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Question
Mal is shopping for a school trip. He buys \(50\) tins of beans and \(20\) packets of cereal. The total cost is \(260\) Australian dollars (\({\text{AUD}}\)).
The triangular faces of a square based pyramid, \({\text{ABCDE}}\), are all inclined at \({70^ \circ }\) to the base. The edges of the base \({\text{ABCD}}\) are all \(10{\text{ cm}}\) and \({\text{M}}\) is the centre. \({\text{G}}\) is the mid-point of \({\text{CD}}\).
Write down an equation showing this information, taking \(b\) to be the cost of one tin of beans and \(c\) to be the cost of one packet of cereal in \({\text{AUD}}\).[1]
Stephen thinks that Mal has not bought enough so he buys \(12\) more tins of beans and \(6\) more packets of cereal. He pays \(66{\text{ AUD}}\).
Write down another equation to represent this information.[1]
Stephen thinks that Mal has not bought enough so he buys \(12\) more tins of beans and \(6\) more packets of cereal. He pays \(66{\text{ AUD}}\).
Find the cost of one tin of beans.[2]
(i) Sketch the graphs of the two equations from parts (a) and (b).
(ii) Write down the coordinates of the point of intersection of the two graphs.[4]
Using the letters on the diagram draw a triangle showing the position of a \({70^ \circ }\) angle.[1]
Show that the height of the pyramid is \(13.7{\text{ cm}}\), to 3 significant figures.[2]
Calculate
(i) the length of \({\text{EG}}\);
(ii) the size of angle \({\text{DEC}}\).[4]
Find the total surface area of the pyramid.[2]
Find the volume of the pyramid.[2]
Answer/Explanation
Markscheme
\(50b + 20c = 260\) (A1)[1 mark]
\(12b + 6c = 66\) (A1)[1 mark]
Solve to get \(b = 4\) (M1)(A1)(ft)(G2)
Note: (M1) for attempting to solve the equations simultaneously.[2 marks]
(i)
(A1)(A1)(A1)
Notes: Award (A1) for labels and some idea of scale, (A1)(ft)(A1)(ft) for each line.
The axis can be reversed.
(ii) \((4,3)\) or \((3,4)\) (A1)(ft)
Note: Accept \(b = 4\), \(c = 3\)[4 marks]
(A1)[1 mark]
\(\tan 70 = \frac{h}{5}\) (M1)
\(h = 5\tan 70 = 13.74\) (A1)
\(h = 13.7{\text{ cm}}\) (AG)[2 marks]
Unit penalty (UP) is applicable in this part of the question where indicated in the left hand column.
(i) \({\text{E}}{{\text{G}}^2} = {5^2} + {13.7^2}\) OR \({5^2} + {(5\tan 70)^2}\) (M1)
(UP) \({\text{EG}} = 14.6{\text{ cm}}\) (A1)(G2)
(ii) \({\text{DEC}} = 2 \times {\tan ^{ – 1}}\left( {\frac{5}{{14.6}}} \right)\) (M1)
\( = {37.8^ \circ }\) (A1)(ft)(G2)[4 marks]
Unit penalty (UP) is applicable in this part of the question where indicated in the left hand column.
\({\text{Area}} = 10 \times 10 + 4 \times 0.5 \times 10 \times 14.619\) (M1)
(UP) \( = 392{\text{ c}}{{\text{m}}^2}\) (A1)(ft)(G2)[2 marks]
Unit penalty (UP) is applicable in this part of the question where indicated in the left hand column.
\({\text{Volume}} = \frac{1}{3} \times 10 \times 10 \times 13.7\) (M1)
(UP) \( = 457{\text{ c}}{{\text{m}}^3}\) (\(458{\text{ c}}{{\text{m}}^3}\)) (A1)(G2)[2 marks]
Question
Sketch the graph of y = 2x for \( – 2 \leqslant x \leqslant 3\). Indicate clearly where the curve intersects the y-axis.[3]
Write down the equation of the asymptote of the graph of y = 2x.[2]
On the same axes sketch the graph of y = 3 + 2x − x2. Indicate clearly where this curve intersects the x and y axes.[3]
Using your graphic display calculator, solve the equation 3 + 2x − x2 = 2x.[2]
Write down the maximum value of the function f (x) = 3 + 2x − x2.[1]
Use Differential Calculus to verify that your answer to (e) is correct.[5]
The curve y = px2 + qx − 4 passes through the point (2, –10).
Use the above information to write down an equation in p and q.[2]
The gradient of the curve \(y = p{x^2} + qx – 4\) at the point (2, –10) is 1.
Find \(\frac{{{\text{d}}y}}{{{\text{d}}x}}\).[2]
The gradient of the curve \(y = p{x^2} + qx – 4\) at the point (2, –10) is 1.
Hence, find a second equation in p and q.[1]
The gradient of the curve \(y = p{x^2} + qx – 4\) at the point (2, –10) is 1.
Solve the equations to find the value of p and of q.[3]
Answer/Explanation
Markscheme
(A1)(A1)(A1)
Note: Award (A1) for correct domain, (A1) for smooth curve, (A1) for y-intercept clearly indicated.[3 marks]
y = 0 (A1)(A1)
Note: Award (A1) for y = constant, (A1) for 0.[2 marks]
Note: Award (A1) for smooth parabola,
(A1) for vertex (maximum) in correct quadrant.
(A1) for all clearly indicated intercepts x = −1, x = 3 and y = 3.
The final mark is to be applied very strictly. (A1)(A1)(A1)[3 marks]
x = −0.857 x = 1.77 (G1)(G1)
Note: Award a maximum of (G1) if x and y coordinates are both given.[2 marks]
4 (G1)
Note: Award (G0) for (1, 4).[1 mark]
\(f'(x) = 2 – 2x\) (A1)(A1)
Note: Award (A1) for each correct term.
Award at most (A1)(A0) if any extra terms seen.
\(2 – 2x = 0\) (M1)
Note: Award (M1) for equating their gradient function to zero.
\(x = 1\) (A1)(ft)
\(f (1) = 3 + 2(1) – (1)^2 = 4\) (A1)
Note: The final (A1) is for substitution of x = 1 into \(f (x)\) and subsequent correct answer. Working must be seen for final (A1) to be awarded.[5 marks]
22 × p + 2q − 4 = −10 (M1)
Note: Award (M1) for correct substitution in the equation.
4p + 2q = −6 or 2p + q = −3 (A1)
Note: Accept equivalent simplified forms.[2 marks]
\(\frac{{{\text{d}}y}}{{{\text{d}}x}} = 2px + q\) (A1)(A1)
Note: Award (A1) for each correct term.
Award at most (A1)(A0) if any extra terms seen.[2 marks]
4p + q = 1 (A1)(ft)[1 mark]
4p + 2q = −6
4p + q = 1 (M1)
Note: Award (M1) for sensible attempt to solve the equations.
p = 2, q = −7 (A1)(A1)(ft)(G3)[3 marks]
Question
Part A
The Green Park Amphitheatre was built in the form of a horseshoe and has 20 rows. The number of seats in each row increase by a fixed amount, d, compared to the number of seats in the previous row. The number of seats in the sixth row, u6, is 100, and the number of seats in the tenth row, u10, is 124. u1 represents the number of seats in the first row.
Part B
Frank is at the amphitheatre and receives a text message at 12:00. Five minutes later he forwards the text message to three people. Five minutes later, those three people forward the text message to three new people. Assume this pattern continues and each time the text message is sent to people who have not received it before.
The number of new people who receive the text message forms a geometric sequence
1 , 3 , …
(i) Write an equation for u6 in terms of d and u1.
(ii) Write an equation for u10 in terms of d and u1.[2]
Write down the value of
(i) d ;
(ii) u1 .[2]
Find the total number of seats in the amphitheatre.[3]
A few years later, a second level was added to increase the amphitheatre’s capacity by another 1600 seats. Each row has four more seats than the previous row. The first row on this level has 70 seats.
Find the number of rows on the second level of the amphitheatre.[4]
Write down the next two terms of this geometric sequence.[1]
Write down the common ratio of this geometric sequence.[1]
Calculate the number of people who will receive the text message at 12:30.[2]
Calculate the total number of people who will have received the text message by 12:30.[2]
Calculate the exact time at which a total of 29 524 people will have received the text message.[3]
Answer/Explanation
Markscheme
(i) u1 + 5d = 100 (A1)
(ii) u1 + 9d = 124 (A1)[2 marks]
(i) 6 (G1)(ft)
(ii) 70 (G1)(ft)
Notes: Follow through from their equations in parts (a) and (b) even if working not seen. Their answers must be integers. Award (M1)(A0) for an attempt to solve two equations analytically.[2 marks]
\(S_{20} = \frac{20}{2}(2 \times 70 + (20 – 1) \times 6)\) (M1)(A1)(ft)
Note: Award (M1) for substituted sum of AP formula, (A1)(ft) for their correct substituted values.
= 2540 (A1)(ft)(G2)
Note: Follow through from their part (b).[3 marks]
\(\frac{n}{2}(2 \times 70 + (n – 1) \times 4) = 1600\) (M1)(A1)
Note: Award (M1) for substituted sum of AP formula, (A1) for their correct substituted values.
4n2 +136n – 3200 = 0 (M1)
Note: Award (M1) for this equation (or other equivalent expanded quadratic) seen, may be implied if correct final answer seen.
n = 16 (A1)(G3)
Note: Do not award the final (A1) for n = 16, – 50 given as final answer, award (G2) if n = 16, – 50 given as final answer without working.[4 marks]
9, 27 (A1)[1 mark]
3 (A1)[1 mark]
1 × 36 (M1)
= 729 (A1)(ft)(G2)
Note: Award (M1) for correctly substituted GP formula. Follow through from their answer to part (b).[2 marks]
\(\frac{{1(3^7 – 1)}}{(3 – 1)}\) (M1)
Note: Award (M1) for correctly substituted GP formula. Accept sum 1+ 3 + 9 + 27 + … + 729. If lists are used, award (M1) for correct list that includes 1093. (1, 4, 13, 40, 121, 364, 1093, 3280…)
= 1093 (A1)(ft)(G2)
Note: Follow through from their answer to part (b). For consistent use of n = 6 from part (c) (243) to part (d) leading
to an answer of 364, treat as double penalty and award (M1)(A1)(ft) if working is shown.[2 marks]
\(\frac{{1(3^n – 1)}}{(3 – 1)} = 29524\) (M1)
Note: Award (M1) for correctly substituted GP formula. If lists are used, award (M1) for correct list that includes 29524. (1, 4, 13, 40, 121, 364, 1093, 3280, 9841, 29524, 88573…). Accept alternative methods, for example continuation of sum in part (d).
n = 10 (A1)(ft)
Note: Follow through from their answer to part (b).
Exact time = 12:45 (A1)(ft)(G2)[3 marks]
Question
Cedric wants to buy an €8000 car. The car salesman offers him a loan repayment option of a 25 % deposit followed by 12 equal monthly payments of €600 .
Write down the amount of the deposit.[1]
Calculate the total cost of the loan under this repayment scheme.[2]
Cedric’s mother decides to help him by giving him an interest free loan of €8000 to buy the car. She arranges for him to repay the loan by paying her €x in the first month and €y in every following month until the €8000 is repaid.
The total amount that Cedric’s mother receives after 12 months is €3500. This can be written using the equation x +11y = 3500. The total amount that Cedric’s mother receives after 24 months is €7100.
Write down a second equation involving x and y.[1]
Cedric’s mother decides to help him by giving him an interest free loan of €8000 to buy the car. She arranges for him to repay the loan by paying her €x in the first month and €y in every following month until the €8000 is repaid.
The total amount that Cedric’s mother receives after 12 months is €3500. This can be written using the equation x +11y = 3500. The total amount that Cedric’s mother receives after 24 months is €7100.
Write down the value of x and the value of y.[2]
Cedric’s mother decides to help him by giving him an interest free loan of €8000 to buy the car. She arranges for him to repay the loan by paying her €x in the first month and €y in every following month until the €8000 is repaid.
The total amount that Cedric’s mother receives after 12 months is €3500. This can be written using the equation x +11y = 3500. The total amount that Cedric’s mother receives after 24 months is €7100.
Calculate the number of months it will take Cedric’s mother to receive the €8000.[3]
Cedric decides to buy a cheaper car for €6000 and invests the remaining €2000 at his bank. The bank offers two investment options over three years.
Option A: Compound interest at an annual rate of 8 %.
Option B: Compound interest at a nominal annual rate of 7.5 % , compounded monthly.
Express each answer in part (f) to the nearest euro.
Calculate the value of his investment at the end of three years if he chooses
(i) Option A;
(ii) Option B.[5]
Answer/Explanation
Markscheme
2000 (euros) (A1)[1 mark]
\(2000 + 12 \times 600\) (M1)
Note: Award (M1) for addition of two correct terms.
9200 (euros) (A1)(ft)(G2)
Note: Follow through from their part (a).[2 marks]
x + 23y = 7100 (A1)[1 mark]
x = 200, y = 300 (A1)(ft)(A1)(ft)(G2)[2 marks]
\(200 + n \times 300 = 8000\) (M1)
Note: Award (M1) for setting up the equation. Follow through from their x and y found in part (d).
n = 26 (A1)(ft)
26 + 1 = 27 (months) (A1)(ft)(G3)
Notes: Middle line n = 26 may be implied if correct answer given. The final (A1)(ft) is for adding 1 to their value of n (even if it is incorrect). Follow through from their part (d). If the final answer is not a positive integer award at most (M1)(A1)(ft)(A0). Award (G2) for final answer of 26.
OR
\(\frac{{8000 – 7100}}{{300}} + 24\) (M1)(A1)
Note: Award (M1) for division of difference by their value of y, (A1) for 24 seen.
27 (months) (A1)(ft)(G3)
Note: Follow through from their value of y.[3 marks]
(i) \(2000{\left( {1 + \frac{8}{{100}}} \right)^3}\) (M1)
Note: Award (M1) for correct substitution in compound interest formula.
2519 (euros) (A1)(G2)
Note: If the answer is not given to the nearest euro award at most (M1)(A0).(ii) \(2000{\left( {1 + \frac{7.5}{{100 \times 12}}} \right)^{3\times12}}\) (M1)(A1)
Note: Award (M1) for substitution in compound interest formula, (A1) for correct substitutions.
2503 (euros) (A1)(G2)
Note: If the answer is not given to the nearest euro, award at most (M1)(A1)(A0), provided this has not been penalized in part (f)(i).[5 marks]
Question
In a college 450 students were surveyed with the following results
150 have a television
205 have a computer
220 have an iPhone
75 have an iPhone and a computer
60 have a television and a computer
70 have a television and an iPhone
40 have all three.
Draw a Venn diagram to show this information. Use T to represent the set of students who have a television, C the set of students who have a computer and I the set of students who have an iPhone.[4]
Write down the number of students that
(i) have a computer only;
(ii) have an iPhone and a computer but no television.[2]
Write down \(n[T \cap (C \cup I)’]\).[1]
Calculate the number of students who have none of the three.[2]
Two students are chosen at random from the 450 students. Calculate the probability that
(i) neither student has an iPhone;
(ii) only one of the students has an iPhone.[6]
The students are asked to collect money for charity. In the first month, the students collect x dollars and the students collect y dollars in each subsequent month. In the first 6 months, they collect 7650 dollars. This can be represented by the equation x + 5y = 7650.
In the first 10 months they collect 13 050 dollars.
(i) Write down a second equation in x and y to represent this information.
(ii) Write down the value of x and of y .[3]
The students are asked to collect money for charity. In the first month, the students collect x dollars and the students collect y dollars in each subsequent month. In the first 6 months, they collect 7650 dollars. This can be represented by the equation x + 5y = 7650.
In the first 10 months they collect 13 050 dollars.
Calculate the number of months that it will take the students to collect 49 500 dollars.[3]
Answer/Explanation
Markscheme
(A1)(A1)(A1)(A1)
Notes: Award (A1) for labelled sets T, C, and I included inside an enclosed universal set. (Label U is not essential.) Award (A1) for central entry 40. (A1) for 20, 30 and 35 in the other intersecting regions. (A1) for 60, 110 and 115 or T(150), C(205), I(220).[4 marks]
In parts (b), (c) and (d) follow through from their diagram.
(i) 110 (A1)(ft)
(ii) 35 (A1)(ft) [2 marks]
In parts (b), (c) and (d) follow through from their diagram.
60 (A1)(ft)[2 marks]
In parts (b), (c) and (d) follow through from their diagram.
450 − (60 + 20 + 40 + 30 + 115 + 35 + 110) (M1)
Note: Award (M1) for subtracting all their values from 450.
= 40 (A1)(ft)(G2)[2 marks]
(i) \(\frac{{230}}{{450}} \times \frac{{229}}{{449}}\) (A1)(M1)
Note: Award (A1) for correct fractions, (M1) for multiplying their fractions.
\(\frac{{52670}}{{202050}}\left( {\frac{{5267}}{{20205}},{\text{ 0}}{\text{.261, 26}}{\text{.1% }}} \right)(0.26067…)\) (A1)(G2)
Note: Follow through from their Venn diagram in part (a).
(ii) \(\frac{{220}}{{450}} \times \frac{{230}}{{449}} + \frac{{230}}{{450}} \times \frac{{220}}{{449}}\) (A1)(A1)
Note: Award (A1) for addition of their products, (A1) for two correct products.
OR
\(\frac{{230}}{{450}} \times \frac{{220}}{{449}} \times 2\) (A1)(A1)
Notes: Award (A1) for their product of two fractions multiplied by 2, (A1) for correct product of two fractions multiplied by 2. Award (A0)(A0) if correct product is seen not multiplied by 2.
\(\frac{{2024}}{{4041}}(0.501,{\text{ 50}}{\text{.1% )(0}}{\text{.50086}}…{\text{)}}\) (A1)(G2)
Note: Follow through from their Venn diagram in part (a) and/or their 230 used in part (e)(i).
Note: For consistent use of replacement in parts (i) and (ii) award at most (A0)(M1)(A0) in part (i) and (A1)(ft)(A1)(A1)(ft) in part (ii).[6 marks]
(i) x + 9y = 13050 (A1)
(ii) x = 900 (A1)(ft)
y = 1350 (A1)(ft)
Notes: Follow through from their equation in (f)(i). Do not award (A1)(ft) if answer is negative. Award (M1)(A0) for an attempt at solving simultaneous equations algebraically but incorrect answer obtained. [3 marks]
49500 = 900 + 1350n (A1)(ft)
Notes: Award (A1)(ft) for setting up correct equation. Follow through from candidate’s part (f).
n = 36 (A1)(ft)
The total number of months is 37. (A1)(ft)(G2)
Note: Award (G1) for 36 seen as final answer with no working. The value of n must be a positive integer for the last two (A1)(ft) to be awarded.
OR
49500 = 900 + 1350(n − 1) (A2)(ft)
Notes: Award (A2)(ft) for setting up correct equation. Follow through from candidate’s part (f).
n = 37 (A1)(ft)(G2)
Note: The value of n must be a positive integer for the last (A1)(ft) to be awarded.[3 marks]
Question
George leaves a cup of hot coffee to cool and measures its temperature every minute. His results are shown in the table below.
Write down the decrease in the temperature of the coffee
(i) during the first minute (between t = 0 and t =1) ;
(ii) during the second minute;
(iii) during the third minute.[3]
Assuming the pattern in the answers to part (a) continues, show that \(k = 19\).[2]
Use the seven results in the table to draw a graph that shows how the temperature of the coffee changes during the first six minutes.
Use a scale of 2 cm to represent 1 minute on the horizontal axis and 1 cm to represent 10 °C on the vertical axis.[4]
The function that models the change in temperature of the coffee is y = p (2−t )+ q.
(i) Use the values t = 0 and y = 94 to form an equation in p and q.
(ii) Use the values t =1 and y = 54 to form a second equation in p and q.[2]
Solve the equations found in part (d) to find the value of p and the value of q.[2]
The graph of this function has a horizontal asymptote.
Write down the equation of this asymptote.[2]
George decides to model the change in temperature of the coffee with a linear function using correlation and linear regression.
Use the seven results in the table to write down
(i) the correlation coefficient;
(ii) the equation of the regression line y on t.[4]
Use the equation of the regression line to estimate the temperature of the coffee at t = 3.[2]
Find the percentage error in this estimate of the temperature of the coffee at t = 3.[2]
Answer/Explanation
Markscheme
(i) 40
(ii) 20
(iii) 10 (A3)
Notes: Award (A0)(A1)(ft)(A1)(ft) for −40, −20, −10.
Award (A1)(A0)(A1)(ft) for 40, 60, 70 seen.
Award (A0)(A0)(A1)(ft) for −40, −60, −70 seen.
\(24 – k = 5\) or equivalent (A1)(M1)
Note: Award (A1) for 5 seen, (M1) for difference from 24 indicated.
\(k = 19\) (AG)
Note: If 19 is not seen award at most (A1)(M0).
(A1)(A1)(A1)(A1)
Note: Award (A1) for scales and labelled axes (t or “time” and y or “temperature”).
Accept the use of x on the horizontal axis only if “time” is also seen as the label.
Award (A2) for all seven points accurately plotted, award (A1) for 5 or 6 points accurately plotted, award (A0) for 4 points or fewer accurately plotted.
Award (A1) for smooth curve that passes through all points on domain [0, 6].
If graph paper is not used or one or more scales is missing, award a maximum of (A0)(A0)(A0)(A1).
(i) \(94 = p + q\) (A1)
(ii) \(54 = 0.5p + q\) (A1)
Note: The equations need not be simplified; accept, for example \(94 = p(2^{-0}) + q\).
p = 80, q = 14 (G1)(G1)(ft)
Note: If the equations have been incorrectly simplified, follow through even if no working is shown.
y = 14 (A1)(A1)(ft)
Note: Award (A1) for y = a constant, (A1) for their 14. Follow through from part (e) only if their q lies between 0 and 15.25 inclusive.
(i) –0.878 (–0.87787…) (G2)
Note: Award (G1) if –0.877 seen only. If negative sign omitted award a maximum of (A1)(A0).
(ii) y = –11.7t + 71.6 (y = –11.6517…t + 71.6336…) (G1)(G1)
Note: Award (G1) for –11.7t, (G1) for 71.6.
If y = is omitted award at most (G0)(G1).
If the use of x in part (c) has not been penalized (the axis has been labelled “time”) then award at most (G0)(G1).
−11.6517…(3) + 71.6339… (M1)
Note: Award (M1) for correct substitution in their part (g)(ii).
= 36.7 (36.6785…) (A1)(ft)(G2)
Note: Follow through from part (g). Accept 36.5 for use of the 3sf answers from part (g).
\(\frac{{36.6785… – 24}}{{24}} \times 100\) (M1)
Note: Award (M1) for their correct substitution in percentage error formula.
= 52.8% (52.82738…) (A1)(ft)(G2)
Note: Follow through from part (h). Accept 52.1% for use of 36.5.
Accept 52.9 % for use of 36.7. If partial working (\(\times 100\) omitted) is followed by their correct answer award (M1)(A1). If partial working is followed by an incorrect answer award (M0)(A0). The percentage sign is not required.