.
Only the number 15 sits in A outside the intersection.
Answer: \(A \cap B' = \{15\}\)
(b)
Total numbers shown in the diagram: \(1 + 2 + 3 + 4 = 11\) (A only: 1, intersection: 2, B only: 3, outside: 4).
\(A'\) is everything not in A: the three numbers in B only \(\{10, 14, 16\}\) plus the four numbers outside both circles \(\{11, 13, 17, 19\}\).
Substitute \(f = 4\) into \(S = \dfrac{8000}{f^2}\).
\(S = \dfrac{8000}{16} = 500\).
Answer: \(S = 500\)
2.4 marks
In a factory, chemical reactions are carried out in spherical containers. The time, \(T\) minutes, the chemical reaction takes is directly proportional to the square of the radius, \(R\) cm, of the spherical container.
When \(R = 120\), \(T = 32\)
Find the value of \(T\) when \(R = 150\)
Worked solution
\(T\) is directly proportional to \(R^2\), so \(T = kR^2\).
The time, \(T\) seconds, it takes a water heater to boil some water is directly proportional to the mass of water, \(m\) kg, in the water heater. When \(m = 250\), \(T = 600\)
(a)
Find \(T\) when \(m = 400\)
The time, \(T\) seconds, it takes a water heater to boil a constant mass of water is inversely proportional to the power, \(P\) watts, of the water heater. When \(P = 1400\), \(T = 360\)
(b)
Find the value of \(T\) when \(P = 900\)
Worked solution
(a)
\(T\) is directly proportional to \(m\), so \(T = km\).
A ball falls vertically after being dropped. The ball falls a distance \(d\) metres in a time of \(t\) seconds. \(d\) is directly proportional to the square of \(t\). The ball falls 20 metres in a time of 2 seconds.
(a)
Find a formula for \(d\) in terms of \(t\).
(b)
Calculate the distance the ball falls in 3 seconds.
(c)
Calculate the time the ball takes to fall 605 m.
Worked solution
(a)
\(d\) is directly proportional to \(t^2\), so \(d = kt^2\).
Bill says 'This means Hajra's weekly pay last year was £192'.
Bill is wrong.
(a)
Explain why.
(b)
Work out Hajra's weekly pay last year.
Worked solution
(a)
<p>Bill has calculated \(20\%\) of this year's pay (£240) and subtracted it: \(240 - 0.20 \times 240 = 192\).</p>
<p>This is wrong because the \(20\%\) increase was applied to last year's pay, not to this year's pay. So you must divide by the multiplier \(1.20\), not take \(20\%\) off £240.</p>
Answer: Bill took 20% off this year's pay, but the 20% increase was applied to last year's pay, so he should divide £240 by 1.20 instead.
(b)
<p>A \(20\%\) increase gives a multiplier of \(1 + 0.20 = 1.20\).</p>
<p>Let last year's weekly pay be \(L\). Then \(L \times 1.20 = 240\).</p>
