2013 p2 past paper
[263 marks]
The table shows the distance, in km, of eight regional railway stations from a city centre terminus and the price, in $, of a return ticket
from each regional station to the terminus.
1a. Draw a scatter diagram for the above data. Use a scale of 1 cm to represent 10 km on the x-axis and 1 cm to represent $10 on
the y-axis.
Markscheme
(A4)
Notes: Award (A1) for correct scale and labels (accept x and y).
Award (A3) for 7 or 8 points plotted correctly.
Award (A2) for 5 or 6 points plotted correctly.
Award (A1) for 3 or 4 points plotted correctly.
Award at most (A1)(A2) if points are joined up.
If axes are reversed, award at most (A0)(A3).
If graph paper is not used, award at most (A1)(A0).
[4 marks]
[4 marks]
[2 marks]
1b. Use your graphic display calculator to find
(i)
x̄, the mean of the distances;
(ii)
ȳ , the mean of the prices.
Markscheme
(x̄ =) 46
(i)
(ȳ =) 57
(ii)
(G1)
(G1)
[2 marks]
1c. Plot and label the point M (x̄, ȳ ) on your scatter diagram.
[1 mark]
Markscheme
M(46,57) plotted and labelled on the scatter diagram
(A1)(ft)
Notes: Follow through from their part (b).
Accept (x̄, ȳ ) as the label.
[1 mark]
1d. Use your graphic display calculator to find
[3 marks]
(i) the product–moment correlation coefficient, r ;
(ii) the equation of the regression line y on x.
Markscheme
0.986 (0.986322...)
(i)
(G1)
y = 1.01x + 10.3 (y = 1.01431 … x + 10.3412 …)
(ii)
(G1)(G1)
Notes: Award (G1) for 1.01x, (G1) for 10.3.
Award (G1)(G0) if not written in the form of an equation.
OR
(y − 57) = 1.01(x − 46) (y − 57 = 1.01431...(x − 46))
(G1)(G1)(ft)
Note: Award (G1) for 1.01, (G1) for their 57 and 46.
[3 marks]
1e. Draw the regression line y on x on your scatter diagram.
Markscheme
straight line drawn on the scatter diagram
(A1)(ft)(A1)(ft)
Notes: The line must be straight for either of the two marks to be awarded.
Award (A1)(ft) passing through their M plotted in (c).
Award (A1)(ft) for correct y-intercept (between 9 and 12).
Follow through from their y-intercept found in part (d).
If part (d) is used, award (A1)(ft) for their intercept (±1).
[2 marks]
[2 marks]
1f. A ninth regional station is 76 km from the city centre terminus.
[3 marks]
Use the equation of the regression line to estimate the price of a return ticket to the city centre terminus from this regional station. Give
your answer correct to the nearest $.
Markscheme
y = 1.01431... × 76 + 10.3412 …
(M1)
Note: Award (M1) for substitution of 76 into their regression line.
= 87.4295 …
(A1)(ft)
Note: Follow through from part (d). If 3 sf values are used the value is 87.06.
$87
(A1)(ft)(G2)
Notes: The final (A1) is awarded for their answer given correct to the nearest dollar.
Method, followed by the answer of 87 earns (M1)(G2). It is not necessary to see the interim step.
Where the candidate uses their graph instead of the equation, and arrives at an answer other than 87, award, at most, (G1)(ft).
If the candidate uses their graph and arrives at the required answer of 87, award (G2)(ft).
[3 marks]
[1 mark]
1g. Give a reason why it is valid to use your regression line to estimate the price of this return ticket.
Markscheme
76 is within the range of distances given in the data OR the correlation coefficient is close to 1.
(R1)
Notes: Award (R1) if either condition is given.
Sufficient to indicate that 76 is ‘within the data range’ and the correlation is ‘strong’.
Allow r2 close to 1.
Do not accept “within the range of prices”.
[1 mark]
1h. The actual price of the return ticket is $80.
Using your answer to part (f), calculate the percentage error in the estimated price of the ticket.
Markscheme
Percentage error =
87−80
80
× 100
(M1)
Note: Award (M1) for correct substitution into formula.
8.75%
(A1)(ft)(G2)
Notes: Follow through from their answer to part (f).
Accept either the rounded or unrounded answer to part (f).
If no integer value seen in part (f), follow through from their unrounded answer to part (f).
Answer must be positive.
[2 marks]
[2 marks]
A manufacturer has a contract to make 2600 solid blocks of wood. Each block is in the shape of a right triangular prism, ABCDEF,
as shown in the diagram.
^ C = 35∘ .
AB = 30 cm, BC = 24 cm, CD = 25 cm and angle AB
2a. Calculate the length of AC .
[3 marks]
Markscheme
AC 2 = 302 + 242 − 2 × 30 × 24 × cos 35∘
(M1)(A1)
Note: Award (M1) for substituted cosine rule formula,
(A1) for correct substitutions.
AC = 17.2 cm (17.2168 …)
(A1)(G2)
Notes: Use of radians gives 52.7002 … Award (M1)(A1)(A0).
No marks awarded in this part of the question where candidates assume that angle ACB = 90∘ .
[3 marks]
2b. Calculate the area of triangle ABC.
[3 marks]
Markscheme
Units are required in part (b).
Area of triangle ABC =
1
2
× 24 × 30 × sin 35∘
(M1)(A1)
Notes: Award (M1) for substitution into area formula, (A1) for correct substitutions.
Special Case: Where a candidate has assumed that angle ACB = 90∘ in part (a), award (M1)(A1) for a correct alternative
substituted formula for the area of the triangle (ie
= 206 cm2 (206.487 … cm2 )
1
2
× base × height) .
(A1)(G2)
Notes: Use of radians gives negative answer, – 154.145 … Award (M1)(A1)(A0).
Special Case: Award (A1)(ft) where the candidate has arrived at an area which is correct to the standard rounding rules from their
lengths (units required).
[3 marks]
3
[2 marks]
2c. Assuming that no wood is wasted, show that the volume of wood required to make all 2600 blocks is 13 400 000 cm , correct
to three significant figures.
Markscheme
206.487 … × 25 × 2600
(M1)
Note: Award (M1) for multiplication of their answer to part (b) by 25 and 2600.
13 421 688.61
(A1)
Note: Accept unrounded answer of 13 390 000 for use of 206.
13 400 000
(AG)
Note: The final (A1) cannot be awarded unless both the unrounded and rounded answers are seen.
[2 marks]
k
2d. Write 13 400 000 in the form a × 10 where 1 ⩽ a < 10 and k ∈ Z.
[2 marks]
Markscheme
1.34 × 107
(A2)
Notes: Award (A2) for the correct answer.
Award (A1)(A0) for 1.34 and an incorrect index value.
Award (A0)(A0) for any other combination (including answers such as 13.4 × 106 ).
[2 marks]
2
2e. Show that the total surface area of one block is 2190 cm , correct to three significant figures.
[3 marks]
Markscheme
2 × 206.487 … + 24 × 25 + 30 × 25 + 17.2168 … × 25
(M1)(M1)
Note: Award (M1) for multiplication of their answer to part (b) by 2 for area of two triangular ends, (M1) for three correct rectangle
areas using 24, 30 and their 17.2.
2193.26 …
(A1)
Note: Accept 2192 for use of 3 sf answers.
2190
(AG)
Note: The final (A1) cannot be awarded unless both the unrounded and rounded answers are seen.
[3 marks]
2
2f. The blocks are to be painted. One litre of paint will cover 22 m .
Calculate the number of litres required to paint all 2600 blocks.
[3 marks]
Markscheme
2190×2600
22×10 000
(M1)(M1)
Notes: Award (M1) for multiplication by 2600 and division by 22, (M1) for division by 10 000.
The use of 22 may be implied ie division by 2200 would be acceptable.
25.9 litres (25.8818 …)
(A1)(G2)
Note: Accept 26.
[3 marks]
A group of 120 women in the USA were asked whether they had visited the continents of Europe (E) or South America (S) or Asia (
A).
7 had visited all three continents
28 had visited Europe only
22 had visited South America only
16 had visited Asia only
15 had visited Europe and South America but had not visited Asia
x had visited South America and Asia but had not visited Europe
2x had visited Europe and Asia but had not visited South America
20 had not visited any of these continents
3a. Draw a Venn diagram, using sets labelled E, S and A, to show this information.
[5 marks]
Markscheme
(A1)(A1)(A1)(A1)(A1)
Notes: Award (A1) for rectangle and three labelled intersecting circles.
Award (A1) for 7 in correct place.
Award (A1) for 28, 22 and 16 in the correct places.
Award (A1) for 15, x and 2x in the correct places.
Award (A1) for 20 in the correct place.
Accept 4 and 8 instead of x and 2x.
Do not penalize if U is omitted from the diagram.
[5 marks]
3b. Calculate the value of x.
[2 marks]
Markscheme
3x = 120 − (20 + 28 + 15 + 22 + 7 + 16)
(M1)
Note: Award (M1) for setting up a correct equation involving x, the 120 and values from their diagram.
x=4
(A1)(ft)(G2)
Note: Follow through from part (a). For the follow through to be awarded x must be a positive integer.
[2 marks]
′
3c. Explain, in words, the meaning of (E ∪ S) ∩ A .
[2 marks]
Markscheme
(Women who had visited) Europe or South America and (but had) not (visited) Asia
(A1)(A1)
Notes: Award (A1) for “(visited) Europe or South America” (or both).
Award (A1) for “and (but) had not visited Asia”.
E(urope) union S(outh America) intersected with not A(sia) earns no marks, (A0).
[2 marks]
3d. Write down n ((E ∪ S ∪ A) ) .
′
[1 mark]
Markscheme
20
(A1)
Note: Award (A0) for the embedded answer of n(20).
[1 mark]
3e. Find the probability that a woman selected at random from the group had visited Europe.
[2 marks]
Markscheme
58
120
( 29
, 0.483, 48.3% ) (0.48333 … )
60
(A1)(ft)(A1)(G2)
Note: Award (A1)(ft) for numerator, follow through from their value of x, or their diagram, (A1) for denominator.
[2 marks]
3f. Find the probability that a woman selected at random from the group had visited Europe, given that she had visited Asia.
[2 marks]
Markscheme
( 37 , 0.429, 42.9% ) (0.428571 … )
15
35
(A1)(ft)(A1)(ft)(G2)
Note: Award (A1)(ft) for numerator, (A1)(ft) for denominator, follow through from their value of x or their diagram.
[2 marks]
[3 marks]
3g. Two women from the group are selected at random.
Find the probability that both women selected had visited South America.
Markscheme
48
120
47
× 119
(A1)(ft)(M1)
Notes: Award (A1)(ft) for two correct fractions, follow through from their denominator in part (e), follow through the numerator from
their answer to part (b) or from their diagram, (M1) for multiplication of their two fractions.
=
2256
14 280
94
( 595
, 0.158, 15,8% ) (0.157983 … )
(A1)(ft)(G2)
Notes: Award (A1)(M1)(A1) for correct fractions, correctly multiplied together with an answer of 0.16.
48
48
Award (A0)(M1)(A0) for 120
× 120
= 0.16.
Award (G1) for an answer of 0.16 with no working seen.
[3 marks]
Consider the function f(x) = 34 x4 − x3 − 9x2 + 20.
[2 marks]
4a. Find f(−2).
Markscheme
3
(−2)4
4
− (−2)3 − 9(−2)2 + 20
(M1)
Note: Award (M1) for substituting x = −2 in the function.
=4
(A1)(G2)
Note: If the coordinates (−2, 4) are given as the answer award, at most, (M1)(A0). If no working shown award (G1).
If x = −2, y = 4 seen then award full marks.
[2 marks]
′
4b. Find f (x).
[3 marks]
Markscheme
3x3 − 3x2 − 18x
(A1)(A1)(A1)
Note: Award (A1) for each correct term, award at most (A1)(A1)(A0) if extra terms seen.
[3 marks]
4c. The graph of the function f(x) has a local minimum at the point where x = −2.
[5 marks]
Using your answer to part (b), show that there is a second local minimum at x = 3.
Markscheme
f ′ (3) = 3 × (3)3 − 3 × (3)2 − 18 × 3
(M1)
Note: Award (M1) for substitution in their f ′ (x) of x = 3.
=0
(A1)
OR
3x3 − 3x2 − 18x = 0
(M1)
Note: Award (M1) for equating their f ′ (x) to zero.
x=3
(A1)
f (x1 ) = 3 × (x1 )3 − 3 × (x1 )2 − 18 × x1 < 0 where 0 < x1 < 3
′
(M1)
Note: Award (M1) for substituting a value of x1 in the range 0 < x1 < 3 into their f ′ and showing it is negative (decreasing).
f ′ (x2 ) = 3 × (x2 )3 − 3 × (x2 )2 − 18 × x2 > 0 where x2 > 3
(M1)
Note: Award (M1) for substituting a value of x2 in the range x2 > 3 into their f ′ and showing it is positive (increasing).
OR
With or without a sketch:
Showing f(x1 ) > f(3) where x1 < 3 and x1 is close to 3.
Showing f(x2 ) > f(3) where x2 > 3 and x2 is close to 3.
(M1)
(M1)
Note: If a sketch of f(x) is drawn in this part of the question and x = 3 is identified as a stationary point on the curve, then
(i) award, at most, (M1)(A1)(M1)(M0) if the stationary point has been found;
(ii) award, at most, (M0)(A0)(M1)(M0) if the stationary point has not been previously found.
Since the gradients go from negative (decreasing) through zero to positive (increasing) it is a local minimum
Note: Only award (R1) if the first two marks have been awarded ie f ′ (3) has been shown to be equal to 0.
[5 marks]
f(x)
f(x)
y
−5 ⩽ x ⩽ 5
x = −2
−40 ⩽ y ⩽ 50
(R1)(AG)
y
Markscheme
(A1)(A1)(A1)(A1)
Notes: Award (A1) for labelled axes and indication of scale on both axes.
Award (A1) for smooth curve with correct shape.
Award (A1) for local minima in 2nd and 4th quadrants.
Award (A1) for y intercept (0,20) seen and labelled. Accept 20 on y-axis.
Do not award the third (A1) mark if there is a turning point on the x-axis.
If the derivative function is sketched then award, at most, (A1)(A0)(A0)(A0).
For a smooth curve (with correct shape) there should be ONE continuous thin line, no part of which is straight and no (one to
many) mappings of x.
[4 marks]
4e. The graph of the function f(x) has a local minimum at the point where x = −2.
[2 marks]
Write down the coordinates of the local maximum.
Markscheme
(0,20)
(G1)(G1)
Note: If parentheses are omitted award (G0)(G1).
OR
x = 0, y = 20
(G1)(G1)
Note: If the derivative function is sketched in part (d), award (G1)(ft)(G1)(ft) for (– 1.12,12.2).
[2 marks]
4f. Let T be the tangent to the graph of the function f(x) at the point (2,– 12).
Find the gradient of T .
Markscheme
f ′ (2) = 3(2)3 − 3(2)2 − 18(2)
(M1)
Notes: Award (M1) for substituting x = 2 into their f ′ (x).
= −24
[2 marks]
(A1)(ft)(G2)
[2 marks]
4g. The line L passes through the point (2,−12) and is perpendicular to T .
[5 marks]
L has equation x + by + c = 0, where b and c ∈ Z.
Find
(i)
the gradient of L;
(ii)
the value of b and the value of c.
Markscheme
Gradient of perpendicular =
(i)
1
24
(0.0417,0.041666 …)
(A1)(ft)(G1)
Note: Follow through from part (f).
y + 12 =
(ii)
1
(x − 2)
24
(M1)(M1)
Note: Award (M1) for correct substitution of (2,– 12), (M1) for correct substitution of their perpendicular gradient into equation of
line.
OR
1
× 2 + d (M1)
24
145
− 12
1
x − 145
(M1)
24
12
−12 =
d=
y=
Note: Award (M1) for correct substitution of (2,– 12) and gradient into equation of a straight line, (M1) for correct substitution of the
perpendicular gradient and correct substitution of dinto equation of line.
b = −24, c = −290
(A1)(ft)(A1)(ft)(G3)
Note: Follow through from parts (f) and g(i).
To award (ft) marks, b and c must be integers.
Where candidate has used 0.042 from g(i), award (A1)(ft) for – 288.
[5 marks]
Give all answers in this question correct to two decimal places.
Arthur lives in London. On 1st August 2008 Arthur paid 37 500 euros (EUR) for a new car from Germany. The price of the same car
in London was 34 075 British pounds (GBP).
The exchange rate on 1st August 2008 was 1 EUR = 0.7234GBP.
5a. Calculate, in GBP, the price that Arthur paid for the car.
[2 marks]
Markscheme
The first answer not given to two decimal places is not awarded the final (A1). Incorrect rounding is not penalized thereafter.
37 500 × 0.7234
= 27 127.50
[2 marks]
5b.
(M1)
(A1)(G2)
GBP
Markscheme
The first answer not given to two decimal places is not awarded the final (A1). Incorrect rounding is not penalized thereafter.
6947.50 (A1)(ft)(G1)
Note: Follow through from part (a) irrespective of whether working is seen.
[1 mark]
st
st
5c. Between 1 August 2008 and 1 August 2012 Arthur’s car depreciated at an annual rate of 9% of its current value.
[3 marks]
st
Calculate the value, in GBP, of Arthur’s car on 1 August 2009.
Markscheme
The first answer not given to two decimal places is not awarded the final (A1). Incorrect rounding is not penalized thereafter.
27 127.50 × 0.91 (A1)(M1)
Note: Award (A1) for 0.91 seen or equivalent, (M1) for their 27 127.50 multiplied by 0.91
OR
27 127.50 − 0.09 × 27 127.50
(A1)(M1)
Note: Award (A1) for 0.09 × 27 127.50 seen, and (M1) for 27 127.50 − 0.09 × 27 127.50.
= 24 686.03
(A1)(ft)(G2)
Note: Follow through from part (a).
[3 marks]
st
st
5d. Between 1 August 2008 and 1 August 2012 Arthur’s car depreciated at an annual rate of 9% of its current value.
st
Show that the value of Arthur’s car on 1 August 2012 was 18 600 GBP, correct to the nearest 100 GBP.
[3 marks]
Markscheme
The first answer not given to two decimal places is not awarded the final (A1). Incorrect rounding is not penalized thereafter.
9
27 127.50 × (1 − 100
)
4
(M1)(A1)(ft)
Notes: Award (M1) for substituted compound interest formula, (A1)(ft) for correct substitution.
Follow through from part (a).
OR
27 127.50 × (0.91)4
(M1)(A1)(ft)
Notes: Award (M1) for substituted geometric sequence formula, (A1)(ft) for correct substitution.
Follow through from part (a).
OR (lists (i))
24 686.03, 22 464.28..., 20 442.50..., 18 602.67...
(M1)(A1)(ft)
Notes: Award (M1) for at least the 2nd term correct (calculated from their (a) × 0.91). Award (A1)(ft) for four correct terms (rounded
or unrounded).
Follow through from part (a).
Accept list containing the last three terms only (24 686.03 may be implied).
OR (lists(ii))
27 127.50 − (2441.47... + 2221.74... + 2021.79... + 1839.82 …)
(M1)(A1)(ft)
Notes: Award (M1) for subtraction of four terms from 27 127.50.
Award (A1) for four correct terms (rounded or unrounded).
Follow through from part (a).
= 18 602.67
= 18 600
(A1)
(AG)
Note: The final (A1) is not awarded unless both the unrounded and rounded answers are seen.
[3 marks]
100 students at IB College were asked whether they study Music (M), Chemistry (C), or Economics (E) with the following results.
10 study all three
15 study Music and Chemistry
17 study Music and Economics
12 study Chemistry and Economics
11 study Music only
6 study Chemistry only
6a.
Draw a Venn diagram to represent the information above.
[4 marks]
Markscheme
(A1) for rectangle and three labelled circles (U need not be seen)
(A1) for 10 in the correct region
(A1) for 2, 7 and 5 in the correct regions
(A1) for 6 and 11 in the correct regions
6b.
(A4)
Write down the number of students who study Music but not Economics.
[1 mark]
Markscheme
16
(A1)(ft)
Note: Follow through from their Venn diagram.
6c. There are 22 Economics students in total.
(i) Calculate the number of students who study Economics only.
(ii) Find the number of students who study none of these three subjects.
[4 marks]
Markscheme
(i) 10 + 7 + 2
(M1)
Note: Award (M1) for summing their 10, 7 and 2.
22 − 19
=3
(A1)(ft)(G2)
Note: Follow through from their diagram. Award (M1)(A1)(ft) for answers consistent with their diagram irrespective of whether
working seen. Award a maximum of (M1)(A0) for a negative answer.
(ii) 22 + 11 + 5 + 6
(M1)
Note: Award (M1) for summing 22, and their 11, 5 and 6.
100 − 44
= 56
(A1)(ft)(G2)
Note: Follow through from their diagram. Award (M1)(A1)(ft) for answers consistent with their diagram and the use of 22
irrespective of whether working seen. If negative values are used or implied award (M0)(A0).
6d.
A student is chosen at random from the 100 that were asked above.
Find the probability that this student
(i) studies Economics;
(ii) studies Music and Chemistry but not Economics;
(iii) does not study either Music or Economics;
(iv) does not study Music given that the student does not study Economics.
Markscheme
22
(i) 100
( 11
,0.22,22%)
50
5
(ii) 100
( 201 ,0.05,5%)
(A1)(G1)
(A1)(ft)(A1)(G2)
Note: Award (A1)(ft) for their 5 in numerator, (A1) for denominator.
Follow through from their diagram.
62
(iii) 100
( 31
,0.62,62%)
50
(A1)(ft)(A1)(G2)
Note: Award (A1)(ft) for 100 − (22 + 11 + their 5), (A1) for denominator.
Follow through from their diagram.
(iv)
62
78
( 31
,0.795,79.5%) (0.794871...)
39
(A1)(ft)(A1)(G2)
Note: Award (A1)(ft) for numerator, (A1) for denominator. Follow
through from part (d)(iii) for numerator.
[7 marks]
George leaves a cup of hot coffee to cool and measures its temperature every minute. His results are shown in the table below.
7a.
Write down the decrease in the temperature of the coffee
[3 marks]
(i) during the first minute (between t = 0 and t =1) ;
(ii) during the second minute;
(iii) during the third minute.
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.
7b.
Assuming the pattern in the answers to part (a) continues, show that k = 19.
[2 marks]
Markscheme
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).
7c. 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 marks]
Markscheme
(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).
7d.
The function that models the change in temperature of the coffee is y = p (2−t )+ q.
[2 marks]
(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.
Markscheme
(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.
7e.
Solve the equations found in part (d) to find the value of p and the value of q.
[2 marks]
Markscheme
p = 80, q = 14
(G1)(G1)(ft)
Note: If the equations have been incorrectly simplified, follow through even if no working is shown.
7f.
The graph of this function has a horizontal asymptote.
[2 marks]
Write down the equation of this asymptote.
Markscheme
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.
7g.
George decides to model the change in temperature of the coffee with a linear function using correlation and linear regression. [4 marks]
Use the seven results in the table to write down
(i) the correlation coefficient;
(ii) the equation of the regression line y on t.
Markscheme
(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).
7h.
Use the equation of the regression line to estimate the temperature of the coffee at t = 3.
[2 marks]
Markscheme
−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).
7i.
Find the percentage error in this estimate of the temperature of the coffee at t = 3.
[2 marks]
Markscheme
36.6785...−24
24
× 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 (×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.
An agricultural cooperative uses three brands of fertilizer, A, B and C, on 120 different crops. The crop yields are classified as High,
Medium or Low.
The data collected are organized in the table below.
The agricultural cooperative decides to conduct a chi-squared test at the 1 % significance level using the data.
8a.
State the null hypothesis, H 0, for the test.
[2 marks]
Markscheme
The (crop) yield is independent of the (type of) fertilizer used.
(A1)(A1)
Note: Award (A1) for (crop) yield and (type of) fertilizer, (A1) for “independent” or “not dependent” or “not associated”.
Do not accept “not correlated” or “not related” or “not connected” or “does not depend on”.
8b.
Write down the number of degrees of freedom.
[1 mark]
Markscheme
4
8c.
(A1)
Write down the critical value for the test.
[1 mark]
Markscheme
13.277
(A1)(ft)
Note: Accept 13.3. Follow through from part (b).
8d.
Show that the expected number of Medium Yield crops using Fertilizer C is 17, correct to the nearest integer.
[2 marks]
Markscheme
50
120
40
× 120
× 120 or
50×40
120
(M1)
Note: Award (M1) for correct substitution in the expected value formula.
= 16.6666... (A1)
= 17 (AG)
Note: Both unrounded and rounded answers must be seen to award (A1).
8e.
Use your graphic display calculator to find for the data
[3 marks]
(i) the χ2 calculated value, χ2calc ;
(ii) the p-value.
Markscheme
(i) χ2calc = 3.86(3.86133...)
(G2)
(ii) p-value = 0.425 (0.425097...)
8f.
(G1)
State the conclusion of the test. Give a reason for your decision.
[2 marks]
Markscheme
Since χ2calc < Critical Value
(R1)
Accept (do not reject) the Null Hypothesis.
(A1)(ft)
Note: Accept decision based on p-value with comparison to 1 % (0.425097... > 0.01) . Do not award (R0)(A1). Follow through from
parts (c) and (e). Numerical answers must be present in the question for a valid comparison to be made.
The graph of the function f(x) =
9a.
Calculate f(1).
14
x
+ x − 6, for 1 ≤ x ≤ 7 is given below.
[2 marks]
Markscheme
14
(1)
+ (1) − 6
(M1)
Note: Award (M1) for substituting x = 1 into f.
=9
9b.
(A1)(G2)
Find f'(x).
[3 marks]
Markscheme
− 142 + 1
x
(A3)
Note: Award (A1) for −14, (A1) for
14
x2
or for x−2 , (A1) for 1.
Award at most (A2) if any extra terms are present.
9c. Use your answer to part (b) to show that the x-coordinate of the local minimum point of the graph of f is 3.7 correct to 2
[3 marks]
significant figures.
Markscheme
− 14
+ 1 = 0 or f ′ (x) = 0
x2
(M1)
Note: Award (M1) for equating their derivative in part (b) to 0.
14
x2
= 1 or x2 = 14 or equivalent
(M1)
Note: Award (M1) for correct rearrangement of their equation.
−−
x = 3.74165...(√14 )
x = 3.7
(A1)
(AG)
Notes: Both the unrounded and rounded answers must be seen to award the (A1). This is a “show that” question; appeals to their
GDC are not accepted –award a maximum of (M1)(M0)(A0).
Specifically, − 142 + 1 = 0 followed by x = 3.74165...,x = 3.7 is awarded (M1)(M0)(A0).
x
9d.
Find the range of f.
[3 marks]
Markscheme
1.48 ⩽ y ⩽ 9
(A1)(A1)(ft)(A1)
−−
Note: Accept alternative notations, for example [1.48,9]. (x = √14 leads to answer 1.48331...)
Note: Award (A1) for 1.48331…seen, accept 1.48378… from using the given answer x = 3.7, (A1)(ft) for their 9 from part (a) seen,
(A1) for the correct notation for their interval (accept ⩽ y ⩽ or ⩽ f ⩽ ).
9e.
Points A and B lie on the graph of f. The x-coordinates of A and B are 1 and 7 respectively.
[1 mark]
Write down the y-coordinate of B.
Markscheme
(A1)
3
Note: Do not accept a coordinate pair.
9f.
Points A and B lie on the graph of f . The x-coordinates of A and B are 1 and 7 respectively.
[2 marks]
Find the gradient of the straight line passing through A and B.
Markscheme
3−9
7−1
(M1)
Note: Award (M1) for their correct substitution into the gradient formula.
= −1
(A1)(ft)(G2)
Note: Follow through from their answers to parts (a) and (e).
9g.
M is the midpoint of the line segment AB.
[2 marks]
Write down the coordinates of M.
Markscheme
(4, 6)
(A1)(ft)(A1)
Note: Accept x = 4, y = 6. Award at most (A1)(A0) if parentheses not seen.
If coordinates reversed award (A0)(A1)(ft).
Follow through from their answers to parts (a) and (e).
9h.
L is the tangent to the graph of the function y = f(x), at the point on the graph with the same x-coordinate as M.
[2 marks]
Find the gradient of L.
Markscheme
− 142 + 1
4
(M1)
Note: Award (M1) for substitution into their gradient function.
Follow through from their answers to parts (b) and (g).
= 18 (0.125)
9i.
(A1)(ft)(G2)
Find the equation of L. Give your answer in the form y = mx + c.
[3 marks]
Markscheme
y − 1.5 = 18 (x − 4)
(M1)(ft)(M1)
Note: Award (M1) for substituting their (4, 1.5) in any straight line formula,
(M1) for substituting their gradient in any straight line formula.
y = x8 + 4
(A1)(ft)(G2)
Note: The form of the line has been specified in the question.
Forty families were surveyed about the places they went to on the weekend. The places were the circus (C), the museum (M) and the
park (P).
16 families went to the circus
22 families went to the museum
14 families went to the park
4 families went to all three places
7 families went to both the circus and the museum, but not the park
3 families went to both the circus and the park, but not the museum
1 family went to the park only
10a.
Draw a Venn diagram to represent the given information using sets labelled C, M and P. Complete the diagram to include the [4 marks]
number of families represented in each region.
Markscheme
(A1)(A1)(A1)(A1)
Award (A1) for 3 intersecting circles and rectangle, (A1) for 1, 3, 4 and 7, (A1) for 2, (A1) for 6 and 5.
10b.
Find the number of families who
(i) went to the circus only;
(ii) went to the museum and the park but not the circus;
(iii) did not go to any of the three places on the weekend.
[4 marks]
Markscheme
(i) 2
(A1)(ft)
(ii) 6
(A1)(ft)
(iii) 40 − (1 + 6 + 2 + 3 + 4 + 7 + 5)
(M1)
Note: Award (M1) for subtracting all their values from 40.
(A1)(ft)(G2)
= 12
Note: Follow through from their Venn diagram for parts (i), (ii) and (iii).
10c.
A family is chosen at random from the group of 40 families. Find the probability that the family went to
[8 marks]
(i) the circus;
(ii) two or more places;
(iii) the park or the circus, but not the museum;
(iv) the museum, given that they also went to the circus.
Markscheme
16
40
(i)
( 25 ,0.4,40%)
(A1)(A1)(G2)
Note: Award (A1) for numerator, (A1) for denominator. Answer must be less than 1 otherwise award (A0)(A0). Award (A0)(A0) if
answer is given as incorrect reduced fraction without working.
(ii)
20
40
( 12 ,0.5,50%)
(A1)(ft) (A1) (G2)
Note: Award (A1)(ft) for numerator, (A1) for denominator. Follow through from their Venn diagram. Answer must be less than 1
otherwise award (A0)(A0). Award (A0)(A0) if answer is given as incorrect reduced fraction without working.
(iii)
6
40
( 203 ,0.15,15%)
(A1)(ft)(A1)(G2)
Note: Award (A1)(ft) for numerator, (A1) for denominator. Follow through from their Venn diagram. Answer must be less than 1
otherwise award (A0)(A0). Award (A0)(A0) if answer is given as incorrect reduced fraction without working.
(iv)
11
16
(0.6875,68.75%)
(A1)(ft)(A1)(G2)
Note: Award (A1)(ft) for numerator, (A1) for denominator. Follow through from their Venn diagram. Answer must be less than 1
otherwise award (A0)(A0). Award (A0)(A0) if answer is given as incorrect reduced fraction without working.
10d.
Two families are chosen at random from the group of 40 families.
Find the probability that both families went to the circus.
[3 marks]
Markscheme
16
40
× 15
39
(A1)(A1)(ft)
Note: Award (A1) for multiplication of their probabilities, (A1)(ft) for their correct probabilities.
240
1560
( 132 ,0.153846...,15.4%)
(A1)(ft)(G2)
Note: Follow through from their answer to part (c)(i). Answer must be less than 1 otherwise award at most (A1)(A1)(A0)(ft).
Francesca is a chef in a restaurant. She cooks eight chickens and records their masses and cooking times. The mass m of each
chicken, in kg, and its cooking time t, in minutes, are shown in the following table.
11a.
Draw a scatter diagram to show the relationship between the mass of a chicken and its cooking time. Use 2 cm to represent
0.5 kg on the horizontal axis and 1 cm to represent 10 minutes on the vertical axis.
[4 marks]
Markscheme
(A1) for correct scales and labels (mass or m on the horizontals axis, time or t on the vertical axis)
(A3) for 7 or 8 correctly placed data points
(A2) for 5 or 6 correctly placed data points
(A1) for 3 or 4 correctly placed data points, (A0) otherwise.
(A4)
Note: If axes reversed award at most (A0)(A3)(ft). If graph paper not used, award at most (A1)(A0).
11b. Write down for this set of data
[2 marks]
(i) the mean mass, m̄ ;
(ii) the mean cooking time, t̄ .
Markscheme
(i) 1.91 (kg) (1.9125 kg)
(ii) 83 (minutes)
(G1)
(G1)
11c. Label the point M(m̄, t̄ ) on the scatter diagram.
[1 mark]
Markscheme
Their mean point labelled.
(A1)(ft)
Note: Follow through from part (b). Accept any clear indication of the mean point. For example: circle around point, (m, t), M , etc.
11d.
Draw the line of best fit on the scatter diagram.
[2 marks]
Markscheme
Line of best fit drawn on scatter diagram.
(A1)(ft)(A1)(ft)
Notes:Award (A1)(ft) for straight line through their mean point, (A1)(ft) for line of best fit with intercept 9(±2) . The second (A1)(ft)
can be awarded even if the line does not reach the t-axis but, if extended, the t-intercept is correct.
11e.
Using your line of best fit, estimate the cooking time, in minutes, for a 1.7 kg chicken.
[2 marks]
Markscheme
75
(M1)(A1)(ft)(G2)
Notes: Accept 74.77 from the regression line equation. Award (M1) for indication of the use of their graph to get an estimate OR for
correct substitution of 1.7 in the correct regression line equation t = 38.5m + 9.32.
11f.
Write down the Pearson’s product–moment correlation coefficient, r .
[2 marks]
Markscheme
0.960 (0.959614...)
(G2)
Note: Award (G0)(G1)(ft) for 0.95, 0.959
11g. Using your value for r , comment on the correlation.
[2 marks]
Markscheme
Strong and positive
(A1)(ft)(A1)(ft)
Note: Follow through from their correlation coefficient in part (f).
11h.
The cooking time of an additional 2.0 kg chicken is recorded. If the mass and cooking time of this chicken is included in the [2 marks]
data, the correlation is weak.
(i) Explain how the cooking time of this additional chicken might differ from that of the other eight chickens.
(ii) Explain how a new line of best fit might differ from that drawn in part (d).
Markscheme
(i) Cooking time is much larger (or smaller) than the other eight
(A1)
(ii) The gradient of the new line of best fit will be larger (or smaller)
(A1)
Note: Some acceptable explanations may include but are not limited to:
The line of best fit may be further away from the plotted points
It may be steeper than the previous line (as the mean would change)
The t-intercept of the new line is smaller (larger)
Do not accept vague explanations, like:
The new line would vary
It would not go through all points
It would not fit the patterns
The line may be slightly tilted
A greenhouse ABCDPQ is constructed on a rectangular concrete base ABCD and is made of glass. Its shape is a right prism, with
cross section, ABQ, an isosceles triangle. The length of BC is 50 m, the length of AB is 10 m and the size of angle QBA is 35°.
12a.
Write down the size of angle AQB.
[1 mark]
Markscheme
110°
12b.
(A1)
Calculate the length of AQ.
[3 marks]
Markscheme
AQ
sin 35∘
=
10
sin 110∘
(M1)(A1)
Note: Award (M1) for substituted sine rule formula, (A1) for their correct substitutions.
OR
AQ = cos535∘
(A1)(M1)
Note: Award (A1) for 5 seen, (M1) for correctly substituted trigonometric ratio.
AQ = 6.10 (6.10387...)
(A1)(ft)(G2)
Notes: Follow through from their answer to part (a).
12c. Calculate the length of AC.
[2 marks]
Markscheme
AC 2 = 102 + 502
(M1)
Note: Award (M1) for correctly substituted Pythagoras formula.
−−−−
AC = 51.0(√2600,50.9901...)
12d.
(A1)(G2)
Show that the length of CQ is 50.37 m, correct to 4 significant figures.
[2 marks]
Markscheme
QC 2 = (6.10387...)2 + (50)2
(M1)
Note: Award (M1) for correctly substituted Pythagoras formula.
QC = 50.3711...
= 50.37
(A1)
(AG)
Note: Both the unrounded and rounded answers must be seen to award (A1).
If 6.10 is used then 50.3707... is the unrounded answer.
For an incorrect follow through from part (b) award a maximum of (M1)(A0) – the given answer must be reached to award the
final (A1)(AG).
12e.
Find the size of the angle AQC.
[3 marks]
Markscheme
cos AQC =
(6.10387...)2+ (50.3711...)2− (50.9901...)2
2(6.10387...)(50.3711...)
(M1)(A1)(ft)
Note: Award (M1) for substituted cosine rule formula, (A1)(ft) for their correct substitutions.
= 92.4° (92.3753...∘ )
(A1)(ft)(G2)
Notes: Follow through from their answers to parts (b), (c) and (d). Accept 92.2 if the 3 sf answers to parts (b), (c) and (d) are used.
Accept 92.5° (92.4858...∘ ) if the 3 sf answers to parts (b), (c) and 4 sf answers to part (d) used.
12f.
Calculate the total area of the glass needed to construct
[5 marks]
(i) the two rectangular faces of the greenhouse;
(ii) the two triangular faces of the greenhouse.
Markscheme
(i) 2(50 × 6.10387...)
(M1)
Note: Award (M1) for their correctly substituted rectangular area formula, the area of one rectangle is not sufficient.
= 610 m2 (610.387...)
(A1)(ft)(G2)
Notes: Follow through from their answer to part (b).
The answer is 610 m2. The units are required.
(ii) Area of triangular face = 12 × 10 × 6.10387... × sin 35∘
(M1)(A1)(ft)
OR
Area of triangular face = 12 × 6.10387... × 6.10387... × sin 110∘
(M1)(A1)(ft)
= 17.5051...
Note: Award (M1) for substituted triangle area formula, (A1)(ft) for correct substitutions.
OR
(Height of triangle) = (6.10387...)2 − 52
= 3.50103...
Area of triangular face = 12 × 10 × their height
= 17.5051...
Note: Award (M1) for substituted triangle area formula, (A1)(ft) for correctly substituted area formula. If 6.1 is used, the height is
3.49428... and the area of both triangular faces 34.9 m2
Area of both triangular faces = 35.0 m2 (35.0103...)
(A1)(ft)(G2)
Notes: The answer is 35.0 m2. The units are required. Do not penalize if already penalized in part (f)(i). Follow through from their
part (b).
12g. The cost of one square metre of glass used to construct the greenhouse is 4.80 USD.
[3 marks]
Calculate the cost of glass to make the greenhouse. Give your answer correct to the nearest 100 USD.
Markscheme
(610.387... + 35.0103...) × 4.80
(M1)
(A1)(ft)
= 3097.90...
Notes: Follow through from their answers to parts (f)(i) and (f)(ii).
Accept 3096 if the 3 sf answers to part (f) are used.
= 3100
(A1)(ft)(G2)
Notes: Follow through from their unrounded answer, irrespective of whether it is correct. Award (M1)(A2) if working is shown and
3100 seen without the unrounded answer being given.
A tent is in the shape of a triangular right prism as shown in the diagram below.
The tent has a rectangular base PQRS .
PTS and QVR are isosceles triangles such that PT = TS and QV = VR .
PS is 3.2 m , SR is 4.7 m and the angle TSP is 35​°.
13a.
Show that the length of side ST is 1.95 m, correct to 3 significant figures.
[3 marks]
Markscheme
ST = cos1.635∘
(M1)(A1)
Note: Award (M1) for correctly substituted trig equation, (A1) for 1.6 seen.
OR
ST
sin 35∘
=
3.2
sin 110∘
(M1)(A1)
Note: Award (M1) for substituted sine rule equation, (A1) for correct substitutions.
ST = 1.95323...
= 1.95 (m)
(A1)
(AG)
Notes: Both unrounded and rounded answer must be seen for final (A1) to be awarded.
13b.
Calculate the area of the triangle PTS.
[3 marks]
Markscheme
1
2
× 3.2 × 1.95323... × sin 35∘ or 12 × 1.95323... × 1.95323... × sin 110∘
(M1)(A1)
Note: Award (M1) for substituted area formula, (A1) for correct substitutions. Do not award follow through marks.
= 1.79 m2 (1.79253...m2)
(A1)(G2)
Notes: The answer is 1.79 m2, units are required. Accept 1.78955... from using 1.95.
OR
1
2
× 3.2 × 1.12033...
(A1) (M1)
−−−−−−−−−−−−−−−
Note: Award (A1) for the correct value for TM (1.12033...) OR correct expression for TM (i.e. 1.6tan35°, √(1.95323...)2 − 1.62 ),
(M1) for correctly substituted formula for triangle area.
= 1.79 m2 (1.79253...m2)
(A1)(G2)
Notes: The answer is 1.79 m2, units are required. Accept 1.78 m2 from using 1.95.
13c. Write down the area of the rectangle STVR.
[1 mark]
Markscheme
9.18 m2 (9.18022 m2)
(A1)(G1)
Notes: The answer is 9.18 m2, units are required. Do not penalize if lack of units was already penalized in (b). Do not award follow
through marks here. Accept 9.17 m2 (9.165 m2) from using 1.95.
13d. Calculate the total surface area of the tent, including the base.
[3 marks]
Markscheme
2 × 1.79253... + 2 × 9.18022... + 4.7 × 3.2
(M1)(A1)(ft)
Note: Award (M1) for addition of three products, (A1)(ft) for three correct products.
= 37.0 m2 (36.9855...m2)
(A1)(ft)(G2)
Notes: The answer is 37.0 m2, units are required. Accept 36.98 m2 from using 3sf answers. Follow through from their answers to
(b) and (c). Do not penalize if lack of units was penalized earlier in the question.
13e.
Calculate the volume of the tent.
[2 marks]
Markscheme
1.79253... × 4.7
(M1)
Note: Award (M1) for their correctly substituted volume formula.
= 8.42 m3 (8.42489...m3)
(A1)(ft)(G2)
Notes: The answer is 8.42 m3, units are required. Accept 8.41 m3 from use of 1.79. An answer of 8.35, from use of TM = 1.11,
will receive follow-through marks if working is shown. Follow through from their answer to part (b). Do not penalize if lack of units
was penalized earlier in the question.
13f. A pole is placed from V to M, the midpoint of PS.
[4 marks]
Find in metres,
(i) the height of the tent, TM;
(ii) the length of the pole, VM.
Markscheme
(i) TM = 1.6 tan 35∘
(M1)
Notes: Award (M1) for their correct substitution in trig ratio.
OR
−−−−−−−−−−−−−−−
TM = √(1.95323...)2 − 1.62
(M1)
Note: Award (M1) for correct substitution in Pythagoras’ theorem.
OR
3.2×TM
2
= 1.79253...
(M1)
Note: Award (M1) for their correct substitution in area of triangle formula.
= 1.12 (m) (1.12033...)
(A1)(ft)(G2)
Notes: Follow through from their answer to (b) if area of triangle is used. Accept 1.11 (1.11467) from use of ST = 1.95.
−−−−−−−−−−−−−
(ii) VM = √1.12033...2 + 4.72
(M1)
Note: Award (M1) for their correct substitution in Pythagoras’ theorem.
= 4.83 (m) (4.83168 )
(A1)(ft)(G2)
Notes: Follow through from (f)(i).
13g. Calculate the angle between VM and the base of the tent.
[2 marks]
Markscheme
sin−1 ( 1.12033...
)
4.83168...
(M1)
OR
4.7
cos−1 ( 4.83168...
)
(M1)
OR
tan−1 ( 1.12033...
)
4.7
(M1)
Note: Award (M1) for correctly substituted trig equation.
OR
cos−1 (
4.72+ (4.83168...)2− (1.12033...)2
)
2×4.7×4.83168...
(M1)
Note: Award (M1) for correctly substituted cosine formula.
= 13.4° (13.4073...)
(A1)(ft)(G2)
Notes: Accept 13.3°. Follow through from part (f).
On Monday Paco goes to a running track to train. He runs the first lap of the track in 120 seconds. Each lap Paco runs takes him 10
seconds longer than his previous lap.
14a. Find the time, in seconds, Paco takes to run his fifth lap.
[3 marks]
Markscheme
120 + 10 × 4
(M1)(A1)
Notes: Award (M1) for substituted AP formula, (A1) for correct substitutions. Accept a list of 4 correct terms.
= 160
14b.
(A1)(G3)
Paco runs his last lap in 260 seconds.
[3 marks]
Find how many laps he has run on Monday.
Markscheme
120 + (n − 1) × 10 = 260
(M1)(M1)
Notes: Award (M1) for correctly substituted AP formula, (M1) for equating to 260. Accept a list of correct terms showing at least the
14th and 15th terms.
= 15
(A1)(G2)
14c. Find the total time, in minutes, run by Paco on Monday.
[4 marks]
Markscheme
15
(120 + 260)
2
or
15
(2 × 120 + (15 − 1) × 10)
2
(M1)(A1)(ft)
Notes: Award (M1) for substituted AP sum formula, (A1)(ft) for correct substitutions. Accept a sum of a list of 15 correct terms.
Follow through from their answer to part (b).
(A1)(ft)(G2)
2850 seconds
Note: Award (G2) for 2850 seen with no working shown.
47.5 minutes
(A1)(ft)(G3)
Notes: A final (A1)(ft) can be awarded for correct conversion from seconds into minutes of their incorrect answer. Follow through
from their answer to part (b).
14d.
On Wednesday Paco takes Lola to train. They both run the first lap of the track in 120 seconds. Each lap Lola runs takes 1.06 [3 marks]
times as long as her previous lap.
Find the time, in seconds, Lola takes to run her third lap.
Markscheme
120 × 1.063−1
(M1)(A1)
Notes: Award (M1) for substituted GP formula, (A1) for correct substitutions. Accept a list of 3 correct terms.
= 135 (134.832)
(A1)(G2)
14e. Find the total time, in seconds, Lola takes to run her first four laps.
[3 marks]
Markscheme
S4 =
120(1.06 4−1)
(1.06−1)
(M1)(A1)
Notes: Award (M1) for substituted GP sum formula, (A1) for correct substitutions. Accept a sum of a list of 4 correct terms.
= 525 (524.953...)
(A1)(G2)
14f. Each lap Paco runs again takes him 10 seconds longer than his previous lap. After a certain number of laps Paco takes less
time per lap than Lola.
Find the number of the lap when this happens.
[3 marks]
Markscheme
120 + (n − 1) × 10 < 120 × 1.06n−1
(M1)(M1)
Notes: Award (M1) for correct left hand side, (M1) for correct right hand side. Accept an equation. Follow through from their
expressions given in parts (a) and (d).
OR
List of at least 2 terms for both sequences (120, 130, … and 120, 127.2, …)
(M1)
List of correct 12th and 13th terms for both sequences (..., 230, 240 and …, 227.8, 241.5)
(M1)
OR
A sketch with a line and an exponential curve,
An indication of the correct intersection point
13th lap
(M1)
(M1)
(A1)(ft)(G2)
Note: Do not award the final (A1)(ft) if final answer is not a positive integer.
The diagram shows an aerial view of a bicycle track. The track can be modelled by the quadratic function
y=
−x2
10
+ 272 x, where x ⩾ 0, y ⩾ 0
(x , y) are the coordinates of a point x metres east and y metres north of O , where O is the origin (0, 0) . B is a point on the bicycle
track with coordinates (100, 350) .
15a. The coordinates of point A are (75, 450). Determine whether point A is on the bicycle track. Give a reason for your answer.
Markscheme
2
y = − 75
+ 272 × 75
10
(M1)
Note: Award (M1) for substitution of 75 in the formula of the function.
= 450
(A1)
Yes, point A is on the bike track. (A1)
Note: Do not award the final (A1) if correct working is not seen.
=
−
2
+
[3 marks]
15b. Find the derivative of y =
−x2
10
+ 272 x.
[2 marks]
Markscheme
dy
dx
= − 2x
+ 272 ( dx = −0.2x + 13.5)
10
dy
(A1)(A1)
Notes: Award (A1) for each correct term. If extra terms are seen award at most (A1)(A0). Accept equivalent forms.
15c.
Use the answer in part (b) to determine if A (75, 450) is the point furthest north on the track between O and B. Give a reason [4 marks]
for your answer.
Markscheme
− 2x
+ 272 = 0
10
(M1)
Note: Award (M1) for equating their derivative from part (b) to zero.
x = 67.5
(A1)(ft)
Note: Follow through from their derivative from part (b).
(Their) 67.5 ≠ 75
(R1)
Note: Award (R1) for a comparison of their 67.5 with 75. Comparison may be implied (eg 67.5 is the x-coordinate of the furthest
north point).
OR
dy
dx
=−
2×(75)
10
+ 272
(M1)
Note: Award (M1) for substitution of 75 into their derivative from part (b).
= −1.5
(A1)(ft)
Note: Follow through from their derivative from part (b).
(Their) − 1.5 ≠ 0
(R1)
Note: Award (R1) for a comparison of their –1.5 with 0. Comparison may be implied (eg The gradient of the parabola at the furthest
north point (vertex) is 0).
Hence A is not the furthest north point.
(A1)(ft)
Note: Do not award (R0)(A1)(ft). Follow through from their derivative from part (b).
15d. (i) Write down the midpoint of the line segment OB.
(ii) Find the gradient of the line segment OB.
[3 marks]
Markscheme
(i) M(50,175)
(A1)
Note: If parentheses are omitted award (A0). Accept x = 50, y = 175.
(ii) 350−0
100−0
(M1)
Note: Award (M1) for correct substitution in gradient formula.
= 3.5 ( 350
, 7)
100 2
(A1)(ft)(G2)
Note: Follow through from (d)(i) if midpoint is used to calculate gradient. Award (G1)(G0) for answer 3.5x without working.
15e. Scott starts from a point C(0,150) . He hikes along a straight road towards the bicycle track, parallel to the line segment OB.
[3 marks]
Find the equation of Scott’s road. Express your answer in the form ax + by = c, where a,b and c ∈ R ​.
Markscheme
y = 3.5x + 150
(A1)(ft)(A1)(ft)
Note: Award (A1)(ft) for using their gradient from part (d), (A1)(ft) for correct equation of line.
3.5x − y = −150 or 7x − 2y = −300 (or equivalent)
(A1)(ft)
Note: Award (A1)(ft) for expressing their equation in the form ax + by = c.
15f.
Use your graphic display calculator to find the coordinates of the point where Scott first crosses the bicycle track.
[2 marks]
Markscheme
(18.4, 214) (18.3772..., 214.320...)
(A1)(ft)(A1)(ft)(G2)(ft)
Notes: Follow through from their equation in (e). Coordinates must be positive for follow through marks to be awarded. If
parentheses are omitted and not already penalized in (d)(i) award at most (A0)(A1)(ft). If coordinates of the two intersection points are
given award (A0)(A1)(ft). Accept x = 18.4, y = 214.
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