Mathematics Paper 2 November 2012

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1. A$60\%$, from taking 240 instead of 250 as the number of pupils who failed
2. B$62,5\%$, because $400 - 150 = 250$ pupils failed out of the 400 who sat
3. C$37,5\%$, which is the percentage that passed rather than the percentage that failed
4. D$40\%$, from dividing 150 by 400 and then rounding the answer upwards

1. A$(x + 3)(x - 1)$, which is only the first of the two expressions itself
2. B$(x + 3)(x - 1)(x + 1)$, every distinct factor of the two taken once
3. C$(x + 3)(x - 1)^2(x + 1)$, with the shared factor $(x - 1)$ counted twice over
4. D$(x - 1)$, which is the highest common factor rather than the L.C.M here

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Answer this when you sit the paper.

1. A$\{3; 9\}$, taking 3 as a perfect square because $3 \times 3 = 9$
2. B$\{3; 6; 9\}$, which is the set B on its own and not the overlap
3. C$\{1; 4; 9\}$, which is the set A on its own and not the overlap
4. D$\{9\}$, the one number that is both a square and a multiple of 3

1. A\$450, since the commission on \$10 000 is \$100 and $550 - 100 = 450$
2. B\$500, the mean of the two salaries \$550 and \$600 taken together
3. C\$50, the difference between the two salaries treated as the constant
4. D\$550, taking the smaller of the two given salaries as the constant part

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1. A$6m^3$, because $m^0 = 1$ and the numbers multiply to give 6
2. B$5m^3$, from adding the coefficients 2 and 3 instead of multiplying
3. C$6m^0$, from taking the smaller of the two indices for the answer
4. D$6$, from treating $m^3 \times m^0$ as cancelling to leave a number

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Answer this when you sit the paper.

1. A$x = -2$, from quoting the index on the right as the answer itself
2. B$x = \frac{1}{2}$, from solving $2x - 1 = 0$ and ignoring the right side
3. C$x = \frac{3}{2}$, from reading $\frac{1}{9}$ as $3^{2}$ rather than as $3^{-2}$
4. D$x = -\frac{1}{2}$, since $\frac{1}{9} = 3^{-2}$, so $2x - 1 = -2$

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1. A$53^\circ$, half of the $106^\circ$ angle at the centre on the same arc
2. B$212^\circ$, from doubling the angle at the centre instead of halving it
3. C$74^\circ$, from subtracting the centre angle from $180^\circ$ instead
4. D$106^\circ$, from taking the angle at D as equal to the one at O

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1. A$53^\circ$, from halving $106^\circ$ as though the angle were at the centre
2. B$37^\circ$, since triangle OAB is isosceles with two radii for its sides
3. C$74^\circ$, from taking the two equal base angles together as one angle
4. D$106^\circ$, from copying the angle at O onto the base of the triangle

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1. ATriangle ABT, which shares the side AT but has no matching pair of angles
2. BTriangle CDT, which shares the vertex T but sits on unrelated arcs entirely
3. CTriangle BCT, since its angles at B and C stand on the same arcs
4. DTriangle ABD, which is not formed by the two chords crossing at the point T

1. A$1 \le x < 4$, from solving both parts with the inequality signs reversed
2. B$-4 \le x < 3$, from dividing $3x < 9$ instead of $3x < 3$ on the left part
3. C$-4 \le x < 1$, from $x < 1$ on the left part and $x \ge -4$ on the right
4. D$-1 \le x < 4$, from swapping the two bounds round when writing them down

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Answer this when you sit the paper.

1. A$23,4$ cm$^2$, from leaving out the factor $\sin 60^\circ$ in the formula
2. B$20,3$ cm$^2$, from $\frac{1}{2} \times 8 \times 5,86 \times \sin 60^\circ$
3. C$11,7$ cm$^2$, from using $\sin 30^\circ$ in place of $\sin 60^\circ$ here
4. D$40,6$ cm$^2$, from leaving out the factor of one half in the area formula

1. AThe set of points that are the same distance from the two vertices P and Q
2. BThe set of points lying at a fixed distance from the single vertex point R
3. CThe set of points equidistant from the lines RP and RQ
4. DThe set of points the same distance from the line PQ as from the point R

Answer this when you sit the paper.

1. A$-2,57$, from attaching the sign of the positive root to the wrong value
2. B$-0,91$, from $\frac{5 - \sqrt{109}}{6} = \frac{-5,4403}{6}$ worked out
3. C$-1,81$, from dividing by $a = 3$ rather than by $2a = 6$ in the formula
4. D$-0,45$, from halving the correct root by mistake at the final step

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1. A\$0,17, from using 178 units in place of the correct 148 units used
2. B\$0,21, from $31,08 \div (55\,926 - 55\,778) = 31,08 \div 148$
3. C\$0,56, from dividing \$31,08 by 55 rather than by the 148 units used
4. D\$4,76, from dividing the 148 units by the \$31,08 charged for them

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1. AA reflection in the line $x = -3$, since each $x$ becomes $-6 - x$
2. BA translation of 4 units to the right, which fails for the point L
3. CA reflection in the line $y = -3$, which would change the $y$ coordinates
4. DA reflection in the $y$-axis, which would send $(-5; 3)$ to $(5; 3)$ instead

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Answer this when you sit the paper.

1. A$\begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix}$, the shear that leaves the $x$-axis invariant
2. B$\begin{pmatrix} 1 & 0 \\ 2 & 1 \end{pmatrix}$, which sends $(x; y)$ to $(x; y + 2x)$
3. C$\begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix}$, an enlargement of scale factor 2 instead
4. D$\begin{pmatrix} 1 & 0 \\ 0 & 2 \end{pmatrix}$, a stretch parallel to the $y$-axis instead

1. A$(-5; 13)$, from adding $2 \times 5$ to the $y$ coordinate of the point
2. B$(-10; 3)$, from doubling the $x$ coordinate and leaving $y$ alone here
3. C$(1; 3)$, from adding $2 \times 3$ to the $x$ coordinate of the point K
4. D$(-5; -7)$, since $x$ is unchanged and $y$ becomes $3 + 2(-5) = -7$

1. A$y \ge x$, which would also allow equal numbers of chairs and desks
2. B$x > y$, which states that there are more desks than chairs instead
3. C$y > x$, since the chair count strictly exceeds the desk count here
4. D$x + y > 0$, which only says that something is bought altogether

1. A$25x + 17,5y = 5000$, which forces every dollar to be spent to the cent
2. B$17,5x + 25y \le 5000$, with the two unit prices attached the wrong way
3. C$25x + 17,5y \le 5000$, the total cost being at most the money available
4. D$25x + 17,5y \ge 5000$, which would force the school to overspend instead

1. A$x \ge 75$ and $y \ge 100$, since at least means greater than or equal to
2. B$x > 75$ and $y > 100$, which would rule out buying exactly 75 or 100
3. C$x \le 75$ and $y \le 100$, which caps the numbers instead of setting a floor
4. D$x \ge 100$ and $y \ge 75$, with the two figures attached the wrong way round

Answer this when you sit the paper.

Answer this when you sit the paper.

Answer this when you sit the paper.

1. A$402$ m, from adding 1 m to each of the two straights of the track
2. B$412,6$ m, from taking the radius of lane 2 as 37 m instead of 36 m
3. C$800$ m, from doubling the 400 m inner boundary of the first lane
4. D$406$ m, using a radius of 36 m for the two semi-circular ends

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1. A$1\,961$ m$^2$, counting only the two curved strips at the ends
2. B$13\,551$ m$^2$, which is the whole outer area with the field left in it
3. C$1\,440$ m$^2$, counting only the two straight strips of the running track
4. D$3\,401$ m$^2$, from $90 \times 16 + \frac{22}{7}(43^2 - 35^2)$

1. A\$288 000, from costing only the two straight strips of the track
2. B\$17,01, from dividing the area by \$200 rather than multiplying by it
3. C\$680 228,57, from $3\,401,14 \times 200$ worked out in full
4. D\$2 710 228,57, from costing the whole outer area including the field

1. A$\mathbf{b} - \mathbf{a}$, since $\vec{AD} = \vec{AO} + \vec{OD}$
2. B$\mathbf{a} - \mathbf{b}$, which is $\vec{DA}$ and points the other way
3. C$\frac{1}{2}(\mathbf{a} + \mathbf{b})$, which would reach the midpoint of AD
4. D$\mathbf{a} + \mathbf{b}$, from adding the two given vectors together

1. A$3\mathbf{b}$, from multiplying by 3 instead of dividing by 3 here
2. B$\frac{3}{2}\mathbf{b}$, which would place X beyond D on the line OD
3. C$\frac{2}{3}\mathbf{b}$, since $\vec{XD} = \frac{1}{3}\mathbf{b}$ leaves two thirds
4. D$\frac{1}{3}\mathbf{b}$, which is $\vec{XD}$ rather than $\vec{OX}$ itself

1. A$h\mathbf{a} + \mathbf{b}$, which is $\vec{OC}$ itself and not its mid-point
2. B$\frac{h}{2}\mathbf{a} + \mathbf{b}$, halving only the first of the two terms
3. C$\frac{1}{2}\mathbf{a} + \frac{h}{2}\mathbf{b}$, with $h$ attached to the wrong vector
4. D$\frac{h}{2}\mathbf{a} + \frac{1}{2}\mathbf{b}$, half of $\vec{OC} = h\mathbf{a} + \mathbf{b}$

1. A$\frac{h}{2}\mathbf{a} - \frac{1}{6}\mathbf{b}$, which is $\vec{XM}$ and points back
2. B$-\frac{h}{2}\mathbf{a} - \frac{1}{6}\mathbf{b}$, from subtracting $\vec{OX}$ instead
3. C$\frac{h}{2}\mathbf{a} + \frac{7}{6}\mathbf{b}$, from adding $\vec{OM}$ to $\vec{OX}$
4. D$-\frac{h}{2}\mathbf{a} + \frac{1}{6}\mathbf{b}$, since $\vec{MX} = \vec{MO} + \vec{OX}$

Answer this when you sit the paper.

Answer this when you sit the paper.

1. A$x = 0,5$ and $x = -3$, with both of the signs taken the wrong way round
2. B$x = -3$ and $x = 1$, from factorising the expression as $x^2 + 2x - 3$
3. C$x = -2$ and $x = 4$, the two ends of the range the curve is drawn over
4. D$x = -0,5$ and $x = 3$, the roots of $(2x + 1)(x - 3) = 0$

1. A$-1$, the value of $4x - 5$ at $x = 1$, matching a tangent drawn there
2. B$4$, the coefficient found by doubling the 2 in front of $x^2$ here
3. C$-6$, which is the value of $y$ at $x = 1$ and not the gradient there
4. D$-5$, the coefficient of $x$ in the equation of the curve itself

1. A$x = 0,22$ and $x = 2,28$, the roots of $2x^2 - 5x + 1 = 0$
2. B$x = -0,50$ and $x = 3,00$, the values where the curve meets the x-axis
3. C$x = 0,25$ and $x = 2,25$, from halving the coefficients before solving
4. D$x = -0,22$ and $x = -2,28$, with both of the signs taken the wrong way

Answer this when you sit the paper.

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Answer this when you sit the paper.

1. A$18\%$, from dividing the 36 people by 200 rather than by the 180
2. B$36\%$, from quoting the combined count of 36 people as a percentage
3. C$20\%$, since $\frac{19 + 17}{180} \times 100 = \frac{36}{180} \times 100$
4. D$10,6\%$, from taking only the 19 who live in Australia into account

1. A$0,0509$, from $\frac{41}{180} \times \frac{40}{179}$ without replacement
2. B$0,228$, the chance that just the first of the two lives in Botswana
3. C$0,0519$, from squaring $\frac{41}{180}$ as if the first were replaced
4. D$0,451$, from adding $\frac{41}{180}$ to $\frac{40}{179}$ rather than multiplying

Answer this when you sit the paper.

Answer this when you sit the paper.