Question 1
Topic-6.2 – Statistical measures
1 Here are eight numbers written in order of size
$h \quad 6 \quad 7 \quad 8 \quad j \quad 16 \quad k \quad k$
where $h, j$ and $k$ are integers.
The median of the eight numbers is 10
The mode of the eight numbers is 18
The range of the eight numbers is 13
Work out the value of $h$, the value of $j$ and the value of $k$
▶️Answer/Explanation
Solution :-
for $k=18$
or
$eg (8+j) \div 2 = 10$ or $(j=) 10 \times 2 – 8$
or $8+j = 2 \times 10$ or $j=12$
or
$eg k-h = 13$ or “18” $- h = 13$ or $h=5$
Correct answer scores full marks (unless from obvious incorrect working)
$h=5$
$j=12$
$k=18$
Question 2
(a) Topic-3.3 – Graphs
(b) Topic-3.3 – Graphs
(a) On the grid, draw the straight line with equation
(i) $y = 2$
(ii) $x = 6$
(iii) $y = x + 1$
Label each line with its equation.
(b) Show, by shading on the grid, the region that satisfies all three of the inequalities
\[y \ge 2\] \qquad \[x \le 6\] \qquad \[y \le x+1\]
Label the region $\mathbf{R}$
▶️Answer/Explanation
Solution :-
(a) (i) $y=2$ drawn
(ii) $x=6$ drawn
(iii) $y=x+1$ drawn
(b)
Question 3
Topic-1.10 – Applying number
A plane takes 9 hours 36 minutes to fly from New Delhi to Perth.
The plane flies at an average speed of 820 km h.
Work out the total distance the plane flies.
▶️Answer/Explanation
Solution :-
For $9 hrs 36 mins = 9.6 (hrs) or (9\frac{36}{60}) (hrs) or
(9\frac{3}{5}) (hrs) oe or 576 (mins)$
$eg~820\times^{46}9.6^{\prime\prime} or
[820\times\frac{576}{60}]
or
(576\times\frac{820}{60}) or (576\times\frac{41}{3}) allow 13.7 for (\frac{41}{3})) oe$
Correct answer scores full marks (unless from
obvious incorrect working)
7872
Question 4
Topic-1.2 – Fractions
Show that $2\frac{4}{7} \times 3\frac{1}{9} = 8$
▶️Answer/Explanation
Solution :-
$\frac{18}{7}, \frac{28}{9}$
$\frac{18^2}{7^1} \times \frac{28^4}{9^1}$ or $\frac{18}{7} \times \frac{28}{9} = \frac{504}{63}$ oe eg $\frac{18^2}{7} \times \frac{28}{9^1} = \frac{56}{7}$
or $\left(\frac{18}{7} \times \frac{28}{9} = \right) \frac{162}{63} \times \frac{196}{63} = \frac{31752}{3969}$ oe
eg $\frac{18^2}{7^1} \times \frac{28^4}{9^1} = 8$ or $\frac{18^2}{7^1} \times \frac{28^4}{9^1} = 2 \times 4 = 8$
eg $\frac{18}{7} \times \frac{28}{9} = \frac{504}{63} = 8$ oe or
eg $\left(\frac{18}{7} \times \frac{28}{9} = \right) \frac{162}{63} \times \frac{196}{63} = \frac{31752}{3969} \left( = \frac{8}{1} \right) = 8$
working required
Question 5
Topic-4.8 – Trigonometry and Pythagora
The diagram shows triangle $ABC$
Work out the value of $x$
Give your answer correct to one decimal place.
▶️Answer/Explanation
Solution :-
$\sin 34 = \frac{x}{6.5}$ or $\frac{x}{\sin 34} = \frac{6.5}{\sin 90}$ oe
$6.5^2 – (6.5 \times \cos 34)^2$ or
$\cos 56 = \frac{x}{6.5}$ oe
$(x=) 6.5 \times \sin 34$ or $x = \frac{6.5 \times \sin 34}{\sin 90}$
or
$(x=) \sqrt{6.5^2 – (6.5 \times \cos 34)^2}$
or
$(x=) 6.5 \times \cos 56$ oe
Correct answer scores full marks (unless from obvious incorrect working) & 3.6
Question 6
Topic-1.10 – Applying number
Change a speed of $w$ metres per second to a speed in kilometres per hour.
Give your answer in terms of $w$ in its simplest form.
▶️Answer/Explanation
Solution :-
For one of
$w \div 1000$ or $w \div 10^3$ or $w \times 10^{-3}$ or $0.001w$ oe
$(w \times 60 \times 60)$ oe or
$w \times 3600$ or $w \div \frac{1}{3600}$ oe
$\frac{w \times 60 \times 60}{1000}$ oe eg $w \times \frac{3600}{1000}$
Correct answer scores full marks (unless from obvious incorrect working) & $3.6w$
Question 7
Topic-4.9 – Mensuration
The diagram shows a 6-sided shape $ABCDEF$
$AF = 21$ cm $CD = 15$ cm $AB = FE = 13$ cm
The perpendicular height of the shape is $h$ cm
$CD$ is parallel to $AF$
The area of the shape is $390 \text{ cm}^2$
Work out the value of $h$
▶️Answer/Explanation
Solution :-
eg $13\times21(=273)$ or $21 \times h(=21h)$
or $0.5(15+21)\times y$ or $15(h-13)$ or $2\times\frac{1}{2}(3(h-13))$
or $\frac{1}{2}(13+h)\times3(=19.5+1.5h)$ or $15 \times h(=15h)$
eg $390-^{6}273^{\prime\prime}(=117)$ or $13\times21$ and $0.5(15+21)(h-13)$
or
$13\times21$ and $0.5(15+21)y$ oe
or
$21h$ and $2\times\frac{1}{2}(3(h-13))$ oe
or
$13\times21$ and $15(h-13)$ and $2\times\frac{1}{2}(3(h-13))$ oe
or
$2\times\frac{1}{2}(13+h)\times3$ and $15\times h$
“117” $\div(0.5\times(15+21))(=6.5)$ or
$\frac{1}{2}(15+21)\times y=^{\prime\prime}117^{\prime\prime}$
or
$273+18(h-13)=390$
or
$15(h-13) + 2\times\frac{1}{2}(3(h-13)) =^{\prime\prime}117^{\prime\prime}$ oe
or
$2\times\frac{1}{2}(13+h)\times3+15h=390$
Typical equations here simplify to :
$18y=117$, $18h-234=117$, $18h+39=390$, $18h=351$
Correct answer scores full marks (unless from obvious incorrect working) & 19.5
Question 8
Topic-1.7 – Ratio and proportion
Ishir plants 600 bulbs in a garden.
He plants tulip bulbs, crocus bulbs and daffodil bulbs so that
number of tulip bulbs : number of crocus bulbs : number of daffodil bulbs = 9 : 4 : 2
45% of the tulip bulbs are for yellow flowers.
$\frac{5}{8}$ of the crocus bulbs are for yellow flowers.
All of the daffodil bulbs are for yellow flowers.
Work out the number of bulbs that are for yellow flowers.
▶️Answer/Explanation
Solution :-
$600 \div (9+4+2) (=40)$
or
tulip: $0.45 \times 600 (=270)$
or
crocus: $\frac{5}{8} \times 600 (=375)$
Daffodils: “$40” \times 2 (=$
$80) \frac{2}{15} \times 600 (=80)$
(implies 1st M1)
Tulip:
$0.45 \times (9 \times “40”) (=162)$
or
$0.45 \times 600 \times \frac{9}{15} (=162)$
(implies 1st M1)
Crocus:
$\frac{5}{8} \times (4 \times “40”) (=100)$
or
$\frac{5}{8} \times 600 \times \frac{4}{15} (=100)$
(implies 1st M1)
Correct answer scores full marks (unless from obvious incorrect working) & 342
Tulips:
$0.45 \times 9 (=4.05)$
or
$0.45 \times \frac{9}{15} (=\frac{27}{100} (=0.27))$ oe
Crocus:
$\frac{5}{8} \times 4 (=2.5)$
or
$\frac{5}{8} \times \frac{4}{15} (=\frac{1}{6} (=0.16))$ oe
Total of parts
$4.05 + 2.5 + 2 (=8.55)$
or
$\frac{27}{100} + \frac{1}{6} + \frac{2}{15} (=\frac{57}{100})$ oe
(implies 1st and 2nd M marks)
$\frac{8.55}{9+4+2} \times 600$ oe
or
$\frac{57}{100} \times 600$ oe
(implies all previous M marks)
Question 9
Topic-1.6 – Percentages
Giovanni invests 4500 koruna in a savings account for 4 years.
He gets 2.4% per year compound interest.
Work out how much money Giovanni will have in the savings account at the end of 4 years.
Give your answer correct to the nearest koruna.
▶️Answer/Explanation
Solution :-
$4500 \times 1.024 (= 4608)$ oe or
$4500 \times 0.024 (= 108)$
“$4608” \times 1.024 (=4718.592)$ and
“$4718.592” \times 1.024 (=4831.838…)$ and
“$4831.838” \times 1.024 (=4947.80…)$
Correct answer scores full marks (unless from obvious incorrect working) & 4948
Question 10
Topic-2.6 – Simultaneous linear equations
Solve the simultaneous equations
$6x + 4y = 1$
$3x + 5y = 8$
Show clear algebraic working.
▶️Answer/Explanation
Solution :-
$6x+4y=1$ $30x+20y=5$
eg $6x+10y=16$ or $12x+20y=32$ oe
$(6y=15)$ $(18x=-27)$
or eg $6x+4(\frac{8-3x}{5})=1$ or $3(\frac{1-4y}{6})+5y=8$
eg
working required & $x=-1.5$, $y=2.5$
Question 11
(i) Topic-2.7 – Quadratic equations
(ii) Topic-2.7 – Quadratic equations
(i) Factorise $x^2 + 9x – 22$
(ii) Hence, solve $x^2 + 9x – 22 = 0$
▶️Answer/Explanation
Solution :-
(i) $(x \pm 2)(x \pm 11)$
Correct answer scores full marks (unless from obvious incorrect working) $(x-2)(x+11)$
(ii) $2, -11$
Question 12
Topic-6.2 – Statistical measures
Ali uses a fitness tracker to count the number of steps he walks each day for 7 days.
For the first 4 days, his mean number of steps is 11 800
For the next 3 days, his mean number of steps is 13 207
Work out his mean number of steps for the 7 days.
▶️Answer/Explanation
Solution :-
$4 \times 11800 (= 47200)$ or $3 \times 13207 (= 39621)$ or 86821
$\frac{“47200” + “39621”}{7} (=\frac{86821}{7})$
Correct answer scores full marks (unless from obvious incorrect working) & 12403
Question 13
(a) Topic-6.1 – Graphical representation of data
(b) Topic-6.1 – Graphical representation of data
(c) Topic-6.2 – Statistical measures
(d) Topic-6.1 – Graphical representation of data
The table gives information about the distances, in km, that 70 teachers travel to school.
(a) Complete the cumulative frequency table.
(b) On the grid opposite, draw a cumulative frequency graph for your table.
(c) Use your graph to find an estimate for the interquartile range of the distances.
(d) Use your graph to find an estimate for the number of teachers who travel more than 46 km.
▶️Answer/Explanation
Solution :-
$(a) 7, 24, 42, 56, 66, 70$
(b) USE OVERLAY 6 points plotted at ends of intervals and joined with curve or line segments
(NB: a ‘bar chart’ type graph scores zero marks)
(ignore any part of the graph before (10, 7))
(c) NB: readings are 16 – 18 and 36 – 38 (but for this M1 these do not have to be correct if correct working is shown – eg lines or marks indicating use of CF 17.5 and CF 52.5 with an indication on the distance axis at the correct points (or they can just show the correct readings)) 18 – 22
If a graph is drawn and answer is in the given range, then award the marks – unless from obvious incorrect working
$(d) 6 or 7 or 8 or 9$
If a graph is drawn and answer is in the given range, then award the marks – unless from obvious incorrect working
Question 14
(a) Topic-2.2 – Algebraic manipulation
(b) Topic-2.2 – Algebraic manipulation
(a) Show that $3y(2y+5)(y+7)$ can be written in the form $ay^3 + by^2 + cy$ where $a$, $b$ and $c$ are integers.
(b) Solve $\frac{2x+3}{5} + \frac{6x-5}{4} = \frac{163}{100}$
Show clear algebraic working.
▶️Answer/Explanation
Solution :-
(a) $(3y)(2y+5) = 6y^2 + 15y$
$(3y)(y+7) = 3y^2 + 21y$
$(2y+5)(y+7) = 2y^2 + 14y + 5y + 35$
$(=2y^2 + 19y + 35)$
$(6y^2 + 15y)(y+7) = 6y^3 + 42y^2 + 15y^2 + 105y$
$(3y^2 + 21y)(2y+5) = 6y^3 + 15y^2 + 42y^2 + 105y$
$3y(2y^2 + 19y + 35) = 6y^3 + 57y^2 + 105y$
working required $6y^3 + 57y^2 + 105y$
(b) eg $\frac{4(2x+3)+5(6x-5)}{20} (=1.63)$ oe or
$\frac{40x+60}{100} (+) \frac{150x-125}{100} (=\frac{163}{100})$ oe
$4(2x+3)+5(6x-5)=1.63 \times 5 \times 4$ oe
eg $8x+12+30x-25=32.6$
or
$40x+60+150x-125=163$
or
$\frac{190x-65}{100} = \frac{163}{100}$
or
$\frac{38x-13}{20} = \frac{163}{100}$ oe
$8x+30x = 32.6-12+25$
or
oe eg $38x=45.6$ or $190x=228$
working required & 1.2
Question 15
(a) Topic-2.3 – Expressions and formulae
(b) Topic-2.8 – Inequalities
(a) Make $g$ the subject of $e = \sqrt{\frac{7g+5}{11+2g}}$
(b) Solve the inequality $3y^2 + 4y – 32 > 0$
Show your working clearly.
▶️Answer/Explanation
Solution :-
(a) $e^2 = \frac{7g+5}{11+2g}$
$11e^2 + 2e^2g = 7g+5$
eg $2e^2g – 7g = 5 – 11e^2$ or $11e^2 – 5 = 7g – 2e^2g$ oe
Correct answer scores full marks (unless from obvious incorrect working) $g = \frac{5 – 11e^2}{2e^2 – 7}$
(b) $(3y-8)(y+4)$
$y = \frac{8}{3}, y = -4$
working required $y < -4, y > \frac{8}{3}$
Question 16
60 art students were asked if they would like to attend workshops for knitting (K),
for photography (P) or for embroidery (E)
Of these students
9 chose knitting, photography and embroidery
17 chose knitting and photography
16 chose photography and embroidery
20 chose knitting and embroidery
28 chose photography
39 chose embroidery
2 chose none of the workshops
(a) Using this information, complete the Venn diagram to show the numbers of students
in each subset.
One of the students is chosen at random.
Given that this student chose photography,
(b) find the probability that this student also chose knitting.
(c) Find $\quad n(P \cap K’)$
(d) Find $\quad n((P \cup E) \cap K)$
▶️Answer/Explanation
Solution :-
(a)
(b) can either ft their Venn diagram or use values given in text & $\frac{17}{28}$
(c) can either ft their Venn diagram or use values given in text & 11
(d) can either ft their Venn diagram or use values given in text & 28
Question 17
$Q$ is directly proportional to the square root of $d$
$Q = 4.5$ when $d = 324$
Find a formula for $Q$ in terms of $d$
▶️Answer/Explanation
Solution :-
$Q=k\sqrt{d}$ oe or $kQ=\sqrt{d}$ or $Q=\sqrt{kd}$
eg $4.5=k\times\sqrt{324}$ or $k=0.25$ oe
Correct answer scores full marks (unless from obvious incorrect working) & $Q=0.25\sqrt{d}$
Question 18
The straight line $\mathbf{P}$ has equation $5y + 2x = 7$
Find the gradient of a straight line that is perpendicular to $\mathbf{P}$
▶️Answer/Explanation
Solution :-
Gradient of $\mathbf{P} = -\frac{2}{5}$ or
$y = \frac{7-2x}{5}$ oe or $y = -0.4x + ….$ or
$\frac{5}{2} x$ or $y = \frac{5}{2} x (+….)$
Correct answer scores full marks (unless from obvious incorrect working) & $\frac{5}{2}$
Question 19
$G = \frac{c}{2f – 3h}$
$c = 8$ correct to the nearest whole number
$f = 6.62$ correct to 2 decimal places
$h = 1.2$ correct to 1 decimal place
Work out the lower bound for the value of $G$
Give your answer correct to 3 decimal places.
Show your working clearly.
▶️Answer/Explanation
Solution :-
7.5, 8.5, 6.615, 6.625, 1.15, 1.25
$(G=) \frac{7.5}{2 \times 6.625 – 3 \times 1.15} \left( = \frac{7.5}{13.25 – 3.45} = \frac{7.5}{9.8} = \frac{75}{98} \right)$
working required & 0.765
Question 20
Given that $k = x – y$ and $x = \frac{1}{4y}$
express $\frac{5k}{x+2}$ in the form $\frac{a-by^2}{c+dy}$ where $a, b, c$ and $d$ are integers.
▶️Answer/Explanation
Solution :-
$\frac{5(\frac{1}{4y}-y)}{\frac{1}{4y}+2}(=\frac{\frac{5}{4y}-5y}{\frac{1}{4y}+2})$ oe or
$\frac{4y(5x-5y)}{8y+1}$ oe
$\frac{\frac{5}{4y}\times4y-5y\times4y}{\frac{1}{4y}\times4y+2\times4y}$ or $\frac{\frac{5-20y^2}{4y}}{\frac{1+8y}{4y}}$ oe or
$\frac{4y(5x-5y)}{8y+1} = \frac{20xy-20y^2}{8y+1}$ oe
Correct answer scores full marks (unless from obvious incorrect working) & $\frac{5-20y^2}{1+8y}$
Question 21
The diagram shows a square ABCD and a circle.
The sides of the square are tangents to the circle.
The total area of the shaded regions is 80 cm^2
Work out the length of AC
Give your answer correct to 3 significant figures.
▶️Answer/Explanation
Solution :-
$(2r)^2 – \pi r^2$ oe or
$x^2 – \pi \times (0.5x)^2$
$4r^2 – \pi r^2 = 80$ oe eg $r^2 – 0.25\pi r^2 = 20$ or $x^2 – 0.25\pi x^2 = 80$ or
$4x^2 – \pi x^2 = 320$ oe
$r^2 = \frac{80}{4-\pi} (=93.19…)$ or $r = \sqrt{\frac{80}{4-\pi}} (=9.65…)$
$x^2 = \frac{80}{1-0.25\pi} (=372.78..)$ or $x = \sqrt{\frac{80}{1-0.25\pi}} (19.307…)$ oe eg
$\sqrt{\frac{320}{4-\pi}}$
$(AC=) \sqrt{(2 \times “9.65”)^2 + (2 \times “9.65”)^2}$ oe or
$(AC=) 2 \times \sqrt{“9.65″^2 + “9.65”^2}$
$(AC=) \sqrt{“19.307″^2 + “19.307…”^2}$ oe eg $\sqrt{8 \times \frac{80}{4-\pi}}$ oe or
$(AC=) \frac{2 \times “9.65”}{\sin 45}$ or $\frac{2 \times “9.65”}{\cos 45}$ oe
Correct answer scores full marks (unless from obvious incorrect working) & 27.3
Question 22
The straight line L has equation $x + y = 5$
The curve C has equation $2x^2 + 3y^2 = 210$
Find the coordinates of the points where L and C intersect.
Show clear algebraic working.
▶️Answer/Explanation
Solution :-
eg $2(5-y)^2 + 3y^2 = 210$ & Eg $2x^2 + 3(5-x)^2 = 210$
$\sqrt{\frac{210-3y^2}{2}} = 5-y$ oe & $\sqrt{\frac{210-2x^2}{3}} = 5-x$ oe
eg $5y^2 – 20y – 160 (=0)$ & eg $5x^2 – 30x – 135 (=0)$
or & or
$y^2 – 4y – 32 (=0)$ & $x^2 – 6x – 27 (=0)$
eg $(y-8)(y+4) (=0)$ & eg $(x-9)(x+3) (=0)$
$y = \frac{–4 \pm \sqrt{(-4)^2 – 4 \times 1 \times -32}}{2 \times 1}$ & $x = \frac{–6 \pm \sqrt{(-6)^2 – 4 \times 1 \times -27}}{2 \times 1}$
eg & eg
$(y-2)^2 – 2^2 = -32$ & $(x-3)^2 – 3^2 – 27 = 0$
(allow incorrect labels for $x/y$) & (allow incorrect labels for $x/y$)
eg $x+8 = 5$ and $x+-4 = 5$ & eg $y = 5-9$ and $y = 5–3$
(correct labels for $x/y$) & (correct labels for $x/y$)
working required & $(9, -4)$
& $(-3, 8)$
Question 23
Simplify $\frac{30 \times 25^{2x+7}}{\sqrt{180} \times (\sqrt{5})^{4x+9}}$
Give your answer in the form $5^w$ where w is an expression in terms of x
Show each stage of your working clearly.
▶️Answer/Explanation
Solution :-
For 2 of
$30 = 5 \times 6$ or $30 = 5 \times 2 \times 3$ oe (for numerator)
or
$\sqrt{180} = 6\sqrt{5}$ oe or $\sqrt{180} = 2 \times 3 \times \sqrt{5}$ (for denominator)
or
$25^{2x+7} = (5^2)^{2x+7}$ or $5^{2(2x+7)}$ oe
or
$(\sqrt{5})^{4x+9} = (5^{\frac{1}{2}})^{4x+9}$ or $5^{\frac{1}{2}(4x+9)}$ oe
or
$\frac{30}{\sqrt{180}} = \sqrt{5}$ oe
or $5^{4x+15}$ as the numerator
or $5^{2x+5}$ as the denominator
$\frac{6 \times 5 \times 5^{4x+14}}{6 \times 5^{0.5} \times 5^{2x+4.5}}$ oe eg $\frac{5^{4x+15}}{5^{2x+5}}$ or $\frac{\sqrt{5}^{8x+30}}{\sqrt{5}^{4x+10}}$ oe or $\frac{25^{2x+7.5}}{25^{x+2.5}}$ oe
working required & $5^{2x+10}$
Question 24
The diagram shows a quadrilateral $OACB$ in which
$\vec{OA} = 4\mathbf{a} \qquad \vec{OB} = 3\mathbf{b} \qquad \vec{BC} = 2\mathbf{a} + \mathbf{b}$
(a) Find $\vec{AC}$ in terms of $\mathbf{a}$ and $\mathbf{b}$
Give your answer in its simplest form.
The point $P$ lies on $AC$ such that $AP:PC = 3:2$
The point $Q$ is such that $OPQ$ and $BCQ$ are straight lines.
(b) Using a vector method, find $\vec{OQ}$ in terms of $\mathbf{a}$ and $\mathbf{b}$
Give your answer in its simplest form.
Show your working clearly.
▶️Answer/Explanation
Solution :-
For 2 of
$30 = 5 \times 6$ or $30 = 5 \times 2 \times 3$ oe (for numerator)
or
$\sqrt{180} = 6\sqrt{5}$ oe or $\sqrt{180} = 2 \times 3 \times \sqrt{5}$ (for denominator)
or
$25^{2x+7} = (5^2)^{2x+7}$ or $5^{2(2x+7)}$ oe
or
$(\sqrt{5})^{4x+9} = (5^{\frac{1}{2}})^{4x+9}$ or $5^{\frac{1}{2}(4x+9)}$ oe
or
$\frac{30}{\sqrt{180}} = \sqrt{5}$ oe
or $5^{4x+15}$ as the numerator
or $5^{2x+5}$ as the denominator
$\frac{6 \times 5 \times 5^{4x+14}}{6 \times 5^{0.5} \times 5^{2x+4.5}}$ oe eg $\frac{5^{4x+15}}{5^{2x+5}}$ or $\frac{\sqrt{5}^{8x+30}}{\sqrt{5}^{4x+10}}$ oe or $\frac{25^{2x+7.5}}{25^{x+2.5}}$ oe
working required & $5^{2x+10}$
Question 25
The diagram shows a sketch of the graph of $y = 2\sin(x+60)^\circ$
(i) Find the coordinates of the point $A$
(ii) Find the coordinates of the point $B$
▶️Answer/Explanation
Solution :-
(i) (30, 2)
(ii) (300, 0)