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SL Paper 2

Eddie decides to construct a path across his rectangular grass lawn using pairs of tiles.

Each tile is $10 cm$ wide and $20 cm$ long. The following diagrams show the path after Eddie has laid one pair and three pairs of tiles. This pattern continues until Eddie reaches the other side of his lawn. When $n$ pairs of tiles are laid, the path has a width of $w_{n}$ centimetres and a length $l_{n}$ centimetres.

The following diagrams show this pattern for one pair of tiles and for three pairs of tiles, where the white space around each diagram represents Eddie’s lawn.

The following table shows the values of $w_{n}$ and $l_{n}$ for the first three values of $n$ .

Find the value of

Write down an expression in terms of $n$ for

Eddie’s lawn has a length $740 cm$ .

The tiles cost $$ 24.50$ per square metre and are sold in packs of five tiles.

To allow for breakages Eddie wants to have at least $8 %$ more tiles than he needs.

There is a fixed delivery cost of $$ 35$ .

$a$ .

[1]

a.i.

$b$ .

[1]

a.ii.

$w_{n}$ .

[2]

b.i.

$l_{n}$ .

[1]

b.ii.

Show that Eddie needs $144$ tiles.

[2]

c.i.

Find the value of $w_{n}$ for this path.

[1]

c.ii.

Find the total area of the tiles in Eddie’s path. Give your answer in the form $a \times 10^{k}$ where $1 \leq a < 10$ and $k$ is an integer.

[3]

Find the cost of a single pack of five tiles.

[3]

Find the minimum number of packs of tiles Eddie will need to order.

[3]

Find the total cost for Eddie’s order.

[2]

Markscheme

$30$ A1

[1 mark]

a.i.

$40$ A1

[1 mark]

a.ii.

arithmetic formula chosen (M1)

$w_{n} = 20 + (n - 1) 10 (= 10 + 10 n)$ A1

[2 marks]

b.i.

arithmetic formula chosen

$l_{n} = 30 + (n - 1) 10 (= 20 + 10 n)$ A1

[1 mark]

b.ii.

$740 = 30 + (n - 1) 10$ OR $740 = 20 - 10 n$ M1

$n = 72$ A1

$144$ tiles AG

Note: The AG line must be stated for the final A1 to be awarded.

[2 marks]

c.i.

$w_{72} = 730$ A1

[1 mark]

c.ii.

$(10 \times 20) \times 144$ (M1)

$= 28800$ (A1)

$2.88 \times 10^{4} {cm}^{2}$ A1

Note: Follow through within the question for correctly converting their intermediate value into standard form (but only if the pre-conversion value is seen).

[3 marks]

EITHER

$1$ square metre $= 100 cm \times 100 cm$ (M1)

(so, $50$ tiles) and hence $10$ packs of tiles in a square metre (A1)

(so each pack is $\frac{$ 24.50}{10 packs}$ )

area covered by one pack of tiles is $(0.2 m \times 0.1 m \times 5 =) 0.1 m^{2}$ (A1)

$24.5 \times 0.1$ (M1)

THEN

$$ 2.45$ per pack (of $5$ tiles) A1

[3 marks]

$\frac{1.08 \times 144}{5} (= 31.104)$ (M1)(M1)

Note: Award M1 for correct numerator, M1 for correct denominator.

$32$ (packs of tiles) A1

[3 marks]

$35 + (32 \times 2.45)$ (M1)

$$ 113 (113.4)$ A1

[2 marks]

Examiners report

In part (a), most candidates were able to find the correct values for $a$ and $b$ .

In part (b), most candidates were able to write down the correct expressions for $w_{n}$ and $l_{n}$ .

In part (c)(i), candidates continued to struggle with “show that” questions. Some substituted 144 for $n$ and worked backwards, however this is never the intention of the question; candidates should progress towards (not “away from”) the given result. Part (ii) was well answered by many candidates.

Part (d) was poorly answered with many candidates multiplying 720 with 730 instead of 10 with 20. However, the majority managed to convert their answer correctly to standard form which gained them a mark for that particular skill.

Part (e) saw very few candidates find the cost of one packet of tiles. The main reason was the failure to convert cm² to m².

In part (f), about half of the candidates managed to find the correct number of packets. Some gained a mark for finding 8% or dividing by 5.

In part (g), most candidates could use their answers to parts (e) and (f) to score “follow through” marks and find the total cost of their order.

a.i.

[N/A]

a.ii.

[N/A]

b.i.

[N/A]

b.ii.

[N/A]

c.i.

[N/A]

c.ii.

[N/A]

The marks obtained by nine Mathematical Studies SL students in their projects (x) and their final IB examination scores (y) were recorded. These data were used to determine whether the project mark is a good predictor of the examination score. The results are shown in the table.

The equation of the regression line y on x is y = mx + c.

A tenth student, Jerome, obtained a project mark of 17.

Use your graphic display calculator to write down ${\bar y}$ , the mean examination score.

[1]

a.ii.

Use your graphic display calculator to write down r , Pearson’s product–moment correlation coefficient.

[2]

a.iii.

Find the exact value of m and of c for these data.

[2]

b.i.

Use the regression line y on x to estimate Jerome’s examination score.

[2]

c.i.

Justify whether it is valid to use the regression line y on x to estimate Jerome’s examination score.

[2]

c.ii.

Markscheme

54 (G1)

[1 mark]

a.ii.

0.5 (G2)

[2 marks]

a.iii.

m = 0.875, c = 41.75 $\left( {m = \frac{7}{8}{\text{,}}\,\,c = \frac{{167}}{4}} \right)$ (A1)(A1)

Note: Award (A1) for 0.875 seen. Award (A1) for 41.75 seen. If 41.75 is rounded to 41.8 do not award (A1).

[2 marks]

b.i.

y = 0.875(17) + 41.75 (M1)

Note: Award (M1) for correct substitution into their regression line.

= 56.6 (56.625) (A1)(ft)(G2)

Note: Follow through from part (b)(i).

[2 marks]

c.i.

the estimate is valid (A1)

since this is interpolation and the correlation coefficient is large enough (R1)

the estimate is not valid (A1)

since the correlation coefficient is not large enough (R1)

Note: Do not award (A1)(R0). The (R1) may be awarded for reasoning based on strength of correlation, but do not accept “correlation coefficient is not strong enough” or “correlation is not large enough”.

Award (A0)(R0) for this method if no numerical answer to part (a)(iii) is seen.

[2 marks]

c.ii.

Examiners report

[N/A]

a.ii.

[N/A]

a.iii.

[N/A]

b.i.

[N/A]

c.i.

[N/A]

c.ii.

Scott purchases food for his dog in large bags and feeds the dog the same amount of dog food each day. The amount of dog food left in the bag at the end of each day can be modelled by an arithmetic sequence.

On a particular day, Scott opened a new bag of dog food and fed his dog. By the end of the third day there were $115.5$ cups of dog food remaining in the bag and at the end of the eighth day there were $108$ cups of dog food remaining in the bag.

Find the number of cups of dog food

In $2021$ , Scott spent $$ 625$ on dog food. Scott expects that the amount he spends on dog food will increase at an annual rate of $6.4 %$ .

fed to the dog per day.

[3]

a.i.

remaining in the bag at the end of the first day.

[1]

a.ii.

Calculate the number of days that Scott can feed his dog with one bag of food.

[2]

Determine the amount that Scott expects to spend on dog food in $2025$ . Round your answer to the nearest dollar.

[3]

Calculate the value of $Σ_{n = 1}^{10} (625 \times 1 . 064^{(n - 1)})$ .

[1]

d.i.

Describe what the value in part (d)(i) represents in this context.

[2]

d.ii.

Comment on the appropriateness of modelling this scenario with a geometric sequence.

[1]

Markscheme

EITHER

$115.5 = u_{1} + (3 - 1) \times d (115.5 = u_{1} + 2 d)$

$108 = u_{1} + (8 - 1) \times d (108 = u_{1} + 7 d)$ (M1)(A1)

Note: Award M1 for attempting to use the arithmetic sequence term formula, A1 for both equations correct. Working for M1 and A1 can be found in parts (i) or (ii).

$(d = - 1.5)$

$1.5$ (cups/day) A1

Note: Answer must be written as a positive value to award A1.

$(d =) \frac{115.5 - 108}{5}$ (M1)(A1)

Note: Award M1 for attempting a calculation using the difference between term $3$ and term $8$ ; A1 for a correct substitution.

$(d =) 1.5$ (cups/day) A1

[3 marks]

a.i.

$(u_{1} =) 118.5$ (cups) A1

[1 mark]

a.ii.

attempting to substitute their values into the term formula for arithmetic sequence equated to zero (M1)

$0 = 118.5 + (n - 1) \times (- 1.5)$

$(n =) 80$ days A1

Note: Follow through from part (a) only if their answer is positive.

[2 marks]

$(t_{5} =) 625 \times 1 . 064^{(5 - 1)}$ (M1)(A1)

Note: Award M1 for attempting to use the geometric sequence term formula; A1 for a correct substitution

$$ 801$ A1

Note: The answer must be rounded to a whole number to award the final A1.

[3 marks]

$(S_{10} =) ($) 8390 (8394.39 \dots)$ A1

[1 mark]

d.i.

EITHER

the total cost (of dog food) R1

for $10$ years beginning in $2021$ OR $10$ years before $2031$ R1

the total cost (of dog food) R1

from $2021$ to $2030$ (inclusive) OR from $2021$ to (the start of) $2031$ R1

[2 marks]

d.ii.

EITHER
According to the model, the cost of dog food per year will eventually be too high to keep a dog.

OR
The model does not necessarily consider changes in inflation rate.

OR
The model is appropriate as long as inflation increases at a similar rate.

OR
The model does not account for changes in the amount of food the dog eats as it ages/becomes ill/stops growing.

OR
The model is appropriate since dog food bags can only be bought in discrete quantities. R1

Note: Accept reasonable answers commenting on the appropriateness of the model for the specific scenario. There should be a reference to the given context. A reference to the geometric model must be clear: either “model” is mentioned specifically, or other mathematical terms such as “increasing” or “discrete quantities” are seen. Do not accept a contextual argument in isolation, e.g. “The dog will eventually die”.

[1 mark]

Examiners report

Parts (a) and (b) were mostly well answered, but some candidates ignored the context and did not give the number of dog food cups per day as a positive number. Most candidates considered geometric sequence in part (c) correctly, and used the correct formula for the nth term, although they used an incorrect value for n at times. Some candidates used the finance application incorrectly. The sum in part (d) was calculated correctly by some candidates, although many seemed unfamiliar with sigma notation and with calculating summations using GDC. In part (d), most candidates interpreted correctly that the sum represented the cost of dog food for 10 years but did not identify the specific 10-year period. Part (e) was not answered well – often candidates made very general and abstract statements devoid of any contextual references.

a.i.

[N/A]

a.ii.

[N/A]

d.i.

[N/A]

d.ii.

[N/A]

Sophie is planning to buy a house. She needs to take out a mortgage for $120000. She is considering two possible options.

Option 1: Repay the mortgage over 20 years, at an annual interest rate of 5%, compounded annually.

Option 2: Pay $1000 every month, at an annual interest rate of 6%, compounded annually, until the loan is fully repaid.

Give a reason why Sophie might choose

Sophie decides to choose option 1. At the end of 10 years, the interest rate is changed to 7%, compounded annually.

Calculate the monthly repayment using option 1.

[2]

a.i.

Calculate the total amount Sophie would pay, using option 1.

[2]

a.ii.

Calculate the number of months it will take to repay the mortgage using option 2.

[3]

b.i.

Calculate the total amount Sophie would pay, using option 2.

[2]

b.ii.

option 1.

[1]

c.i.

option 2.

[1]

c.ii.

Use your answer to part (a)(i) to calculate the amount remaining on her mortgage after the first 10 years.

[2]

d.i.

Hence calculate her monthly repayment for the final 10 years.

[2]

d.ii.

Markscheme

evidence of using Finance solver on GDC M1

Monthly payment = $785 ($784.60) A1

[2 marks]

a.i.

$240 \times 785 = {\text{\$}}188000$ M1A1

[2 marks]

a.ii.

$N = 180.7$ M1A1

It will take 181 months A1

[3 marks]

b.i.

$181 \times 1000 = {\text{\$ }}181000$ M1A1

[2 marks]

b.ii.

The monthly repayment is lower, she might not be able to afford $1000 per month. R1

[1 mark]

c.i.

the total amount to repay is lower. R1

[1 mark]

c.ii.

$74400 (accept $74300) M1A1

[2 marks]

d.i.

Use of finance solver with N =120, PV = $74400, I = 7% A1

$855 (accept $854 − $856) A1

[2 marks]

d.ii.

Examiners report

[N/A]

a.i.

[N/A]

a.ii.

[N/A]

b.i.

[N/A]

b.ii.

[N/A]

c.i.

[N/A]

c.ii.

[N/A]

d.i.

[N/A]

d.ii.

Consider the function $f\left( x \right) = \frac{{27}}{{{x^2}}} - 16x,\,\,\,x \ne 0$ .

Sketch the graph of y = f (x), for −4 ≤ x ≤ 3 and −50 ≤ y ≤ 100.

[4]

Use your graphic display calculator to find the zero of f (x).

[1]

b.i.

Use your graphic display calculator to find the coordinates of the local minimum point.

[2]

b.ii.

Use your graphic display calculator to find the equation of the tangent to the graph of y = f (x) at the point (–2, 38.75).

Give your answer in the form y = mx + c.

[2]

b.iii.

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

(A1)(A1)(A1)(A1)

Note: Award (A1) for axis labels and some indication of scale; accept y or f(x).

Use of graph paper is not required. If no scale is given, assume the given window for zero and minimum point.

Award (A1) for smooth curve with correct general shape.

Award (A1) for x-intercept closer to y-axis than to end of sketch.

Award (A1) for correct local minimum with x-coordinate closer to y-axis than end of sketch and y-coordinate less than half way to top of sketch.

Award at most (A1)(A0)(A1)(A1) if the sketch intersects the y-axis or if the sketch curves away from the y-axis as x approaches zero.

[4 marks]

1.19 (1.19055…) (A1)

Note: Accept an answer of (1.19, 0).

Do not follow through from an incorrect sketch.

[1 mark]

b.i.

(−1.5, 36) (A1)(A1)

Note: Award (A0)(A1) if parentheses are omitted.

Accept x = −1.5, y = 36.

[2 marks]

b.ii.

y = −9.25x + 20.3 (y = −9.25x + 20.25) (A1)(A1)

Note: Award (A1) for −9.25x, award (A1) for +20.25, award a maximum of (A0)(A1) if answer is not an equation.

[2 marks]

b.iii.

Examiners report

[N/A]

b.i.

[N/A]

b.ii.

[N/A]

b.iii.

In this question, give all answers to two decimal places.

Bryan decides to purchase a new car with a price of €14 000, but cannot afford the full amount. The car dealership offers two options to finance a loan.

Finance option A:

A 6 year loan at a nominal annual interest rate of 14 % compounded quarterly. No deposit required and repayments are made each quarter.

Finance option B:

A 6 year loan at a nominal annual interest rate of $r$ % compounded monthly. Terms of the loan require a 10 % deposit and monthly repayments of €250.

Find the repayment made each quarter.

[3]

a.i.

Find the total amount paid for the car.

[2]

a.ii.

Find the interest paid on the loan.

[2]

a.iii.

Find the amount to be borrowed for this option.

[2]

b.i.

Find the annual interest rate, $r$ .

[3]

b.ii.

State which option Bryan should choose. Justify your answer.

[2]

Bryan’s car depreciates at an annual rate of 25 % per year.

Find the value of Bryan’s car six years after it is purchased.

[3]

Markscheme

N = 24
I % = 14
PV = −14000
FV = 0
P/Y = 4
C/Y = 4 (M1)(A1)

Note: Award M1 for an attempt to use a financial app in their technology, award A1 for all entries correct. Accept PV = 14000.

(€)871.82 A1

[3 marks]

a.i.

4 × 6 × 871.82 (M1)

(€) 20923.68 A1

[2 marks]

a.ii.

20923.68 − 14000 (M1)

(€) 6923.68 A1

[2 marks]

a.iii.

0.9 × 14000 (= 14000 − 0.10 × 14000) M1

(€) 12600.00 A1

[2 marks]

b.i.

N = 72

PV = 12600

PMT = −250

FV = 0

P/Y = 12

C/Y = 12 (M1)(A1)

Note: Award M1 for an attempt to use a financial app in their technology, award A1 for all entries correct. Accept PV = −12600 provided PMT = 250.

12.56(%) A1

[3 marks]

b.ii.

EITHER

Bryan should choose Option A A1

no deposit is required R1

Note: Award R1 for stating that no deposit is required. Award A1 for the correct choice from that fact. Do not award R0A1.

Bryan should choose Option B A1

cost of Option A (6923.69) > cost of Option B (72 × 250 − 12600 = 5400) R1

Note: Award R1 for a correct comparison of costs. Award A1 for the correct choice from that comparison. Do not award R0A1.

[2 marks]

$14\,000{\left( {1 - \frac{{25}}{{100}}} \right)^6}$ (M1)(A1)

Note: Award M1 for substitution into compound interest formula.
Award A1 for correct substitutions.

= (€)2491.70 A1

N = 6

I% = −25

PV = ±14 000

P/Y = 1

C/Y = 1 (A1)(M1)

Note: Award A1 for PV = ±14 000, M1 for other entries correct.

(€)2491.70 A1

[3 marks]

Examiners report

[N/A]

a.i.

[N/A]

a.ii.

[N/A]

a.iii.

[N/A]

b.i.

[N/A]

b.ii.

[N/A]

Paul wants to buy a car. He needs to take out a loan for $7000. The car salesman offers him a loan with an interest rate of 8%, compounded annually. Paul considers two options to repay the loan.

Option 1: Pay $200 each month, until the loan is fully repaid

Option 2: Make 24 equal monthly payments.

Use option 1 to calculate

Use option 2 to calculate

Give a reason why Paul might choose

the number of months it will take for Paul to repay the loan.

[3]

a.i.

the total amount that Paul has to pay.

[2]

a.ii.

the amount Paul pays each month.

[2]

b.i.

the total amount that Paul has to pay.

[2]

b.ii.

option 1.

[1]

c.i.

option 2.

[1]

c.ii.

Markscheme

evidence of using Finance solver on GDC M1

$N = 39.8$ A1

It will take 40 months A1

[3 marks]

a.i.

$40 \times 200 = {\text{\$}}8000$ M1A1

[2 marks]

a.ii.

Monthly payment = $316 ($315.70) M1A1

[2 marks]

b.i.

$24 \times 315.7 = {\text{\$}}7580{\text{ }}\left( {{\text{\$}}7576.80} \right)$ M1A1

[2 marks]

b.ii.

The monthly repayment is lower, he might not be able to afford $316 per month. R1

[1 mark]

c.i.

the total amount to repay is lower. R1

[1 mark]

c.ii.

Examiners report

[N/A]

a.i.

[N/A]

a.ii.

[N/A]

b.i.

[N/A]

b.ii.

[N/A]

c.i.

[N/A]

c.ii.

A new café opened and during the first week their profit was $60.

The café’s profit increases by $10 every week.

A new tea-shop opened at the same time as the café. During the first week their profit was also $60.

The tea-shop’s profit increases by 10 % every week.

Find the café’s profit during the 11th week.

[3]

Calculate the café’s total profit for the first 12 weeks.

[3]

Find the tea-shop’s profit during the 11th week.

[3]

Calculate the tea-shop’s total profit for the first 12 weeks.

[3]

In the mth week the tea-shop’s total profit exceeds the café’s total profit, for the first time since they both opened.

Find the value of m.

[4]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

60 + 10 × 10 (M1)(A1)

Note: Award (M1) for substitution into the arithmetic sequence formula, (A1) for correct substitution.

= ($) 160 (A1)(G3)

[3 marks]

$\frac{{12}}{2}\left( {2 \times 60 + 11 \times 10} \right)$ (M1)(A1)(ft)

Note: Award (M1) for substituting the arithmetic series formula, (A1)(ft) for correct substitution. Follow through from their first term and common difference in part (a).

= ($) 1380 (A1)(ft)(G2)

[3 marks]

60 × 1.1¹⁰ (M1)(A1)

Note: Award (M1) for substituting the geometric progression nth term formula, (A1) for correct substitution.

= ($) 156 (155.624…) (A1)(G3)

Note: Accept the answer if it rounds correctly to 3 sf, as per the accuracy instructions.

[3 marks]

$\frac{{60\left( {{{1.1}^{12}} - 1} \right)}}{{1.1 - 1}}$ (M1)(A1)(ft)

Note: Award (M1) for substituting the geometric series formula, (A1)(ft) for correct substitution. Follow through from part (c) for their first term and common ratio.

= ($)1280 (1283.05…) (A1)(ft)(G2)

[3 marks]

$\frac{{60\left( {{{1.1}^n} - 1} \right)}}{{1.1 - 1}} > \frac{n}{2}\left( {2 \times 60 + \left( {n - 1} \right) \times 10} \right)$ (M1)(M1)

Note: Award (M1) for correctly substituted geometric and arithmetic series formula with n (accept other variable for “n”), (M1) for comparing their expressions consistent with their part (b) and part (d).

(M1)(M1)

Note: Award (M1) for two curves with approximately correct shape drawn in the first quadrant, (M1) for one point of intersection with approximate correct position.

Accept alternative correct sketches, such as

Award (M1) for a curve with approximate correct shape drawn in the 1^st (or 4^th) quadrant and all above (or below) the x-axis, (M1) for one point of intersection with the x-axis with approximate correct position.

17 (A2)(ft)(G3)

Note: Follow through from parts (b) and (d).
An answer of 16 is incorrect. Award at most (M1)(M1)(A0)(A0) with working seen. Award (G0) if final answer is 16 without working seen.

[4 marks]

Examiners report

[N/A]

The admissions team at a new university are trying to predict the number of student applications they will receive each year.

Let $n$ be the number of years that the university has been open. The admissions team collect the following data for the first two years.

It is assumed that the number of students that apply to the university each year will follow a geometric sequence, $u_{n}$ .

In the first year there were $10 380$ places at the university available for applicants. The admissions team announce that the number of places available will increase by $600$ every year.

Let $v_{n}$ represent the number of places available at the university in year $n$ .

For the first $10$ years that the university is open, all places are filled. Students who receive a place each pay an $$ 80$ acceptance fee.

When $n = k$ , the number of places available will, for the first time, exceed the number of students applying.

Calculate the percentage increase in applications from the first year to the second year.

[2]

Write down the common ratio of the sequence.

[1]

b.i.

Find an expression for $u_{n}$ .

[1]

b.ii.

Find the number of student applications the university expects to receive when $n = 11$ . Express your answer to the nearest integer.

[2]

b.iii.

Write down an expression for $v_{n}$ .

[2]

Calculate the total amount of acceptance fees paid to the university in the first $10$ years.

[3]

Find $k$ .

[3]

State whether, for all $n > k$ , the university will have places available for all applicants. Justify your answer.

[2]

Markscheme

$\frac{12 669 - 12 300}{12 300} \times 100$ (M1)

$3 %$ A1

[2 marks]

$1.03$ A1

Note: Follow through from part (a).

[1 mark]

b.i.

$(u_{n} =) 12 300 \times 1 . 03^{n - 1}$ A1

[1 mark]

b.ii.

$(u_{11} =) 12 300 \times 1 . 03^{10}$ (M1)

$16530$ A1

Note: Answer must be to the nearest integer. Do not accept $16500$ .

[2 marks]

b.iii.

$(v_{n} =) 10380 + 600 (n - 1)$ OR $600 n + 9780$ M1A1

Note: Award M1 for substituting into arithmetic sequence formula, A1 for correct substitution.

[2 marks]

$80 \times \frac{10}{2} (2 (10380) + 9 (600))$ (M1)(M1)

Note: Award (M1) for multiplying by $80$ and (M1) for substitution into sum of arithmetic sequence formula.

$$ 10 500 000 ($ 10 464 000)$ A1

[3 marks]

$12 300 \times 1 . 03^{n - 1} < 10 380 + 600 (n - 1)$ or equivalent (M1)

Note: Award (M1) for equating their expressions from parts (b) and (c).

EITHER

graph showing $y = 12 300 \times 1 . 03^{n - 1}$ and $y = 10 380 + 600 (n - 1)$ (M1)

graph showing $y = 12 300 \times 1 . 03^{n - 1} - (10 380 + 600 (n - 1))$ (M1)

list of values including, $(u_{n =}) 17537$ and $(v_{n =}) 17580$ (M1)

$12.4953 \dots$ from graphical method or solving numerical equality (M1)

Note: Award (M1) for a valid attempt to solve.

THEN

$(k =) 13$ A1

[3 marks]

this will not guarantee enough places. A1

EITHER

A written statement that $u_{n} > v_{n}$ , with range of $n$ . R1

Example: “when $n = 24$ (or greater), the number of applications will exceed the number of places again” (“ $u_{n} > v_{n}, n \geq 24$ ”).

exponential growth will always exceed linear growth R1

Note: Accept an equivalent sketch. Do not award A1R0.

[2 marks]

Examiners report

The percentage increase proved more difficult than anticipated, with many using an incorrect denominator. Part (b) was accessible with many candidates earning at least three marks. The expression for the general term of the arithmetic sequence was found in part (c). A surprising number of candidates found the acceptance fees paid in the tenth year, rather than the required total acceptance fees found in the first ten years. Most candidates were able to earn one mark for multiplying their value by $80$ . In part (e), candidates were usually able to find the first point of intersection of $u_{n}$ and $v_{n}$ , but did not always realize the answer must be an integer. Part (f) required candidates to support their answer through a comparison of growth rates, or by finding a range of values/single data point where $u_{n} > v_{n}$ for $n > k$ . Though the answer " $u_{n}$ is geometric, $v_{n}$ is arithmetic", inferred some understanding, this was insufficient justification and required a description that went a little bit further. It is recommended that teachers provide opportunities for candidates to explore the more advanced features of the GDC. An inappropriate choice of calculator window made it difficult for candidates to assess and appreciate the behaviour of $u_{n}$ and $v_{n}$ .

b.i.

b.ii.

b.iii.

Boris recorded the number of daylight hours on the first day of each month in a northern hemisphere town.

This data was plotted onto a scatter diagram. The points were then joined by a smooth curve, with minimum point $(0, 8)$ and maximum point $(6, 16)$ as shown in the following diagram.

Let the curve in the diagram be $y = f (t)$ , where $t$ is the time, measured in months, since Boris first recorded these values.

Boris thinks that $f (t)$ might be modelled by a quadratic function.

Paula thinks that a better model is $f (t) = a \cos (b t) + d$ , $t \geq 0$ , for specific values of $a, b$ and $d$ .

For Paula’s model, use the diagram to write down

The true maximum number of daylight hours was $16$ hours and $14$ minutes.

Write down one reason why a quadratic function would not be a good model for the number of hours of daylight per day, across a number of years.

[1]

the amplitude.

[1]

b.i.

the period.

[1]

b.ii.

the equation of the principal axis.

[2]

b.iii.

Hence or otherwise find the equation of this model in the form:

$f (t) = a \cos (b t) + d$

[3]

For the first year of the model, find the length of time when there are more than $10$ hours and $30$ minutes of daylight per day.

[4]

Calculate the percentage error in the maximum number of daylight hours Boris recorded in the diagram.

[3]

Markscheme

EITHER
annual cycle for daylight length R1

OR
there is a minimum length for daylight (cannot be negative) R1

OR
a quadratic could not have a maximum and a minimum or equivalent R1

Note: Do not accept “Paula's model is better”.

[1 mark]

$4$ A1

[1 mark]

b.i.

$12$ A1

[1 mark]

b.ii.

$y = 12$ A1A1

Note: Award A1 “ $y =$ (a constant)” and A1 for that constant being $12$ .

[2 marks]

b.iii.

$f (t) = - 4 \cos (30 t) + 12$ OR $f (t) = - 4 \cos (- 30 t) + 12$ A1A1A1

Note: Award A1 for $b = 30$ (or $b = - 30$ ), A1 for $a = - 4$ , and A1 for $d = 12$ . Award at most A1A1A0 if extra terms are seen or form is incorrect. Award at most A1A1A0 if $x$ is used instead of $t$ .

[3 marks]

$10.5 = - 4 \cos (30 t) + 12$ (M1)

EITHER

$t_{1} = 2.26585 \dots, t_{2} = 9.73414 \dots$ (A1)(A1)

$t_{1} = \frac{1}{30} \cos^{- 1} \frac{3}{8}$ (A1)

$t_{2} = 12 - t_{1}$ (A1)

THEN

$9.73414 \dots - 2.26585 \dots$

$7.47 (7.46828 \dots)$ months ( $0.622356 \dots$ years) A1

Note: Award M1A1A1A0 for an unsupported answer of $7.46$ . If there is only one intersection point, award M1A1A0A0.

[4 marks]

$|\frac{16 - (16 + \frac{14}{60})}{16 + \frac{14}{60}}| \times 100 %$ (M1)(M1)

Note: Award M1 for correct values and absolute value signs, M1 for $\times 100$ .

$= 1.44 % (1.43737 \dots %)$ A1

[3 marks]

Examiners report

Part (a) indicated a lack of understanding of quadratic functions and the cyclical nature of daylight hours. Some candidates seemed to understand the limitations of a quadratic model but were not always able to use appropriate mathematical language to explain the limitations clearly.

In part (b), many candidates struggled to write down the amplitude, period, and equation of the principal axis.

In part (c), very few candidates recognized that it would be a negative cosine graph here and most did not know how to find the “b” value even if they had originally found the period in part (b). Some candidates used the regression features in their GDC to find the equation of the model; this is outside the SL syllabus but is a valid approach and earned full credit.

In part (d), very few candidates were awarded “follow through” marks in this part. Some substituted 10.5 into their equation rather that equate their equation to 10.5 and attempt to solve it using their GDC to graph the equations or using the “solver” function.

Part (e) was perhaps the best answered part in this question. However, due to premature rounding, many candidates did not gain full marks. A common error was to write the true number of daylight hours as 16.14.

[N/A]

b.i.

[N/A]

b.ii.

[N/A]

b.iii.

[N/A]

The following table shows values of ln x and ln y.

The relationship between ln x and ln y can be modelled by the regression equation ln y = a ln x + b.

Find the value of a and of b.

[3]

Use the regression equation to estimate the value of y when x = 3.57.

[3]

The relationship between x and y can be modelled using the formula y = kxⁿ, where k ≠ 0 , n ≠ 0 , n ≠ 1.

By expressing ln y in terms of ln x, find the value of n and of k.

[7]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

valid approach (M1)

eg one correct value

−0.453620, 6.14210

a = −0.454, b = 6.14 A1A1 N3

[3 marks]

correct substitution (A1)

eg −0.454 ln 3.57 + 6.14

correct working (A1)

eg ln y = 5.56484

261.083 (260.409 from 3 sf)

y = 261, (y = 260 from 3sf) A1 N3

Note: If no working shown, award N1 for 5.56484.
If no working shown, award N2 for ln y = 5.56484.

[3 marks]

METHOD 1

valid approach for expressing ln y in terms of ln x (M1)

eg ${\text{ln}}\,y = {\text{ln}}\,\left( {k{x^n}} \right),\,\,{\text{ln}}\,\left( {k{x^n}} \right) = a\,{\text{ln}}\,x + b$

correct application of addition rule for logs (A1)

eg ${\text{ln}}\,k + {\text{ln}}\,\left( {{x^n}} \right)$

correct application of exponent rule for logs A1

eg ${\text{ln}}\,k + n\,{\text{ln}}\,x$

comparing one term with regression equation (check FT) (M1)

eg $n = a,\,\,b = {\text{ln}}\,k$

correct working for k (A1)

eg ${\text{ln}}\,k = 6.14210,\,\,\,k = {e^{6.14210}}$

465.030

$n = - 0.454,\,\,k = 465$ (464 from 3sf) A1A1 N2N2

METHOD 2

valid approach (M1)

eg ${e^{{\text{ln}}\,y}} = {e^{a\,{\text{ln}}\,x + b}}$

correct use of exponent laws for ${e^{a\,{\text{ln}}\,x + b}}$ (A1)

eg ${e^{a\,{\text{ln}}\,x}} \times {e^b}$

correct application of exponent rule for $a\,{\text{ln}}\,x$ (A1)

eg ${\text{ln}}\,{x^a}$

correct equation in y A1

eg $y = {x^a} \times {e^b}$

comparing one term with equation of model (check FT) (M1)

eg $k = {e^b},\,\,n = a$

465.030

$n = - 0.454,\,\,k = 465$ (464 from 3sf) A1A1 N2N2

METHOD 3

valid approach for expressing ln y in terms of ln x (seen anywhere) (M1)

eg ${\text{ln}}\,y = {\text{ln}}\,\left( {k{x^n}} \right),\,\,{\text{ln}}\,\left( {k{x^n}} \right) = a\,{\text{ln}}\,x + b$

correct application of exponent rule for logs (seen anywhere) (A1)

eg ${\text{ln}}\,\left( {{x^a}} \right) + b$

correct working for b (seen anywhere) (A1)

eg $b = {\text{ln}}\,\left( {{e^b}} \right)$

correct application of addition rule for logs A1

eg ${\text{ln}}\,\left( {{e^b}{x^a}} \right)$

comparing one term with equation of model (check FT) (M1)

eg $k = {e^b},\,\,n = a$

465.030

$n = - 0.454,\,\,k = 465$ (464 from 3sf) A1A1 N2N2

[7 marks]

Examiners report

[N/A]

Abdallah owns a plot of land, near the river Nile, in the form of a quadrilateral ABCD.

The lengths of the sides are ${\text{AB}} = {\text{40 m, BC}} = {\text{115 m, CD}} = {\text{60 m, AD}} = {\text{84 m}}$ and angle ${\rm{B\hat AD}} = 90^\circ$ .

This information is shown on the diagram.

N17/5/MATSD/SP2/ENG/TZ0/03

The formula that the ancient Egyptians used to estimate the area of a quadrilateral ABCD is

${\text{area}} = \frac{{({\text{AB}} + {\text{CD}})({\text{AD}} + {\text{BC}})}}{4}$ .

Abdallah uses this formula to estimate the area of his plot of land.

Show that ${\text{BD}} = 93{\text{ m}}$ correct to the nearest metre.

[2]

Calculate angle ${\rm{B\hat CD}}$ .

[3]

Find the area of ABCD.

[4]

Calculate Abdallah’s estimate for the area.

[2]

d.i.

Find the percentage error in Abdallah’s estimate.

[2]

d.ii.

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

${\text{B}}{{\text{D}}^2} = {40^2} + {84^2}$ (M1)

Note: Award (M1) for correct substitution into Pythagoras.

Accept correct substitution into cosine rule.

${\text{BD}} = 93.0376 \ldots$ (A1)

$= 93$ (AG)

Note: Both the rounded and unrounded value must be seen for the (A1) to be awarded.

[2 marks]

$\cos C = \frac{{{{115}^2} + {{60}^2} - {{93}^2}}}{{2 \times 115 \times 60}}{\text{ }}({93^2} = {115^2} + {60^2} - 2 \times 115 \times 60 \times \cos C)$ (M1)(A1)

Note: Award (M1) for substitution into cosine formula, (A1) for correct substitutions.

$= 53.7^\circ {\text{ }}(53.6679 \ldots ^\circ )$ (A1)(G2)

[3 marks]

$\frac{1}{2}(40)(84) + \frac{1}{2}(115)(60)\sin (53.6679 \ldots )$ (M1)(M1)(A1)(ft)

Note: Award (M1) for correct substitution into right-angle triangle area. Award (M1) for substitution into area of triangle formula and (A1)(ft) for correct substitution.

$= 4460{\text{ }}{{\text{m}}^2}{\text{ }}(4459.30 \ldots {\text{ }}{{\text{m}}^2})$ (A1)(ft)(G3)

Notes: Follow through from part (b).

[4 marks]

$\frac{{(40 + 60)(84 + 115)}}{4}$ (M1)

Note: Award (M1) for correct substitution in the area formula used by ‘Ancient Egyptians’.

$= 4980{\text{ }}{{\text{m}}^2}{\text{ }}(4975{\text{ }}{{\text{m}}^2})$ (A1)(G2)

[2 marks]

d.i.

$\left| {\frac{{4975 - 4459.30 \ldots }}{{4459.30 \ldots }}} \right| \times 100$ (M1)

Notes: Award (M1) for correct substitution into percentage error formula.

$= 11.6{\text{ }}(\% ){\text{ }}(11.5645 \ldots )$ (A1)(ft)(G2)

Notes: Follow through from parts (c) and (d)(i).

[2 marks]

d.ii.

Examiners report

[N/A]

d.i.

[N/A]

d.ii.

On her first day in a hospital, Kiri receives ${u_1}$ milligrams (mg) of a therapeutic drug. The amount of the drug Kiri receives increases by the same amount, $d$ , each day. On the seventh day, she receives 21 mg of the drug, and on the eleventh day she receives 29 mg.

Kiri receives the drug for 30 days.

Ted is also in a hospital and on his first day he receives a 20 mg antibiotic injection. The amount of the antibiotic Ted receives decreases by 50 % each day. On the second day, Ted receives a 10 mg antibiotic injection, on the third day he receives 5 mg, and so on.

Write down an equation, in terms of ${u_1}$ and $d$ , for the amount of the drug that she receives on the seventh day.

[1]

a.i.

Write down an equation, in terms of ${u_1}$ and $d$ , for the amount of the drug that she receives on the eleventh day.

[1]

a.ii.

Write down the value of $d$ and the value of ${u_1}$ .

[2]

Calculate the total amount of the drug, in mg, that she receives.

[3]

Find the amount of antibiotic, in mg, that Ted receives on the fifth day.

[3]

d.i.

The daily amount of antibiotic Ted receives will first be less than 0.06 mg on the $k$ th day. Find the value of $k$ .

[3]

d.ii.

Hence find the total amount of antibiotic, in mg, that Ted receives during the first $k$ days.

[3]

d.iii.

Markscheme

(amount taken in the 7th day): ${u_1} + 6d = 21$ (A1)

Note: Accept ${u_1} + \left( {7 - 1} \right)d = 21$ . The equations do not need to be simplified. They should be given in terms of ${u_1}$ and $d$ for the marks to be awarded.

[1 mark]

a.i.

(amount taken in the 11th day): ${u_1} + 10d = 29$ (A1)

Note: Accept ${u_1} + \left( {11 - 1} \right)d = 29$ . The equations do not need to be simplified. They should be given in terms of ${u_1}$ and $d$ for the marks to be awarded.

[1 mark]

a.ii.

( ${u_1}$ =) 9 (A1)(ft)

( $d$ =) 2 (A1)(ft)

Note: Follow through from part (a), but only if values are positive and ${u_1}$ < 21.

[2 marks]

$\left( {{S_{30}} = } \right)\frac{{30}}{2}\left( {2 \times 9 + \left( {30 - 1} \right) \times 2} \right)$ (M1)(A1)(ft)

Note: Award (M1) for substitution in the sum of an arithmetic sequence formula; (A1)(ft) for their correct substitution.

1140 (mg) (A1)(ft)(G3)

Note: Follow through from their ${u_1}$ and $d$ from part (b).

[3 marks]

20 × (0.5)⁴ (M1)(A1)

Note: Award (M1) for substitution into the geometric sequence formula, (A1) for correct substitution.

1.25 (mg) (A1)(G3)

[3 marks]

d.i.

$20 \times {\left( {0.5} \right)^{k - 1}} < 0.06$ (M1)(M1)

Note: Award (M1) for correct substitution into the geometric sequence formula; (M1) for comparing their expression to 0.06. Accept an equation instead of inequality.

( $k$ =) 10 (10th day) (A1)(ft)(G3)

Note: Follow through from part (d)(i), if 0 < $r$ < 1. Follow through answers must be rounded up for final mark.

[3 marks]

d.ii.

$\frac{{20\left( {1 - {{0.5}^{10}}} \right)}}{{1 - 0.5}}$ (M1)(A1)(ft)

Note: Award (M1) for substitution into sum of a geometric sequence formula, (A1)(ft) for correct substitution.
Follow through from their ${u_1}$ and $r$ in part (d)(i), if 0 < $r$ < 1. Follow through from their $k$ in part (d)(ii) but only if $k$ is a positive integer.

40.0 (39.9609…) (mg) (A1)(ft)(G2)

[3 marks]

d.iii.

Examiners report

[N/A]

a.i.

[N/A]

a.ii.

[N/A]

d.i.

[N/A]

d.ii.

[N/A]

d.iii.

A water container is made in the shape of a cylinder with internal height $h$ cm and internal base radius $r$ cm.

N16/5/MATSD/SP2/ENG/TZ0/06

The water container has no top. The inner surfaces of the container are to be coated with a water-resistant material.

The volume of the water container is $0.5{\text{ }}{{\text{m}}^3}$ .

The water container is designed so that the area to be coated is minimized.

One can of water-resistant material coats a surface area of $2000{\text{ c}}{{\text{m}}^2}$ .

Write down a formula for $A$ , the surface area to be coated.

[2]

Express this volume in ${\text{c}}{{\text{m}}^3}$ .

[1]

Write down, in terms of $r$ and $h$ , an equation for the volume of this water container.

[1]

Show that $A = \pi {r^2} + \frac{{1\,000\,000}}{r}$ .

[2]

Find $\frac{{{\text{d}}A}}{{{\text{d}}r}}$ .

[3]

Using your answer to part (e), find the value of $r$ which minimizes $A$ .

[3]

Find the value of this minimum area.

[2]

Find the least number of cans of water-resistant material that will coat the area in part (g).

[3]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

$(A = ){\text{ }}\pi {r^2} + 2\pi rh$ (A1)(A1)

Note: Award (A1) for either $\pi {r^2}$ OR $2\pi rh$ seen. Award (A1) for two correct terms added together.

[2 marks]

$500\,000$ (A1)

Notes: Units not required.

[1 mark]

$500\,000 = \pi {r^2}h$ (A1)(ft)

Notes: Award (A1)(ft) for $\pi {r^2}h$ equating to their part (b).

Do not accept unless $V = \pi {r^2}h$ is explicitly defined as their part (b).

[1 mark]

$A = \pi {r^2} + 2\pi r\left( {\frac{{500\,000}}{{\pi {r^2}}}} \right)$ (A1)(ft)(M1)

Note: Award (A1)(ft) for their ${\frac{{500\,000}}{{\pi {r^2}}}}$ seen.

Award (M1) for correctly substituting only ${\frac{{500\,000}}{{\pi {r^2}}}}$ into a correct part (a).

Award (A1)(ft)(M1) for rearranging part (c) to $\pi rh = \frac{{500\,000}}{r}$ and substituting for $\pi rh$ in expression for $A$ .

$A = \pi {r^2} + \frac{{1\,000\,000}}{r}$ (AG)

Notes: The conclusion, $A = \pi {r^2} + \frac{{1\,000\,000}}{r}$ , must be consistent with their working seen for the (A1) to be awarded.

Accept ${10^6}$ as equivalent to ${1\,000\,000}$ .

[2 marks]

$2\pi r - \frac{{{\text{1}}\,{\text{000}}\,{\text{000}}}}{{{r^2}}}$ (A1)(A1)(A1)

Note: Award (A1) for $2\pi r$ , (A1) for $\frac{1}{{{r^2}}}$ or ${r^{ - 2}}$ , (A1) for $- {\text{1}}\,{\text{000}}\,{\text{000}}$ .

[3 marks]

$2\pi r - \frac{{1\,000\,000}}{{{r^2}}} = 0$ (M1)

Note: Award (M1) for equating their part (e) to zero.

${r^3} = \frac{{1\,000\,000}}{{2\pi }}$ OR $r = \sqrt[3]{{\frac{{1\,000\,000}}{{2\pi }}}}$ (M1)

Note: Award (M1) for isolating $r$ .

sketch of derivative function (M1)

with its zero indicated (M1)

$(r = ){\text{ }}54.2{\text{ }}({\text{cm}}){\text{ }}(54.1926 \ldots )$ (A1)(ft)(G2)

[3 marks]

$\pi {(54.1926 \ldots )^2} + \frac{{1\,000\,000}}{{(54.1926 \ldots )}}$ (M1)

Note: Award (M1) for correct substitution of their part (f) into the given equation.

$= 27\,700{\text{ }}({\text{c}}{{\text{m}}^2}){\text{ }}(27\,679.0 \ldots )$ (A1)(ft)(G2)

[2 marks]

$\frac{{27\,679.0 \ldots }}{{2000}}$ (M1)

Note: Award (M1) for dividing their part (g) by 2000.

$= 13.8395 \ldots$ (A1)(ft)

Notes: Follow through from part (g).

14 (cans) (A1)(ft)(G3)

Notes: Final (A1) awarded for rounding up their $13.8395 \ldots$ to the next integer.

[3 marks]

Examiners report

[N/A]

The Tower of Pisa is well known worldwide for how it leans.

Giovanni visits the Tower and wants to investigate how much it is leaning. He draws a diagram showing a non-right triangle, ABC.

On Giovanni’s diagram the length of AB is 56 m, the length of BC is 37 m, and angle ACB is 60°. AX is the perpendicular height from A to BC.

Giovanni’s tourist guidebook says that the actual horizontal displacement of the Tower, BX, is 3.9 metres.

Use Giovanni’s diagram to show that angle ABC, the angle at which the Tower is leaning relative to the
horizontal, is 85° to the nearest degree.

[5]

a.i.

Use Giovanni's diagram to calculate the length of AX.

[2]

a.ii.

Use Giovanni's diagram to find the length of BX, the horizontal displacement of the Tower.

[2]

a.iii.

Find the percentage error on Giovanni’s diagram.

[2]

Giovanni adds a point D to his diagram, such that BD = 45 m, and another triangle is formed.

Find the angle of elevation of A from D.

[3]

Markscheme

$\frac{{{\text{sin BAC}}}}{{37}} = \frac{{{\text{sin 60}}}}{{56}}$ (M1)(A1)

Note: Award (M1) for substituting the sine rule formula, (A1) for correct substitution.

angle ${\text{B}}\mathop {\text{A}}\limits^ \wedge {\text{C}}$ = 34.9034…° (A1)

Note: Award (A0) if unrounded answer does not round to 35. Award (G2) if 34.9034… seen without working.

angle ${\text{A}}\mathop {\text{B}}\limits^ \wedge {\text{C}}$ = 180 − (34.9034… + 60) (M1)

Note: Award (M1) for subtracting their angle BAC + 60 from 180.

85.0965…° (A1)

85° (AG)

Note: Both the unrounded and rounded value must be seen for the final (A1) to be awarded. If the candidate rounds 34.9034...° to 35° while substituting to find angle ${\text{A}}\mathop {\text{B}}\limits^ \wedge {\text{C}}$ , the final (A1) can be awarded but only if both 34.9034...° and 35° are seen.
If 85 is used as part of the workings, award at most (M1)(A0)(A0)(M0)(A0)(AG). This is the reverse process and not accepted.

a.i.

sin 85… × 56 (M1)

= 55.8 (55.7869…) (m) (A1)(G2)

Note: Award (M1) for correct substitution in trigonometric ratio.

a.ii.

$\sqrt {{{56}^2} - 55.7869{ \ldots ^2}}$ (M1)

Note: Award (M1) for correct substitution in the Pythagoras theorem formula. Follow through from part (a)(ii).

cos(85) × 56 (M1)

Note: Award (M1) for correct substitution in trigonometric ratio.

= 4.88 (4.88072…) (m) (A1)(ft)(G2)

Note: Accept 4.73 (4.72863…) (m) from using their 3 s.f answer. Accept equivalent methods.

[2 marks]

a.iii.

$\left| {\frac{{4.88 - 3.9}}{{3.9}}} \right| \times 100$ (M1)

Note: Award (M1) for correct substitution into the percentage error formula.

= 25.1 (25.1282) (%) (A1)(ft)(G2)

Note: Follow through from part (a)(iii).

[2 marks]

${\text{ta}}{{\text{n}}^{ - 1}}\left( {\frac{{55.7869 \ldots }}{{40.11927 \ldots }}} \right)$ (A1)(ft)(M1)

Note: Award (A1)(ft) for their 40.11927… seen. Award (M1) for correct substitution into trigonometric ratio.

(37 − 4.88072…)² + 55.7869…²

(AC =) 64.3725…

64.3726…² + 8² − 2 × 8 × 64.3726… × cos120

(AD =) 68.7226…

$\frac{{{\text{sin 120}}}}{{68.7226 \ldots }} = \frac{{{\text{sin A}}\mathop {\text{D}}\limits^ \wedge {\text{C}}}}{{64.3725 \ldots }}$ (A1)(ft)(M1)

Note: Award (A1)(ft) for their correct values seen, (M1) for correct substitution into the sine formula.

= 54.3° (54.2781…°) (A1)(ft)(G2)

Note: Follow through from part (a). Accept equivalent methods.

[3 marks]

Examiners report

[N/A]

a.i.

[N/A]

a.ii.

[N/A]

a.iii.

[N/A]

The following table shows the average body weight, $x$ , and the average weight of the brain, $y$ , of seven species of mammal. Both measured in kilograms (kg).

M17/5/MATSD/SP2/ENG/TZ1/01

The average body weight of grey wolves is 36 kg.

In fact, the average weight of the brain of grey wolves is 0.120 kg.

Find the range of the average body weights for these seven species of mammal.

[2]

For the data from these seven species calculate $r$ , the Pearson’s product–moment correlation coefficient;

[2]

b.i.

For the data from these seven species describe the correlation between the average body weight and the average weight of the brain.

[2]

b.ii.

Write down the equation of the regression line $y$ on $x$ , in the form $y = mx + c$ .

[2]

Use your regression line to estimate the average weight of the brain of grey wolves.

[2]

Find the percentage error in your estimate in part (d).

[2]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

$529 - 3$ (M1)

$= 526{\text{ (kg)}}$ (A1)(G2)

[2 marks]

$0.922{\text{ }}(0.921857 \ldots )$ (G2)

[2 marks]

b.i.

(very) strong, positive (A1)(ft)(A1)(ft)

Note: Follow through from part (b)(i).

[2 marks]

b.ii.

$y = 0.000986x + 0.0923{\text{ }}(y = 0.000985837 \ldots x + 0.0923391…)$ (A1)(A1)

Note: Award (A1) for $0.000986x$ , (A1) for 0.0923.

Award a maximum of (A1)(A0) if the answer is not an equation in the form $y = mx + c$ .

[2 marks]

$0.000985837 \ldots (36) + 0.0923391 \ldots$ (M1)

Note: Award (M1) for substituting 36 into their equation.

$0.128{\text{ (kg) }}\left( {0.127829 \ldots {\text{ (kg)}}} \right)$ (A1)(ft)(G2)

Note: Follow through from part (c). The final (A1) is awarded only if their answer is positive.

[2 marks]

$\left| {\frac{{0.127829 \ldots - 0.120}}{{0.120}}} \right| \times 100$ (M1)

Note: Award (M1) for their correct substitution into percentage error formula.

$6.52{\text{ }}(\% ){\text{ }}\left( {6.52442...{\text{ }}(\% )} \right)$ (A1)(ft)(G2)

Note: Follow through from part (d). Do not accept a negative answer.

[2 marks]

Examiners report

[N/A]

b.i.

[N/A]

b.ii.

[N/A]

Let $f\left( x \right) = {{\text{e}}^{2\,{\text{sin}}\left( {\frac{{\pi x}}{2}} \right)}}$ , for x > 0.

The k th maximum point on the graph of f has x-coordinate x_k where $k \in {\mathbb{Z}^ + }$ .

Given that x_k_{+ 1} = x_k + a, find a.

[4]

Hence find the value of n such that $\sum\limits_{k = 1}^n {{x_k} = 861}$ .

[4]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

valid approach to find maxima (M1)

eg one correct value of x_k, sketch of f

any two correct consecutive values of x_k (A1)(A1)

eg x₁ = 1, x₂ = 5

a = 4 A1 N3

[4 marks]

recognizing the sequence x_1, x₂_, x_{3, …,} x_n is arithmetic (M1)

eg d = 4

correct expression for sum (A1)

eg $\frac{n}{2}\left( {2\left( 1 \right) + 4\left( {n - 1} \right)} \right)$

valid attempt to solve for n (M1)

eg graph, 2n² − n − 861 = 0

n = 21 A1 N2

[4 marks]

Examiners report

[N/A]

Give your answers to parts (b), (c) and (d) to the nearest whole number.

Harinder has 14 000 US Dollars (USD) to invest for a period of five years. He has two options of how to invest the money.

Option A: Invest the full amount, in USD, in a fixed deposit account in an American bank.

The account pays a nominal annual interest rate of r % , compounded yearly, for the five years. The bank manager says that this will give Harinder a return of 17 500 USD.

Option B: Invest the full amount, in Indian Rupees (INR), in a fixed deposit account in an Indian bank. The money must be converted from USD to INR before it is invested.

The exchange rate is 1 USD = 66.91 INR.

The account in the Indian bank pays a nominal annual interest rate of 5.2 % compounded monthly.

Calculate the value of r.

[3]

Calculate 14 000 USD in INR.

[2]

Calculate the amount of this investment, in INR, in this account after five years.

[3]

Harinder chose option B. At the end of five years, Harinder converted this investment back to USD. The exchange rate, at that time, was 1 USD = 67.16 INR.

Calculate how much more money, in USD, Harinder earned by choosing option B instead of option A.

[3]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

$17500 = 14000{\left( {1 + \frac{r}{{100}}} \right)^5}$ (M1)(A1)

Note: Award (M1) for substitution into the compound interest formula, (A1) for correct substitution. Award at most (M1)(A0) if not equated to 17500.

N = 5

PV = ±14000

FV = $\mp$ 17500

P/Y = 1

C/Y = 1 (A1)(M1)

Note: Award (A1) for C/Y = 1 seen, (M1) for all other correct entries. FV and PV must have opposite signs.

= 4.56 (%) (4.56395… (%)) (A1) (G3)

[3 marks]

14000 × 66.91 (M1)

Note: Award (M1) for multiplying 14000 by 66.91.

936740 (INR) (A1) (G2)

Note: Answer must be given to the nearest whole number.

[2 marks]

$936740 \times {\left( {1 + \frac{{5.2}}{{12 \times 100}}} \right)^{12 \times 5}}$ (M1)(A1)(ft)

Note: Award (M1) for substitution into the compound interest formula, (A1)(ft) for their correct substitution.

N = 60

I% = 5.2

PV = ±936740

P/Y= 12

C/Y= 12 (A1)(M1)

Note: Award (A1) for C/Y = 12 seen, (M1) for all other correct entries.

N = 5

I% = 5.2

PV = ±936740

P/Y= 1

C/Y= 12 (A1)(M1)

Note: Award (A1) for C/Y = 12 seen, (M1) for all other correct entries

= 1214204 (INR) (A1)(ft) (G3)

Note: Follow through from part (b). Answer must be given to the nearest whole number.

[3 marks]

$\frac{{1214204}}{{67.16}}$ (M1)

Note: Award (M1) for dividing their (c) by 67.16.

$\left( {\frac{{1214204}}{{67.16}}} \right) - 17500 = 579$ (USD) (M1)(A1)(ft) (G3)

Note: Award (M1) for finding the difference between their conversion and 17500. Answer must be given to the nearest whole number. Follow through from part (c).

[3 marks]

Examiners report

[N/A]

Give your answers in parts (a), (d)(i), (e) and (f) to the nearest dollar.

Daisy invested $37 000$ Australian dollars ( $AUD$ ) in a fixed deposit account with an annual interest rate of $6.4 %$ compounded quarterly.

After $m$ months, the amount of money in the fixed deposit account has appreciated to more than $50 000 AUD$ .

Daisy is saving to purchase a new apartment. The price of the apartment is $200 000 AUD$ .

Daisy makes an initial payment of $25 %$ and takes out a loan to pay the rest.

The loan is for $10$ years, compounded monthly, with equal monthly payments of $1700 AUD$ made by Daisy at the end of each month.

For this loan, find

After $5$ years of paying off this loan, Daisy decides to pay the remainder in one final payment.

Calculate the value of Daisy’s investment after $2$ years.

[3]

Find the minimum value of $m$ , where $m \in ℕ$ .

[4]

Write down the amount of the loan.

[1]

the amount of interest paid by Daisy.

[2]

d.i.

the annual interest rate of the loan.

[3]

d.ii.

Find the amount of Daisy’s final payment.

[3]

Find how much money Daisy saved by making one final payment after $5$ years.

[3]

Markscheme

EITHER

$N = 2$
$P V = - 37 000$
$I % = 6.4$
$P / Y = 1$
$C / Y = 4$ (M1)(A1)

Note: Award M1 for an attempt to use a financial app in their technology, award A1 for all entries correct.

OR

$N = 8$
$P V = - 37 000$
$I % = 6.4$
$P / Y = 4$
$C / Y = 4$ (M1)(A1)

Note: Award M1 for an attempt to use a financial app in their technology, award A1 for all entries correct.

$F V = 37 000 \times {(1 + \frac{6.4}{100 \times 4})}^{4 \times 2}$ (M1)(A1)

Note: Award M1 for substitution into compound interest formula, (A1) for correct substitution.

$= 42 010 AUD$ A1

Note: Award (M1)(A1)A0 for unsupported $42009.87$ .

[3 marks]

EITHER

$P V = - 37 000$
$F V = 50 000$
$I % = 6.4$
$P / Y = 1$
$C / Y = 4$ (M1)(A1)

Note: Award M1 for an attempt to use a financial app in their technology, award A1 for all entries correct. The final mark can still be awarded for the correct number of months (multiple of $3$ ).

OR

$P V = - 37 000$
$F V = 50 000$
$I % = 6.4$
$P / Y = 4$
$C / Y = 4$ (M1)(A1)

Note: Award M1 for an attempt to use a financial app in their technology, award A1 for all entries correct.

$50 000 < 37 000 \times {(1 + \frac{6.4}{100 \times 4})}^{4 \times n}$ OR $50 000 < 37 000 \times {(1 + \frac{6.4}{100 \times 4})}^{n}$ (M1)(A1)

Note: Award M1 for the correct inequality, $50 000$ and substituted compound interest formula. Allow an equation. Award A1 for correct substitution.

THEN

$N = 4.74 (years) (4.74230 \dots)$ OR $N = 18.9692 \dots (quarters)$ (A1)

$m = 57$ months A1

Note: Award A1 for rounding their $m$ to the correct number of months. The final answer must be a multiple of $3$ . Follow through within this part.

[4 marks]

$150 000 AUD$ A1

[1 mark]

$120 \times 1700 - 150 000$ (M1)

$= 54 000 AUD$ A1

[2 marks]

d.i.

$N = 120$
$P V = - 150 000$
$P M T = 1700$
$F V = 0$
$P / Y = 12$
$C / Y = 12$ (M1)(A1)

Note: Award M1 for an attempt to use a financial app in their technology or an attempt to use an annuity formula or $F V = 0$ seen. If a compound interest formula is equated to zero, award M1, otherwise award M0 for a substituted compound interest formula.
Award A1 for all entries correct in financial app or correct substitution in annuity formula, but award A0 for a substituted compound interest formula. Follow through marks in part (d)(ii) are contingent on working seen.

$r = 6.46 (%) (6.45779 \dots)$ A1

[3 marks]

d.ii.

$N = 60$
$I = 6.46 (6.45779 \dots)$
$P V = - 150 000$
$P M T = 1700$
$P / Y = 12$
$C / Y = 12$ (M1)(A1)

Note: Award M1 for an attempt to use a financial app in their technology or an attempt to use an annuity formula. Award (M0) for a substituted compound interest formula. Award A1 for all entries correct. Follow through marks in part (e) are contingent on working seen.

$F V = 86973 AUD$ A1

[3 marks]

$204 000 - (60 \times 1700 + 86973)$ OR $204 000 - 188 973$ (M1)(A1)

Note: Award M1 for $60 \times 1700$ . Award M1 for subtracting their $(60 \times 1700 + 86973)$ from their $(204 000)$ . Award at most M1M0 for their $204 000 - (60 \times 1700)$ or M0M0 for their $204 000 - (86973)$ . Follow through from parts (d)(i) and (e). Follow through marks in part (f) are contingent on working seen.

$= 15 027 AUD$ A1

[3 marks]

Examiners report

[N/A]

d.i.

[N/A]

d.ii.

[N/A]

The first term of an infinite geometric sequence is 4. The sum of the infinite sequence is 200.

Find the common ratio.

[2]

Find the sum of the first 8 terms.

[2]

Find the least value of n for which S_n > 163.

[3]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

correct substitution into infinite sum (A1)
eg $200 = \frac{4}{{1 - r}}$

r = 0.98 (exact) A1 N2

[2 marks]

correct substitution (A1)

$\frac{{4\left( {1 - {{0.98}^8}} \right)}}{{1 - 0.98}}$

29.8473

29.8 A1 N2

[2 marks]

attempt to set up inequality (accept equation) (M1)
eg $\frac{{4\left( {1 - {{0.98}^n}} \right)}}{{1 - 0.98}} > 163,\,\,\frac{{4\left( {1 - {{0.98}^n}} \right)}}{{1 - 0.98}} = 163$

correct inequality for n (accept equation) or crossover values (A1)
eg n > 83.5234, n = 83.5234, S₈₃ = 162.606 and S₈₄ = 163.354

n = 84 A1 N1

[3 marks]

Examiners report

[N/A]

Rosa joins a club to prepare to run a marathon. During the first training session Rosa runs a distance of 3000 metres. Each training session she increases the distance she runs by 400 metres.

A marathon is 42.195 kilometres.

In the $k$ th training session Rosa will run further than a marathon for the first time.

Carlos joins the club to lose weight. He runs 7500 metres during the first month. The distance he runs increases by 20% each month.

Write down the distance Rosa runs in the third training session;

[1]

a.i.

Write down the distance Rosa runs in the $n$ th training session.

[2]

a.ii.

Find the value of $k$ .

[2]

Calculate the total distance, in kilometres, Rosa runs in the first 50 training sessions.

[4]

Find the distance Carlos runs in the fifth month of training.

[3]

Calculate the total distance Carlos runs in the first year.

[3]

Markscheme

3800 m (A1)

[1 mark]

a.i.

$3000 + (n - 1)400{\text{ m}}$ $\,\,\,$ OR $\,\,\,$ $2600 + 400n{\text{ m}}$ (M1)(A1)

Note: Award (M1) for substitution into arithmetic sequence formula, (A1) for correct substitution.

[2 marks]

a.ii.

$3000 + (k - 1)400 > 42195$ (M1)

Notes: Award (M1) for their correct inequality. Accept $3 + (k - 1)0.4 > 42.195$ .

Accept $=$ OR $\geqslant$ . Award (M0) for $3000 + (k - 1)400 > 42.195$ .

$(k = ){\text{ }}99$ (A1)(ft)(G2)

Note: Follow through from part (a)(ii), but only if $k$ is a positive integer.

[2 marks]

$\frac{{50}}{2}\left( {2 \times 3000 + (50 - 1)(400)} \right)$ (M1)(A1)(ft)

Note: Award (M1) for substitution into sum of an arithmetic series formula, (A1)(ft) for correct substitution.

$640\,000{\text{ m}}$ (A1)

Note: Award (A1) for their $640\,000$ seen.

$= 640{\text{ km}}$ (A1)(ft)(G3)

Note: Award (A1)(ft) for correctly converting their answer in metres to km; this can be awarded independently from previous marks.

$\frac{{50}}{2}\left( {2 \times 3 + (50 - 1)(0.4)} \right)$ (M1)(A1)(ft)(A1)

Note: Award (M1) for substitution into sum of an arithmetic series formula, (A1)(ft) for correct substitution, (A1) for correctly converting 3000 m and 400 m into km.

$= 640{\text{ km}}$ (A1)(G3)

[4 marks]

$7500 \times {1.2^{5 - 1}}$ (M1)(A1)

Note: Award (M1) for substitution into geometric series formula, (A1) for correct substitutions.

$= 15\,600{\text{ m }}(15\,552{\text{ m}})$ (A1)(G3)

$7.5 \times {1.2^{5 - 1}}$ (M1)(A1)

Note: Award (M1) for substitution into geometric series formula, (A1) for correct substitutions.

$= 15.6{\text{ km}}$ (A1)(G3)

[3 marks]

$\frac{{7500({{1.2}^{12}} - 1)}}{{1.2 - 1}}$ (M1)(A1)

Notes: Award (M1) for substitution into sum of a geometric series formula, (A1) for correct substitutions. Follow through from their ratio ( $r$ ) in part (d). If $r < 1$ (distance does not increase) or the final answer is unrealistic (eg $r = 20$ ), do not award the final (A1).

$= 297\,000{\text{ m }}(296\,853 \ldots {\text{ m}},{\text{ }}297{\text{ km}})$ (A1)(G2)

[3 marks]

Examiners report

[N/A]

a.i.

[N/A]

a.ii.

[N/A]

John purchases a new bicycle for 880 US dollars (USD) and pays for it with a Canadian credit card. There is a transaction fee of 4.2 % charged to John by the credit card company to convert this purchase into Canadian dollars (CAD).

The exchange rate is 1 USD = 1.25 CAD.

John insures his bicycle with a US company. The insurance company produces the following table for the bicycle’s value during each year.

The values of the bicycle form a geometric sequence.

During the 1st year John pays 120 USD to insure his bicycle. Each year the amount he pays to insure his bicycle is reduced by 3.50 USD.

Calculate, in CAD, the total amount John pays for the bicycle.

[3]

Find the value of the bicycle during the 5th year. Give your answer to two decimal places.

[3]

Calculate, in years, when the bicycle value will be less than 50 USD.

[2]

Find the total amount John has paid to insure his bicycle for the first 5 years.

[3]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

1.042 × 880 × 1.25 OR (880 + 0.042 × 880) × 1.25 (M1)(M1)

Note: Award (M1) for multiplying 880 by 1.042 and (M1) for multiplying 880 by 1.25.

1150 (CAD) (1146.20 (CAD)) (A1)(G2)

Note: Accept 1146.2 (CAD)

[3 marks]

$\frac{{704}}{{880}}$ OR $\frac{{563.20}}{{704}}$ (M1)

Note: Award (M1) for correctly dividing sequential terms to find the common ratio, or 0.8 seen.

880(0.8)⁵⁻¹ (M1)

Note: Award (M1) for correct substitution into geometric sequence formula.

360.45 (USD) (A1)(G3)

Note: Do not award the final (A1) if the answer is not correct to 2 decimal places. Award at most (M0)(M1)(A0) if $r = 1.25$ .

[3 marks]

$880{\left( {0.8} \right)^{n - 1}} < 50$ (M1)

Note: Award (M1) for correct substitution into geometric sequence formula and (in)equating to 50. Accept weak or strict inequalities. Accept an equation. Follow through from their common ratio in part (b). Accept a sketch of their GP with $y = 50$ as a valid method.

${u_{13}} = 60.473$ AND ${u_{14}} = 48.379$ (M1)

Note: Award (M1) for their ${u_{13}}$ and ${u_{14}}$ both seen. If the student states ${u_{14}} = 48.379 < 50$ , without ${u_{13}} = 60.473$ seen, this is not sufficient to award (M1).

14 or “14th year” or “after the 13th year” (A1)(ft)(G2)

Note: The context of the question requires the final answer to be an integer. Award at most (M1)(A0) for a final answer of 13.9 years. Follow through from their 0.8 in part (b).

[2 marks]

$\frac{5}{2}\left( {\left( {2 \times 120} \right) + \left( { - 3.5\left( {5 - 1} \right)} \right)} \right)$ (M1)(A1)

Note: Award (M1) for substitution into arithmetic series formula, (A1) for correct substitution.

565 (USD) (A1)(G2)

[3 marks]

Examiners report

[N/A]

A large underground tank is constructed at Mills Airport to store fuel. The tank is in the shape of an isosceles trapezoidal prism, $ABCDEFGH$ .

$AB = 70 m$ , $AF = 200 m$ , $AD = 40 m$ , $BC = 40 m$ and $CD = 110 m$ . Angle $ADC = 60 °$ and angle $BCD = 60 °$ . The tank is illustrated below.

Once construction was complete, a fuel pump was used to pump fuel into the empty tank. The amount of fuel pumped into the tank by this pump each hour decreases as an arithmetic sequence with terms $u_{1}, u_{2}, u_{3}, \dots, u_{n}$ .

Part of this sequence is shown in the table.

At the end of the $2 nd$ hour, the total volume of fuel in the tank was $88 200 m^{3}$ .

Find $h$ , the height of the tank.

[2]

Show that the volume of the tank is $624 000 m^{3}$ , correct to three significant figures.

[3]

Write down the common difference, $d$ .

[1]

Find the amount of fuel pumped into the tank in the $13th$ hour.

[2]

Find the value of $n$ such that $u_{n} = 0$ .

[2]

e.i.

Write down the number of hours that the pump was pumping fuel into the tank.

[1]

e.ii.

Find the total amount of fuel pumped into the tank in the first $8$ hours.

[2]

Show that the tank will never be completely filled using this pump.

[3]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

$\sin 60 ° = \frac{h}{40}$ OR $\tan 60 ° = \frac{h}{20}$ (M1)

Note: Award (M1) for correct substitutions in trig ratio.

$20^{2} + h^{2} = 40^{2} (\sqrt{40^{2} - 20^{2}})$ (M1)

Note: Award (M1) for correct substitutions in Pythagoras’ theorem.

$(h =) 34.6 (m) (\sqrt{1200}, 20 \sqrt{3}, 34.6410 \dots)$ (A1)(G2)

[2 marks]

$\frac{1}{2} (70 + 110) (34.6410 \dots) \times 200$ (M1)(M1)

Note: Award (M1) for their correctly substituted area of trapezium formula, provided all substitutions are positive. Award (M1) for multiplying by $200$ . Follow through from part (a).

$(2 \times \frac{1}{2} \times 20 \times 34.6410 \dots + 70 \times 34.6410 \dots) \times 200$ (M1)(M1)

Note: Award (M1) for the addition of correct areas for two triangles and one rectangle. Award (M1) for multiplying by $200$ . Follow through from part (a).

$70 \times 34.6410 \dots \times 200 + 2 \times \frac{1}{2} \times 34.6410 \dots \times 20 \times 200$ (M1)(M1)

Note: Award (M1) for their correct substitution in volume of cuboid formula. Award (M1) for correctly substituted volume of triangular prism(s). Follow through from part (a).

$623538 \dots$ (A1)

$624000 (m^{3})$ (AG)

Note: Both an unrounded answer that rounds to the given answer and the rounded value must be seen for the (A1) to be awarded.

[3 marks]

$(d =) - 1800$ (A1)

[1 mark]

$(u_{13} =) 45000 + (13 - 1) (- 1800)$ (M1)

Note: Award (M1) for correct substitutions in arithmetic sequence formula.
OR
Award (M1) for a correct $4^{th}$ term seen as part of list.

$23400 (m^{3})$ (A1)(ft)(G2)

Note: Follow through from part (c) for their value of $d$ .

[2 marks]

$0 = 45000 + (n - 1) (- 1800)$ (M1)

Note: Award (M1) for their correct substitution into arithmetic sequence formula, equated to zero.

$(n =) 26$ (A1)(ft)(G2)

Note: Follow through from part (c). Award at most (M1)(A0) if their $n$ is not a positive integer.

[2 marks]

e.i.

$25$ (A1)(ft)

Note: Follow through from part (e)(i), but only if their final answer in (e)(i) is positive. If their $n$ in part (e)(i) is not an integer, award (A1)(ft) for the nearest lower integer.

[1 mark]

e.ii.

$(S_{8} =) \frac{8}{2} (2 \times 45000 + (8 - 1) \times (- 1800))$ (M1)

Note: Award (M1) for their correct substitutions in arithmetic series formula. If a list method is used, award (M1) for the addition of their $8$ correct terms.

$310 000 (m^{3}) (309 600)$ (A1)(ft)(G2)

Note: Follow through from part (c). Award at most (M1)(A0) if their final answer is greater than $624 000$ .

[2 marks]

$(S_{25} =) \frac{25}{2} (2 \times 45000 + (25 - 1) \times (- 1800)), (S_{25} =) \frac{25}{2} (45000 + 1800)$ (M1)

Note: Award (M1) for their correct substitutions into arithmetic series formula.

$S_{25} = 585000 (m^{3})$ (A1)(ft)(G1)

Note: Award (M1)(A1) for correctly finding $S_{26} = 585000 (m^{3})$ , provided working is shown e.g. $(S_{26} =) \frac{26}{2} (2 \times 45000 + (26 - 1) \times (- 1800))$ , $(S_{26} =) \frac{26}{2} (45000 + 0)$ . Follow through from part (c) and either their (e)(i) or (e)(ii). If $d < 0$ and their final answer is greater than $624 000$ , award at most (M1)(A1)(ft)(R0). If $d > 0$ , there is no maximum, award at most (M1)(A0)(R0). Award no marks if their number of terms is not a positive integer.

$585000 (m^{3}) < 624000 (m^{3})$ (R1)

Hence it will never be filled (AG)

Note: The (AG) line must be seen. If it is omitted do not award the final (R1). Do not follow through within the part.
For unsupported $(S_{25}) = 585000$ seen, award at most (G1)(R1)(AG). Working must be seen to follow through from parts (c) and (e)(i) or (e)(ii).

$(S_{n} =) \frac{n}{2} (2 \times 45000 + (n - 1) \times (- 1800))$ (M1)

Note: Award (M1) for their correct substitution into arithmetic series formula, with $n$ .

Maximum of this function $585225 (m^{3})$ (A1)

Note: Follow through from part (c). Award at most (M1)(A1)(ft)(R0) if their final answer is greater than $624 000$ . Award at most (M1)(A0)(R0) if their common difference is not $- 1800$ . Award at most (M1)(A0)(R0) if $585 225$ is not explicitly identified as the maximum of the function.

$585225 (m^{3}) < 624000 (m^{3})$ (R1)

Hence it will never be filled (AG)

Note: The (AG) line must be seen. If it is omitted do not award the final (R1). Do not follow through within the part.

sketch with concave down curve and labelled $624000$ horizontal line (M1)

Note: Accept a label of “tank volume” instead of a numerical value. Award (M0) if the line and the curve intersect.

curve explicitly labelled as $(S_{n} =) \frac{n}{2} (2 \times 45000 + (n - 1) \times (- 1800))$ or equivalent (A1)

Note: Award (A1) for a written explanation interpreting the sketch. Accept a comparison of values, e.g $585225 (m^{3}) < 624000 (m^{3})$ , where $585225$ is the graphical maximum. Award at most (M1)(A0)(R0) if their common difference is not $- 1800$ .

the line and the curve do not intersect (R1)

hence it will never be filled (AG)

Note: The (AG) line must be seen. If it is omitted do not award the final (R1). Do not follow through within the part.

$624000 = \frac{n}{2} (2 \times 45000 + (n - 1) \times (- 1800))$ (M1)

Note: Award (M1) for their correctly substituted arithmetic series formula equated to $624000 (623538)$ .

Demonstrates there is no solution (A1)

Note: Award (A1) for a correct working that the discriminant is less than zero OR correct working indicating there is no real solution in the quadratic formula.

There is no (real) solution (to this equation) (R1)

hence it will never be filled (AG)

Note: At most (M1)(A0)(R0) for their correctly substituted arithmetic series formula $= 624000, 623538$ or $622800$ with a statement "no solution". Follow through from their part (b).

[3 marks]

Examiners report

[N/A]

e.i.

[N/A]

e.ii.

[N/A]

The sum of an infinite geometric sequence is 33.25. The second term of the sequence is 7.98. Find the possible values of $r$ .

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

correct substitution into formula for infinite geometric series (A1)

eg $33.25 = \frac{{{u_1}}}{{1 - r}}$

correct substitution into formula for ${u_n}$ (seen anywhere) (A1)

eg $7.98 = {u_1}r$

attempt to express ${u_1}$ in terms of $r$ (or vice-versa) (M1)

eg ${u_1} = \frac{{7.98}}{r}$ , ${u_1} = 33.25\left( {1 - r} \right)$ , $r = \frac{{7.98}}{{{u_1}}}$ , $r = \frac{{33.25 - {u_1}}}{{33.25}}$

correct working (A1)

eg $\frac{{\left( {\frac{{7.98}}{r}} \right)}}{{1 - r}} = 33.25$ , $33.25\left( {1 - r} \right) = \frac{{7.98}}{r}$ , (0.4, 19.95), (0.6, 13.3), $\frac{{{u_1}}}{{1 - \frac{{7.98}}{{{u_1}}}}} = 33.25$

$r = 0.4\,\,\,\,\left( { = \frac{2}{5}} \right)$ , $r = 0.6\,\,\,\,\left( { = \frac{3}{5}} \right)$ A1A1 N3

[6 marks]

Examiners report

[N/A]

Maegan designs a decorative glass face for a new Fine Arts Centre. The glass face is made up of small triangular panes. The first three levels of the glass face are illustrated in the following diagram.

The $1 st$ level, at the bottom of the glass face, has $5$ triangular panes. The $2 nd$ level has $7$ triangular panes, and the $3 rd$ level has $9$ triangular panes. Each additional level has $2$ more triangular panes than the level below it.

Maegan has $1000$ triangular panes to build the decorative glass face and does not want it to have any incomplete levels.

Find the number of triangular panes in the $12 th$ level.

[3]

Show that the total number of triangular panes, $S_{n}$ , in the first $n$ levels is given by:

$S_{n} = n^{2} + 4 n$ .

[3]

Hence, find the total number of panes in a glass face with $18$ levels.

[2]

Find the maximum number of complete levels that Maegan can build.

[3]

Each triangular pane has an area of $1.84 m^{2}$ .

Find the total area of the decorative glass face, if the maximum number of complete levels were built. Express your area to the nearest $m^{2}$ .

[4]

Markscheme

$u_{12} = 5 + (12 - 1) \times (2)$ (M1)(A1)

Note: Award (M1) for substituted arithmetic sequence formula, (A1) for correct substitutions.

$27$ (A1)(G3)

[3 marks]

$S_{n} = \frac{n}{2} (2 \times 5 + (n - 1) (2))$ (M1)(A1)

Note: Award (M1) for substituted arithmetic sequence formula, (A1) for correct substitutions.

$S_{n} = \frac{n}{2} (8 + 2 n)$ OR $S_{n} = n (5 + n - 1)$ (M1)

Note: Award (M1) for evidence of expansion and simplification, or division by $2$ leading to the final answer.

$S_{n} = n^{2} + 4 n$ (AG)

Note: The final line must be seen, with no incorrect working, for the final (M1) to be awarded.

[3 marks]

$(S_{18} =) 18^{2} + 4 \times 18$ (M1)

Note: Award (M1) for correctly substituted formula for $S_{n}$ .

$(S_{18} =) 396$ (A1)

Note: The use of “hence” in the question paper means that the $S_{n}$ formula (from part (b)) must be used.

[2 marks]

$1000 = n^{2} + 4 n$ OR $1000 = \frac{n}{2} (10 + (n - 1) 2)$ (or equivalent) (M1)

Note: Award (M1) for equating $S_{n}$ to $1000$ or for equating the correctly substituted sum of arithmetic sequence formula to $1000$ .

a sketch of the graphs $S_{n} = n^{2} + 4 n$ and $S_{n} = 1000$ intersecting (M1)

Note: Award (M1) for a sketch of a quadratic and a horizontal line with at least one point of intersection.

a sketch of $n^{2} + 4 n - 1000$ intersecting the $x$ -axis (M1)

Note: Award (M1) for a sketch of $n^{2} + 4 n - 1000$ with at least one $x$ -intercept.

$(n =) 29.6859 \dots$ OR $- 2 + 2 \sqrt{251}$ (A1)

Note: Award (A1) for $29.6859 \dots$ or $- 2 + 2 \sqrt{251}$ seen. Can be implied by a correct final answer.

$(n =) 29$ (A1)(ft)(G2)

Note: Do not accept $30$ . Award a maximum of (M1)(A1)(A0) if two final answers are given. Follow though from their unrounded answer.

$S_{30} = 1020$ and $S_{29} = 957$ (A2)

Note: Award (A2) for both “crossover” values seen. Do not split this (A2) mark.

$(n =) 29$ (A1)(G2)

[3 marks]

$(A =) (29^{2} + 4 \times 29) \times (1.84)$ (M1)(M1)

Note: Award (M1) for their correct substitution to find the total number of triangular panes. Award (M1) for multiplying their number of panes by $1.84$ .

$(A =) 957 \times 1.84$ (A1)(ft)(M1)

Note: Award (A1)(ft) for their $957$ seen. Award (M1) for multiplying their number of panes by $1.84$ . Follow through from part (d).

$(A =) 1760.88 (m^{2})$ (A1)(ft)(G2)

$(A =) 1761 (m^{2})$ (A1)(ft)(G3)

[4 marks]

Examiners report

[N/A]

An infinite geometric series has first term $u_{1} = a$ and second term $u_{2} = \frac{1}{4} a^{2} - 3 a$ , where $a > 0$ .

Find the common ratio in terms of $a$ .

[2]

Find the values of $a$ for which the sum to infinity of the series exists.

[3]

Find the value of $a$ when $S_{\infty} = 76$ .

[3]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

evidence of dividing terms (in any order) (M1)

eg $\frac{u_{1}}{u_{2}}, \frac{\frac{1}{4} a^{2} - 3 a}{a}$

$r = \frac{1}{4} a - 3$ A1 N2

[2 marks]

recognizing $|r| < 1$ (must be in terms of $a$ ) (M1)

eg $|\frac{1}{4} a - 3| < 1, - 1 \leq \frac{1}{4} a - 3 \leq 1, - 4 < a - 12 < 4$

$8 < a < 16$ A2 N3

[3 marks]

correct equation (A1)

eg $\frac{a}{1 - (\frac{1}{4} a - 3)} = 76, a = 76 (4 - \frac{1}{4} a)$

$a = \frac{76}{5} (= 15.2)$ (exact) A2 N3

[3 marks]

Examiners report

[N/A]

In an arithmetic sequence, ${u_1} = 1.3$ , ${u_2} = 1.4$ and ${u_k} = 31.2$ .

Consider the terms, ${u_n}$ , of this sequence such that $n$ ≤ $k$ .

Let $F$ be the sum of the terms for which $n$ is not a multiple of 3.

Find the value of $k$ .

[4]

Find the exact value of ${S_k}$ .

[2]

Show that $F = 3240$ .

[5]

An infinite geometric series is given as ${S_\infty } = a + \frac{a}{{\sqrt 2 }} + \frac{a}{2} + \ldots$ , $a \in {\mathbb{Z}^ + }$ .

Find the largest value of $a$ such that ${S_\infty } < F$ .

[5]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

attempt to find $d$ (M1)

eg 1.4 − 1.3 , ${u_1} - {u_2}$ , $1.4 = 1.3 + \left( {2 - 1} \right)d$

$d = 0.1$ (may be seen in expression for ${u_n}$ ) (A1)

correct equation (A1)

eg $1.3 + \left( {k - 1} \right) \times 0.1 = 31.2$ , $0.1k = 30$

$k = 300$ A1 N3

[4 marks]

correct substitution (A1)

eg $\frac{{300}}{2}\left( {1.3 + 31.2} \right)$ , $\frac{{300}}{2}\left[ {2\left( {1.3} \right) + \left( {300 - 1} \right)\left( {0.1} \right)} \right]$ , $\frac{{300}}{2}\left[ {2.6 + 299\left( {0.1} \right)} \right]$

${S_k} = 4875$ A1 N2

[2 marks]

recognizing need to find the sequence of multiples of 3 (seen anywhere) (M1)

eg first term is ${u_3}$ (= 1.5) (accept notation ${u_1} = 1.5$ ) ,

$d = 0.1 \times 3$ (= 0.3) , 100 terms (accept $n = 100$ ), last term is 31.2

(accept notation ${u_{100}} = 31.2$ ) , ${u_3} + {u_6} + {u_9} + \ldots$ (accept $F = {u_3} + {u_6} + {u_9} + \ldots$ )

correct working for sum of sequence where n is a multiple of 3 A2

$\frac{{100}}{2}\left( {1.5 + 31.2} \right)$ , $50\left( {2 \times 1.5 + 99 \times 0.3} \right)$ , 1635

valid approach (seen anywhere) (M1)

eg ${S_k} - \left( {{u_3} + {u_6} + \ldots } \right)$ , ${S_k} - \frac{{100}}{2}\left( {1.5 + 31.2} \right)$ , ${S_k} -$ (their sum for $\left( {{u_3} + {u_6} + \ldots } \right)$ )

correct working (seen anywhere) A1

eg ${S_k} - 1635$ , 4875 − 1635

$F = 3240$ AG N0

[5 marks]

attempt to find $r$ (M1)

eg dividing consecutive terms

correct value of $r$ (seen anywhere, including in formula)

eg $\frac{1}{{\sqrt 2 }}$ , 0.707106… , $\frac{a}{{0.293 \ldots }}$

correct working (accept equation) (A1)

eg $\frac{a}{{1 - \frac{1}{{\sqrt 2 }}}} < 3240$

correct working A1

METHOD 1 (analytical)

eg $3240 \times \left( {1 - \frac{1}{{\sqrt 2 }}} \right)$ , $a < 948.974$ , 948.974

METHOD 2 (using table, must find both ${S_\infty }$ values)

eg when $a = 948$ , ${S_\infty } = 3236.67 \ldots$ AND when $a = 949$ , ${S_\infty } = 3240.08 \ldots$

$a = 948$ A1 N2

[5 marks]

Examiners report

[N/A]

A new concert hall was built with $14$ seats in the first row. Each subsequent row of the hall has two more seats than the previous row. The hall has a total of $20$ rows.

Find:

The concert hall opened in $2019$ . The average number of visitors per concert during that year was $584$ . In $2020$ , the average number of visitors per concert increased by $1.2 %$ .

The concert organizers use this data to model future numbers of visitors. It is assumed that the average number of visitors per concert will continue to increase each year by $1.2 %$ .

the number of seats in the last row.

[3]

a.i.

the total number of seats in the concert hall.

[2]

a.ii.

Find the average number of visitors per concert in $2020$ .

[2]

Determine the first year in which this model predicts the average number of visitors per concert will exceed the total seating capacity of the concert hall.

[5]

It is assumed that the concert hall will host $50$ concerts each year.

Use the average number of visitors per concert per year to predict the total number of people expected to attend the concert hall from when it opens until the end of $2025$ .

[4]

Markscheme

recognition of arithmetic sequence with common difference $2$ (M1)

use of arithmetic sequence formula (M1)

$14 + 2 (20 - 1)$

$52$ A1

[3 marks]

a.i.

use of arithmetic series formula (M1)

$\frac{14 + 52}{2} \times 20$

$660$ A1

[2 marks]

a.ii.

$584 + (584 \times 0.012)$ OR $584 \times {(1.012)}^{1}$ (M1)

$591 (591.008)$ A1

Note: Award M0A0 if incorrect $r$ used in part (b), and FT with their $r$ in parts (c) and (d).

[2 marks]

recognition of geometric sequence (M1)

equating their $n$ th geometric sequence term to their $660$ (M1)

Note: Accept inequality.

METHOD 1

EITHER

$600 = 584 \times {(1.012)}^{x - 1}$ A1

$(x - 1 =) 10.3 (10.2559 \dots)$

$x = 11.3 (11.2559 \dots)$ A1

$2030$ A1

$600 = 584 \times {(1.012)}^{x}$ A1

$x = 10.3 (10.2559 \dots)$ A1

$2030$ A1

METHOD 2

11^th term $658 (657.987 \dots)$ (M1)A1

12^th term $666 (666.883 \dots)$ (M1)A1

$2030$ A1

Note: The last mark can be awarded if both their 11^th and 12^th correct terms are seen.

[5 marks]

$7$ seen (A1)

EITHER

$584 (\frac{1 . 012^{7} - 1}{1.012 - 1})$ (M1)

multiplying their sum by $50$ (M1)

sum of the number of visitors for their $r$ and their seven years (M1)

multiplying their sum by $50$ (M1)

$29 200 (\frac{1 . 012^{7} - 1}{1.012 - 1})$ (M1)(M1)

THEN

$212000 (211907.3 \dots)$ A1

Note: Follow though from their $r$ from part (b).

[4 marks]

Examiners report

[N/A]

a.i.

[N/A]

a.ii.

[N/A]

The first terms of an infinite geometric sequence, ${u_n}$ , are 2, 6, 18, 54, …

The first terms of a second infinite geometric sequence, ${v_n}$ , are 2, −6, 18, −54, …

The terms of a third sequence, ${w_n}$ , are defined as ${w_n} = {u_n} + {v_n}$ .

The finite series, $\sum\limits_{k = 1}^{225} {{w_k}}$ , can also be written in the form $\sum\limits_{k = 0}^m {4{r^k}}$ .

Write down the first three non-zero terms of ${w_n}$ .

[3]

Find the value of $r$ .

[2]

b.i.

Find the value of $m$ .

[2]

b.ii.

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

attempt to add corresponding terms (M1)

eg $2 + 2{\text{,}}\,\,6 + \left( { - 6} \right){\text{,}}\,\,2{\left( 3 \right)^{n - 1}} + \,2{\left( { - 3} \right)^{n - 1}}$

correct value for ${w_5}$ (A1)

eg 324

4, 36, 324 (accept 4 + 36 + 324) A1 N3

[3 marks]

valid approach (M1)

eg $4 \times {r^1} = 36$ , $4 \times {9^{n - 1}}$

$r = 9$ (accept $\sum\limits_{k = 0}^m {4 \times {9^k}}$ ; $m$ may be incorrect) A1 N2

[2 marks]

b.i.

recognition that 225 terms of ${w_n}$ consists of 113 non-zero terms (M1)

eg $\sum\limits_1^{113} {}$ , $\sum\limits_0^{112} {}$ , 113

$m = 112$ (accept $\sum\limits_{k = 0}^112 {4 \times {r^k}}$ ; $r$ may be incorrect) A1 N2

[2 marks]

b.ii.

Examiners report

[N/A]

b.i.

[N/A]

b.ii.

Consider a geometric sequence where the first term is 768 and the second term is 576.

Find the least value of $n$ such that the $n$ th term of the sequence is less than 7.

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

attempt to find $r$ (M1)

eg $\,\,\,\,\,$ $\frac{{576}}{{768}},{\text{ }}\frac{{768}}{{576}},{\text{ }}0.75$

correct expression for ${u_n}$ (A1)

eg $\,\,\,\,\,$ $768{(0.75)^{n - 1}}$

EITHER (solving inequality)

valid approach (accept equation) (M1)

eg $\,\,\,\,\,$ ${u_n} < 7$

valid approach to find $n$ M1

eg $\,\,\,\,\,$ $768{(0.75)^{n - 1}} = 7,{\text{ }}n - 1 > {\log _{0.75}}\left( {\frac{7}{{768}}} \right)$ , sketch

correct value

eg $\,\,\,\,\,$ $n = 17.3301$ (A1)

$n = 18$ (must be an integer) A1 N2

OR (table of values)

valid approach (M1)

eg $\,\,\,\,\,$ ${u_n} > 7$ , one correct crossover value

both crossover values, ${u_{17}} = 7.69735$ and ${u_{18}} = 5.77301$ A2

$n = 18$ (must be an integer) A1 N2

OR (sketch of functions)

valid approach M1

eg $\,\,\,\,\,$ sketch of appropriate functions

valid approach (M1)

eg $\,\,\,\,\,$ finding intersections or roots (depending on function sketched)

correct value

eg $\,\,\,\,\,$ $n = 17.3301$ (A1)

$n = 18$ (must be an integer) A1 N2

[6 marks]

Examiners report

[N/A]